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In This Chapter

► Optimizing space Relating rates

Getting up to speed with position, velocity, and acceleration Going off on a tangent »- Doing /37 in your head

71 low that you’re an expert at finding derivatives, I’m sure you can’t wait to put your / V Expertise to use solving some practical problems. In this section, you find problems that actually come up in the real world — problems like how should a cat rancher use 200 feet of fencing to build a three-sided corral next to a river (he only needs three sides because the river makes the fourth side and cats hate water) to maximize the grazing area for his cats?

Optimization Problems: From Soup to Nuts

Optimization problems are one of the most practical types of calculus problems. You use the techniques shown below whenever you want to maximize or minimize something: like maximizing profit or area or volume or minimizing cost or energy consumption, and so on.

Q. A rancher has 400 feet of fencing and wants

To build a corral that’s divided into three

Equal rectangles. See the following figure. What length and width will maximize the

Area?

XXX

A. 100 feet by 50 feet with an area of 5000 square feet.

1. Draw a diagram and label with variables.

2. a. Express the thing you want maxi-

Mized, the area, as a function of the

Variables.

Area = Length X Width A = 3x ■ y

B. Use the given information to relate

The two variables to each other.

6x + 4y = 400 3x + 2y = 200

C. Solve for one variable and substitute into the equation from Step 2a to

Create a function of a single variable.

2y = 200 – 3x y = 100 – 1.5x A = 3x ■ y A ( x ) = 3x (100 – 1.5x ) = 300X – 4.5×2

3. Determine the domain of the function.

You can’t have a negative length of fence, so X Can’t be negative. And if you build the ridiculous corral with no width, all

400 feet of fencing would equal 6x. So x > 0 and 6x – 400 x < 200

4. Find the critical numbers of A (X).

A ( x) = 300x – 4.5×2 A ( x ) = 300 – 9x 0 = 300 – 9x 9x = 300

X = 100 X 3

A (x ) is defined everywhere, so 100 is

The only critical number.

5. Evaluate A (x) at the critical number

And at the endpoints of the domain.

A (0) = 0

A (100) = 300 (100) – 4.5 (100 )2

5000

A

200) = 0

The first and third results above should be obvious because they represent corrals with zero length and zero width.

You’re done.

100

: -

Will maximize the

Area. Plug that into y = 100 – 1.5x and you get y = 50. So the largest corral is

100

3

‘-, or 100 feet long, 50 feet wide,

And has an area of 5000 square feet.

1. What are the dimensions of the soup can of

Greatest volume that can be made with 50

Square inches of tin? (The entire can, including the top and bottom, are made

Of tin.) And what’s its volume?

2. A Norman window is in the shape of a

Semicircle above a rectangle. If the straight

Edges of the frame cost $20 per linear foot and the circular frame costs $25 per linear

Foot and you want a window with an area of 20 square feet, what dimensions will minimize the cost of the frame?

Solve It

3. A right triangle is placed in the first quadrant with its legs on the x and y axes. Given

That its hypotenuse must pass through the

Point (2, 5), what are the dimensions and

Area of the smallest such triangle?

Solve It

6. You’re designing an open-top cardboard box for a purveyor of nuts. The top will be made of clear plastic, but the plastic-box-top designer is handling that. The box must have a square base and two cardboard pieces that divide the box into four

Sections for the almonds, cashews, pecans, and walnuts. See the following figure. Given that you want a box with a volume of 72 cubic inches, what dimensions will minimize the total cardboard area and thus minimize the cost of the cardboard? What’s the total area of cardboard?

/

/

R

Solve It

Problematic Relationships: Related Rates

Related rates problems are the Waterloo for many a calculus student. But they’re not that bad after you get the basic technique down. The best way to learn them is by working through examples, so get started!

Q.

After working each problem, ask yourself whether the answer makes sense. Asking this question is one of the best things you can do to increase your success in mathematics and science. And while it’s not always possible to decide whether a math answer is

Reasonable, when it’s possible, this inquiry should be a quick, extra step of every problem you do.

A homeowner decides to paint his home. He

Picks up a home improvement book, which recommends that a ladder should be placed against a wall such that the distance from the foot of the ladder to the bottom of the wall is one third the length of the ladder. Not being the sharpest tool in the shed, the homeowner gets mixed up and thinks that

It’s the distance from the Top Of the ladder to

The base of the wall that should be a third of

The ladder’s length. He sets up his 18 foot

Ladder accordingly, and — despite this unstable ladder placement — he manages to

Climb the ladder and start painting. (Perhaps the foot of the ladder is caught on

A tree root or something.) His luck doesn’t

Last long, and the ladder begins to slide rapidly down the wall. One foot before the top

Of the ladder hits the ground, it’s falling at a

Rate of 20 feet/second. At this moment, how

Fast is the foot of the ladder moving away from the wall?

A. Roughly 1.11 Feet/second.

1. Draw a diagram, labeling it with any

Unchanging Measurements and assigning variables to any Changing Things.

See the following figure.

You don’t have to draw the house — the

Basic triangle is enough. But I’ve sketched a fuller picture of this scenario to make clear what a bonehead this guy is.

2. List all given rates and the rate you’re asked to figure out. Write these rates as

Derivates with respect to time.

You’re told that the ladder is Falling At a rate of 20 Ft/ sec. Going down is Negative, So

Dh 20 Db ? ~dt = – 20 ~dt = ?

(hIs for the distance from the top of the

Ladder to the bottom of the wall; B Is for

The distance from the base of the ladder to the wall.)

3. Write down the formula that connects

The variables in the problem, H And b.

That’s the Pythagorean Theorem, of

Course:

A2 + B2 = C2, thus H2 + B2 = 182

B

Continued

4.Differentiate with respect to time.

This is a lot like implicit differentiation because you’re differentiating with respect to T But the equation is in terms of H And B.

2hDhh-2h dt

2b dt

182 0

5.Substitute known values for the rates and variables in the equation from Step4, and then solve for the thing you’re asked to determine.

You’re trying to determine Dbb, so you

Have to plug numbers into everything else. But, as often happens, you don’t have a

Number for B, So use a formula to get the number you need. This will usually be the same formula you already used.

H2 + B2 = 182 12 + B2 = 182

B = +/323 ~ +17.97 feet (Obviously, you can reject the negative

Answer.)

Now you’ve got what you need to finish the problem.

2hdhh; + 2b = 0 Dt dt

2 20) + 2 (17.97) ^ = 0

= ~ 1.11 Feet/sec

6. Ask yourself whether your answer is

Reasonable.

Yes, it does make sense. Lean a yardstick

Against a wall so the bottom of it is about

4 or 5 inches from the wall. Then push the bottom of the yardstick the 4 or 5

Inches to the wall. You’ll see that the top

Would barely move up. Right triangles

With a fixed hypotenuse like this one always work like that. If one leg is much shorter than the other, the short leg can change a lot while the long leg barely changes. It’s a by-product of the Pythagorean Theorem.

5. A farmer’s hog trough is 10 feet long, and

Its cross-section is an isosceles triangle

With a base of 2 feet and a height of 2 feet 6

Inches (with the vertex at the bottom, naturally). The farmer is pouring swill into the

Trough at a rate of 1 Cubic foot per minute. Just as the swill reaches the brim, her three hogs start violently sucking down the swill at a rate of Vi cubic foot per minute For

Each hog. They’re going at it so vigorously

That another X cubic foot of swill is being

Splashed out of the trough each minute. The farmer keeps pouring in swill, but she’s no match for his hogs. When the depth of the swill falls to 1 foot 8 inches, how fast is the swill level falling?

Solve It

6. A pitcher delivers a fastball, which the batter pops up — it goes straight up above

Home plate. When it reaches a height of 60 feet, it’s moving up at a rate of 50 Ft/sec.

Atthis point, how fast is the distance from

The ball to 2nd base growing? Note: The

Distance between the bases of a baseball

Diamond is 90 feet.

Solve It

7. A six-foot tall man looking over his

Shoulder sees his shadow that’s cast by a

15-foot tall lamp post in front of him. The

Shadow frightens him so he starts running away from it — toward the lamp post.

Unfortunately, this only makes matters

Worse, as it causes the frightening head of the shadow to gain on him. He starts to

Panic and runs even faster. Five feet before

He crashes into the lamp post, he’s running

At a speed of 15 Miles/hour. At this point,

How fast is the tip of the shadow moving?

8. Salt is being unloaded onto a conical pile at

A rate of 200 Cubic feet per minute. If the

Height of the cone-shaped pile is always equal to the radius of the cone’s base, how fast is the height of the pile increasing

When it’s 18 feet tall?

Solve It

A Day at the Races: Position, Velocity, and Acceleration

The most important thing to know about this type of problem is that velocity is the derivative of position and acceleration is the derivative of velocity. The following points about position, velocity, and acceleration with regard to the chariot race in Figure 8-1 provide some keys to approaching these problems.

\/> The finish is 100 palameters from the start As the crow flies, So 100 palameters is the total Displacement. (A Palameter Is a little-known unit of distance used in ancient Rome equal to the length of Julius Caesar’s palace — roughly 380 feet.) Say the start is at (0, 0) on a coordinate system and the finish is at (100, 0). It’s 100 from 0 to 100, of course, so 100 is the total displacement.

\S Distance Is different. You can see that the charioteers backtrack 50 palameters in the middle of the race. Because there are two extra 50-palameter legs, the total length of the race is 200 palameters — that’s the distance. Distance is always positive or zero.

\S Displacement to the Left Is Negative (in other problems, down would be negative). When, say, Maximus passes Atlas, his Position Is 75 palameters from the start.

AtAphrodite, he’s back to only 25 palameters from the start as the crow flies. Displacement equals final position minus initial position, so from Atlas to

Aphrodite is a displacement of 25 – 75, or -50 palameters.

\S Velocity Is related to displacement, not distance traveled. Velocity has a special meaning in calculus and physics so forget the everyday meaning of it. Like displacement, if you’re going Left (or down), that’s a Negative Velocity. And here’s a critical point: When you switch directions, your velocity is Zero. Think of a ball

Thrown straight up. At its peak, for an infinitesimal moment, it is motionless, so its velocity is zero.

V Average velocity Is defined as Total displacement Divided by Total time. Say Glutius

Completes the race in 1 hour. Because he traveled 200 palameters, his Average

Speed Would be 200 Palameters per hour. But because the total displacement is only 100, his average velocity would be a mere 100 Palameters per hour (roughly

7 Miles per hour).

\S Speed Is regular old speed, and, unlike velocity, it’s always positive (or zero). If Maximus picks up speed to make the jump over the lion pit, his speed, naturally, would be increasing. Note: His velocity would be Decreasing — even though we would see him speeding up — because his velocity would be negative and would

Be becoming a larger and larger negative.

The meaning of Acceleration, Like velocity, (in calculus and physics) agrees with the way we use it in everyday life only for movement to the right (or up). But, going left (or down), it’s strange. When Glutius speeds up to jump over the lion

Pit, we would say that he’s accelerating. But because his velocity is becoming a

Larger and larger negative and is thus decreasing, he is technically Decelerating.

9. For problems 9, 10, and 11, a duck-billed

Platypus is swimming back and forth along

The side of your boat, blithely unaware that

He’s the subject for calculus problems in rectilinear motion. The back of your boat is at the zero position, and the front of your boat is in the positive direction (see the following figure). S(t) Gives his position (in feet) as a function of time (seconds). Find his a) position, b) velocity, c) speed, and d)

Acceleration, at T = 2 seconds.

|——–.

0 +

S (t ) = 5t2 + 4

10. S (t) = 3t4 – 5t3 +1 – 6

Solve It

11. S (t) = I + Ј – 3

Solve It

12. For problems 12, 13, and 14, a three-toed

Sloth is hanging onto a tree branch and moving right and left along the branch.

(The tree trunk is at zero and the positive direction goes out from the trunk.) s(t)

Gives his position (in feet) as a function of

Time (seconds). Between T = 0 and T = 5, for

Each problem, find a) the intervals when he’s moving away from the trunk, the intervals when he’s moving toward the trunk,

And when and where he turns around; b)

His total distance moved and his average speed; and c) his total displacement and his average velocity.

S (t) = 2t3 – T2 + 8t – 5

Solve It

13. S(t)=t4+t2-t

Solve It

U. s (t) = t

Solve It

T+1

Make Sure \lou Kno© \lour Lines: Tangents and Normals

In everyday life, it’s perfectly normal to go off on a tangent now and then. In calculus, on the other hand, there is nothing at all normal about a tangent. You need only note a couple points before you’re ready to try some problems:

At its point of tangency, a Tangent Line has the same slope as the curve it’s tangent to. In calculus, whenever a problem involves slope, you should immediately think derivative. The derivative is the key to all tangent line problems.

\S At its point of intersection to a curve, a Normal Line is Perpendicular To the tangent line drawn at that same point. When any problem involves perpendicular lines,

You use the rule that perpendicular lines have slopes that are opposite reciprocals. So all you do is use the derivative to get the slope of the tangent line, and then the opposite reciprocal of that gives you the slope of the normal line.

4

Ready to try a few problems? Say, that reminds me. I once had this problem with my carburetor. I took my car into the shop, and the mechanic told me the problem would be easy to fix, but when I went back to pick up my car. . . Wait a minute. Where was I?

Q. Find all lines through (1, -4) either tangent to or normal to Y = x3. For each tangent

Line, give the point of tangency and the

Equation of the line; for the normal lines, give only the points of normalcy.

A. Point of tangency is (2, 8); equation of

Tangent line is § = 12x - 16. Points of normalcy are approximately (-1.539, -3.645), (-.335, -.038), and (.250, .016).

1. Find the derivative.

Y=x3 y’ = 3x2

2. For the tangent lines, set the slope from

The general point (x, X3) to (1, -4)

Equal to the derivative and solve.

-4 – X3

3×2

1 – X - 4 – X3 = 3×2 – 3×3 2×3 – 3×2 – 4 = 0

X = 2 (I used my calculator:)

3.Plug this solution into the original function to find the point of tangency.

The point is (2, 8).

4. Get your algebra fix by finding the

Equation of the tangent line that passes

Through (1, -4) and (2, 8).

You can use either the point-slope form or the two-point form to arrive at

Y = 12x – 16.

5. For the normal lines, set the slope from

The general point (x x3) to (1, -4)

Equal to the opposite reciprocal of the derivative and solve.

- 4 – X3 = -1 1 – X 3x2 -12×2 – 3×5 = X – 1 3×5 + 12×2 + X - 1 = 0

X « -1.539, – .335,

Or.250 (Use your

Calculator.)

6. Plug these solutions into the original function to find the points of normalcy.

Plugging the points into Y = X3 gives you the three points: (-1.539, -3.645), (-.335, -.038), and (.250, .016).

15. Two lines through the point (1, -3) are tangent to the parabola Y = x2. Determine the

Points of tangency.

Solve It

16.

The Earth has a radius of 4,000 miles. Say

You’re standing on the shore and your eyes

Are 5′ 3.36" above the surface of the water.

How far out can you see to the horizon before the Earth’s curvature makes the water dip below the horizon? See the following figure.

.Your line of site to horizon

X2 + y2 = 40002

Note: not drawn to scale — you would be ©ay Smaller.

Solve It

17. Find all lines through (0, 1) normal to the

Curve Y x4. The results may surprise you. Before you begin solving this, graph Y x4

And put the cursor at (0, 1). Now guess

Where normal lines will be and whether they represent shortest paths or longest

Paths from (0, 1) to Y = x4. Note: Do ZoomSqr To get the distances on the graph

To appear in their proper proportion.

Solve It

18. An ill-prepared adventurer has run out of water on a hot sunny day in the desert. He’s 30 miles due north and 7 miles due

East of his camp. His map shows a winding river — that by some odd coincidence happens to flow according to the function

Y = 10sin1x0 + 10cos5 + X (where his

Camp lies at the origin). See the following figure. What point along the river is closest to him? He figures that he and his camel can just barely make it another 15 miles or so.

Y 1 K

50>

Solve It

Looking Smart ©ith Linear Approximation

Linear approximation is easy to do, and once_you get the hang of it, you can impress your friends by approximating things like 3/70 in your head — like this: Bingo! 4.125.

How did I do it? Look at Figure 8-2 and then at the example to see how I did it.

Figure 8-2:

The line tangent to the curve at (64, 4) can be used to approximate cube roots of numbers near 64.

<2

-1010 2030405060708090100110120

64

Y

Q. Use linear approximation to estimate 3Jl0.

A. 4.125

1. Find a perfect cube root near 3/70.

You notice that 3J70 Is near a no-brainer, 3/64. That gives you the point (64, 4) on the graph of Y = 3Jx.

2. Find the slope of § = 3/x at x = 64.

Y‘= i X-2/3 So the slope at 64 is )4.

This tells you that you add (or subtract)

>4 to 4 for each increase (decrease) of

One from 64. For example, the cube root

Of 65 is 4K8 And the cube root of 66 is 4^8,

Or 4>k

3. Use the point-slope form to write the equation of the tangent line at (64, 4).

Y - 4 = 48(X - 64)

Y = 48(X - 64) + 4

4. Because this tangent line runs so close to the function § = 3/x near x = 64, use

It to estimate cube roots of numbers near 64, namely at x = 70.

Y = 418(70 – 64) + 4

By the way, in your calc text, the above simple point-slope form from algebra

Is probably rewritten in highfalutin

Calculus terms — like this:

L (x) = F (x 0) + F’ (x 0)(X - X 0 )

Don’t be intimidated by this equation. It’s just your friendly old algebra equation in disguise! Look carefully at it term by term and you’ll see that it’s

Mathematically identical to the point-slope equation tweaked like this:

Y = Y 0 + M (x - X 0)

19. Estimate the 4th root of 17.

Solve It

20. Approximate 3.015

Solve It

21. Estimate siny|j, that’s one degree of

Course.

Solve It

22. Approximate ln(

Solve It

Solutions to differentiation Problem Solving

MM What are the dimensions of the soup can of greatest volume that can be made with 50 square inches of tin? What’s its volume? The dimensions are 3% inches wide and 3% inches tall. The volume is 27.14 cubic inches.

1. Draw your diagram (see the following figure).

•Soup For Dummies

2. a. Write a formula for the thing you want to maximize, the volume: V= nr2 h B. Use the given information to relate R And h.

Top and bottom lateral area

Surface Area = 2nr2 + 2nrh 50 = 2nr2+2nrh 25 = nr2+nrh

C. Solve for h and substitute to create a function of one variable.

Nrh = 25 – Nr2 V=nr2h

H = 25..

V(r) = nr2[^-r

25r-nr"

3. Figure the domain.

R >0 is obvious

H > 0 is also obvious

And because 25 = Nr2+ nrh (from Step 2b), when H About 2.82 inches.

4. Find the critical numbers of V(r).

V(r) = 25r-nr*

V (r) = 25-3nr2

0 = 25-3rcr2

R2= 25 3Tc

0, r =

/25

So r must be less than

^25

Or

= 1.63 inches (You can reject the negative answer because it’s outside the domain.) 5. Evaluate the volume at the critical number. V(1.63) = 25- 1.63-Tc(1.63 )3 = 27.14 cubic inches

R

D

That’s about 15 ounces. The can will be 2 ■ 1.63 or about 3X inches wide and

25

, „„ 1.63 or K ■ 1.63

About 3X inches tall. Isn’t that nice? The largest can has the same width and height and would thus fit perfectly into a cube. Geometric optimization problems frequently have results where the dimensions have some nice, simple mathematical relation to each other.

. . . What dimensions will minimize the cost of the frame? The dimensions are 4’3" wide and 5’1" high. The minimum cost is $373.

1. Draw a diagram with variables (see the following figure).

2x

2. a. Express the thing you want to minimize, the cost.

Cost = (length of curved frame) ■ (cost per linear foot) (length of straight frame) ■ (cost per linear foot)

= (tcx)(25) + (2x+2y)(20) = 25rcx+40x+40y

B. Relate the two variables to each other. Area = Semicircle + Rectangle

20 = 4|1 + 2xy

C. Solve for Y And substitute.

2xy = 20 Y

20 2x

10

Tcx

2

_ 7ix2

" 4x Nx ‘ 4

Cost = 25Tcx + 40x + 40y C(X) = 25Tcx+40X+40

10

■■ 25kx + 40x -■ 15nx+ 40x -

400

X

400

X

Nx 4

IOtcx

3. Find the domain.

X > 0 is obvious. And when X Gets large enough, the entire window of 20 square feet in area

Will be one big semicircle, so

20

Tcx

40 = nx2 X 2 40

X = J~ft

Thus, X Must be less than or equal to 3.57.

Y

4. Find the critical numbers of C(x).

Omit -2.143 because it’s outside the domain. So 2.143 is the only critical number.

5. Evaluate the cost at the critical number and at the endpoints.

C(x) = 15Tcx+40X+ C (0) = undefined C (2.143) ^ $373 C (3.57) ^ $423

So, the least expensive frame for a 20-square-foot window will cost about $373 and will be

2 x 2.143, or about 4.286 feet or 4’3" wide at the base. Because Y = – the height of the

Rectangular lower part of the window will be 2.98, or about 3′ tall. The total height will thus be 2.98 plus 2.14, or about 5T\

D . . . Given that a right triangle’s hypotenuse must pass through the point (2, 5), what are the dimensions and area of the smallest such triangle? The hypotenuse meets the §-axis at (0, 10) and the x-axis at (4, 0) and the triangle’s area is 20.

1. Draw a diagram (see the following figure).

2. a. Write a formula for the thing you want to minimize, the area: A = 2 bh B. Use the given constraints to relate B And h.

This is a bit tricky — Hint: Consider similar triangles. If you draw a horizontal line from (0, 5)

To (2, 5), you create a little triangle in the upper-left corner that’s similar to the whole triangle.

(You can prove their similarity with AA — remember your geometry? — both triangles have a

Right angle and both share the top angle.)

Because the triangles are similar, their sides are proportional:

Heightbig height^

Base

Big triangle

BaseS,

H – 5

Y

H

2

X

B

C. Solve for one variable in terms of the other — take your pick — and substitute into your formula to create a function of a single variable.

2h = b(h – 5) 2H = bh – 5b h (2 – b) =-5b

5B

A = 2 bh A(b) = Ib ■ (b5b2 5b2

B-2

2b – 4

3. Find the domain.

B must be greater than 2 — do you see why? And there’s no maximum value for B

4. Find the critical numbers.

A(b)

A (b ) =

5B2

2B-4

(5b2)‘(2b – 4) – (5b2)(2b – 4)’

(2b – 4)2 10b(2b – 4) – 10b2

(2b – 4)2 10b2 - 40b (2b – 4)2

10b2 – 40b

0

(2b – 4)’

10b2 -40b=0 10b(b-4)=0

B=0or 4

Zero is outside the domain, so 4 is the only critical number. The smallest triangle must occur at b = 4 because near the endpoints you get triangles with astronomical areas.

5. Finish.

B=4

H

5b B – 2 5 ■ 4

So

10;

4-2

Andthetriangle’sareaisthus 20.

D . . . Given that you want a box with a volume of 72 cubic inches, what dimensions will minimize the total cardboard area and thus minimize the cost of the cardboard? The minimizing dimensions are 6-by-6-by-2, made with 108 square inches of cardboard.

1. Draw a diagram and label with variables (see the following figure).

Mixed Nuts For Dummies _V_

X

2. a. Express the thing you want to minimize, the cardboard area, as a function of the variables.

Square Tan fcur Sides,©,„ dividers

Cardboard area = x2 + 4xy + 2x§ A = x2 + 6xy

H

_

B. Use the given constraint to relate x to y. Vol = I ■ © ■ h

72 = x ■ x ■ y

C. Solve for y and substitute in equation from Step 2a to create a function of one variable. 72

: -

Y =—2

X2

A = x2 + 6xy A (X) = x2 + 6x (g)

432

3. Find the domain. x > 0 is obvious y > 0 is also obvious

And if you make y small enough, say the height of a proton — great box, eh? — x would have to be astronomically big to make the volume 72 cubic inches. Technically, there is no maximum value for x

4. Find the critical numbers.

You know this number has to be a minimum because near the endpoints, say when x = .0001 or y = .0001, you get absurd boxes — either thin and tall like a mile-high toothpick or short and flat like a square piece of cardboard as big as a city block with a microscopic lip. Both of these would have Enormous Area and would be of interest only to calculus professors.

5. Finish.

X = 6, so the total area is

A (6) = 62 + 432 Because y = p

36+72 y 2

= 108

That’s it — a 6-by-6-by-2 box made with 108 square inches of cardboard.

D . . . When the depth of the swill falls to 1 foot 8 inches, how fast is the swill level falling? It’s falling at a rate of %> Inches per minute.

1. Draw a diagram, labeling the diagram with any Unchanging Measurements and assigning variables to any Changing Things. See the following figure.

2′ 6"

Fc(l’8">

Note that the figure shows the Unchanging Dimensions of the trough, 2 feet by 2 feet 6 inches by 10 feet, and these dimensions are Not Labeled with variable names like L (for length), © (for width), or H (for height). Also note that the Changing Things — the height (or depth) of the

Swill and the width of the surface of the swill (which gets narrower as the swill level falls) —

Do Have variable names, H For height and B For base (I realize it’s at the top, but it’s the base of the upside-down triangle shape made by the swill). Finally note that the height of 1’8" — which is the height only at one particular point in time — is in parentheses to distinguish it from the other Unchanging Dimensions.

2. List all given rates and the rate you’re asked to figure out. Express these rates as derivatives with respect to time. Give yourself a high-five if you realized that the thing that matters about the changing volume of swill is the Net Rate of change of volume.

Swill is coming in at 1 Cubic foot per minute And is going out at 3 ■ 2 Cubic feet per minute (for

The three hogs) plus another X Cubic feet per minute (the splashing). So the net is 1 Cubic foot per minute Going out — that’s a Negative Rate of change. In calculus language, you write:

Dv ~dt

Dv =-1 Cubic foot per minute.

You’re asked to determine how fast the height is changing, so write:

Dh ~dt

?

3. a. Write down a formula that involves the variables in the problem — V, h, And B.

The technical name for the shape of the trough is a Right prism. And the shape of the swill in the trough — what you care about here — has the same shape. Imagine tipping this up so it

Stands vertically. Any shape that has a flat base and a flat top and that goes straight up from base to top has the same volume formula: Volume = areabase ■ height

Note that this "base" is the entire swill triangle and totally different from B In the figure; also this "height" is totally different from the swill height, H

The area of the triangular base equals 2 Bh And the height of the prism is 10 feet, so here’s your formula: V = 1 bh ■ 10 = 5bh

Because B Doesn’t appear in your list of derivatives in Step 2, you want to get rid of it.

ANG/

B. Find an equation that relates your unwanted variable, b, to some other variable in the problem so you can make a substitution and be left with an equation involving only V And H. The triangular face of the swill is the same shape of the triangular side of the trough. If you remember geometry, you know that such similar shapes have proportional sides. So,

B = _h_ 2 2.5 2.5b = 2h

B = .8h

Similar triangles often come up in related rate problems involving triangles, triangular prisms,

And cones.

Now substitute.8h for b in the formula from Step 3a:

V 5bh

= 5 ■ .8h ■ h

= 4h2

4. Differentiate with respect to T.

Dv St

8hF

In all related rates problems, make sure you differentiate (like you do here in Step 4) Before You substitute the values of the variables into the equation (like you do below when you plug 1’8"

Into H In Step 5).

5. Substitute all known quantities into this equation and solve for Dhh.

You were given that h = 1’8" (you must convert this to feet) and you figured out in Step 2

That TJf

= -1, so

,2 Dh 1 3 ‘ dt

Dh St

-1

:-3 Ft/min - 9

-j-7j Inches/min

Thus, when the swill level drops to a depth of 1’8", it’s falling at a rate of % inches per minute. Mmm, mmm, good!

6. Ask whether this answer makes sense.

Unlike the example problem, it’s not easy to come up with a common-sense explanation of why this answer is or is not reasonable. But there’s another type of check that works here and

In many other related rates problems.

Take a very small increment of time — something much less than the time unit of the rates used in the problem. This problem involves rates per Minute, So use 1 second for your time

Increment. Now ask yourself what happens in this problem in 1 second. The swill is leaving

The trough at 1 Cubic foot / minute; So in 1 second, I4 cubic foot will leave the trough. What does that do to the swill height? Because of the similar triangles mentioned in Step 3b, when the swill falls to a depth of 1’8», which is 23 Of the height of the trough, the width of the surface of the swill must be 23 Of the width of the trough — and that comes to 1X feet. So the surface

Area of the swill is 13 X 10 feet.

Assuming the trough walls are straight (this type of simplification always works in this type of

Checking process), the swill that leaves the trough would form the shape of a Very, very Short box ("box" sounds funny because this shape is so thin; maybe "thin piece of plywood" is a

Better image).

The volume of a box equals Length ■ width ■ height, thus 60 = 10 ^ ^ ^ Height

Height .00125

This tells you that in 1 second, the height should fall.00125 feet or something very close to it.

(This process sometimes produces an exact answer and sometimes an answer with a very

Small error.) Now, finally, see whether this number agrees with the answer. Your answer was -%> inches/minute. Convert this to Feet/second:

-10 + 12 + 60 =-.00125. It checks.

D . . . When it reaches a height of 60 feet, it’s moving up at a rate of 50 Ft/sec. At this point, how fast is the distance from 2nd base to the ball growing? The distance is growing 21.3 Feet/second.

1. Draw your diagram and label it. See the following figure.

2nd

H(60)

-Vertical right triangle

9(WT

2. List all given rates and the rate you’re asked to figure out.

Dhh = 50 Ft/sec

Dd ? ~dt = ?

3. Write a formula that involves the variables: H2

4. Differentiate with respect to time: 2hdhh = 2dddd

(90/2)2

Like in the example, you’re missing a needed value, d. So use the Pythagorean Theorem to get it:

H2 + (90/2) = D2

2

602 + (90/2) = D2

D « +140.7 feet (You can reject the negative answer.)

Now do the substitutions:

Dh dd 2hdt = 2d~dt

2 ■ 60 ■ 50 = 2 ■ 140.7 Dd dt

Dd

2 ■ 60■50 2 ■ 140.7

21.3 Ft/sec

5. Check whether this answer makes sense.

For this one, you’re on your own. Hint: Use the Pythagorean Theorem to calculate D 34 second after the critical moment. Do you see why I picked this time increment?

. . . Five feet before the man crashes into the lamp post, he’s running at a speed of 15 Miles/hour. At this point, how fast is the tip of the shadow moving? It’s moving at 25 Miles/hour.

1. The diagram thing: See the following figure.

Initial Position

Critical Position

15 ft.

D

2. List the known and unknown rates.

Dcc =-15 Miles/hour (This is negative because C Is shrinking.) Dbb =

3. Write a formula that connects the variables.

This is another similar triangle situation, so

HeightM8 triangle FtflSeMG triangle

- — -

Height„ttle triangle BOSe^ Triangle

15 = B 6 B – c 15b – 15c = 6b

9b = 15c

3B 5C

4. Differentiate with respect to t: 3 Dbb = 5 Dcc

5. Substitute known values.

3 db = 5 (-15)

Dt

Dbb =-25 Miles/hour

Thus, the top of the shadow is moving toward the lamp post at 25 Miles/hour — and is thus gaining on the man at a rate of 10 Miles/hour.

A somewhat unusual twist in this problem is that you never had to plug in the given distance of 5 ft. This is because the speed of the shadow is independent of the man’s position.

. . . If the height of the cone-shaped pile is always equal to the radius of the cone’s base, how fast is the height of the pile increasing when it’s 18 feet tall? It’s increasing at 2^ Inches/min.

1. Draw your diagram: See the following figure.

C

C

?

2. List the rates: ^ = 200 Cubic ft/min dhh

Dt

3. a. The formula thing: Vc, ne= ^nr’h

B. Write an equation relating R And H So that you can get rid of R:r = h What could be simpler? Now get rid of R: V= I Nh2h = i rc/i3

4. Differentiate: =

5. Substitute and solve for Dh.

Dt

200 = Tc- 182-^7 At

F « .196 Ft/min « 2i Inches/min

6. Check whether this answer makes sense.

Calculate the increase in the height of the cone from the critical moment (h = 18) to 32oo Minute

After the critical moment. When H = 18, V= I Tc (18)3, or about 6107.256 cubic feet. LAm Minute

Later, the volume (which grows at a rate of 200 Cubic feet per minute) Will increase by 1 cubic foot to about 6108.256 cubic feet. Now solve for h:

6108.256 = - nh3

6108.256

1 „

3"

^ 18.000982

Thus, in 3200 Minute, the height would grow from 18 feet to 18.000982 feet. That’s a change of.000982 feet. Multiply that by 200 to get the change in 1 minute: .000982 ■ 200 ^ .196

It checks.

S (t ) = 5t2 + 4

A. At – = 2, the platypus’s position is s (t) = 24 feet from the back of your boat.

B. V (t) = S’ (t) = 10t, so at t = 2, the platypus’s velocity is s’ (2) = 20 Feet/second (20 is positive so that’s toward the front of the boat).

C. Speed is the absolute value of velocity, so the speed is also 20 ft/sec.

D. Acceleration, A (t), equals v’ (t) = S" (T) = 10. That’s a constant, so the platypus’s acceleration

Is 10

Feet/second Second

At all times.

R

M S (t) = 3t4 – 5t3 + – - 6

A. s (2) gives the platypus’s position at – = 2; that’s 3 ■ 24 - 5 ■ 23 + 2 - 6, or 4 feet, from the back of the boat.

B. V (t) = S’ (t) = 12t3 – 15t2 + 1. At – = 2, the velocity is thus 37 Feet per second.

C. Speed is also 37 feet per second.

Ra

M

Feet/ second Second.

D. A (t) = v (t) = s" (t) = 36t2 – 30t. A (2) equals 84

S (t) = I + Ј – 3

A. At T = 2, s (2) equals 1 + 1 - 3, or -1^2 Feet. This means that the platypus is 1Vi Feet Behind

The back of the boat. 2

B. V (2) = S’ (2) = -2-

= -1 - 24 4 16

24(2)

= -14 Feet/second

A negative velocity means that the platypus is swimming "backwards," in other words away from the back of the boat.

C.

Speed = \Velocity\, so the platypus’s speed is 133 Feet/second.

D. A (t) = v (t) = s (t) = 2t-

31 Feet/ second Or 34 Second .

96t-

Or P + 7^ A (2) is therefore | + ||,

Give yourself a pat on the back if you figured out that this positive acceleration with a negative velocity means the platypus is actually slowing down.

S (t) = 2t3- t2 + 8t – 5

A. Find the zeros of the velocity:

V (t) = S’ (t) = 6t2 – 2t + 8 0 = 6t2 – 2t + 8 = 3t2 -1 + 4

No solutions because the discriminant is negative. The discriminant equals B2 – 4ac.

The fact that the velocity is never zero means that the sloth never turns around. At T = 0, V (t) = 8 Ft/sec Which is positive, so the sloth moves away from the trunk for the entire

Interval T = 0 to T = 5.

B. and c. Because there are no turnaround points and because the motion is in the positive direction, the total distance and total displacement are the same: 265 feet.

S (5) – s (0) = 260 – (-5) = 265 Whenever the total displacement equals the total distance, average velocity also equals

Average speed: 53 Ft/sec.

Total displacement = S (5) – s (0) = 265 = 53 ft/ Total time = 5 – 0 = 5 =53 /sec

2T I

M S (t)=t4+12-1

A. Find the zeros of V(t): v(t)=s’(t)=4T3

You’ll need your calculator for this:

3 1 and locate the x-intercepts. There’s just one: x ~

Graph y = 4×3 zero of S’(t)=

+2X

V(t).

.385. That’s the only

Don’t forget that a zero of a derivative can be a horizontal inflection as well as a local extremum. You get a turnaround point only at the local extrema.

Because v (0) =-1 (a leftward velocity) and v (1) = 5 (a rightward velocity), s (.385) must be a turnaround point (and it’s also a local min on the position graph). Does the first derivative test ring a bell?

Thus, the sloth is going left From T = 0 sec to T = .385 sec and right from.385 to 5 sec. He turns around, obviously, at .385 sec when he is at s (.385) = .3854 + .3852- .385 or -.215 meters — that’s.215 meters to the left of the trunk. I presume you figured out that there

Must be another branch on the tree on the other side of the trunk to allow the sloth to go left to a negative position.

0 till t = .385, then right from t = .385

B. There are two legs of the sloth’s trip. He goes left from t till t = 5. Just add up the Positive Lengths of the two legs.

LengthEgi = |s (.385) – s (0)| = |-.215 – 0| =.215 meters

Lengthy = |s(5) – s(.385)|

= |54 + 52- 5 – (-.215)|

=645.215 meters

The total distance is thus.215 + 645.215, or 645.43 meters. That’s one big tree! The branch is over 2000 feet long.

His average speed is 645.43 / 5, or about 129.1 Meters/second. That’s one fast sloth! Almost 300 Miles/hour!

C. Total displacement is s (5) - S (0), that’s 645 – 0 = 645 meters. Lastly, his average velocity

Is simply total displacement divided by total time — that’s 645/5, or 129 Meters/second.

M S (t)

T+1

T2+4

A. Find the zeros of V(t):

S’ (t) = v (t)=^12+4)- (t +1)(12+4)’

(Tl+4)

T2 + 4 – (2t2 + 2t)

(t+4)

T2 2T+4

(t+4)

Set this equal to zero and solve: -12- 2t +4 = 0

(t2 + 4 )2 "

T2 +2T 4=0

-2 + /4 – (-16) 2

~ – 3.236 or 1.236

T=

_

Reject the negative solution because it’s outside the interval of interest: t = 0 to t = 5. So, the only zero velocity occurs at t = 1.236 seconds.

Because v (0) = .25 Meters/second And v (5) ^ -.037, the first derivative test tells you that s (1.236) must be a local max and therefore a turnaround point.

The sloth thus goes right from T = 0 till T = 1.236 seconds, then turns around at S(1.236), or about.406 meters to the right of the trunk, and goes left till T = 5.

B. His total distance is the sum of the lengths of the two legs:

Going right = |s (1.236) – s (0)| = |.405 – .25|

Going left = |s (5) – s (1.236)|

Total distance is therefore.155 + .198 = .353 meters. His average speed is thus.353/5, or.071 Meters/second. That’s roughly a sixth of a Mile/hour — Much More like it for a sloth.

That’s pretty darn slow, but how quick do you think you’d be with only three toes?

C. Total displacement is defined as final position minus initial position, so that’s

S (5) – s (0) = ^ – 4

~ -.043 meters

And thus his average velocity is -.043/5, or -.0086 Meters/second. You’re done.

± Two lines through the point (1, -3) are tangent to the parabola y = x2. Determine the points of tangency. The points of tangency are (-1, 1} And (3, 9).

1. Express a point on the parabola in terms of x.

The equation of the parabola is y = x2, so you can take a general point on the parabola (x, y)

And substitute x2 for y. So your point is (x, x2).

2. Take the derivative of the parabola.

Y=x2 y’ = 2x

3. Using the slope formula, M = jXI-Јi, set the slope of the tangent line from (1, - 3) To (X, x2) Equal to the derivative. Then solve for X.

X2 ( 3)

–—- – 2x

X – 1

X2+3=2×2 2x x2 2x 3=0 (x+1)(x 3)=0

X= 1or 3

4. Plug these x-coordinates into Y = X2 To get the y-coordinates.

Y = (-1)2 = 1

Y=32=9

So there’s one line through (1, -3) that’s tangent to the parabola at (-1, 1) and another through (1, -3) that’s tangent at (3, 9). You may want to confirm these answers by graphing

The parabola and your two tangent lines:

Y= 2(x+1)=1

Y = 6 (x – 3) + 9

° . . . How far out can you see to the horizon before the Earth’s curvature makes the water dip

Below the horizon? You can see out 2.83 miles.

1. Write the equation of the Earth’s circumference as a function of y (see the figure in the

Problem).

X2 + y2 = 40002

Y = +/40 002 – x2

You can disregard the negative half of this circle because your line of sight will obviously be tangent to the upper half of the Earth.

2. Express a point on the circle in terms of x: (x, /40002-x2).

3. Take the derivative of the circle.

Y = /40 002 – x2

Y’ = i(40002 – x2)-1’2(-2x) (Chain Rule)

= - x /40 002 – x2

4. Using the slope formula, set the slope of the tangent line from your eyes to (x, /40002 – x2)

Equal to the derivative and then solve for x.

Your eyes are 5′ 3.36" above the top of the Earth at the point (0, 4000) on the circle. Convert your height to miles, that’s exactly.001 miles (What an amazing coincidence!). So the coordinates of your eyes are (0, 4000.001).

M

- x

/40 002 – x2

(40 002- x2) – 4000.001/40002- x2 (Cross multiplication) – 4000.001/40002 – x2 (Use your calculator, of course) /40002 – x2 (Now square both sides)

40002 x2 8

2/2 ~ 2.83 miles

Many people are surprised that the horizon is so close. What do you think?

IB Find all lines through (0, 1) normal to the curve y = x4. Five normal lines can be drawn to Y = X4 from (0, 1). The points of normalcy are (-.915, .702}, (-.519, .073}, (0, 0), (.519, .073), and (.915, .702).

1. Express a point on the curve in terms of x: A general point is (x, x4).

2. Take the derivative.

4

Y=x

Y’ = 4x3

3. Set the slope from (0, 1) to (x, x4) equal to the opposite reciprocal of the derivative and

Solve.

X4 - 1 = -1 X – 0 4x3 4×7 4×3+x=0 x (4×6 – 4×2 + 1) = 0 x = 0 or 4x6 - 4×2 + 1 = 0

X 2 - y 1 =

/40002 – x2 - 4000.001 = x – 0 =

X2= 40002= 3999.999= 15999992=

X2=

X=

Unless you have a special gift for solving 6th degree equations, you better use your calculator — just graph y = 4×6 – 4×2 + 1 and find all of the x-intercepts. There are x-intercepts at —.915, —.519, -.519, and -.915. Dig those palindromic numbers!

4. Plug these four solutions into y =

No-brainer.

X4 to get the y-coordinates. And there’s also the x = 0

(

-.519)’ -.915)4

(.519)4 (.915)4

-.073 -.702

You’re done. Five normal lines can be drawn to y = x4 from (0, 1). The points of normalcy are (-.915, .702), (-.519, .073), (0, 0), (.519, .073), and (.915, .702).

I find this result interesting. First, because there are so many normal lines, and second, because the normal lines from (0, 1) to (-.915, .702), (0, 0), and (.915, .702) are all Shortest

Paths (compared to other points in their respective vicinities). The other two normals are

Longest paths. This is curious because y = x4 is everywhere concave Toward (0, 1). When a curve is concave away from a point, a normal to the curve can only be a local shortest path. But when a curve is concave toward a point, you can get either a local shortest or a local

Longest path.

I played slightly fast and loose with the math for the x

Doesn’t work if you plug it back into the equation, become zero? However

-

■■ 0 solution. Did you notice that x = 0

X—^ = – i because both denominators

X – 0 4X3

Promise not to leak this to your calculus teacher — this is okay here

Because both sides of the equation become

5 = 2

Non-zero number Zero.

(Actually, they’re both

But something like 0 = 0 would also work.) Non-zero over zero means a vertical line with undefined slope. So the -1 = tells you that you’ve got a vertical normal line at x = 0.

. . . What point along the river is closest to the adventurer? The closest point is (6.11, 15.26), which is 14.77 miles away.

1. Express a point on the curve in terms of x: (x, 10sin10 .

2. Take the derivative.

10cos |–

Y

Y’

10sin1x0

10cos x -

10cos’ 10)’ 10

10sin’ 5)’ 5

Cos 10 – 2 sin 5-

1

3. Set the slope from (7, 30) to the general point equal to the opposite reciprocal of the derivative and solve.

ANG/

30

10sin1x0 + 10cos ^5 + x

Cos

10

2 sin

Unless you wear a pocket protector, don’t even think about solving this equation without a calculator.

Solve on your calculator by graphing the following equation and finding the x-intercepts

30

10sin1x0 + 10cos ^5 + x

Cos 10 – 2 sin y -

Y

It’s a bit tricky to find the x-intercepts for this hairy function. You have to play around with the Window Settings a bit. And don’t forget that your calculator will draw vertical asymptotes that look like zeros of the function, but are not. To see the first zero, set Xmin = -1, Xmax = 10, Xscl = 1, Ymin = -5, Ymax = 25, and Yscl = 5. To see the other two zeros, set Xmin = 10, Xmax = 30, Xscl = 1, Ymin = -2, Ymax = 10, and Yscl = 1. The zeros are at roughly 6.11, 13.75, and 20.58.

4. Plug the zeros into the original function to obtain the Y-coordinates. You get the following

Points of normalcy: (6.11, 15.26), (13.75, 14.32), (20.58, 23.80).

5. Use the distance formula, D = J(x2 - X 1)2 + (Y2 - Y 1)2, to find the distance from our parched

Adventurer to the three points of normalcy.

The distances are 14.77 miles to (6.11, 15.26), 17.07 miles to (13.75, 14.32), and 14.93 miles to (20.58, 23.80). Using his trusty compass, he heads mostly south and a little east to (6.11, 15.26). An added benefit of this route is that it’s in the direction of his camp.

D Estimate the 4th root of 17. The approximation is 2.03125.

1. Write a function based on the thing you’re trying to estimate: F (X) =

2. Find a "round" number near 17 where the 4th root is very easy to get: that’s 16, of course.

And you know 4/16 = 2. So (16, 2) is on F.

3. Determine the slope at your point.

F (x) = 4/x

F’(x) = ± X- 3’4

F‘(i6\- 1 f (16)= 32

4. Use the point-slope form of a line to write the equation of the tangent line at (16, 2).

Y – 2 = ^(X – 16)

5. Plug your number into the tangent line and you’ve got your approximation.

Y=32 (17 -16)+2

= 2132 or 2.03125

The exact answer is about 2.03054. Your estimate is only ^ of 1 percent too big! Not too shabby. Extra credit question (solve this or we may have to vote you off the island): No matter what 4th

Root you estimate with linear approximation, your answer will be too big. Do you see why?

— Approximate 3.015. The approximation is 247.05.

1. Write your function: G (X) = x5

2. Find your round number. That’s 3, well duhh. So your point is (3, 243).

3. Find the slope at your point.

G ( X) = 5X4

G ( 3) = 405

4. Tangent line equation.

Y – y 1 = M (x – x 1) Y – 243 = 405 (X – 3)

5. Get your approximation: Y = 405 (3.01 – 3) + 243 = 247.05 Only X00 Of a percent off.

EQ Estimate sin Y|g, that’s one degree of course. The approximation is J^tj.

You know the routine

° (x) = Sinx Y – Y1 = M (x – X1)

° (0) = 0 — (0,0) is the point Y – 0 = 1 (X – 0) °’ (X) = cos X y = x

°’ (0) = 1 — 1 is the slope

Your number is ygjj, so you get Y = ygjj.

Which shows that for very small angles, the sine of the angle and the angle itself are approximately equal. (The same is true of the tangent of an angle, by the way.) is only LAm% Too bi

EH Approximate ln (e10 + 5). The approximation is 10 + »50.

Just imagine all the situations where such an approximation will come in handy!

Q (x) = ln(x) Y – y1= m (x – X1)

Q (e10) = 10 — (e10, 10) is the point Y – 10 = -X{x – E10) — The tangent line

Q‘(X) = Y Y = ^((e10 + 5) – e10) + 10

Q’ (e10) = — Is the slope = 10 + _5_

Hold on to your hat. This answer is a mere 0.00000026% too big.

In This Chapter

> Boning up on basic derivative rules

Producing your quota of product and quotient problems

► Joining the chain rule gang Achieving higher order differentiation

Hapter 5 gives you the meaning of the derivative. In this chapter, you practice rules for finding derivatives. But before you practice the following rules, you may want to go to the Cheat Sheet in Calculus For Dummies Or to your calc text to review the basic derivatives. For example, you need to know that the derivative of sine is cosine.

C

Rules for Beginners

Okay, now that you’ve got the memorization stuff taken care, you now begin some rules that involve more than just memorizing the answer.

First there’s the derivative of a constant rule: The derivative of a constant is zero. Alright —

This one’s also just memorization.

And then there’s the power rule: To find the derivative of a variable raised to a power, bring

The power in front — multiplying it by the coefficient, if there is one — then reduce the power by one.

Q. What’s the derivative of 5×3?

A. 15×2

1. Bring the power in front, multiplying it by the coefficient.

So far you’ve got 15x\ Note that this does not equal 5×3 and so you should

Notput an equal sign in front of it.

In fact, there’s no reason to write this interim step down at all. I do it simply to make the process clear.

2. Reduce the power by one.

This gives you the final answer of 15×2.

1. What’s the derivative of °(x) = 8?

Solve It

2. What’s the derivative of g ( X )

Solve It

3. What’s the derivative of

G(x) = & sin-| cos 2Ji, Where K Is a constant?

Solve It

Ft. For ° ( x ) = 5x 4, °’ ( x ) = ?

Solve It

5. For g ( x ) = – Xp what’s g’ ( x )?

Solve It

6. Find y’ if Y = JX~5 (x > 0)

Solve It

J. What’s the derivative of s (?) = 7?6 + – + 10?

Solve It

8. Find the derivative for y = (x3- 6)2.

Solve It

Giving It Up for the Product and Quotient Rules

Now that you’ve got the easy stuff down, I’m sure you’re dying to get some practice with advanced differentiation rules. The Product rule And the Quotient rule Give you the derivatives for the product of two functions and the quotient of two functions, respectively and obviously.

The Product rule Is a snap. The derivative of a product of two functions, (firstj(second), Equals (First)’(second) + (first)( second)’.

The Quotient rule Is also a piece of cake. The derivative of a quotient of two functions,

( first)

( second)

Equals

(first) (second) – (first)(second)

(second)

Here’s a good way to remember the quotient rule. Notice that the numerator of the quotient rule looks exactly like the product rule, except that there’s a minus sign instead of a plus sign. And note that when you read a product, you read from left to right, and when you read a quotient, you read from top to bottom. So just remember

That the quotient rule, like the product rule, works in the natural order in which you read, beginning with the derivative of the first thing you read.

For some mysterious reason, many textbooks give the quotient rule in a different form that’s harder to remember. Learn it the way I’ve written it above, beginning with (First) ‘ That’s the easiest way to remember it.

8- DL

D

A. Dx

D f X 2sin x )

: (x2) ‘(sinx) + (X2)(sinx)’ 2x sin x + x2 Cos X

8.

A.

D

D (sin )

D x 2 (x2)(sinx) – (x2)(sinx)’ Dx (Sin X) =

(sin ) 2 sin – 2cos

Sin

One more thing: I’ve purposely designed

This example to resemble the product rule example, so you can see the similarity between the quotient rule numerator and the product rule.

Solve It

Cos )

10. (Sinxtanx) =

Solve It

11. Dx(5x3lnx) = ?

Solve It

*12. ^(x2 Ex Lnx)

Solve It

13.

D_ X3 Dx cosx

=?

Solve It

1k.

D Cosx

Dx ex

=?

Solve It

15.

D 3×2+3

D Arctan

*16.

D_ Sinx

Dx X3lnx

Solve It

Solve It

Linking Up ©ith the Chain Rule

The chain rule is probably the trickiest among the advanced rules, but it’s really not that bad at all if you focus clearly on what’s going on — I promise! Most of the basic derivative rules have aplain old X As the argument (or input variable) of the function. For example, F (x) = JX, sin x, and Y = ex All have just X As the argument.

When the argument of a function is anything other than a plain old x, such as Y = Sinx2 or lnlOx, you’ve got a chain rule problem.

Here’s what you do. You simply apply the derivative rule that’s appropriate to the outer function, temporarily ignoring the not-a-plain-old-x argument. Then multiply that result by the derivative of the argument. That’s all there is to it.

Q. What’s the derivative of Y = Sinx3?

A. § ‘= 3x 2cos x3

1. Temporarily think of the argument, x3, as a Glob.

So, you’ve got Y = Sin(glob).

2. Use the regular derivative rule.

Y = Sin (Glob), So Y ‘ = Cos (Glob)

(This is only a provisional answer, so the "=" sign is false — egad! The math police

Are going to pull me over.)

3. Multiply this by the derivative of the

Argument.

Y‘ = Cos (glob) ■ glob’

4. Get rid of the Glob.

The Glob Equals 3 so Glob Equals 3 2.

Y’= Cos(x3) ■ 3×2 =3 2cos 3

Q. What’s the derivative of sin4x3? You’ve got

To use the chain rule twice for this one.

A. The answer is 12×2 sin3 x3 ■ cos x3

1. Rewrite sin4 x 3to show what it really means: (Sin x3)

2. The outermost function is the 4th power, so use the derivative rule for Stuff4 — that’s 4stuff3; then multiply that by the derivative of the inside Stuff, Sinx3.

4 (sinx3)3 ^ (sinx3)’

With chain rule problems, always work

From the outside, in.

3. To get the derivative of sin x3, use the derivative rule for sin (glob), And then

Multiply that by Glob .

4 (sinX3 )3 ^ Cos (X3) ^ (X3) ‘

4. The derivative of X3 Is 3X2, So you’ve got

4 (sinX3 )3 ■ Cos (X3) ■ 3×2

5. To simplify, rewrite the Sin Power and

Move the 3X2 To the front.

= l2x2sin3X3^ CosX3

Can you remember when you’ve had so

Much fun?

With chain rule problems, never use more than one derivative rule per step. In other words, when you do the derivative rule for the outermost function, don’t

Touch the inside stuff! Only in the next step do you multiply the outside derivative by the derivative of the inside stuff.

17.

F( )=sin 2

F ( )=?

Solve It

18.

G( )=sin3

G ( )=?

Solve It

19.

S(t)=tan(lnT)

S (t)=?

Solve It

20.

Solve It

*21.

F( )= 4sin3

F ( )=?

Solve It

*22.

G( ) g ( )

(Ln X) 5x – 4 ?

Solve It

*23.

Y=cos34 2

Y =?

Solve It

*2k. Dx (tan

Solve It

What to Do ©ith Implicit Differentiation

You use implicit differentiation when your equation isn’t in "y =" form, such as sinY2 = X3 + 5y3, and it’s impossible to solve for Y. If you can solve for Y, Implicit differentiation will still work, but it’s not necessary.

Implicit differentiation Problems are chain rule problems in disguise. Here’s what I

Mean. You know that the derivative of sin is cos, and that according to the chain

Rule, the derivative of sinx3 is cos (X3) ■ (X3) ‘. You would finish that problem by doing

The derivative of 3, but I have a reason for leaving the problem unfinished.

To do implicit differentiation, all you do (sort of) is every time you see a "y" in a problem, you treat it as if it were X3. Thus, because the derivative of sinx3 is cos (X3) ■ (X3) ‘, the derivative of sinY Is cosY ■ y’. Then, after doing the differentiation, you just rearrange terms so that you get Y’ = Something.

Atn By the way, I used "y" in the explanation above, but that’s not the whole story.

Consider that Y’ = 2Ox3 is the same as -f – = 2Ox3. It’s the variable on the top that you

Apply implicit differentiation to. This is typically Y, But it could be any other variable. And it’s the variable on the bottom that you treat the ordinary way. This is typically X, But it could also be any other variable.

Q. A.

3. Factor out § ‘.

Y (3Y2-cosY)=cos -3

= Cos X – 3x2 3§2 – cos §

4. Divide.

1.Take the derivative of all four terms,

Using the chain rule for terms containing § And using the ordinary method

For terms containing X.

That’s your answer. Note that this

Derivative — unlike ordinary derivatives — contains YS as well as s.

Cos -3

3y2 – cosY

3y2 ■ y’+ 3x2 = cos Y ■ y’+ Cos X

2.Move all terms containing § to the left

Side and all other terms to the right

Side.

3y2 ■ y’ – CosY ■ y ‘= Cosx – 3x

25. If

Y3 – x2 = x + y, find ^ by implicit

Differentiation. D

Solve It

26. If 3y + lny = 4ex, find

Solve It

27. For X2y = y3x + 5y + x, find & by implicit

Differentiation. D

Solve It

*28. If Y + Cos2 Y3 = sin5x2, find the slope of the curve at

Solve It

Getting High on Calculus: Highe®Order Derivatives

You often need to take the derivative of a derivative, or the derivative of a derivative

Of a derivative, and so on. In the next two chapters, you see a few applications. For example, a second derivative will tell you the acceleration of a moving body. To find a higher order derivative, you just treat the first derivative as a new function and take its

Derivative in the ordinary way. You can keep doing this indefinitely.

29. For Y = x4, find the 1st through 6th

Derivatives.

Solve It

30. For Y = x5 + lOx3, find the 1st, 2nd, 3rd, and

4th derivatives.

Solve It

31. For Y = sin X + cos x, find the 1st through

6th derivatives.

Solve It

32. For Y = cos X2, find the 1st, 2nd, and 3rd

Derivatives.

Solve It

Solutions for Differentiation Problems

MM

MM g(x) = N>; g(x) = 0

MM

MM

MM

MM

F (x) = 8; F (X) = 0

The derivative of any constant is zero.

Don’t forget that even though N Sort of looks like a variable (and even though other Greek letters like 9, a, And 00 are Variables), N Is a number (roughly 3.14) and behaves like any other number. The same is true of E ~ 2.718. And when doing derivatives, constants like C And K Also behave like ordinary numbers.

G(x) = &sin-|cos2;t (where K Is a constant); G (x) = 0

If you feel bored because this first page of problems was so easy, just enjoy it; it won’t last. F (x) = 5x4; F‘ (X) = 20×3

Bring the 4 in front and multiply it by the 5, and at the same time reduce the power by 1, from 4 to 3: F’ (x) = 2Ox3. Notice that the coefficient 5 has no effect on the derivative in the following sense: You can ignore the 5 temporarily, do the derivative of x4 (which is 4×3), and then put the 5 back where it was and multiply it by 4.

D G (x)

1 o ; 8 (X) Io X2

You can just write the derivative without any work: ically, it works like this:

-3×2 10

But if you want to do it more method -

1. Rewrite so you can see an ordinary coefficient:

2. Bring the 3 in front, multiply, and reduce the power by 1.

1 X3

G )

2 (which is the same, of course, as

.)

1O 1O

Y = JX~5 (x> 0); §’=- JX-7/2

Rewrite with an exponent and finish like problem 5 (/P5 = x-:

To write your answer without a negative power, you write Y =

Your answer without a fraction power, to wit: Y’ = 5 You say "po-tay-to," I say "po-tah-to."

2/x7 or 2/x7

2×7′

5

Or

Or

-5

2×7/2

Or you can write

Or

5

2(jx\ 2 (jx\

S (t) = It6 + T + 10; S ‘ (T) = 42-5 + 1

Note that the derivative of plain old T Or plain old (or any other variable) is simply 1. In a sense, this is the simplest of all derivative rules, not counting the derivative of a constant. Yet

For some reason, many people get it wrong. This is simply an example of the power rule: X Is the same as x1, so you bring the 1 in front and reduce the power by 1, from 1 to 0. That gives you 1×0. But because anything to the 0 power equals 1, you’ve got 1 times 1, which of course is 1.

Y = (x3-6)2; §‘= 6×5-36x2 FOIL and then take the derivative.

=( 3 6)( 3 6)

= x6- 12×3 + 36 Y =6 5 36 2

■■•ft

-fX (x3 cosx) = 3×2 cos X-x3 sin X

Remember that cosx = – sinx. For a great mnemonic to remember the derivatives of trig functions, check out Chapter 15.

-fx (x 3cosx) = (X 3) ‘(cosx) + (X 3 )(cosx) ‘ =3 2cos + 3 sin ) =3 2cos 3sin -fix (sinxtanx) = Cos X Tan X + Sec X Tan X

A helpful rule: Fx tanx = Sec2 x.

Fx (Sinxtanx) = (sinx)’(tanx) + (sinx)(tanx) ‘

=cos tan +sin sec2 =cos tan +sec tan

Ix 5x3lnx = 5x 2( 3 In X + 1)

Another helpful rule: Fx

Ln

1

As for the 5, you can deal with it in two ways. First, you can

Ignore it temporarily (because coefficients have no effect on differentiation), and put it back when you’re done differentiating. If you do it this way, don’t forget that the "5" multiplies the

Entire derivative, not just the first term. The second way is probably easier and better: just

Make the "5" part of the first function. To wit:

Ix (5x3lnx) = (5×3) ‘(Lnx) + (5×3 )(Lnx)’

= 15x2lnx + 5×3 ■ — =15 2ln +5 2or =5 2 3ln +1); takeyourpick.

Fx X2 Ex lnx = Ex In X (x2 + 2x) + Xex

This is a challenge problem because, as you’ve probably noticed, there are three functions in this product instead of two. But it’s a piece o’ cake. Just make it two functions: either (x2e*)(lnx) or (x2)(Ex Lnx). Take your pick.

A handy rule: -^ex = ex (Note that Ex And its multiples (like 4ex) Are the only functions that are

Their own derivatives.)

1. Rewrite this "triple function" as the product of two functions.

2. Apply the product rule.

Ix (x2 e* ))mx ) = (X2 ^ ) + (X2 ^ X1^ )’

3. Apply the product rule separately to (x2Ex) ‘, then substitute the answer back where it

Belongs.

(x2 Ex) ‘ = (x2) ‘(Ex) + (x2)(Ex) ‘ =2 E + 2e

M

4. Complete the problem as shown in Step 2.

( 2e ) ln )+( 2e ) ln ) =(2 E + 2e ) ln )+ 2e

1

X

-2xexLnx + x2ExLnx + Xex Or 2e Ln +2 E Ln + E Or E Ln +2ln +1)or E Ln ( 2+2 )+ E

You say "pa-/a-mas," I say "po-/ah-mas." I 33X2cosX+x3sinX

I Cos

I

X

Cos X

( 3) cos )

( 3) cos )

Ix Cosx

Cos ) 3 2cos 3 sin )

Cos2x 3x2cosx + x3sinx Cos2x

I Cosx

Ix ex

SinX CosX

E

I Cosx = (cosx) (E ) (cosx)(E ) E Sin E Cos

Sin cos

I 3×2 + 3 6x arctanx - 3

Ix Arctanx

Arctan2x

A handy-dandy rule: Ix Arctanx =

1 + 2

I 3×2

. 3 (3×2 + 3) (arctanx) – (3×2 + 3)(arctanx) ‘

Ix Arctanx

(arctanx)

6 arctan (3 2+3)

1

X2 + 1

Arctan2 = 6x arctanx – 3 Arctan2 X

To remember the derivatives of the inverse trig functions, notice that the derivative of each co -

Function (arccosine, arccotangent, And Arccosecanf) Is the negative of its corresponding function. So, you really only need to memorize the derivatives of Arcsin, arctan, And Arcsec. These three have a 1 in the numerator. The two that contain the letter "s," Arcsin And Arcsec, Contain a SQuare root in the denominator. Arctan Has no "s" and no square root.

_

1

I Sinx _ X CosX InX - 3 sinX InX - SinX

Ix X3lnx X * (in X )2

I Sin sin ) ( 3ln ) sin )( 3ln )

I 3ln

3ln )

Product Rule

(cosx)(x3 lnx) – (sinx)((x3) (lnx) + (x3)(lnx) x6( lnx )2

X3cosxlnx – (sinx)(3x2lnx + (x3)(—

6 Ln )

= X3cosxlnx – 3×2 sinx lnx – x2 sinx

X6 (lnx )2 = Xcosxlnx- 3 sinx lnx – sinx 4 Ln )

M F(x) = sinx2; F‘(X) = 2XCosX2

Because the argument of the sine function is something other than a plain old , This is a chain rule problem. Just use the rule for the derivative of sine, not touching the inside stuff ( 2), then multiply your result by the derivative of 2.

F’ (x) = cos(x2) ■ 2x =2 cos 2

E3 G (x) = sin3x G‘ (X) = 3sin2 XCosX

Rewrite sin3 as sin )3 So that it’s clear that the outermost function is the cubing function.

By the chain rule, the derivative of Stuff3 Is 3 Stuff2 ■ stuff. The Stuff Here is sin X And thus Stuff Is cos x. So your final answer is 3 (sinx) ^ cosx, or 3sin2XCosx.

EH S (T) = tan(lnT); S ‘ (T) = Sec2(inT) ■ -

The derivative of tan is sec2, so the derivative of tan(lump) is sec2( Lump) ■ lump’. You better know by now that the derivative of ln T Is 1, so your final result is sec2(lnT) ■ 1.

— § = E4X3; § ‘= 12×2 E *x 3

The derivative of E Is ex, so by the chain rule, the derivative of Eglob Is Eglob■ glob’. So §’= E4X3■ 12×2, or 12×2E4x\

*E1 F(x) = x4sin3x; F‘(X) = 4x3sin3X + 3x*sin2XCosx

This problem involves both the product rule and the chain rule. Which do you do first? Note

That the chain rule part of this problem, sin3 , Is one of the two things being multiplied, so it is

Part of — or sort of Insiie — the product. And, like with pure chain rule problems, with problems involving more than one rule, you work from outside, in. So here you begin with the product rule. Here’s another way to look at it:

If you’re not sure about the order of the rules in a complicated derivative problem, imagine that

You plugged a number into in the original function and had to compute the answer. Your Last

Computation tells you where to start. If, for example, you plugged 2 into x4sin3x, you would compute 24, then sin 2, then you’d cube that to get sin3 2, and, finally, you’d multiply 24 by sin3 2. Because your final step was Multiplication, You begin with the Proiuct Rule.

F )= 4sin3

=( 4)(sin3 ) F’ (x) = (x4) (sin3x) + (x4)(sin3X) (product rule)

Use the chain rule to solve (sin3 x) ‘, then go back and finish the problem. sin3x means (sinx)3 and that’s Stuff3. The derivative of Stuff3 Is 3Stuff2■ stuff, So the derivative of (sinx)3 is 3 (sinx )2 ■ cosx. Now continue the solution.

F’ (x) = (x4) ‘(sin3 x) + (x4)(sin3 X) ‘ = 4x3sin3x + x4^ 3 (sinx)2cosx =4 3sin3 +3 4sin2 cos

(x) = 0^; „‘ (X) = 2inX__5(InX)2

G(x) 5x-4; G (X) X(5x-4) (5x-4)2

Here you’ve got the chain rule inside the quotient rule. Start with the quotient rule:

((lnx)2)’(5x – 4) – (lnx)2 (5x-4) ‘

G‘(x) =–75—7?-

5 4)

Next, take care of the chain rule solution for ((lnx )2)’. You want the derivative of Glob2 — that’s 2Glob■ glob’. So the derivative of (lnx)2 is 2 (lnx)(1 ]. Now you can finish:

G )

2 (lnx)(X )(5x-4)- (lnx)’(5)

5 4)

(10x-8) Lnx-5x (lnx )2

5 4)

2ln 5 ln )2

5 4) 5 4)2

Y = cos34x2; §‘= -24xcos24x2sin4x2 Triply nested!

23

Y=(cos4x )

The derivative of Stuff3 Is 3 Stuff2 ■ stuff, So, you’ve got,

Y ‘= 3 (cos4x2 )2 ■ (cos4x2) ‘ Now you do the derivative of cos (glob), which is – sin (glob) ■ Glob’. Two down, one to go:

Y‘= 3 (cos4x2)2(-sin4x2) ■ (4×2) ‘ = 3cos24x2(-sin4x2) ■ 8x = -24xcos24x2sin4x2

*E1 J* Tan3 E*

6xex tan2 ex sec2 ex

Holy quadrupely nested quadruple nestedness, Batman! This is one for the Riddler.

-J (Tane* 1

= 3 (tane*’) ■ (tanE*^

= 3tan2 E* Sec2 E* ■ (e

= 3tan2 e*sec2 e*2 ■ e*

(because j*stuff3 = 3stuff2■ stuff’^j (because J* tan (glob) = sec2(glob) ■ glob

X’

D*

D_{

D*

(because 4-.elump = elump■ lump’

3tan2e Sec2e ■e ■2 – 6*e*2 tan2 E’2 sec2 E’2

§ If §

* = * + §, y

2x + 1

3y2 - 1

1. Take the derivative of all four terms, using the chain rule (sort of) for all terms containing a §.

3§2 §’- 2* = 1 + §‘

2. Move all terms containing §’ To the left, all other terms to the right, and factor out

3§2§ -§ =1+2

§ ‘ (3§2 – 1) = 1 + 2*

3. Divide and voila! 2* + 1

§

3§2-1

EL If 3§ + Ln§ = 4e*, Y ‘=

4ye’

3Y+1

Follow steps for problem 25.

4E 4E

3 + 7 4§e* 3§ + 1

EL For *2 § = §3 * + 5§ + *, find J*§ By implicit differentiation. Y’= -3§y-X2′y.+

This time you’ve got two Products To deal with, so use the product rule for the two products and

The regular rules for the other two terms.

(*2) ‘§ + *2 § ‘= (§3) ‘* + §3 *’+ 5§‘+ 1 2*§ + *2 § ‘= 3§2 § ‘* + §3 + 5§’+ 1 *2 §’ – 3§2 § ‘* – 5§’ = §3 + 1 – 2*§ §’ (*2 - 3§2* – 5) = §3 - 2*§ + 1

§ ‘ = §3 - 2*§ + 1 § -3§2* + *2- 5

!E If § + cos2 §3 = sin5*2, find the slope of the curve at (J^, 0j. The slope is zero. You need a slope, so you need the derivative.

§’ + 2cos§3 ■ (-sin§3 )(§3)’ = cos( 5*2 )(10*)

Implicit ’ Chain Rule ’ ‘ Chain" Rule

Differentiation . .

I twice nested 1

§’+ 2cos §3 (-sin§3 )(3§2 §’) = 10* cos5*2

Implicit

Differentiation

§’ (1 – 6§2cos§3sin§3) = 10*cos5*2

10 cos5 2

§

1 6§2cos§3sin§3

You need the slope at X = Jfir, y = 0, so plug those numbers in to the derivative. Actually, you

Can save yourself a lot of work if you notice that the numerator will equal zero (because cos»5 Jj0 ) = 0) and the denominator will equal 1 (because § = 0). And thus the slope of the

Curve at this point is zero. A tangent line with a zero slope is horizontal, and because this tangent line touches the curve where § = 0, the tangent line is the *-axis.

EL For § = *4, find the 1st through 6th derivatives.

Y’= 4×3

Y =12X2

Y ‘ ‘ ‘= 24x y » = 24

YC5) = O yC6) = O Extra credit: §(20055) = 0

EL For § = *5 + 10*3 find the 1st, 2nd, 3rd, and 4th derivatives.

Y ‘ = 5×4 + 30×2

Y ‘ ‘= 20×3 + 60x y ‘ ‘ ‘= 60×2 + 60 yC4) = 120x

EH For § = sin * + cos *, find the 1st through 6th derivatives. Y =cosx sinx y = Sinx cos x y = Cosx+sin x yC4) = Sin x + Cos x yC5) = Cos x - Sin x yC6> = – Sin x - Cos x

Notice that the 4th derivative equals the original function; that the 5th derivative equals the

1st, and so on. This cycle of four functions repeats ad infinitum.

E3 For § = cos *2, find the 1st, 2nd, and 3rd derivatives.

§=cos 2

Y‘ =- 2x sinx2 (Chain rule)

§»= (-2*) ‘(sinx2) + (-2x)(sinx2) (product rule) = -2sin*2- 2*cos(x2) 2x (chain rule) = 2sinx2 4x2cosx2

-2cos (*2) 2* – [(4*2) ‘(cos *2) + (4*2)(cos*2) ‘ |

§

4 cos 2

8* cos * + 4*2 y – sin (*

- – 4* cos *2 – 8* cos *2 + 8*3 sin*2 8×3 sin x2 - 12x cos x2

) 2*)

Chapter 7

The 5th Wave _BV Rich Tennant

"I looked over your equations, Mrs. T)undt. Your concavity and inflection points are clean and there’s nothing A^rong – with your velocity and acceleration. It might be your diЈЈerentiation, but Ivqon’t be able to look at it until Thursday."

In this part…

/n this part, you begin calculus in earnest. Differentiation is the first of the two big ideas in calculus (integration is the second). With all the mystique surrounding calculus, you may think that differentiation is as difficult a subject as string theory or ancient Sanskrit. But Differentiation Is

Simply the fancy calculus term for slope, which, as you

Remember from algebra I, is just the steepness of a line. Whenever possible, remind yourself that every problem involving differentiation is really just a slope problem. In Part III, you get differentiation basics, differentiation rules, techniques for using the derivative to analyze the shape of

Curves, and methods for solving practical problems with the derivative.

Chapter 5

In This Chapter

► The ups and downs of finding slope and rate

► The difference quotient: the other DQ

Power

(ifferentiation is the process of finding Derivatives. The derivative is one of the most important inventions in the history of mathematics and one of mathematics’ most powerful tools. I’m sure you will feel both a deep privilege as you do the practice problems below — and a keen sense of indebtedness to the great mathematicians of the past. Yeah, yeah, yeah.

The Derivative: A Fancu Calculus Word (or Slope and Rate

A derivative of a function tells you how fast the output variable (like y) is changing compared to the input variable (like x). For example, if y is increasing 3 times as fast as X — like with the line y = 3x + 5 — then you say that the derivative of y with respect to x equals 3, and you

Write d§ = 3. This, of course, is the same as d§ = 3, and that means nothing more than

Saying that the rate of change of y compared to x is 3-to-l, or that the line has a slope of

The following problems emphasize the fact that a derivative is basically just a rate or a slope. So to solve these problems, all you have to do is answer the questions as if they had asked

You to determine a rate or a slope instead of a derivative.

Q. What’s the derivative of y = 4x – 5?

A. The answer is 4. You know, of course, that the slope of y = 4x – 5 is 4, right? No? Egad! Any line of the form y = Mx + b has a slope equal to m. I hope that rings a bell. The

Derivative of a line or curve is the same thing as its slope, so the derivative of this line is 4.

You can think of the derivative

As basically Jj^.

Dy

1. If you leave your home at time = 0, and

Go 60 ^±»r in your car, what’s &, the

Derivative of your position with respect to time?

Solve It

2. Using the information from problem 1,

Write a function that gives your position

As a function of time.

Solve It

3. What’s the slope of the parabola

Y = -1 x2 + ^ X – 85 at the point (7, 9)

In the following figure?

Ft. What’s the derivative of the parabola

Y = – x2 + 5 at the point (0, 5)? Hint: Look

Atits graph.

Solve It

Y

X

Solve It

5.

With your graphing calculator, graph both

The line y = – 4x + 9 and the parabola

Y = 5 – x2. You’ll see that they’re tangent at

The point (2, 1).

A. What is the derivative of y = 5 – x2 when

X = 2?

B. On the parabola, how fast is y changing compared to x when x = 2?

Solve It

6. Draw a function containing three points where — for three different reasons — you

Would not be able to determine the slope and, thus, where you would not be able to find a derivative.

Solve It

The Handy-bandy Difference Quotient

The Difference quotient Is the almost magical tool that gives us the slope of a curve at a single point. To make a long story short, here’s what happens when you use the difference quotient. (If you want an excellent version of the long story, check out Calculus For Dummies.) Look again at the figure in problem 3. You can’t get the slope of the

Parabola at (7,9) with the algebra slope formula, ^ m = y 2 – YX 1 J, because no matter

What other point on the parabola you use with (7,9) in the formula, you’ll get a slope

That’s steeper or less steep than the precise slope at (7,9).

But if your second point on the parabola is Extremely Close to (7,9) — like

(7.001,9.0029996) — your line would be almost exactly as steep as the tangent line.

The difference quotient gives the precise slope of the tangent line by sliding the second point closer and closer to (7,9) until its distance from (7,9) is infinitely small.

Enough of this mumbo jumbo; now for the math. Here’s the definition of the derivative

Based on the difference quotient:

(/ x f (x + h) – f (x) f (X) = Lim —-I-—

Like with most limit problems, plugging the arrow-number in at the beginning of a

Difference quotient problem won’t help because that gives you 00. You have to do a

Little algebraic mojo so that you can cancel the H And then plug in. (The techniques

From Chapter 4 also work here.)

Now for a difference quotient problem.

Q. What’s the slope of the parabola

F (x) = 10 – X2 at X = 3?

A. The slope is -6.

1. Because ° (X) = 10 - X2, then ° (X + H) = 10 - (X + Ft)2, and so the derivative is

F (x) = lim-H-

2. Simplify.

10 – (x2 + 2xh + H2

Lim

H – 0

Lim

H – 0

Lim

H – 0

10 + X2

H

10 – X2 - 2xh – H2 - 10 + X2

H

2Xh-h2

3. Factor out ft.

H (-2x – H) = lim-r-1

H-0 H

4. Cancel.

= lim(-2x – H)

H-0

5. Plug in the arrow-number.

= – 2x – 0

F (x) -2X

6. You want the slope or derivative at x = 3, so plug in 3.

F’ (3) =-2 ■ 3

-6

7. Use the difference quotient to determine

The derivative of the line Y 4X-3.

Solve It

8. Use the difference quotient to find the derivative of the parabola F (x) = 3×2.

Solve It

9. Use the difference quotient to find the

Derivative of the parabola from problem 4,

Y = – X2 + 5.

Solve It

10. Figure the derivative of G(x) = JAx-

Using the difference quotient.

Solve It

11. Use the parabola from problem 8, but make

It a position function, S (t) = 3t2, where T Is

In hours and S(t) Is in miles.

A. What’s the average velocity from T = 4 to T = 5?

B. What’s the average velocity from T = 4 to T = 4.1?

C. What’s the average velocity from T = 4 to T = 4.01?

12. For the position function in problem 11, what’s the Instantaneous Velocity at T = 4? Hint: Use the derivative.

Solve It

Solve It

Solutions for Differentiation Basics

D

D

D

If you leave your home at time = 0, and go 60 Mm0^ In your car, what’s Dpp, the derivative of

Your position with respect to time? The answer is = 60. A derivative is always a rate, and

A rate is always a derivative (assuming we’re talking about instantaneous rates). So, if your

Speed, or rate, is 60 Miles, the derivative, Dp, is also 60. One way to think about a derivative Hour dt

Like

Dp

Df Is that it tells you (in this case) how much your position (p) changes when the time (t) increases by one hour. A rate of 60 means that your position changes 60 miles each time

The number of hours of your trip goes up by 1.

Using the information from problem 1, write a function that gives your position as a function of time. P (t) = 60t Or P = 60t, where t is in hours and P Is in miles. If you plug 1 into T, Your position is 60 miles; plug 2 into T And your position is 120 miles. P = 60t is a line, of course, in the form Y = Mx + B (where B = 0). So the slope is 60 and the derivative is thus 60. And again

You see that a derivative is a slope and a rate.

D What’s the slope of the parabola Y = –3 X2 + 23 X - 85 at the point (7, 9)? The slope is 3.

85

23

12, is tangent to the parabola, Y = -3 X2 + ^3. x - ^ at the

You can see that the line, Y 3X

Point (7, 9). You know from Y = Mx + B That the slope of Y = 3x – 12 is 3. At the point (7, 9), the parabola is exactly as steep as the line, so the derivative (that’s the slope) of the parabola

At (7, 9) is also 3.

Although the slope of the line stays constant, the slope of the parabola changes as you climb

Up from (7, 9), getting less and less steep. Even if you go to the right just 0.001 to X = 7.001, the

Slope will no longer be exactly 3.

What’s the derivative of the parabola Y = -X2 + 5 at the point (0, 5)? The answer is 0.

The point (0, 5) is the very top of the parabola, Y = – X2 + 5. At the top, the parabola is neither

Going up nor down — just like you’re neither going up nor down when you’re walking on top ofa hill. The top of the parabola is flat or level in this sense, and thus the slope and derivative

Both equal zero.

The fact that the derivative is zero at the top of a hill — and at the bottom of a valley critically important point which we’ll return to time and time again.

Is a

-4x + 9 and the parabola Y = 5 – X2

With your graphing calculator, graph both the line Y

You’ll see that they’re tangent at the point (2, 1).

A. What is the derivative of Y = 5 – X2 when X = 2? The answer is -4. The derivative of a curve tells you its slope or steepness. Because the line and the parabola are equally steep at (2, 1), and because you know the slope of the line is -4, the slope of the parabola at (2, 1) is also -4

And so is its derivative.

B. On the parabola, how fast is Y Changing compared to X When X = 2? It’s decreasing 4 times as fast as x increases. A derivative is a rate as well as a slope. Because the derivative of the parabola is -4 at (2, 1), that tells you that Y Is changing 4 times as fast as x, but because the 4 is Negative, y decreases 4 times as fast as X increases. This is the rate of Y Compared to X Only for the one instant at (2, 1) — and thus it’s called an Instantaneous Rate. A split second later say at X = 2.000001 — Y Will be decreasing a bit faster.

D Draw a function containing three points where — for three different reasons — you would not be able to determine the slope and thus where you would not be able to find a derivative.

Your sketch should contain (1) Any type of gap or discontinuity. There’s no slope and thus no derivative at a gap because you can’t draw a tangent line at a gap (try it). (2) A sharp point or cusp. It’s impossible to draw a tangent line at a cusp because a line touching the function at such a sharp point could rock back and forth. So there’s no slope and no derivative at a cusp. (3) A vertical inflection point. Although you can draw a tangent line at a vertical inflection point, since it’s vertical, its slope — and therefore its derivative — is undefined.

D Use the difference quotient to determine the derivative of the line y = 4x – 3. y’ = 4.

Y ‘= lim

4 (x + h) - 3 – (4x – 3)

4x + 4h – 3 – 4x + 3

Lim

H – 0

H – o h lim4

H – 0

Y’=4

You can also figure this out because the slope of y=4x-3is 4.

D Use the difference quotient to find the derivative of the parabola °(x) = 3×2. F’(x) = 6X

° (x) = lim ——

H-0 h

3 ( X2 + 2xh + h2

3×2

H

3×2 + 6xh + 3h2 – 3×2

=lim

H-0

= lim

H-0 h

= lim6xh + 3h 2 (Now, factor out the h)

H-0h

= him h (Cancelthe h)

=l i-m0(6x+3 )(Nowplugin 0) = 6x + 3 ■ 0 °’ (x) = 6x

Use the difference quotient to find the derivative of the parabola from problem 4, y = – x2 + 5. y’ = -2X.

Y =lim

Lim

Lim

(x+ )2+5

(x2 + 2xh + h2

5 + x2 – 5

-x2 – 2xh – h2 + 5 + x2 – 5

H

Lim – 2x– h2 (Now factor)

-0

(-2x – ) lim-1 (And cancel)

=lim(-2x – )

-0

Y =-2x

In problem 4, you see that the top of this parabola (y = – x2 + 5) is at the point (0,5) and that the derivative is zero there because the parabola is neither going up nor down at its peak. That explanation was based on common sense. But now, with the result given by the difference quotient, namely y ‘=- 2x, you have a rigorous confirmation of the derivative’s value at (0,5). Just plug 0 in for x in y’=-2x, and you get y ‘= 0.

E9 Figure the derivative of g (x) = /4x + 5 using the difference quotient. G’ (x) = – t-2-

If you got this one, give yourself a pat on the back. It’s a bit tricky.

G (x) = /4x + 5

./4 (x + h) + 5 – /4x + 5 g (x) = him? h

Lim

Lim

/4x + 4h + 5 – /4x + 5 H

(/4x + 4h + 5 -/4x + 5) (/4x + 4h + 5 + /4x + 5)

Lim

Ft

H

(4x + 4h + 5) – (4x + 5)

(Conjugate multiplication)

Lim

0 h (/4x + 4h + 5 + /4x + 5

4

(/4x + 4h + 5 + /4x + 5) – (Because (a – b)(a + B) = a2- B2, of course)

0 h (/4x + 4h + 5 + /4x + 5)

Lim

4

/4x + 4h + 5 + /4x + 5

(Nowyoucanplugin)

=

4

—/

/4x + 4 ■ 0 + 5

4

4x

=

G (x)

-/

2/4x + 5 2

/4x + 5

Use the parabola from problem 8, but make it a position function, s (-) = 3-2, where – is in hours and s(-) is in miles.

Average velocity equals to-a! ^’f-ance.

A. What’s the average velocity from – = 4 to – = 5? The answer is 27 Miles/hour.

S (5) – s (4)

5 – 4 3 ■ 52 – 3 ■ 42

:27

1

Mi/es Hour

B. What’s the average velocity from – = 4 to – = 4.1? The answer is 24.3 Miles/hour.

S(4.1)- s(4)

Auerage Ve/oc//y4

4.1- 4

3 ■ 4.12 – 3 ■ 42

24. 3

0.1

Mi/es Hour

-0

-0

H

-0

C. What’s the average velocity from – = 4 to – = 4.01? The answer is 24.03 Miles/hour.

S (4.01) – s (4)

Average Ve/ocify

4to 4.01 =

4.01- 4

3 ■ 4.012 – 3 ■ 42

24. 03

0.01

Mi/es Hour

For the position function in problem 11, what’s the instantaneous velocity at – = 4? The answer is 24 Miles/hour. Problem 8 gives you the derivative of this parabola, °’ (x) = 6x. The position function in this problem is the same except for different variables, so its derivative is s’ (-) = 6-.

Plug in 4, and you get s’ (4) = 24 . Notice how the average velocities get closer and closer

To 24 7^ as the total travel time gets less and less and the ending time hones in on – = 4.

That’s precisely how the difference quotient works as h shrinks to zero.

Rhis chapter gives you lots of practice using the derivative to analyze the shape of curves and their significant features and points. Don’t forget: The derivative tells you the slope of a curve, so any problem involving anything about the slope or steepness of a curve is a derivative problem.

The First Derivative Test and Local Extrema

One of the most common applications of the derivative employs the simple idea that at the top of a hill or at the bottom of a valley, you’re neither going up nor down; in other words, there’s no steepness and the slope — and thus the derivative — equals zero. You can therefore use the derivative to locate the top of "hills" and the bottom of "valleys," called Local extrema, On just about any function by setting the derivative of the function equal to zero

And solving.

In This Chapter

* Mum’s the word: Minimum, maximum, extremum Concavity and inflection points

► The nasty mean value theorem

KJ Q. Use the first derivative test to determine & the location of the local extrema of

G (x) = 15×3 – x 5. See the following figure.

A. The local min is at (-3, -162), and the local max is at (3, 162).

1. Find the first derivative of g using the power rule.

G(x)=15X3-x5

G’ (x) = 45×2 – 5×4

2. Set the derivative equal to zero and solve for x to get the critical numbers

Of G.

45X2-5X4=0

5×2 (9 – x2) = 0 5×2(3 – x)(3 + x) = 0

5×2 = 0 or 3 – x = 0 or 3 + x = 0 X=0,3,or -3

If the first derivative were undefined for

Some X-values in the domain of G, There could be more critical numbers, but

Because g’ (x) = 45×2 – 5×4 is defined for all real numbers, 0, 3, -3 is the complete list of critical numbers of g.

Remember: If °is defined at a number c and the derivative at x = c is either zero

Or undefined, then c is a critical number

Of F.

3. Plot the three critical numbers on a number line, noting that they create four regions (see the figure in Step 5).

4. Test a number from each of the four

Regions, noting whether the results are positive or negative.

Note that if you use round numbers like

0, -10, or 10, you can often do the arithmetic in your head.

G’ (-10) = 45 (-10)2 – 5 (-10)4 =-45,500 G (-1) = 45 (-1)2 – 5 (-1)4 = 40 g’ (1) = 45 (1)2 -5 (1)4 = 40

G’(10)=45(10)2-5(10)4=-45,500

5. Draw a "sign graph." Take your number line and label each region — based on your results from Step 4 — positive

(increasing) or negative (decreasing). See the following figure.

Decreasing increasing increasing decreasing

— + + —

-1-1-1-*~

-3 0 3

This sign graph tells you where the function is rising or increasing and where it is falling or decreasing.

6. Use the sign graph to determine whether there’s a local minimum, local

Maximum, or neither at each critical number.

Because g goes down on its way to x = —3 and up after x = —3, it must bottom out at x = —3, so there’s a local min there. Conversely, g peaks at x = 3 because it rises until x = 3, then falls. There is thus a local max at x = 3. And because g climbs on its way to x = 0 and then climbs further, there is neither a min nor a max at

X = 0.

7. Determine the y-values of the local extrema by plugging the x-values into

The original function.

G (-3) = 15 (-3)3 – (-3)5 = -162

G (3) = 15 (3)3 – (3)5

=162

So the local min is at (—3, —162), and the local max is at (3, 162).

1. Use the first derivative to find the local extrema of f (x) = 6×2/3 – 4x + 1. Tip: You

Better write small if you want to do this problem on half a page.

Solve It

2. Find the local extrema of

H (x) = -4= + cosx – in the interval

0, 2Ji With the first derivative test.

Solve It

3. Locate the local extrema of § = (x2

With the first derivative test.

Solve It

I. Using the first derivative test, determine

The local extrema of s =

T4 + 4 - 2t2 .

Solve It

The Second Derivative Test and Local Extrema

With the second derivative test, you use — can you guess? — the Second Derivative to test for local extrema. The second derivative test is based on the absolutely brilliant idea that the crest of a hill has a hump shape (fl) and the bottom of a valley has a trough shape (U).

After you find a function’s critical numbers, you’ve got to decide whether to use the first or the second derivative test to find the extrema. For some functions, the second

Derivative test is the easier of the two because 1) The second derivative is usually

Easy to get, 2) You can often plug the critical numbers into the second derivative and do a quick computation, and 3) you will often get non-zero results and thus get your answers without having to do a sign graph and test regions. On the other hand, testing

Regions on a sign graph (the first derivative test) is also fairly quick and easy, and if the

Second derivative test fails (see the warning), you’ll have to do that anyway. As you do practice problems, you’ll get a feel for when to use each test.

When the second derivative equals zero, the second derivative test fails and you learn nothing about local extrema. You’ve then got to use the first derivative test to determine whether there’s a local Extremum (the singular of extrema) there.

Take the function from the example in the

Previous section, g (x) = 15×3 – x5, but this

Time find its local extrema using the second derivative test.

A. The local min is at X

Max is at X = 3.

-3 and the local

You’ve probably figured out that to use

The second derivative test you need the second derivative:

G (x) = 15×3 – x5 g’ (x) = 45×2 – 5×4 g" (x) = 90x – 20×3

Now all you do is plug in the critical

Numbers of g from the example in the

Last section:

G’ (-3) = 270 g’ (0) = 0 g’ (3)=-270

The fact that g" (-3) is positive tells you that g is concave up (U) there, and thus

That there’s a local min. And the fact that

G" (3) is negative tells you that g is concave do©n (If) at x = 3, and, therefore, that there’s a local max there. And, while it may seem that g" (0) = 0 confirms what you figured out previously (that there’s

Neither a min nor a max there), you actually learn nothing when the second derivative is zero; you have to use the first derivative test.

Tip: If you only have one critical point between a local min and a local max (and no discontinuities), it has to be an inflection point, and if you have a single critical number between two known maxes (see problem

7), the only possibility for the middle critical number is a local min (and vice-versa). So in these cases, it really doesn’t matter if the second derivative test fails with the middle critical number. If this not-by-the -

Book reasoning doesn’t work for your calc

Teacher, you might say (with just a touch of sarcasm in your voice), "Oh, so in other words, you’ve got something against logic and common sense."

5. Use the second derivative test to analyze the critical numbers of the function from

Problem 2, h (x) = -4- + cosx – ^.

Solve It

J2

0. Find the local extrema of

F (x) = – 2×3 + 6×2 + 1 with the

Second derivative test.

Solve It

7. Find the local extrema of § = 2×4 – 3.

With the second derivative test. 3

Solve It

8. Consider the function from problem 3,

, and the function 7t Which of the two func -

§ = (x2 – 21t

S = 8 + 24L 4 tions is easier to analyze with the second

Derivative test, and why? For the function you pick, use the second derivative test to find its local extrema.

Solve It

Finding Mount Everest: Absolute Extrema

The basic idea in this section is quite simple. Instead of finding all local extrema like in the previous sections (all the peaks and all the valleys), you just want to determine the single highest point and single lowest point along a Continuous Function in some Closed Interval. These Absolute extrema Can occur at a peak or valley or at an edge(s) of the interval. (Note: You could have, say, two peaks at the same height so there’d be a tie for the absolute max; but there would still be exactly one Y-value That’s the absolute

Maximum value on the interval.)

Before you practice with some problems, look at Figure 7-1 to see two standard absolute extrema problems (continuous Functions on a Closed Interval) and at Figure 7-2 for four strange functions that don’t have the standard single absolute max and single

Absolute min.

Figure 7-1:

Two standard absolute extrema functions.

Abs max

—i—i—i—h

12 3 4

Mbax

Abs

Min

-I-1-1-h -

1234

Open interval closed interval

Open interval closed

No abs max f

1 2 3 4 1 2 3 4

Asymptote interval

Figure 7-2:

Fournon -

Standard absolute extrema functions.

Mabasx

No abs max or min

Hole

\-1-1-1-*-x

1 2 3 4 1 1 2 3 4

Closed intervalopen interval

Y

X

X

No abs max

Y

Y

X

X

No abs min

Y

Y

X

0. A.

Determine the absolute min and absolute max of F (x) = /\x\-x In the interval [-1,

The absolute max is 2 and the absolute min is 0.

1. Get all the critical numbers:

First, set the derivative equal to zero

(note, first split this function in two to get rid of the absolute value bars):

F (x) = /x - x (x > 0)

F' (x)-

0

1

2/x

1

2/x

2/x = 1

F(x)

F'(x)-

0

2/-x ■■

X - x (x < 0)

1

2 J - x

1

1

~2/-x 1

=-1 Nosolution

Now, determine whether the derivative is undefined anywhere:

The derivative is undefined at X = 0

Because the denominator of the derivative can't equal zero. (If you graph this function (always a good idea) you'll also

See the cusp at X = 0 and thus know immediately that there's no derivative

There.)

The critical numbers are therefore 0

And K

2. Compute the function values (the heights) at all the critical numbers.

F(0)=0

It's just a coincidence, by the way, that in

Both cases the argument equals the

Answer.

3. Compute the function values at the two

Edges of the interval.

F(-1)=2

F (2

/2 -1

I 2

0.207

4. The highest of all the function values from Steps 2 and 3 is the absolute max; the lowest of all the values from Steps 2

And 3 is the absolute min.

Thus, 2 is the absolute max and 0 is the absolute min.

Note that finding absolute extrema involves less work than finding local extrema because you don't have to use

The first or second derivative tests — do you see why?

9. Find the absolute extrema of

F(x) = Sinx+ cosx on the interval [0, 2n\.

Solve It

10. Find the absolute extrema of

G (x) = 2x3 - 3x2 - 5 on the interval [- .5,.5]

Solve It

11. Find the absolute extrema of

° (x) = (x + 1) .5x on the interval

[-2, 31].

Solve It

12. Find the absolute extrema of

Q(x) = 2cosX + 4sinX On the interval N 5Ji

24

Solve It

Smiles and FroUns: Concavity and InFLection Points

Another purpose of the second derivative is to analyze concavity and points of inflection. (For a refresher, look at Figure 7-3: The area between A and B is concave down — like an upside-down spoon or a frown; the areas on the outsides of A and B are concave up — right-side up spoon or a smile; and A and B are inflection points.) A positive

Second derivative means concave up; a negative second derivative means concave

Down. Where the concavity switches from up to down or down to up, you’ve got an

Inflection point, and the second derivative there will be zero (or sometimes undefined).

Figure 7-3:

Concavity and points of inflection.

ANG/

All inflection points have a second derivative of zero (if the second derivative exists), but not all points with a second derivative of zero are inflection points. This is no different from "all ships are boats but not all boats are ships."

However, you can have an inflection point where the second derivative is undefined.

This occurs when the inflection point has a vertical tangent and in some bizarre

Curves that you shouldn’t worry about that have a weird discontinuity in the second

Derivative.

Q. Find the intervals of concavity and the inflection points of F (x) = 3x5 – 5×3 + 10.

Note that the following solution is analogous to the solution for finding local extrema with the first derivative.

A. f Is concave down from - oo To the inflec-

Tion point at

4′ ~11.24

; concave

Upfrom there to the inflection point at

(0, 10); concave down from there to the third inflection point at ~ 8.76 and concave up from there to Oo.

1. Find the second derivative of f.

F (x) = 3x5 – 5×3 + 10 F’ (x) = 15×4 – 15×2 F" (x) = 60×3 – 30x

2. Set the second derivative equal to zero

And solve.

60×3 – 30x = 0 30x (2×2 – 1) = 0

2X2-1=0

30X=0 X=0

Or

2X2

X2

H

2

3. Check whether there are any x-values

Where the second derivative is undefined.

0, and ^ are

There are none, so – ^

The three second derivative "critical

Numbers." (Technically these aren’t called critical numbers, but they could be because they work just like first

Derivative critical numbers.)

4. Plot these "critical numbers" on a

Number line and test the regions.

You can use -1, -.5, .5, and 1 as test

Numbers. The following figure shows thesecond derivative sign graph.

F" (x) = 60×3 – 30x

F"(-1) F"(-.5) F"(.5) F"(1)

Concave down

-30 7.5 -7.5 30

Concave up

Concave

Down

Concave up

-V2

2

V2

2

Because the concavity switches at all

These "critical numbers" and because the second derivative exists at those

Numbers (from Steps 2 and 3), there are

Inflection points at those three X-values.

(If the concavity switches at a point

Where the second derivative is undefined, you have to check one more thing before concluding that you’ve got an inflection point: whether you can draw a tangent line there. This is the case when the first derivative is defined or if there’s

A vertical tangent.) In a nutshell, if the concavity switches at a point where the

Curve is smooth, you’ve got an inflection point.

5. Determine the location of the three

Inflection points.

F (x) = 3×5 – 5×3 + 10 F2 ^

11.24

F

F(0)=10 72

8.76

So F Is concave down from -o To the inflection point at

~ 11.24

Con -

Cave up from there to the inflectio n point

At (0, 10), concave down from there to

The third inflection point at ~ 8.76 and, finally, concave up from there to oo.

0

X

13. Find the intervals of concavity

And the inflection points of

F (x) = – 2x3 + 6×2 – 10x + 5.

Solve It

1}. Find the intervals of concavity and the inflection points of G (x) = x4 – 12×2.

Solve It

15. Find the intervals of concavity and the inflection points of ° (X) = "

-9"

Solve It

16. Find the intervals of concavity and the inflection points of Q (x) = 5/x – 3Jx.

Solve It

The Mean Value Theorem: Go Ahead, Make M§ Dai;

The Mean Value Theorem Is based on an incredibly simple idea. Say you go for a one hour drive and travel 50 miles. Your average speed, of course, would be 50 mph. The Mean Value Theorem says that there must be at least one point during your trip when your speed was exactly 50 mph. But you don’t need a fancy-pants calculus theorem to tell you that. It’s just common sense. If you went slower than 50 mph the whole way, you couldn’t average 50. And if you went faster than 50 the whole way (this assumes you’re going faster than 50 at your starting point), your average speed would be

Greater than 50. The only way to average 50 is to go exactly 50 the whole way or to goslower than 50 some times and faster than 50 at other times. And when you speed up or slow down, you have to pass exactly 50 at some point.

With the Mean Value Theorem, you figure an average rate or slope over an interval and

Then use the first derivative to find one or more points in the interval where the instantaneous rate or slope equals the average rate or slope. Here’s an example:

Q. Given f (x) = X3 – 4×2 – 5x, find all numbers c in the open interval (2, 4) where the

Instantaneous rate equals the average rate over the interval.

A. The only answer is 4 + .

Basically, you’re finding the points along the curve in the interval where the slope

Is the same as the slope from (2, f (2)) to (4, f (4)). Mathematically speaking,

Find all numbers C Where

F’ (c) =

F(4) – f(2) 4 – 2 .

1. Get the first derivative.

F (x) = x3 – 4×2 – 5x f’ (x) = 3×2 – 8x – 5

2. Using the slope formula, m = Jff-§7, figure the slope from (2, ° (2)) to

(4, ° (4)).

F (4) = 43 – 4. 42 – 5 . 4

=-20

F (2) = 23 – 4. 22 – 5 . 2 = -18

F(4)-f(2)

M

4-2 20-(-18)

3. Set the derivative equal to this slope

And solve.

3×2 – 8x – 5 =-1 3×2 – 8x – 4 = 0

_ 8 +/(-8)2 – 4 (3)(-4)

6

8 + 4/7

4 + 2/7 4- 2/7 3 or 3

3.10or

.43

Because – .43 is outside the interval

4 + 2/7

(2, 4), your only answer is

3

By the way, the Mean Value Theorem only works for functions that are differ-entiable over the open interval in question and continuous over the open interval and its endpoints.

_

17. For G (x) = x3 + x2 – x, find all the values c in the interval (-2, 1) that satisfy the Mean

Value Theorem.

Solve It

18. For s (-) = -4/3 – 3-1/3, find all the values c in

The interval (0, 3) that satisfy the Mean Value Theorem.

Solve It

Solutions for derivatives and Shapes of Curves

MM Use the first derivative to find the local extrema of f (x) = 6×2/3 – Ax + 1. Local min at (0, 1); local max at (1, 3).

1. Find the first derivative using the power rule.

F (x) = 6×2/3 – 4x + 1 f’ (x) = 4x-1/3 – 4

2. Find the critical numbers of f

A. Set the derivative equal to zero and solve.

4x-1/3 – 4 = 0 x-1/3=1 x=1

B. Determine the x-values where the derivative is undefined: f’ (x) = 4x 1/3 – 4 = 4- - 4

3x

Because the denominator is not allowed to equal zero, f’ (x) is undefined at x = 0. Thus the critical numbers of fare 0 and 1.

3. Plot the critical numbers on a number line.

I’m going to skip the figure this time because I assume you can imagine a number line with dots at 0 and 1. Don’t disappoint me!

4. Plug a number from each of the three regions into the derivative.

F’ (-1) = 4 (-1)-1/3- 4 =-4 – 4 =-8

-1/3

F’ (1) = 4 (2) – 4 = 4 (2)1/3- 4 = Positive

22

F’(8)=4(8)-1/3-4=2-4=-2

Note, again, how the numbers I picked for the first and third computations made the math easy. With the second computation, you can save a little time and skip the final calculation because all you care about is whether the result is positive or negative — this assumes that you know that the cube root of 2 is more than 1 (you better!).

5. Draw your sign graph (see the following figure).

Decreasing increasing decreasing -*-1-1-*■

0 1

6. Determine whether there’s a local min or max or neither at each critical number.

F Goes down to x = 0 and then up, so there’s a local min at x = 0, and F Goes up to x = 1 and then down, so there’s a local max at x = 1.

7. Figure the y-value of the two local extrema.

F (0) = 6 (0)2/3- 4 (0) + 1 = 1 f (1) = 6 (1)2/3 – 4 (1) + 1 = 3 Thus, there’s a local min at (0, 1) and a local max at (1, 3). Check this answer by looking at a

Graph of fon your graphing calculator.

D Find the local extrema of h (x) = -4- + cosx – in the interval (0, 2k ) with the first derivative ‘* ^\local2minatff,^-/2l.

Test. Local max at

4′ 8

1. Find the first derivative.

X ______ /2

H(x) h’(x)

J2 /2

Cosx -

2. Find the critical numbers of h.

A. Set the derivative equal to zero and solve:

-X – Sinx = 0

/2 sinx 0

Sinx

Ft 2

Or ^ (These are the solutions in the given interval.^

B. Determine the x-values where the derivative is undefined.

The derivative isn’t undefined anywhere, so the critical numbers of H Are J

And

3n

3. Test numbers from each region on your number line.

H’‘6) y2 sin6 h’(f)^-sinf

Ft -1

" 2 2 positive

2 1

= negative

4. Draw a sign graph (see the following figure).

Hh (k) = ^ – sinn

= ^-0 2 0

=positive

Increasing decreasing

Increasing

7t 37t

27t

5. Decide whether there’s a local min, max, or neither at each of the two critical numbers.

Going from left to right along the function, you go up until x = -| and then down, so there’s

Local max at x = 4- It’s vice-versa for x =

3k

4. it s vice-versa for x = —

6. Compute the y-values of these two extrema.

So there’s a local min there.

K_

Ft

Cos I

4/2

KJ2

Ft

2

Ft. ft 2 2

So you’ve got a max at

7t 7t/2

4′ 8

And a min at

3k,-

J2 + COS 4 2

3kJ2 ft ft 8 2 2

—g— v2

3k 3kJ2 k — —— ~ /2

0

A

H

H

D Locate the local extrema of y = (x2 -And (2 Ft, 0); a local max at (0, 4)

With the first derivative test. Local mins at (- 2/2, 0)

Same basic steps as problems 1 and 2, but abbreviated a bit.

1. Find the derivative.

2/3

Y=( x

Y’ = 3(x

(2x)

4x

33/ x 2-8

2. Find the critical numbers.

—TT= = 0

A. 33x 8

X=0

B. The first derivative will be undefined when the denominator is zero, so

33/ x 2-8 = 0 3/ x 2-8 = 0 x2 8=0 x2=8

X = + 2/2

The critical numbers are -2/2, 0, and 2/2.

3. Test the values.

Y’ (-10) = f((-10)-8) (2 ■ (-10))

2 1/3 = 3(Positive) ^ negative

= 3 Positive ^ negative =negative

Y’ (-1) = § ((-1)2-8 )l\ 2 ■ (-1))

2 1/3 = 3 ^ (negative) ^ negative

= ^ Negative ^ negative = Positive

Y (1) = Negative And y’ (10) = Positive (What? I should do all your work?)

4. Make a sign graph (see the following figure).

Decreasing increasing decreasing increasing

-2V2 0 2^2

5. Find the Y-values.

22/3

Y = ((-2/2) -8) = 0 There’s a local min at (-2/2, 0).

Y = (02 – 8)2/3 = (-8)2/3 = 4 There’s a local max at (0, 4).

22/3

Y = ((2/2) – = 0 There’s another local min at (2 Ft, 0). Check out this interesting curve

On your graphing calculator.

T4 + 4

D Using the first derivative test, determine the local extrema of S = Hpf-. Local maxes at -72,-2) and (/2,-2).

1. Do the differentiation thing.

T4+4 2T2

‘ = (t4 + 4)’(-2t 2)-( T4 + 4)(-2t2)’ = (4t3)(-2t2)-(T4 + 4)(-4t) = -T 4 + 4 S= (-2t2)2 = 4t4 = T3

2. Find the critical numbers.

-t4+4 = 0

4 T4 =0 (2 T2)(2+t2)=0 (/2-1 )(/2 +1)(2 +12) = 0

T =/2 or -/2

So – J2 And J2 Are two critical numbers of S.

T = 0 is a third important number because T = 0 makes the derivative’s denominator equal zero, and so you need to include zero on your sign graph in order to define test regions. Note, however, that T = 0 is Not A critical number of S Because S Is undefined at T = 0. And because there is no point on S At T = 0, there cannot be a local extremum at T = 0.

3. Test values: You’re on your own.

4. Make a sign graph (see the following figure).

Increasing decreasing increasing decreasing

-V2

V2

She loves me; she loves me not; she loves me; she loves me not. 5. Find the y-values.

-J2 )4 + 4

72) =

-2 (-J2 )

Local max there.

^i,4- = -2 You climb up the hill to (- /2, -2), then down, so there’s

S(0)

04

2(0)

Undefined (which you already knew). Therefore, there’s no local extremum

At T = 0. Remember that if a problem asks you to identify only the x-values and not the

Y-values of the local extrema, and you only consider the sign graph, you would incorrectly

Conclude — using the current problem as an example — that there’s a local min at T = 0.

Soyou should always check where your function is undefined.

4

J2 + 4

-2/2

-2. Up then down again, so there’s another local max at (J2, -2).

As always, you should check out this function on your graphing calculator.

0

A

D Use the second derivative test to analyze the critical numbers of the function from problem 2,

D

H (x ) =

CosX

2 2

1. Find the second derivative.

^ Local max at x = -; local min at X –

J2

~2

H(x) h ( x)

H"(x)=

72 +cos x-12

-4- – sinx

J2

CosX

2. Plug in the critical numbers (from problem 2).

-cos |

-cos-^p -

J2

You’re done. H Is concave down at X J2

/2

At X=

2~, so there’s a local max there, and H Is concave up

So there’s a local min at that X-value.

(In problem 2, you already determined the y-values for these extrema.) H Is an example of a

Function where the second derivative test is quick and easy.

Find the local extrema of F (X) = -2×3 + 6×2 + 1 with the second derivative test. Local min at (0, 1); local max at (2, 9).

1. Find the critical numbers.

F (X) = -2×3 + 6×2 + 1 F’ (X) = -6×2 + 12x 0 = -6×2 + 12x

0= 6X(x 2) X=0, 2

2. Find the second derivative.

F’ (X) = -6×2 + 12x F"(X)= -12x+12

3. Plug in the critical numbers.

F" (0) = -12 (0) + 12 f" (2):

=12 (concaveup: min)

4. Determine the ycoordinates for the extrema.

F (0) = -2 (0)3 + 6 (0)2 + 1 F (2) = -2 (2)3 + 6 (2)2 + 1 = 1 = 9

So there’s a min at (0, 1) and a max at (2, 9). F Is another function where the second derivative

Test works like a charm.

12(2)+12

12(concavedown: max)

_

D Find the local extrema of y = 2×4 – 1 X6 with the second derivative test. You find local maxes at

X = -2 and X = 2 with the second derivative test; you find a local min at X = 0 with street smarts.

1. Find the critical numbers.

Y = 2×4 – 3 x6

Y’ = 8×3 – 2×5 8×3 – 2×5 = 0 2×3 (4 – x2) = 0 2×3(2 – x)(2 + x) = 0

X = 0, 2, -2

2. Get the second derivative.

Y’ = 8×3 – 2×5 y’ = 24×2 – 10×4

3. Plug in.

Y" (-2) = 24 (-2)2- 10 (-2)4 Y" (0) = 24(0)2 – 10(0)4 Y" (2) = 24 (2)2- 10 (2)4 = 96 – 160 = 0 thus inconclusive = same as y" (-2)

= Negative Thusa max = Negative Thus a max

The second derivative test fails at x = 0, so you’ve got to use the first derivative test for that critical number. And this means, basically, that the second derivative test was a waste of time

For this function.

If — as in the function for this problem — one of the critical numbers is x = 0, and you can see that the second derivative will equal zero at x = 0 (because, for example, all the terms of the second derivative will be simple powers of x), then the second derivative test will fail for x = 0 and thus, will likely be a waste of time. Use the first derivative test instead.

However, because this is a continuous function and because there’s only one critical number between the two maxes you found, the only possibility is that there’s a min at x = 0. Try this

Streetwise logic out on your teacher and let me know if it works.

D Consider the function from problem 3, y = (x2 – 8F \ and the function s = 8 + 21t – ^. Which is easier to analyze with the second derivative test and why? For the function you pick, use the

Second derivative test to find its local extrema. S = 8 + 2|- - ^; local min at (-1, 4.5) and a local max at (-1, 11.5). 4 4

The second derivative test fails where the second derivative is undefined (in addition to failing where the second derivative equals zero).

To pick, look at the first derivative of each function:

Y = (x2 – 8)2/3 s = 8 + 21t – f

Y’ = 2(x2 – 8)-1’3(2x) s’ = 24i – 2112 = 4x 3 (x2 – 8 F

Do you see the trouble you’re going to run into with y(x)? The first derivative is undefined at x – + 2/2. And the second derivative will also be undefined at those x-values, because when you take the second derivative with the quotient rule, squaring the bottom, the denominator will contain that same factor, (x2 - 8). The second derivative test will thus fail at + 2/2, and you’ll have to use the first derivative test. In contrast to y(x), the second derivative test works great with s(-):

1. Get the critical numbers.

S’ – 21 – 21 . 2 S 4 4 -

0-T – T -2

21 ,2_ 21

T – - T +1

S’ is not undefined anywhere, so -1 and 1 are the only critical numbers.

2. Do the second derivative.

S’-21 – 21 .2 S 4 4 -

S— 21 -

3. Plug in the critical numbers.

21

2

S – (1) – -22- (concave down: max)

S – (-1)- 221 (concave up: min)

21

2

4. Get the heights of the extrema.

SU 8 + 21(-1)-4.

S ( 1) 8 + 4 4 4.3

S (1)-8 + - 7Jf – 11.5

You’re done. s has a local min at (-1, 4.5) and a local max at (-1, 11.5).

Find the absolute extrema of F (x) = sinx+ cosx on the interval [0, 2Ji]. Absolute max at ■|, /2~); Absolute min at -Jl

1. Find critical numbers. F (x) – Sinx + cosx F’ (x) – Cosx – sinx 0 – cosx – sinx sinx – cosx (divide both sides by cosx) tanx 1

X= \’ ^ (tne solutions in the given interval)

The derivative is never undefined, so these are the only critical numbers.

If you divide both sides of an equation by something that can equal zero at one or more x-values (like you do above when dividing both sides by cos x), you may miss one or more solutions. You

3Ji T

Have to check whether any of those x-values is a solution. In this problem, cosx = 0 at -| and

And it’s easy to check (in line 4 above) that sinx does not equal cosx at either of those values. If it did, you’d have one or two more solutions and one or two more critical numbers. Note that you have to check these values before dividing out the "dividing thing" (cosx here).

2. Evaluate the function at the critical numbers.

F(f) = sinf + cosf,(^ = sin^ + cos5K

-/2 Y2 - _J2_ /J2

- 2 + 2 - 2 2

-/2 - -J2

3. Evaluate the function at the endpoints of the interval.

F (0) sin0+cos0 1

F( 2n ) = Sin 2n + Cos 2n = 1

4. The largest of the four answers from Steps 2 and 3 is the absolute max; the smallest is the

Absolute min.

The absolute max is at (4, /2). The absolute min is at (54-, -J2 J.

Find the absolute extrema of g (x) – 2×3 – 3×2- 5 on the interval [-.5,.5]. Absolute min at (-.5, -6); absolute max at (0, -5).

1. Find critical numbers.

G (x) – 2×3 – 3×2 – 5 g’ (x) – 6×2 – 6x

0 – 6×2 – 6x

0 – 6x (x – 1)

X 0, 1

X = 1 is neglected because it’s outside the given interval; x = 0 is your only critical number.

2. Evaluate the function at x = 0.

G (0) – 2 (0)3 – 3 (0)2 – 5 –5

3. Do the endpoint thing.

G (-.5) – 2 (-.5)3- 3 (-.5)2- 5 – 2 ■ (-.125) – 3 ■ .25 – 5

-6

G (.5) – 2 ■ .53 – 3 ■ .52 – 5

- 2 ■ .125 – 3 ■ .25 – 5

-5.5

4. Pick the smallest and largest answers from Steps 2 and 3.

The absolute min is at the left endpoint, (-.5, -6). The absolute max is smack dab in the

Middle, (0, -5).

10 Find the absolute extrema of ° (x) – (x + 1)4’5 – .5x on the interval [-2, 31]. Absolute max at (-2, 2); absolute mins at (-1, .5) and (31, .5).

I think you know the steps by now.

° (x) – (x + 1)4’5 – .5x °’ (x) – 5(x + 1)-1’5 – .5

- 4 5 0 – » „5 – .5

.5

2.5 (x + 1)1’5 – 4

X – 9.48576

That’s one critical number, but x = -1 is also one because it produces an undefined derivative.

° (-1) – (-1 + 1)4’5- .5 (-1)

.5

° (9.48576) – (9.48576 + 1)4’5 – .5 (9.48576) ^ 1.81072

Left endpoint: ° (-2) – (-2 + 1)4’5 – .5 (-2) – 2

Right endpoint: ° (31) – (31 + 1)"’5 – .5 (31) – 16 – 15.5 – .5

Your absolute max is at the left endpoint: (-2, 2). There’s a tie for the absolute min: At the cusp:

(-1, .5) and at the right endpoint: (31, .5).

71 5Tc

H Find the absolute extrema of q(x) – 2cos2x + 4sinx on the interval ~^)’ a",so’u*e maxes at 3 J And 3

Q(x) – 2cos2x + 4sinx q’ (x) – -2sin2x^ 2 + 4cosx 0 -4sin2x+4cosx 0 – sin2x – cosx (dividingby -4) 0 2sinxcosx-cosx(trigidentity) 0 cosx(2sinx-1)

2sinx-1 0

Cosx 0

-n re or sinx=i

6 ’6

24

Absolute min at

^jtAEH Technically x = is not one of the critical numbers; being at an endpoint, it is refused raem -

^^TX Bership in the critical number club. It’s a moot point though, because you have to evaluate the JIM I endpoints anyway.

6

Left endpoint: Q[-± 5Tc

Right endpoint: ? I ^) = 2 cos I 2 • ^ I + 4 sin ^

5Tc

5Tc

2 ■ 0 + 4

2 + 4 (-1)

Ft 2

-2.828

Pick your winners: Absolute min at left endpoint: ( --, -6 ) and a tie for absolute max: ( 3 and ^ 3

Find the intervals of concavity and the inflection points of F (x) = - 2X3 + 6x2 - 10x + 5. /"is concave up from - oo To the inflection point at (1, -I), Then concave down from there to Oo.

1. Get the second derivative.

F (x) = - 2x3 + 6x2 - 10x + 5 f' (x) = - 6x2 + 12x - 10 f'(x)= -12x+12

2. Set equal to 0 and solve.

-12x + 12 = 0 x = 1

3. Check for x-values where the second derivative is undefined. None.

■ to the left and to the right of x = 1 — and make your sign graph (see

4. Test your two regions the following figure).

F' (x )= -12x + 12

F' (0) = 12 f' (2) =-12

— i

Concave up

Concave down

Q

Q

1

Because the concavity switches at x = 1 and because f" equals zero there, there's an inflection point at x = 1.

5. Find the height of the inflection point.

F (x ) = - 2x3 + 6x2 - 10x + 5 F (1)=-1

Thus F Is concave up from - oo to the inflection point at (1, -1), then concave down from there to oo. As always, you should check your result on your graphing calculator. Hint: To get a good feel for the look of this function, you'll need a fairly odd graphing window — try something like Xmin = -2, Xmax = 4, Ymin = -20, Ymax = 20.

EL Find the intervals of concavity and the inflection points of G (x) = x4 - 12x2. G Is concave up from - oo To the inflection point at (-JL, -20); then concave down to an inflection point at (/2, -20); then concave up again to Oo.

1. Find the second derivative.

G (x) = x4 - 12x2 G' (x) = 4x3 - 24x G" (x) = 12x2 - 24

2. Set to 0 and solve.

12x2 - 24 = 0 X2=2 X = +/2

3. Is the second derivative undefined anywhere? No.

4. Test the three regions and make a sign graph. See the following figure.

G" (x) = 12x2 - 24

G"(-2) = 24 G" (0)=-24 G" (2) = 24

Concave up

Concave down

Concave up

-V2

V2

Because the concavity switched signs at the two zeros of G", there are inflection points at these two x-values.

5. Find the heights of the inflection points.

G (x) = X4 - 12x2

G (-/2) =- 20 G (J2) =- 20

G Is concave up from - oo to the inflection point at (-J2, - 20), concave down from there to another inflection point at (/2, - 20), then concave up again from there to oo.

± Find the intervals of concavity and the inflection points of P (x) = xx-S-

Ftx + 9 3/^' --j2"

. Concave down from

O To an inflection point at

Then concave down again till the third inflection point at

Up to O .

1. Get the second derivative.

, then concave up till the inflection point at (0, 0),

, and, finally, concave

(x)'(X2 + 9) - (X)(X2 + 9)'

(X2 + 9) X2 + 9 - 2x2 (X2 + 9 )2 9 - X2

(x2+9)

2. Set equal to 0 and solve.

2x (X2 - 27)

0

(x +9)

2x (X2- 27) = 0

3Ft ^3

P"(x)

(x2+9

- 2x3 - 18x – 36x + 4x3

(x2 + 9)3 2x (x2 – 27)

(x2+9

2x = 0 X2 – 27 = 0

Or

X = 0 X + 3/3

3. Check for undefined points of the second derivative. None.

4. Test four regions with the second derivative. You can skip the sign graph. Tip: You can do all of this in your head because all that matters is whether the answers are positive or negative.

P"(x) P"(-10)

P"(1)

2x (X2 – 27)

(x2+9 2(-10)((-10)2 -27)

((-10/ + 9) 2 (N)(P) P3

N P

N

2(1)((1)2-27)

((1)+9)

2(P)(N)

P3

N P

■■N

P"(-1)

2(-1)((-1)-27)

2(N)(N)

P3

P

P

P

P"(10)

2(10)((10) – 27)

((10)+9) 2(P)(P)

P3

P

P

P

The concavity goes N P N P So there’s an inflection point at each of the three zeros of p".

5. Find the heights of the inflection points.

P (x )’■ P (-) =

X2 + 9 - 3/3

- 3/3) + 9 -3/3

P (0) = 0 p (3/3) =

3/3

27 + 9 -/3

(3/3)+9

T2

12

-3A -#

Taking a drive on highway p, you’ll be turning right from negative oo to – 3/3, – 1y – , where

You’ll point straight ahead for an infinitesimal moment, then you’ll be turning left till (0, 0),

Then right again till 3 J3, 13 , and on your final leg to oo, you’ll round a very long bend to the left. I 121

° Find the intervals of concavity and the inflection points of q (x) = FtX – L/x. Concave down from - oo Till an inflection point at about (-.085, -.171), then concave up till a vertical inflection point at (0, 0), then concave down till a third inflection point at about (.085, .171), then concave up

Out to O .

You know the routine.

Q (x) q’ (x) q’ (x)

25×9

9×5/3

Whoops, I guess this algebra’s kind of messy. Let’s get the zeros on our calculators: just graph -4 , 2

Y=

25×9/59×5/

■ and find the x-intercepts. There are two: x;

-0.085 and x ~ 0.085.

So you’ve got two "critical numbers," right? Wrong! Don’t forget to check for undefined points

Of the second derivative. Because q" (x) = – 25 x- 9/

2 x – 5/3

9x,

Q" is undefined at x = 0. Since q(x)

Is defined at x = 0, 0 is another critical number. So you have three critical numbers and four regions. You can test them with -1, -.01, .01, and 1:

Q’ (x) =- 25 x – 9/5

14 q’ (-.01) ^ 158 q’ (.01) ^ -158

2 x – 5/3

9 x

Q‘(1) = IK

Thus the concavity goes: down, up, down, up. Because the second derivative is zero at -0.085 and 0.085 and because the concavity switches there, you can conclude that there are inflection points at those two x-values. But because both the first and second derivatives are undefined

At x = 0, you have to check whether there’s a vertical tangent there. You can see that there is by just looking at the graph, but if want to be rigorous about it, you figure the limit of the first derivative as x approaches zero. Since that equals infinity, you’ve got a vertical tangent at x = 0,

And thus there’s an inflection point there.

X

_

Now plug in -0.085, 0, and 0.085 into q to get the y-values and you’re done.

IS For g (x) = x3 + x2 – x, find all the values c in the interval (- 2, 1) that satisfy the Mean Value

Theorem. The values of c are gv and

1. Find the first derivative.

G (x) = x3 + x2-x g’ (x) = 3×2 + 2x – 1

2. Figure the slope between the endpoints of the interval.

G( 2) g(1)

G (-2) = (-2) + (-2) -(-2) = -2 g(1)=1

2 1

2-1

2-1

1

3. Set the derivative equal to this slope and solve.

-2 + /4- (-24) x 6 _ -2 + 2/7

3×2 + 2x – 1 = 1 3×2 + 2x-2 = 0

6

-1-/7

Or

1+ /7

33

Both are inside the given interval, so you’ve got two answers.

® For s (t) = t4/3- 3t1/3, find all the values c in the interval (0, 3) that satisfy the Mean Value Theorem. The value of c is

1. Find the first derivative.

S (t) = t4/3-3t1/3

S’ (t) = J t1/3 -

T

2. Figure the slope between the endpoints of the interval.

S(3) s(0)

S(0)=0

S (3) = 34/3-3 ■ 31/3

=0

3-0 0-0 3

0

3. Set the derivative equal to the result from Step 2 and solve.

T 2/3 =0

3 t

T-2/3 (31 ^1) = 0

^0 or 4= 0

Or t = 4

Graph s and check that its slope at t = 4

3 is zero.