Master AP Calculus AB & BC

Part III. FOUR PRACTICE TESTS

ANSWER KEY AND EXPLANATIONS

Section I, Part A

1. C

7. B

13. E

19. D

24. B

2. E

8. E

14. C

20. C

25. B

3. E

9. C

15. B

21. B

26. C

4. D

10. C

16. E

22. E

27. C

5. E

11. D

17. C

23. C

28. D

6. A

12. C

18. A

1. The correct answer is (C). Instantaneous rates of change always imply differentiation. To quickly determine this derivative, it is helpful to recognize x3 + 3x2 + 3x + 1 as (x + 1)3. We then simplify the original function to

Now, use the Power and Chain Rules:

f’(x) = 2(x + 1)

To find the instanteous rate of change when x = 2,

f'(2) = 2(2 + 1) = 6

2. The correct answer is (E), To approximate the actual number of cars crossing the bridge, approximate the area under this graph. One way to do this is to divide the interval from t = 0 to t = 12 into 2 equal subintervals. Both of these regions resemble trapezoids. The area of the left one is

The area of the right trapezoid would be

So, the area under the curve would be approximately 825. But we must be careful here. The rate is in cars per minute. Since the x-axis is in hours, we must convert the rate to cars per hour. To do this, we multiply our 825 by 60 (minutes per hour) and get 49,500 cars.

3. The correct answer is (E), Find the definite integral:

4. The correct answer is (D), There is not enough information to determine whether or not choices (A), (B), (C), or (E) are true. Look at choice (D):

Average rate of change

5. The correct answer is (E), We must recognize that This lets us rewrite the integral as

Next, we can evaluate this integral using u-substitution. If we let u = sec x and du = sec x tan x dx, we get

6. The correct answer is (A), We need the derivative of the curve. Since x and y are not separated for us, we must use implicit differentiation. Differentiate everything with respect to x.

Now, group all terms with dy/dx, and solve for dy/dx.

Finally, we just substitute our point (1,1) into our expression for dy/dx and get

7. The correct answer is (B), Before we try to integrate anything here, distribute that √x and change the notation to that of rational exponents. After these two steps, we get

Integrating leads to

8. The correct answer is (E), This is a related-rates problem. We are given dr/dt, the rate at which the radius is increasing, and need to find dV/dt, the rate at which the volume is increasing when A, the surface area, is 9π. Our primary equation is the volume equation for a sphere:

As in all such problems, we differentiate with respect to t.

Knowledge of basic formulas is useful here. 4πr2 is merely the surface area formula for a sphere. We were given that the surface area is equal to 9π and that dr/dt = 2. Substituting these values yields

9. The correct answer is (C). To find the area of a region bounded by two curves, we should apply the following formula:

where a and b are the endpoints of the interval, f(x) represents the top curve, while g(x) represents the bottom curve. Finding a and b is simple enough. Since the region begins at the y-axis, a = 0. To find b, we will determine what value satisfies the following equation:

x3 = x + 6

By inspection, we can see that x = 2. Now, the area of the region would be given by

10. The correct answer is (C). Whenever the average rate of change is requested, we just need to compute the slope of the secant line.

11. The correct answer is (D). Since we know that the second derivative is a line of slope 6, we can say that

f"(x) = 6x + C1

That implies that f'(x) = 3x2 + C1x + C2 which in turn implies that f(x) = x3 + C3x2 + C2x + C4. That is a cubic function.

12. The correct answer is (C). The MVT for integrals says that the average value of a function over a given interval is the area under the curve divided by the length of the interval. So, the average value, f(c), of f over [—4,4] could be found like this:

which is equivalent to

13. The correct answer is (E). Since f has a horizontal tangent at x = 1, we know that f'(1) = 0. By reading the graph, we can see that f(1) > 0. Since the graph is concave down at x = 1, f"(1) < 0. Hence, f"(1) < f'(1) < f(1).

14. The correct answer is (C). The MVT states that at some point c on the interval [a,b]. Since then at some point c on [a,b], f’(c) = ½.

15. The correct answer is (B). This is an application of the Fundamental Theorem of Calculus, Part Two, which states

We must remember that when the upper limit of integration is some function of x, such as 2x2, we must multiply f(x) by the derivative of that function with respect to x. Hence,

16. The correct answer is (E). Since we know that rate of change implies derivative, from the information in the problem, we can write

f’(c) = e3f'(2)

We are also told that f(x) = ex, so f'(x) = ex. So, the above equation becomes

ec = e3 ∙ e2 ∙ e5

So, c = 5.

17. The correct answer is (C). This problem is testing if we can apply the chain rule to functions defined by a table. If h(x) = f(g(x)), then h'(x) = f'(g(x))g'(x). This is why, when x = 3,

So, h(3) = h’(3)

18. The correct answer is (A). Since there are exactly two points on (0,10) where/has a value of 4, the graph of f must cross the line x = 4 twice: Once on the way up and once on the way down. The fact that f is differentiable over the interval insures no cusps or discontinuities. Since the curve turns around somewhere on (0,10), there must be at least one horizontal tangent on (0,10). Horizontal tangents are places where the derivative is equal to zero.

19. The correct answer is (D). Normal lines are perpendicular to tangent lines. The slopes of two perpendicular lines are opposite reciprocals of each other. So, this problem needs us to determine the slope of the curve at x = 2, and then determine the opposite reciprocal of that slope. Since slope of the curve is determined by the value of its derivative,

The slope of the normal line is the opposite reciprocal

20. The correct answer is (C). Evaluate the definite integral and apply the Fundamental Theorem, Part One:

21. The correct answer is (B). The rate of change of y being directly proportional to y is the same statement as

y' = ky.

which we know leads to

y = Nekt

22. The correct answer is (E). The question to answer here is when, if ever, is the derivative of y = 3x3 — 2x2 + 6x — 2 negative? We should try to determine the derivative, find any critical values, and examine a wiggle graph.

y' = 9x2 — 4x + 6 = 0

This is an unfactorable trinomial. Since we are not permitted to use our calculators, we’d better use the quadratic formula. So,

Aha! The radicand, 16 — 216, is less than zero, which would indicate that the equation has no real solutions, which would imply that the derivative of y = 3x3 — 2x2 + 6x — 2 is never zero. Since it is a polynomial function, then it must be continuous; hence, y = 3x3 — 2x2 + 6x — 2 is strictly monotonic. Now, we should determine the value of the derivative at one x value to determine if the derivative is always positive or always negative. Using the equation, let’s determine the value of the derivative when x = 0:

y'(0) = 0 — 0 + 6 = 6 > 0

Since the derivative is always positive, the function y = 3x3 — 2x2 + 6x — 2 is never decreasing.

23. The correct answer is (C). Remember, the MVT guarantees that for some c on [2,5],

Now, we should determine the derivative of set it equal to 3/2, and solve for x.

Since —1 is not on [2,5], we throw that value out and the value on [2,5] that satisfies the MVT is 3.

24. The correct answer is (B). The slope field shows us that f(x) will have a derivative of zero when y = —1 and when x = 1 (since the slopes are horizontal there). The easiest possible differential equation with such characteristics is f’(x) = (x — 1)(y + 1), since plugging in —1 for y or 1 for x makes the slope 0. If you factor choice (B) by grouping, that is exactly what you get

25. The correct answer is (B). This problem is not as clear as it may appear initially. After examining the diagram, we should look for

However, this is not an answer choice. One of the choices must be equivalent to ours. Remember that when you switch the limits of integration, you get the opposite value. Switching our limits gives us

This is still not a choice. What if we treat that negative sign as if it were the constant — 1 and distribute it through the integral? We get

Eureka!

26. The correct answer is (C). The derivative being positive over [0,2] implies that the function is increasing over this interval. The second derivative being negative means that the function is concave down. The curve must look something like the curve drawn below:

Notice that every y-coordinate over the interval (0,2) is less than 2, so f(1) < 2. Notice that the entire curve is above the secant line from (0, —1) to (2,2). (This is true due to the concavity of the curve.) Since the secant line segment passes through (1,1/2), f(1) > 1/2. Therefore, f(1) could be 1.

27. The correct answer is (C). This is another related-rates problem. We know that , where h represents the water level in the barrel. We are looking for dV/dt with V representing the volume of the barrel. Our primary equation is the formula for volume of a cylinder:

Since it is given that the radius, r, is a constant of 10, we can substitute this into the equation and get

V = 100πh

Now, as in any related-rates problem, we should differentiate with respect to t:

Substituting into the equation yields

Therefore, the water is leaving the barrel at 100π in 3/min.

Notice that the information that the volume was 500π cubic inches was unnecessary.

28. The correct answer is (D). All we need here is the derivative of f(x) = arctan u.

Since our function is u = x2, the equation becomes

Now, we substitute x = √3 and get

Section I, Part B

29. D

33. C

37. E

40. D

43. D

30. D

34. A

38. D

41. C

44. E

31. E

35. A

39. C

42. E

45. B

32. B

36. D

29. The correct answer is (D). The particle’s positive velocity indicates that the position function’s graph is increasing. The decreasing velocity indicates that the position function’s graph should be concave down.

30. The correct answer is (D). In order for these two functions to have parallel tangent lines, their derivatives must be equal. So, we should find the derivatives of both functions, set them equal to each other, and solve for x. Since f(x) = 3 ln (2x), f'(x) = 3/x. The derivative of g(x) = x3 + 2x is g'(x) = 3x2 + 2. Now, we will set these two expressions equal and use our calculator to solve for x:

31. The correct answer is (E). The rate of increase of the derivative is the second derivative. So,

f’’(x) = 2

To find an expression for the first derivative, we can find an antiderivative:

f’(x) = 2x + C1

In order to determine C1, we can use the information given to us that f'(2) = 4. We will substitute this and solve for C1:

f'(2) = 4 = 4 + C1

C1 = 0

Substituting this value into the second equation yields

f'(x) = 2x

Now, we will determine f(x) and find an antiderivative of the last equation:

f(x) = x2 + C2

To solve for C2, we can use the fact that f(1) = 2, so

f(1) = 2 = 1 + C2

C2 = 1

Then, f(x) = x2 + 1

Finally, we can determine f(3):

f(3) = 9 + 1 = 10

32. The correct answer is (B). We must remember how to factor the difference of perfect cubes:

a3 — b3 = (a — b)(a2 + ab + b2)

Using this formula, we can simplify the limit:

which we can evaluate by substitution:

33. The correct answer is (C). The best way to attack this problem would be to plot the 4 points given and sketch the 3 rectangles, as shown in the diagram below:

Notice that the heights of the rectangles are determined by the y-value corresponding to the right endpoint of the subintervals. Next, we determine the area of each rectangle and then add them up:

34. The correct answer is (A). Since the graph of the derivative of f crosses the x-axis twice, there will be two relative extrema. There will be one maximum because the derivative changes from positive to negative once. There will also be one minimum since the derivative changes from negative to positive once as well.

35. The correct answer is (A). This is a rather complicated application of the MVT. We will also have to use the Fundamental Theorem of Calculus, Part Two. First, let’s determine the value of f'(c):

Note:

Now, we will determine the derivative of

Next, we will set our value for f'(c) equal to our expression for f’(x) and use our calculator to solve for x:

36. The correct answer is (D). This problem asks where is the slope of the tangent line, which is the instantaneous velocity, equal to the slope of the secant line, which is the average velocity, over [0,3].

To find the slope of the tangent line, find the derivative of the curve:

To determine where the two slopes are the same, we will set msec equal to mtan and solve for x using our calculator:

37. The correct answer is (E). For g to have a local maximum at x = 1, the derivative of g, which is f, must change from positive to negative at x = 1. It does not.

38. The correct answer is (D). This is a tedious example of the trapezoidal rule. Since we have 5 subintervals and the interval is 3 units long, we will be dealing with some messy numbers. Anyway, we still have to remember the trapezoidal rule:

Applying it to this function, we get

39. The correct answer is (C). Since we can immediately say that we are dealing with an exponential function of the following form:

y = Nekt

We know that after five years, the population will be three times what it was initially. If we substitute 3N for y and 5 for t and solve for k, we get

40. The correct answer is (D). In this related-rates problem, we are going to need . To quickly find , let’s use the formula for the circumference, differentiate with respect to t, and solve for :

C = 2πr

We are given that We can now substitute this value into the equation to get a value for :

The question is asking us about the rate at which the area is increasing, , when the circumference is 10π inches. We will take the formula for the area of a circle and differentiate with respect to t:

Notice that we have the expression 2πr. This is just the circumference that we know to be 10π. We can substitute this value and the value 1/5 for to determine :

41. The correct answer is (C). Remember that the formula for the volume of a solid with known cross sections is

where A(x) represents the area of the cross sections. In this problem, we are dealing with cross sections that are equilateral triangles. The formula for the area of an equilateral triangle is

where s is the length of one side. As we can see from the diagram, the interval is from x = 0 to x = 6. Therefore, the volume of this solid is

Our calculator will now do the rest and get

V = 112.237

42. The correct answer is (E). First, we must determine an equation for the tangent line to this curve at x = 2. We need a point on the line and the slope of the line. First the point: Since f(2) = 4, (2,4) is on the line. Now the slope:

f'(x) = 2x + 4

f'(2) = 4 + 4 = 8

The equation for the tangent line is

y - 4 = 8(x - 2)

or

y = 8x - 12

Now, let’s examine our choices.

I. The value of the function at x = -√2 is -4√2 - 6, and the tangent line approximation is -8√2 - 12, which is twice the value of the function. So, I checks out.

II. The function value at x = 1 is — 3, while the tangent line approximation is — 4.

III. The function value at x = √2 is 4√2 - 6, and the tangent line approximation is 8√2 – 12. So, III applies too.

43. The correct answer is (D). In order to determine the number of critical values of the function, we can count the zeros of the derivative. This would require us to graph the derivative on the calculator and count how many times it crosses the x-axis. It crosses four times.

44. The correct answer is (E). This is a tricky Fundamental Theorem of Calculus, Part Two problem. First, we should rewrite it as such:

Once we’ve rewritten the problem like this, it’s not so difficult:

45. The correct answer is (B). In order for the left-hand and right-hand limits to be equal, the function must be continuous. So, we need to find the value of k for which this equation is true:

This is relatively simple to solve:

1 + k = 0

k = -1

Section II, Part A

1. (a) The five subintervals would each be of length 2 and would be [0,2], [2,4], [4,6], [6,8], and [8,10], The midpoints of these subintervals would be 1, 3, 5, 7, and 9, respectively. The Riemann sum that we are looking for is just the sum of five rectangles, each of width 2 and height f(mi), where m; is the midpoint of the ith subinterval. So,

(b) Recall the trapezoidal rule:

(c) The average value, f(c), of the function is the area under the curve divided by the length of the interval. So, we can approximate f(c) like this:

(d) Here, we should determine the exact area under the curve and divide it by the length of the interval:

2. (a) Well start by labeling a fourth point, Q, as the point on the track directly in front of observer O. We will also define some variables: x will be the distance from the horse H to the point Q, y will be the distance from the observer O to the point Q, and z will be the distance between the horse H and the observer O. All of this is shown in the diagram below.

How long is the distance from P to Q? We can use the 30-60-90 triangle theorem to determine that it is 100√3 or 173.2051 feet. Since the horse is running at 45 feet per second, he has run 180 feet after 4 seconds. So, x = 180 — 173.2051 = 6.79492.

Now we can use the Pythagorean theorem to write our primary equation:

x2 + 1002 = z2

Substituting x = 6.79492 into the equation and solving for z gives us

The question asked us for the rate that the distance from the horse to the observer is increasing after four seconds. In other words, what is when t = 4? To answer this, let’s differentiate with respect to t and solve for .

So, when t = 4, the distance from the horse to the observer is increasing at 3.051 feet per second.

(b) This is not a difficult problem. The area of a triangle is

A = 1/2 ∙ base ∙ height

The height of this triangle is a constant, 100 feet, so

A = 50 ∙ base

To determine the rate at which the area of the triangle is changing, let’s differentiate with respect to t:

What is d/dt(base)? That’s the rate at which the base is changing, which is merely the speed of the horse, 45 feet per second. Now we have

So, the area of the triangle formed by P, H, and O is increasing at a constant rate of 2,250 feet2 per second.

(c) We will use x, y, and z to represent the same distances as in part A. It can be determined easily that the horse has galloped 225 feet in the same amount of time that the man ran 50 feet. So, y = 50, x = 225, and z can be determined as such:

To determine , we should differentiate with respect to t:

Solving for gives us

Now, we will substitute the following values into the equation: is negative because y is getting shorter.

The distance from the horse to the observer is increasing at 41.759 feet per second.

3. This problem involves solving the separable differential equation First, we should separate the v’s and t’s:

Integrate both sides:

To determine the value of C, we use the initial condition given to us in the problem. Since v(0) = 197.28, then

and C = 197.28. Now, we have our expression for v in terms of t:

(b) This is a two-part question. First, we should determine how much time it takes the car to stop, and then we should integrate the velocity curve using that value. In order for the car to stop, v(t) = 0.

It takes the car 18 seconds to come to a stop. Now, to determine how far the car travels in those 18 seconds, we should find the area under the velocity curve from t = 0 to t = 18:

The car travels 1,989.36 feet while slowing down.

(c) To write the equation for a line, we need a point on the line and the slope of the line. To determine the y coordinate of the point, we will evaluate v at t = 9:

This tells us that (9116.46) is on our tangent line. Now, determine the slope by evaluating the derivative (which we already know from the problem itself at t = 9:

The tangent line passes through (9116.46) and has a slope of —10.96. We can now write its equation using point-slope form:

v - 116.46 = -10.96(t - 9)

or in slope ≠ intercept form:

v = - 10.96t + 215.1

(d) This problem calls for the average value formula applied to the velocity equation you found in part (A).

You already know the value of the integral from your work in part (B).

Section II, Part B

4. (a)

(b) The two curves intersect at (2,4). The area of the region R can be determined using the following definite integral:

(c) The volume is easiest to determine using the washer method. The outer radius, R(x), is 8 — x2, and the inner radius, r(x), is x2:

(d) The volume of a solid with known cross sections can be determined like this:

The cross sections are semicircles whose area formula is A = 1/2πr2.

Now, we need an expression in terms of x for the radius of one of these semicircles. Because the height of R is the diameter of a semicircle, the radius would be

This leads to the area of a semicircle:

which gives us the volume of the solid:

5. (a) The local maximum occurs at x = 5 because the derivative changes from positive to negative there. This means that the function changes from increasing to decreasing there as well.

(b) f(7) < f(4) < f(5)

Since the function increases over [4,5] and decreases over [5,7], f(5) is the greatest of the three. To determine which is greater, f(4) or f(7), we examine the accumulated area over [4,7]. Since this area is negative, the function has a net decrease over [4,7]. Thus, f(4) > f(7).

(c) f has two points of inflection: one at x = 4 and one at x = 7. Points of inflection are places where the graph changes concavity. The graph changes concavity whenever the derivative changes from increasing to decreasing or from decreasing to increasing. The derivative changes from increasing to decreasing at x = 4 and from decreasing to increasing at x = 7.

6. (a) Since the x and y are not separated, we should differentiate implicitly.

(b) Differentiating implicitly again yields

(c) To determine the y-intercept, we let x = 0 and solve for y:

So, the y-intercept is (0,b).

(d) We just need the slope when x = 0 and y = b. We will substitute these values into our expression for dy/dx:

Now, we write the equation for a line with y-intercept b and slope b2: