AP Calculus Chapter 4 Testbank(Mr. Surowski)

Part I. Multiple-Choice Questions

1. Let f(x) = x3 + 3x2 − 45x+ 4. Then the local extrema of f are

(A) a local minimum of −179 at x = 5 and a local maximum of77 at x = −3.

(B)a local minimum of −77 at x = 3 and a local maximum of179 at x = −5

.

(C) a local minimum of −179 at x = −5 and a local maximumof −77 at x = 3.

(D) a local minimum of −77 at x = 3 and a local maximum of77 at x = 5.

(E) a local minimum of 77 at x = −5 and a local maximum of179 at x = 3.

2. Consider the function f(x) = x3 + 3x2 − 45x+ 4. Then

I. f is decreasing on (−5, 3) and increasing on (−∞,−5)∪(3,∞).

II. f has a local minimum at x = 3.

III. f has a local maximum at x = −5.

(A) I only (B) II only (C) I and II only (D) I and III only (E) I, II, and III

3. For what value of x does the function f(x) = x3− 9x2− 120x+ 6have a local minimum?

(A) 10 (B) 4 (C) 3 (D) −4 (E) −10

4. Use differentials to approximate the change in the volume of asphere when the radius is increased from 10 to 10.02 cm.

(A) 4213.973 (B) 1261.669 (C) 1256.637 (D) 25.233 (E) 25.133

5. The graph of y = x3 − 5x2 + 4x+ 2 has a local minimum at

(A) (0.46, 2.87)

(B) (0.46, 0)

(C) (2.94,−4.05)

(D) (4.06, 2.87)

(E) (1.66,−0.59)

6. Find a positive value, c, for x, that satisfies the conclusion of theMean Value Theorem for Derivatives for f(x) = 3x2 − 5x+ 1 onthe interval [2, 5].

(A) 1 (B)13

6(C)

11

6(D)

23

6(E)

7

2

7. The graph of y = x3 − 2x2 − 5x+ 2 has a local maximum at

(A) (2.120, 0)

(B) (2.120,−8.061)

(C) (−0.786, 0)

(D) (−0.786, 4.209)

(E) (0.666,−1.926)

8. A 20-foot ladder slides down a wall at 5 ft/sec. At what speedis the bottom sliding out (in ft/sec) when the top is 10 feet fromthe floor?

(A) 0.346 (B) 2.887 (C) 0.224 (D) 5.774 (E) 4.472

9. What are the coordinates of the point of inflection on the graphof y = x3 − 15x2 + 33x+ 100?

(A) (9, 0) (B) (5,−48) (C) (1, 119) (D) (9,−89) (E) (5, 15)

10. The graph of y = x4 + 8x3 − 72x2 + 4 is concave down for

(A) −6 < x < 2

(B) x > 2

(C) x < −6

(D) x < −3− 3√

5 or x > −3 + 3√

5

(E) −3− 3√

5 < x < −3 + 3√

5

11. The function f is given by f(x) = x4 + 4x3. On which of thefollowing intervals is f decreasing?

(A) (−3, 0) (B) (0,∞) (C) (−3,∞) (D) (−∞,−3) (E) (−∞, 0)

12. The value of c that satisfies the Mean Value Theorem on the in-terval [0, 5] for the function f(x) = x3 − 6x is

(A) − 5√3

(B) 0 (C) 1 (D)5

3(E)

5√3

13. The graph of the function y = x3 + 12x2 + 15x + 3 has a relativemaximum at x =

(A) −10.613 (B) −.248 (C) −7.317 (D) −1.138 (E) −.683

14. The side of a square is increasing at a constant rate of 0.4 cm/sec.In terms of the perimeter, P , what is the rate of change of thearea of the square, in cm2/sec?

(A) 0.05P (B) 0.2P (C) 0.4P (D) 6.4P (E) 51.2P

15. The second derivative of a function is given by f ′′(x) = x sinx−2. How many points of inflection does f have on the interval(−10, 10)?

(A) Zero (B) Two (C) Four (D) Six (E) Eight

Part II. Free-Response Questions

16. Find the extreme values of the function f(x) = x3 +3x2−36x+4on the interval [0, 4].

The critical x-values are obtained by solving f ′(x) = 0. Thisleads to 3x2 + 6x− 36 = 0⇒ x = −1 +

√13. The extreme values

are obtained by comparing:

f(0) = 4

f(−1 +√

13) = −51.74

f(4) = −28

Therefore the maximum of f is 4 and the minimum is −51.74.

17. Find the extreme values of the function f(x) =x

x2 + 1on the

interval [−3, 3].

From f ′(x) =1− x2

1 + x2 , we see that the critical x-values on the

given interval are x = ±1. We compare:

f(−3) =−3

10

f(−1) = −1

2

f(1) =1

2

f(3) =3

10

Therefore f has a maximum of1

2(at x = 1) and a minimum of

−1

2(at x = −1).

18. Find the extreme values of the function f(x) =x

x2 + 1on the

interval (−∞,∞).

This is very similar to the above problem. As above, that thecritical x-values on the given interval are x = ±1. We compare:

f(−∞) = limx→−∞

x

x2 + 1= 0

f(−1) = −1

2

f(1) =1

2

f(+∞) = limx→+∞

x

x2 + 1= 0

Therefore f has a maximum of1

2(at x = 1) and a minimum of

−1

2(at x = −1).

Drawing the graph of y = f(x) will reveal the relationship be-tween the above two problems.

19. Find the extreme values of the function g(x) = x√

2− x on theinterval [−2, 2].

The critical x-values are those for which either g′(x) = 0 or g′(x)doesn’t exist. From 0 = g′(x) =

√2− x− x

2√

2− xwe get 4−2x =

x⇒ x =4

3.Also, note that g′ does not exist at x = 2 (an endpoint

of the given interval). We now compare:

g(−2) = −4

g

(4

3

)= 4

3

√23

g(2) = 0

Therefore the maximum of g is4

3

√2

3(at x =

4

3), and the mini-

mum of g is −4 (at x = −2).

20. Find all relative extrema of the function h(x) = x2e1/x.

We find the critical x-values:

h′(x) = 2xe1/x − e1/x set= 0 ⇒ x =

1

2. Since h′(x) < 0 when x <

1

2

and h′(x) > 0 when x >1

2, we conclude that h has a relative

minimum of e2/4 at x =1

2.

21. Let f(x) = x√

4− x2.

(a) Determine all relative extrema of f .The critical x-values are obtained by solving f ′(x) = 0. This

leads to the equation√

4− x2 − x2√

4− x2= 0, and so x =

±√

2. Since f(x) < 0 when −2 < x < 0, and since f(x) > 0when 0 < x < 2 we infer that f has a relative minimumof f(−

√2) = −2 (at x = −

√2) and a relative maximum of

f(√

2) = 2 (at x =√

2).

(b) Determine all points of inflection of the graph of y = f(x).

Here, we have (after some work) that f ′′(x) =2x(x2 − 4)

(4− x2)3/2 =

0 ⇒ x = 0 (It’s true that x = ±2 are also solutions of theabove, but we don’t allow points of inflection to be at theendpoints of the domain of definition of a function.) Sincef ′′(x) > 0 when x < 0 and f ′′(x) < 0 when x > 0, we see thatthe inflection changes at x = 0, and so the point (0, f(0)) =(0, 0) is a genuine point of inflection.

22. Determine all relative extrema of the function f(x) = x+ cosx.

We compute the critical x-values:

f ′(x) = 1 − sinxset= 0 ⇒ sinx = 1 ⇒ x =

π

2+ 2kπ, where

k is an arbitrary integer. However, as f ′(x) is ALWAYS non-negative, we see f is always increasing and no there can be NORELATIVE EXTREMA.

23. Let f(x) = 2xe−x, x ≥ 0 and determine the extrema of f(x) on[0,∞).

We have f ′(x) = 2e−x − 2xe−x = (2 − 2x)e−x. From this we seethat the only critical x-value is x = 1. Since f ′(x) > 0 when0 ≤ x < 1 and f ′(x) < 0 when x > 1, we conclude that f has arelative maximum of 2/e (at x = 1).

24. Let f(x) = 2xe−x , x ≥ 0.

(a) Determine where f is increasing and where f is decreasing.Using the work already done above, we conclude that f ↗on the interval 0 ≤ x < 1 and that f ↘ on the interval x > 1.

(b) Determine where f is concave up and where f is concavedown.f ′′(x) = (2x − 4)e−x = 0 when x = 2. Also f ′′(x) < 0 whenx < 2 and f ′′(x) < 0 when x > 2. Therefore there is a changein concavity at x = 2 forcing the point (2, 4/e2) to be theunique point of inflection.

(c) Sketch the graph of y = f(x),

25. Find the points of inflection of the graph y =2

1 + e−x.

There are at least two ways of doing this. The first, and easiest,way is to recognize the function as a logistic function, where wehave already learned that the point of inflection occurs wherey is one-half the carrying capacity. In this case the carrying ca-pacity is y = 2 and so the point of inflection is at (x, 1), where

1 =2

1 + e−x⇒ x = 0. That is the point of inflection is (0, 1).

The more straightforward method is to compute the second deriva-tive y′′ of y. We have

y′ =−2e−x

(1 + e−x)2 ,

and so

y′′ =2e−x(1 + e−x)2 − 4e−2x(1 + e−x)

(1 + e−x)4

=2e−x + 2e−2x − 4e−2x

(1 + e−x)3

=2e−x(1− e−x)

(1 + e−x)3

= 0 when x = 0.

Furthermore, it’s easily checked that a change in concavity oc-curs through x = 0 and so the point of inflection is (0, 1).

26. A particle moves along the x-axis so that its position at time t ≥0 is given by the function x(t) = t3 − 12t2 + 48t, t ≥ 0, where xis measured in meters and t is measured in seconds.

(a) Determine those valuesof t where the particle isnot moving. The particle isnot moving when v(t) =x′(t) = 0. This leads to theequation 3t2 − 24t + 48 =0 ⇒ (t − 4)2 = 0, and sot = 4.

(b) Determine when the parti-cle is moving to the rightand when the particle ismoving to the left. Asv(t) = x′(t) = 3(t − 4)2 ≥ 0for all t, we see that unlesst = 4 the particle is alwaysmoving to the right.

(c) Determine when the par-ticle is accelerating andwhen the particle is decel-erating. We have thata(t) = x′′(t) = 6t − 24 > 0whe t > 4 and x′′(t) < 0when t < 4. That is to say,the particle is acceleratingwhen t > 4 and is deceler-ating with t < 4.

(d) At what value(s) of t doesthe particle have zero ac-celeration? This happenswhen 0 = x′′(t) = 6t − 24,i.e, when t = 4.

27. Find all relative extrema of the function f(x) = e−x sinx, x > 0.

The critical x values are those for which f ′(x) = 0, i.e., wheree−x cosx − e−x sinx = 0. This happens when cosx = sin x, i.e.,when x =

π

4+ kπ, where k is any non-negative integer. Next,

note that f ′′(x) = −e−x cosx − e−x sinx + e−x sinx − e−x cosx =−2e−x cosx. We have

f ′′(π

4+ kπ

)< 0 if x =

π

4+ kπ, k even

> 0 if x =π

4+ kπ, k odd.

and so f has relative maxima of e−π4 +kπ/

√2, k even and ≥ 0 and

relative minima of e−π4 +kπ/

√2, k odd and > 0.

28. Find the critical x-values of the function f(x) = x2 3√x− 1.

These are the solutions of f ′(x) = 0 or where the derivative failsto exist. We have

f ′(x) = 2x 3√x− 1 +

1

3x2(x− 1)−2/3

=6x(x− 1) + x2

3(x− 1)2/3

=x(7x− 6)

3(x− 1)2/3 .

The above shows that the critical x-values are x = 0, 67 (where

f ′(x) = 0) and x = 1 (where f ′(x) is not defined).

29. Find the open intervals on which the function f(x) = 14x

3− 3x isincreasing or decreasing.

We know that f↗ precisely where f ′(x) > 0 and that f ↘precisely where f ′(x) < 0. From f ′(x) = 3

4x2 − 3 = 3

4(x2 − 4).

From this it follows quickly that f ′(x) > 0 on the open intervals(−∞,−2) and (2,∞) and that f ′(x) < 0 on the open interval(−2, 2). Therefore

f ↗ on (−∞,−2) and on (2,∞), and

f ↘ on (−2, 2).

30. Use the second derivative test to find the relative extrema off(x) = x4 − 4x3 + 3. As part of your work, be sure to iden-tify all points of inflection. Sketch a graph, indicating all pointsof inflection.

We have f ′(x) = 4x3 − 12x2 = 4x2(x − 3) revealing critical x-values at x = 0, 3. Next, f ′′(x) = 12x2 − 24x = 12x(x− 2). Notethat f ′′(0) = 0 and so the second derivative test is INCONCLU-SIVE at the critical point x = 0. Next, f ′′(3) = 36·1 > 0 and so we

see that f has a relative minimum of f(3) = 34− 4 · 33 + 3 = −24(at x = 3). Finally note that f ′′(x) > 0 on the intervals (−∞, 0)and (2,∞) and f ′′(x) < 0 on the interval (0, 2). Therefore weobtain points of inflection at x = 0, 2; the corresponding pointsare P and Q, with coordinates P (0, 3) and Q(2,−13)

-�

6

?

x0 x1 x2 x3 x4 x5 x6 x

yy = f(x)

?

6

31. The graph above depicts a function f defined on the interval(−∞,∞). In terms of the notation given in the graph, determine

(i) the critical x values of f ; These are x = x0, x2, x4, and x6.

(ii) the open intervals on which f is increasing and decreas-ing; f ↗ on (−∞, x0), (x2, x4), (x6,∞); f ↘ on (x0, x2), and on(x4, x6).

(iii) the open intervals on which f is concave up and concavedown. f ∪ on the intervals (x1, x3) and on (x5,∞); f ∩ on theintervals (−∞, x1) and on (x3, x5).

(iv) Find the values of x at which f has points of inflection.These are the values x = x1, x3, and x5.

32. Find the production level that produces the maximum profit forhamburgers in a fast-food restaurant whose profit function (Pdollars) in terms of hamburgers (x) is described as

P = 2.44x− x2

20, 000− 5, 000 , 0 ≤ x ≤ 50, 000.

From P ′(x) = 2.44 − x

10, 000set= 0 we infer the critical value

x = 24, 400. We now compare:

P (0) = −5000 (certainly not a maximum!);

P (24400) = 2.44 · 24400− 244002

20000− 5000 = 24768;

P (50000) = 2.44 · 50000− 500002

20000− 5000 = −8000

Therefore the production level which maximizes profit is x =24, 400 hamburgers.

33. Suppose that the semicircle y =√

1− x2 is drawn. Now drawa rectangle whose base is on the x-axis and height is such thatthe rectangle is incribed within the semicircle. What is the max-imum area of this rectangle?

If the base of the rectangle is drawn between the points (−x, 0)and (x, 0), then the height would be

√1− x2. Therefore, the area

to be maximized is given by the functionA(x) = 2x√

1− x2, −1 ≤x ≤ 1. Solving 0 = A′(x) = 2

√1− x2 − 2x2(1 − x2)−1/2 yields

1 − x2 = x2 and so x = ± 1√2

The resulting maximum area is

then A(± 1√

2

)=√

2√

1− 12 = 1.

34. Suppose that on one side of a 2 km-wide river is an electricity-generating plant. On the opposite side, and 10 km down theriver is a small town that will be consuming the electricity. If itcosts $80/m to lay cable under the river and $40/m to lay cableover land, find the most strategic method for laying the cable.

Power Plant•

•Town

HHHHHH

HHHHHH

HH︸ ︷︷ ︸x

︷ ︸︸ ︷10− x2 km

From the above picture, the total cost of running the cable fromthe power plant to the town is

C(x) = 80000√

4 + x2 + 40000(10− x) dollars.

Differentiating and setting equal to 0 give the equation

80000x√4 + x2

− 40000 = 0,

and so

4x2 = 4 + x2 ⇒ x =2√3

km.

Comparing the cost at the critical x-value versus that at the end-points yields

C(0) = 560, 000 (dollars)

C(10) = 815, 843 (dollars)

C(2/√

3) = 538, 564 (dollars)

from which we see that the most economical method is to runthe cable to the opposite side of the river where x = 2/

√3 (km)

and then run the cable along the ground to the town. The result-ing cost of doing this is $538,564.

35. Find the minimum distance from the point (2, 0) to a point onthe parabola whose graph is y = 2x2.

We shall minimize the distance-squared from the point P (2, 0)to the parabola whose equation is y = 2x2. Setting D equal tothis distance-squared, then we have D(x) = (x − 2)2 + (2x2)2.Differentiating and setting equal to 0 produces

0 = D′(x) = 2(x− 2) + 8x3.

This cannot be solved (easily) in closed form; using a graphicscalculator gives the approximate solution x ≈ 0.689 Therefore,the minimum distance is approximated as

√D(0.689) =

√(0.689− 2)2 + 4(0.689)4 ≈ 1.62

36. Write the linearization of the function f(x) = 3√

1 + x valid nearx = 0. Use this linearization to approximate 3

√1.1.

We have that f(x) ≈ f(0) + f ′(0)x = 1 + 13x.

From this we get the approximation

3√

1.1 = f(.1) ≈ 1 +1

3(0.1) ≈ 1.033.

(Note that this is pretty good, as a calculator yields 3√

1.1 ≈1.032.)

37. Write the linearization of the function g(x) = cosx valid nearx = 0. Use this to approximate cos(.01). (Are you surprised bythis linearization?)

Setting f(x) = cos x we have cosx ≈ f(0) + f ′(0)x = 1 + 0, sinceddx cosx = − sinx and sin 0 = 0. This forces the approximationcos(.01) ≈ 1.

38. Write the linearization of f(x) = ln x valid near x = 1. Use thisapproximate ln(1.2).

We have f(x) ≈ f(1)+f ′(1)(x−1) = 0+(x−1) = x−1. Therefore,ln(1.2) = f(1.2) ≈ 1.2− 1 = .2.

39. Assume that y = x2 lnx.

(a) Compute dy in terms of x and dx.

Fromdy

dx= 2x lnx+ x we get dy = (2x lnx+ x) dx.

(b) Compute dy given that x = 1 and that dx = 0.1.dy = dx = 0.1

40. Let V be the volume of a sphere of radius r and let dV be its dif-ferential value (in terms of r and dr). For fixed dr, what wouldyou say about dV :

(a) that it is an increasing function of r,

(b) that it is a decreasing function of r, or

(c) that it doesn’t depend on r?

From V = 43πr

3 we get dV = 4πr2dr. Therefore, for fixed dr, dVis an increasing function of r .

41. Let Q be a quantity that depends on a measurable quantity, x,by the rule Q =

√1 + x2. Suppose that we measure x to be 5.2±

0.015.

(a) Estimate the resulting range in computed values for Q.

We have dQ =x dx√1 + x2

; we take x = 5.2 and dx = 0.015. The

resulting range of values of Q is approximately Q ± dQ =√

1 + 5.22 ± 5.2 · 0.015√1 + (5.2)2

= 5.30± 0.015.

(b) Estimate the resulting relative error in the computed value

for Q. The relative error isdQ

Q= 0.015/5.30 ≈ 0.003, or a .3%

relative error.

42. Assume that ball bearings are to be made whose volume must bewithin a 1% error. Use differentials to determine the necessarytolerance of the radius of the given ball bearings.

We require thatdV

V≤ .01. Since dV = 4πr2dr, we get

dV

V=

4πr2dr

(4/3)πr3 =3dr

r. Therefore the relative error on the measure-

ments of the radius cannot exceed 1/3 of 1%.

43. Suppose that a sphere is expanding at a rate given by dr/dt =5m/sec. If S denotes the surface area of this sphere and V de-notes its volume, compute dV/dt and dS/dt when r = 2. Do thesame for r = 4. Does this make sense?

Using the chain rule and the facts that V = 43πr

3, S = 4πr2, we

getdV

dt=dV

dr· drdt

= (4πr2) · (5 m/sec) r=2= 80πm3/sec. Similarly

dS

dt=dS

dr· drdt

= (8πr) · (5 m/sec) r=2= 80πm2/sec.

The corresponding results for r = 4 are similarly obtained.

44. Suppose that you are standing exactly 1km from the Pearl Towerin Shanghai. Now suppose that someone releases a helium-filled balloon from the top of the Pearl Tower, and that this bal-loon is rising at a constant rate of 1m/sec. Let D be the distancebetween you (standing on the ground) and the balloon. Let h(t)be the height of the balloon as a function of time, where we as-sume that h(0) = 400m.

(a) Compute dD/dtwhen h = 1000m. We have thatD2 = h2+1,

which implies that 2DdD

dt= 2h

dh

dt. This implies that

dD

dt=

h

D

dh

dt=h

D=

1000√1 + 10002

≈ 1m/sec

(b) Compute dD/dt when h = 2000m.

Here,dD

dt=h

D

dh

dt=h

D=

2000√1 + 20002

≈ 1m/sec.

(c) Compute limh→∞

dD/dt. (Does this make sense?)

The above work shows thatdD

dt=

h

D

dh

dt=

h

D=

h√1 + h2

≈

1m/sec→ 1 as h→∞.This makes sense for as the balloon continues to rise, it be-comes almost directly overhead, rising at very nearly 1 m/secrelative to the observer.

(d) Show that dD/dt is a increasing function of both h and t.

This is almost obvious for the function F (h) =h√

1 + h2sat-

isfies F (h) < 1, but that F (h)→ 1 as h→∞. Therefore,dD

dtis an increasing function of h. As h increases as t increases,dD

dtis also an increasing function of t.

Of course, if you’re nervous about this, you can show thatF (h) is increasing by computing its derivative with respectto h and showing that it’s positive.

45. Let h be a function defined for all x 6= 0 such that h(4) = −3 and

the derivative of h is given by h′(x) =x2 − 2

xfor all x 6= 0

(a) Find all values of x for which the graph of h has a horizontaltangent, and determine whether h has a local maximum, alocal minimum, or neither at each of these values. Justifyyour answers. From what’s given we see immediately that

h′(x) = 0 when x = ±√

2. We have that h′(x) = x − 2

xand

so h′′(x) = 1 +2

x2 . Therefore, h′′(±√

2) = 1 + 1 = 2 > 0,and so h will have a relative minimum at each of the valuesx = ±

√2.

(b) On what intervals, if any, is the graph of h concave up? Jus-tify your answer. Since h′′(x) > 0 for all values of x in thedomain of h, we infer that h is concave up on the intervals(−∞, 0) and (0,∞).

(c) Write an equation for the line tangent to the graph of h at

x = 4. We have that h′(4) =42 − 2

4=

7

2. Therefore, since

we’re already given that h(4) = −3, the equation of the linetangent to the graph of y = h(x) at the point (4,−3) is

y + 3 =7

2(x− 4).

(d) Does the line tangent to the graph of h at x = 4 lie above orbelow the graph of h for x > 4. Why? The line tangent to thegraph of h at x = 4 will lie below the graph of h since thegraph of h is everywhere concave up.

46. A cubic polynomial function g is defined by

f(x) = x3 + ax2 + bx+ c

where a, b, and c are constants. The function has a critical valueat x = −1, and the graph of f has a point of inflection at thepoint (−2, 0). Find a, b, and c.

We have that f ′(x) = 3x2+2ax+b and that 0 = f ′(−1) = 3−2a+b.Next, f ′′(x) = 6x + 2a and so 0 = f ′′(−2) = −12 + 2a. Solvingthis latter equation results in a = 6 . From this together with theequation 3− 2a+ b = 0 we obtain b = 9 . Finally, since the point(−2, 0) is on the graph, we know that f(−2) = 0, which forces0 = (−2)3 + a(−2)2 + b(−2) + c = −8 + 6 · 4 + 9 · (−2) + c =−2 + c⇒ c = 2 .

x 0 0 < x < 1 1 1 < x < 2 2 2 < x < 3 3 3 < x < 4

f(x) −1 Negative 0 Positive 2 Positive 0 Negative

f ′(x) 4 Positive 0 Positive DNE Negative −3 Negative

f ′′(x) −2 Negative 0 Positive DNE Negative 0 Positive

47. Let f be a function that is continuous on the interval [0, 4). Thefunction f is twice differentiable except at x = 2. The functionf and its derivatives have the properties indicated in the tableabove, where DNE indicates that the derivatives of f do not existat x = 2.

(a) For 0 < x < 4, find all values of x at which f has a relativeextremum. Determine whether f has a relative maximumor a relative minimum at each of these values. Justify youranswer. f has a relative maximum at x = 2 for the deriva-tive of f changes from positive to negative through x = 2.(The fact that the derivative of f does not exist at x = 2 isunimportant.) There are no other relative extrema.

(b) On the axes provided, sketch the graph of a function thathas all the characteristics of f .

-

6

x

y

�

�

48. An airplane is flying at a constant speed at a constant altitude of3 km in a straight line that will take it directly over an observerat ground level. At a given instant the observer notes that theangle θ is π/3 and is increasing at 1/20 radians per second. Findthe speed of the airplane, in kilometers per hour.

����

�����

���

���

����

θ•

Observer

• Airplane

3 km

︷ ︸︸ ︷x

We start with the equation cot θ =x

3, where x is as indicated in

the diagram above. Note that the speed of the airplane is∣∣∣dxdt

∣∣∣.Differentiating the above equation gives

− csc2 θdθ

dt=

1

3

dx

dt.

This implies thatdx

dt= −3

dθ

dtcsc2 θ = − 3

20csc2 θ, so that when

θ =π

3, csc2 θ =

4

3, all of which implies that

dx

dt= −

(3

20

)(4

3

)=

1

5km/sec. Therefore, we see that the speed of the plane is

1

5×

3600 = 720 km/hr.

49. Continue that we have, as in the above problem, an airplaneflying toward the observer at an altitude of 3 km. This time,

however, assume that angular rate of change,dθ

dt=

1

20, and is

constant.

(i) What is the speed of the airplane when it is directly over-head? Again, we have the equation cot θ =

x

3, which im-

plies thatdx

dt= −3

dθ

dtcsc2 θ = − 3

20csc2 θ, so that when the

plane is directly overhead we have θ =π

2, and so

dx

dt= − 3

20csc2 θ = − 3

20.

This says that the speed of the plane is

speed =∣∣∣dxdt

∣∣∣ =3

20× 3600 = 540 km/hr.

(ii) Is the airplane slowing down or speeding up as it approachesthe observer? Justify your conclusion.

From∣∣∣dxdt

∣∣∣ =3

20csc2 θ, we conclude that as θ increases to

π

2, csc2 θwill decrease to 1 from values greater than 1. There-

fore the speed of the plane is decreasing .