Ap Calculus Ab 2014 Asdfds 1j3y34

AP Calculus AB 2014 Solutions∗ J. Sloan May 11, 2014

∗

I have no affiliation with The College Board

© and cannot guarantee the correctness of these solutions 1

Problem 1 Grass clippings are placed in a bin, where they decompose. For 0 ≤ t ≤ 30, the amount of grass clippings remaining in the bin is modeled by A(t) = 6.687(0.931)t , where A(t) is measured in pounds and t is measured in days. a) Find the average rate of change of A(t) over the interval 0 ≤ t ≤ 30. Indicate units of measure. b) Find the value of A0 (15) Using correct units, interpret the meaning of the value in the context of the problem. c) Find the time t for which the amount of grass clippings in the bin is equal to the average amount of grass clippings in the bin over the interval 0 ≤ t ≤ 30. d) For t > 30, L(t), the linear approximation to A at t = 30, is a better model for the amount of grass clippings remaining in the bin. Use L(t) to predict the time at which there will be 0.5 pound of grass clippings remaining in the bin. Show the work that leads to your answer.

a) The average value of A(t) on this interval is just the slope of the tangent line between endpoints on the interval. This is given as Rate of Changeavg =

A(30) − A(0) = −0.197 pounds/day. 30 − 0

b) Use a graphing calculator to evaluate d 0 A (15) = A(t) = −0.164 pounds/day. dt t=15 This means that at time t = 15 the weight of grass clippings in the bin is instantaneously changing at a rate of −0.164 pounds per day. c) The average value of the total mass of grass clipping in the bin on the given interval is found by integrating A(t) over the interval, and then dividing by the interval width of 30 days. This is Z 30 1 A(t) dt = 2.75264 pounds. A(t)avg = 30 0 The problem asks when the function A(t) is equal to this average value. Use a calulator to solve 6.687(0.931)t = 2.75264, it comes out to t = 12.415 seconds. d) In order to make a linear approximation (a tangent line to a certain point), we need a point, and a slope. The point of interest is just (30, A(30)) = (30, 0.782928) found by calculator. The slope at this point is d 0 A (30) = A(t) = −0.0559762 pounds/day. dt t=30 2

We will call the above value m. Then the tangent line approximation is given as y − A(30) = m(t − 30). To find the value of t for which A(t) = 0.5, when there is a half pound of grass clippings in the bin, set y = 0.5 and solve for t with algebraic manipulation and a calculator. The result is t = 35.054 seconds.

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Problem 2

Let R be the region encolsed by the graph of f (x) = x4 − 2.3x3 + 4 and the horizontal line y = 4, as shown in the figure above. a) Find the volume of the solid generated when R is rotated about the horizontal line y = −2. b) Region R is the base of a solid. For this solid, each cross section perpendicular to the x-axis is an isoceles right triangle with a leg in R. Find the volume of the solid. c) The vertical line x = k divides R into two regions with equal areas. Write, but do not solve, an equation involving integral expressions whose solution gives the value k.

a) First find the where the two functions intersect using a calulator. Let a be the leftmost intersection x value, and b be the rightmost intersection x value. We will use a and b for the remainder of the problem to avoid numerical crowding. Using a calculator to set the functions equal to each other and find the intersection points gives a = 0, b = 2.3 In order to for an axis of rotation of y = −2, it is necessary to shift both of the functions up by 2 units, which effectively places the new axis of rotation on the x axis. Then the volume is just the difference between the outer and inner portions of the object. That is Z b V olume = π (4 + 2)2 − (f (x) + 2)2 dx = 98.868. a

b) The isosceles right triangles to which the problem refers look like this:

w w 4

Both sides are denoted w since the triangleis isosceles. The area of the triangle is then 1 Areatriangle = w2 . 2 Since the width in this problem is specified by the distance between the two functions, w = 4 − f (x). Integrating across the whole x interval gives 1 V olume = 2

Z

b

(4 − f (x))2 dx = 8.518.

a

c) The integral equation is Z

k

Z (4 − f (x)) dx =

0

b

(4 − f (x)) dx, k

since each integral gives the area of a section of the region, and the problem stipulates that they be equal to one another.

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Problem 3

The function f is defined on the closed interval [−5, 4]. The graph of f consists of three line Z segments and is shown in the figure above. Let g be the function defined by g(x) = x

f (t) dt. −3

a) Find g(3). b) On what open intervals contained in −5 < x < 4 is the graph of g both increasing and concave down? Give a reason for your answer. g(x) . Find h0 (3). 5x d) The function p is defined by p(x) = f x2 − x . Find the slope of the line tangent to the graph of p at the point where x = −1. c) The function h is defined by h(x) =

a) By definition, Z

3

g(3) =

f (t) dt = −3

12 8 2 + − = 9. 2 2 2

The numerical values are just triangular areas between f and the x-axis. The last value in the sequence of 3 is negative because the function is below the x-axis. b) Since by definition, Zx g(x) =

f (t) dt, −3

we can use the Second Fundamental Theorem of Calculus to conclude that g 0 (x) = f (x). Furthermore, taking the derivative of both sides again gives the relation that g 00 (x) = f 0 (x). 6

If we want to find when g is increasing and concave down, we want to know when g 0 (x) > 0, and g 00 (x) < 0. This is equivalent, via the relations above, to inquiring about when f (x) > 0, and f 0 (x) < 0. So we want f to be above the x-axis, and sloping downward. This happens on the intervals (−5, −3) and (0, 2). c) By the quotient rule, h0 (x) =

5xg 0 (x) − 5g(x) . 25x2

So 5(3)g 0 (3) − 5g(3) 25(3)2 15f (3) − 5(9) = 225 15(−2) − 5(9) = 225 1 = . 3

h0 (3) =

Note that g 0 (3) = f (3) due to the relationship established in (b). Also note that no simplification is required once all values in the expression are numerical. d) Taking the derivative of p(x) with the chain rule gives p0 (x) = f 0 (x2 − x)(2x − 1). So p0 (−1) = f 0 (2)(−3) = (−2)(−3) = 6. Note that f 0 (2) = −2 because the slope of f can be geomtrically found to be −2.

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Problem 4

Train A runs back and forth on an east-west section of railroad track. Train A’s velocity, measured in meters per minute, is given by a differentiable function vA (t), where the time t is measured in minutes. Selected values for vA (t) are given in the table above. a) Find the average acceleration of train A over the interval 2 ≤ t ≤ 8. b) Do the data in the table the conclusion that train A’s velocity is −100 meters per minute at some time t with 5 ≤ t ≤ 8. Give a reason for your answer. c) At time t = 2, train A’s position is 300 meters east of the Origin Station, and the train is moving to the east. Write an expression involving an integral that gives the position of train A, in meters from the Origin Station, at time t = 12. Use a trapezoidal sum with three subintervals indicated by the table to approximate the position of the train at time t = 12. d) A second train, train B, travels north from the Origin Station. At time t the velocity of train B is given by vB (t) = −5t2 + 60t + 25, and at time t = 2 the train is 400 meters north of the station. Find the rate, in meters per minute, at which the distance between train A and train B is changing at time t = 2. a) The average acceleration is the average rate of change of the velocity function on the specified interval. That is, aavg =

−120 − 100 110 vA (8) − vA (2) = =− meters/min. 8−2 6 3

b) Yes, we can guarantee that there exists a time for which v = −100 on the given interval because vA (5) is positive and vA (8) is negative. Because the function is differentiable, and thus continuous, the velocity function must through the x-axis, where v = 0. c) The position of the train at time t = 12 can be found by taking the initial position given in the problem, and then adding the change in position found by integrating the velocity function. That is, Z 12

xA (12) = 300 +

vA (t) dt. 2

The right trapezoidal sum is given as Z 12 100 + 40 40 − 120 −120 − 150 vA (t) dt ≈ 3 +3 +4 = −450. 2 2 2 2 Adding the initial value of 300 gives a final result of −150 meters. 8

d) A picture is in order. N

s = 500 (3-4-5 triangle)

xB (t) = 400

E

xA (t) = 300 By the Pythagorean Theorem, we conclude that xA (t)2 + xB (t)2 = s2 ,

where s is the distance between the trains. Taking the derivative of both sides implcitly with respect to time gives 2xA (t)vA (t) + 2xB (t)vB (t) = 2s

ds . dt

Then ds 300(100) + 400(125) 300vA (2) + 400vB (2) = = 160 meters/min. = dt t=2 500 500 Note that vA (2) is taken directly from the table, and vB (2) is found to 125 by substituting t = 2 into the given function.

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Problem 5

The twice-differentiable functions f and g are defined for all real numbers x. Values of f , f 0 , g, and g 0 for various values of x are given in the table above. a) Find the x-coordinate of each relative minimum of f on the interval [−2, 3]. Justify your answers. b) Explain why there must be a value c, for −1 < c < 1, such that f 00 (c) = 0. c) The function h is defined by h(x) = ln(f (x)). Find h0 (3). Show the computations that lead to your answer. Z 3 d) Evaluate f 0 (g(x))g 0 (x) dx. −2

a) The function f can have a relative minimum only when f 0 (x) goes from negative to positive. This occurs at x = 1. b) Find the slope of the secant line on f 0 (x) beetween the endpoints of the interval. This is slope =

0 f 0 (1) − f 0 (−1) = = 0. 1 − (−1) 2

Since this secant slope is zero, and the functions in question are differentiable, Rolles’ Theorem, a special case of the Mean Value Theorem, guarantees that there exists a c in ther interval such that f 00 (c) is equal to the secant slope of zero found above. c) By the chain rule, h0 (x) =

1 0 f 0 (x) f (x) = . f (x) f (x)

h0 (3) =

f 0 (3) 1/2 1 = = . f (3) 7 14

Thus

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d) This is a seemingly impossible integral which requires the use of a clever trick. Use u substitution, letting u = g(x), then du = g 0 (x) dx, which conveniently exists already. Also, the boundaries must be changed from −2 and 3 to respectively g(−2) and g(3) in accordance with the substitution. Together, this is Z

3 0

Z

0

g(3)

f (g(x))g (x) dx = −2

f 0 (u) du.

g(−2)

The integral of the derivative of f 0 (u) is just f (u). So Z

g(3)

g(−2)

1 g(3) = f (u) = f (1) − f (−1) = 2 − (−8) = 10. f (u) du = f (u) 0

g(−2)

−1

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Problem 6 dy Consider a differential equation = (3 − y) cos x. Let y = f (x) be the particular solution dx to the differential equation with the initial condition f (0) = 1. The function f is defined for all real numbers. a) A portion of the slope field of the differential equation is given below. Sketch the solution curve through the point (0, 1).

b) Write an equation for the line tangent to the solution curve in part (a) at the point (0, 1). Use the equation to approximate f (0.2). c) Find y = f (x), the particular solution to the differential equation with the initial condition f (0) = 1.

a) The sketch of the solution curve is shown above. b) For a tangent line we need a point and a slope. The point given is (0, 1). By the derivative given, dy = (3 − 1) cos(0) = 2. dx (0,1) Thus the equation of the tangent line, through slope form, is y − 1 = 2x, y = 2x + 1. Then y(0.2) = 2(0.2) + 1 = 1.4.

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c) Seperating variables, we get Z

dy = 3−y

Z cos(x) dx.

The integration and algebraic solve is fairly straightforward. Just don’t forget that the left side picks up a negative sign once integrated due to the 3 − y in the denominator. Z Z dy = cos(x) dx 3−y − ln |3 − y| = sin(x) + C ln |3 − y| = − sin(x) + C2 |3 − y| = C3 e− sin(x) 3 − y = C4 e− sin(x) y = 3 − C4 e− sin(x) .

Then taking into the initial condition, we let x = 0 and y = 1 and solve for C4 . 1 = 3 − C4 e− sin(0) . 1 = 3 − C4 e 0 1 = 3 − C4 (1) 1 = 3 − C4 C4 = 2. Which leads to the final conclusion that y = f (x) = 3 − 2e− sin(x) .

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