Ap Calc Ab 2017 Practice Exam

Let's delve into a comprehensive practice exam for the AP Calculus AB 2017, designed to mirror the format and difficulty level you can expect on the actual test. This practice exam covers a wide range of calculus topics, from limits and derivatives to integrals and applications of calculus. By working through these problems and understanding the solutions, you'll be well-prepared to tackle the AP Calculus AB exam with confidence.

Section I: Multiple Choice

This section consists of two parts: Part A, where you can't use a calculator, and Part B, where a graphing calculator is required.

Part A: No Calculator

1. Limits and Continuity

• Problem: Find the limit, if it exists: lim (x→2) (x^2 - 4) / (x - 2)

  • Solution: We can factor the numerator: (x^2 - 4) = (x - 2)(x + 2). Thus, the expression simplifies to (x + 2) when x ≠ 2. Therefore, lim (x→2) (x^2 - 4) / (x - 2) = lim (x→2) (x + 2) = 2 + 2 = 4.

• Problem: Which of the following functions is continuous at x = 0?

  I. f(x) = |x| II. g(x) = 1/x III. h(x) = sin(x)/x

  • Solution:

    • I. f(x) = |x| is continuous everywhere, including x = 0.
    • II. g(x) = 1/x is not continuous at x = 0 because it's undefined.
    • III. h(x) = sin(x)/x has a removable discontinuity at x = 0, but the limit as x approaches 0 exists and is equal to 1. Therefore, if we define h(0) = 1, the function becomes continuous.

    Therefore, I and III are continuous (with a suitable definition for h(0)).

2. Derivatives

• Problem: Find the derivative of f(x) = x^3 - 4x^2 + 5x - 2.

  • Solution: Using the power rule, f'(x) = 3x^2 - 8x + 5.

• Problem: If y = sin(2x), find dy/dx.

  • Solution: Using the chain rule, dy/dx = cos(2x) * 2 = 2cos(2x).

• Problem: Find the derivative of f(x) = ln(x^2 + 1).

  • Solution: Using the chain rule, f'(x) = (1/(x^2 + 1)) * 2x = 2x / (x^2 + 1).

• Problem: Given f(x) = e^(3x), find f''(x).

  • Solution: First, f'(x) = 3e^(3x). Then, f''(x) = 9e^(3x).

3. Applications of Derivatives

• Problem: Find the equation of the tangent line to the curve y = x^2 - 3x + 2 at x = 1.

  • Solution: First, find the y-coordinate at x = 1: y = (1)^2 - 3(1) + 2 = 0. So, the point is (1, 0). Next, find the derivative: dy/dx = 2x - 3. Evaluate the derivative at x = 1: dy/dx |_(x=1) = 2(1) - 3 = -1. This is the slope of the tangent line. Using the point-slope form, the equation of the tangent line is y - 0 = -1(x - 1), or y = -x + 1.

• Problem: A particle moves along the x-axis such that its position at time t is given by x(t) = t^3 - 6t^2 + 9t + 1. Find the time(s) when the particle is at rest.

  • Solution: The particle is at rest when its velocity is zero. First, find the velocity function: v(t) = x'(t) = 3t^2 - 12t + 9. Set v(t) = 0: 3t^2 - 12t + 9 = 0. Divide by 3: t^2 - 4t + 3 = 0. Factor: (t - 1)(t - 3) = 0. Therefore, the particle is at rest at t = 1 and t = 3.

• Problem: Find the critical points of f(x) = x^3 - 3x.

  • Solution: First, find the derivative: f'(x) = 3x^2 - 3. Set f'(x) = 0: 3x^2 - 3 = 0. Divide by 3: x^2 - 1 = 0. Factor: (x - 1)(x + 1) = 0. Therefore, the critical points are x = 1 and x = -1.

4. Integrals

• Problem: Evaluate the definite integral: ∫(from 0 to 1) (x^2 + 1) dx.

  • Solution: Find the antiderivative: ∫(x^2 + 1) dx = (1/3)x^3 + x + C. Evaluate at the limits: [(1/3)(1)^3 + 1] - [(1/3)(0)^3 + 0] = (1/3 + 1) - 0 = 4/3.

• Problem: Find the indefinite integral: ∫cos(3x) dx.

  • Solution: Using u-substitution, let u = 3x, so du = 3dx, and dx = (1/3)du. Then, ∫cos(3x) dx = ∫cos(u) (1/3)du = (1/3)∫cos(u) du = (1/3)sin(u) + C = (1/3)sin(3x) + C.

• Problem: Evaluate the definite integral: ∫(from 1 to e) (1/x) dx.

  • Solution: The antiderivative of 1/x is ln|x|. So, ∫(from 1 to e) (1/x) dx = ln|e| - ln|1| = 1 - 0 = 1.

Part B: Calculator Allowed

5. Applications of Integrals

• Problem: The velocity of a particle moving along the x-axis is given by v(t) = t^2 - 4t + 3. Find the total distance traveled by the particle from t = 0 to t = 3.

  • Solution: The total distance traveled is given by ∫(from 0 to 3) |v(t)| dt. First, find when v(t) = 0: t^2 - 4t + 3 = 0. (t - 1)(t - 3) = 0. So, v(t) = 0 at t = 1 and t = 3. We need to split the integral into intervals where v(t) is positive and negative.

    • From t = 0 to t = 1, v(t) > 0.
    • From t = 1 to t = 3, v(t) < 0.

    So, the total distance is ∫(from 0 to 1) (t^2 - 4t + 3) dt - ∫(from 1 to 3) (t^2 - 4t + 3) dt.

    • ∫(from 0 to 1) (t^2 - 4t + 3) dt = [(1/3)t^3 - 2t^2 + 3t] |_(from 0 to 1) = (1/3 - 2 + 3) - 0 = 4/3.
    • ∫(from 1 to 3) (t^2 - 4t + 3) dt = [(1/3)t^3 - 2t^2 + 3t] |_(from 1 to 3) = (9 - 18 + 9) - (1/3 - 2 + 3) = 0 - (4/3) = -4/3.

    Therefore, the total distance traveled is |4/3| + |-4/3| = 8/3. Use your calculator to evaluate the definite integrals directly to save time.

• Problem: The region R is bounded by the curves y = x^2 and y = 4. Find the volume of the solid generated when R is revolved about the x-axis.

  • Solution: We'll use the washer method. The outer radius is R(x) = 4 and the inner radius is r(x) = x^2. The limits of integration are the x-values where the curves intersect: x^2 = 4, so x = -2 and x = 2.

    The volume is given by V = π∫(from -2 to 2) [R(x)^2 - r(x)^2] dx = π∫(from -2 to 2) (16 - x^4) dx. Using symmetry, we can rewrite this as V = 2π∫(from 0 to 2) (16 - x^4) dx.

    V = 2π [16x - (1/5)x^5] |_(from 0 to 2) = 2π [32 - 32/5] = 2π [128/5] = 256π/5. Use your calculator to evaluate the definite integral directly.

6. Differential Equations

• Problem: Solve the differential equation dy/dx = x/y, with the initial condition y(1) = 2.

  • Solution: This is a separable differential equation. Separate the variables: y dy = x dx. Integrate both sides: ∫y dy = ∫x dx. (1/2)y^2 = (1/2)x^2 + C. Multiply by 2: y^2 = x^2 + 2C.

    Use the initial condition y(1) = 2: (2)^2 = (1)^2 + 2C. 4 = 1 + 2C. 2C = 3. So, y^2 = x^2 + 3. Therefore, y = √(x^2 + 3) (we take the positive root because y(1) = 2 is positive).

• Problem: The rate of change of the population P of a town is proportional to the population. Which of the following differential equations best models this situation?

  • Solution: The rate of change of population is dP/dt. If it's proportional to P, then dP/dt = kP, where k is a constant of proportionality.

7. Other Topics

• Problem: Use Euler's method with a step size of 0.1 to approximate y(1.2) given dy/dx = x + y and y(1) = 0.

  • Solution:

    • y(1) = 0 (given)
    • y'(1) = 1 + 0 = 1
    • y(1.1) ≈ y(1) + y'(1) * 0.1 = 0 + 1 * 0.1 = 0.1
    • y'(1.1) ≈ 1.1 + 0.1 = 1.2
    • y(1.2) ≈ y(1.1) + y'(1.1) * 0.1 = 0.1 + 1.2 * 0.1 = 0.1 + 0.12 = 0.22

    Therefore, y(1.2) ≈ 0.22.

• Problem: Find the average value of the function f(x) = x^2 on the interval [0, 2].

  • Solution: The average value is given by (1/(b - a)) ∫(from a to b) f(x) dx. In this case, it's (1/(2 - 0)) ∫(from 0 to 2) x^2 dx = (1/2) ∫(from 0 to 2) x^2 dx.

    ∫(from 0 to 2) x^2 dx = (1/3)x^3 |_(from 0 to 2) = (1/3)(2)^3 - 0 = 8/3. So, the average value is (1/2) * (8/3) = 4/3.

Section II: Free Response

This section consists of 6 free-response questions that cover a wide range of calculus topics. Show all your work clearly and completely.

Question 1: Area and Volume

The region R is enclosed by the graphs of y = x^2 and y = 2x.

(a) Find the area of region R.

(b) Find the volume of the solid generated when region R is revolved about the x-axis.

(c) The region R is the base of a solid. For this solid, the cross-sections perpendicular to the x-axis are squares. Find the volume of this solid.

• Solution:

  (a) First, find the intersection points: x^2 = 2x. x^2 - 2x = 0. x(x - 2) = 0. So, x = 0 and x = 2. The area is given by ∫(from 0 to 2) (2x - x^2) dx.

  ∫(from 0 to 2) (2x - x^2) dx = [x^2 - (1/3)x^3] |_(from 0 to 2) = (4 - 8/3) - 0 = 4/3.  Therefore, the area of region R is 4/3.
  

  (b) We'll use the washer method. The outer radius is R(x) = 2x and the inner radius is r(x) = x^2. The limits of integration are x = 0 and x = 2.

  The volume is given by V = π∫(from 0 to 2) [R(x)^2 - r(x)^2] dx = π∫(from 0 to 2) (4x^2 - x^4) dx.
  
  V = π [(4/3)x^3 - (1/5)x^5] |_(from 0 to 2) = π [(32/3 - 32/5)] = π [160/15 - 96/15] = π [64/15] = 64π/15.
  

  (c) The side length of the square cross-section is (2x - x^2). The area of the square is (2x - x^2)^2 = 4x^2 - 4x^3 + x^4. The volume is given by ∫(from 0 to 2) (4x^2 - 4x^3 + x^4) dx.

  V = [(4/3)x^3 - x^4 + (1/5)x^5] |_(from 0 to 2) = (32/3 - 16 + 32/5) - 0 = (160/15 - 240/15 + 96/15) = 16/15.
  

Question 2: Rates of Change

Water is leaking out of a tank at a rate proportional to the square root of the volume of water remaining in the tank. Initially, the tank contains 100 liters of water. After 16 days, there are only 25 liters remaining.

(a) Write a differential equation that models the volume of water in the tank at time t.

(b) Solve the differential equation you wrote in part (a) to find V(t), the volume of water in the tank at time t.

(c) When will the tank be empty?

• Solution:

  (a) dV/dt = -k√V, where V is the volume and k is a positive constant (since the volume is decreasing).

  (b) Separate variables: dV/√V = -k dt. Integrate both sides: ∫V^(-1/2) dV = ∫-k dt. 2√V = -kt + C.

  Use the initial condition V(0) = 100: 2√100 = -k(0) + C.  20 = C. So, 2√V = -kt + 20.
  
  Use the condition V(16) = 25: 2√25 = -k(16) + 20.  10 = -16k + 20.  -10 = -16k.  k = 10/16 = 5/8.
  
  Therefore, 2√V = -(5/8)t + 20.  √V = -(5/16)t + 10.  V(t) = (-(5/16)t + 10)^2.
  

  (c) The tank is empty when V(t) = 0. (-(5/16)t + 10)^2 = 0. -(5/16)t + 10 = 0. (5/16)t = 10. t = 10 * (16/5) = 32. The tank will be empty after 32 days.

Question 3: Related Rates

A ladder 10 feet long rests against a vertical wall. If the bottom of the ladder slides away from the wall at a rate of 1 foot per second, how fast is the top of the ladder sliding down the wall when the bottom of the ladder is 6 feet from the wall?

• Solution: Let x be the distance from the bottom of the ladder to the wall, and let y be the distance from the top of the ladder to the ground. We have x^2 + y^2 = 10^2 = 100.

  We are given dx/dt = 1 ft/s, and we want to find dy/dt when x = 6.

  Differentiate both sides of x^2 + y^2 = 100 with respect to t: 2x(dx/dt) + 2y(dy/dt) = 0.

  When x = 6, we have 6^2 + y^2 = 100. 36 + y^2 = 100. y^2 = 64. y = 8 (since y must be positive).

  Plug in the values: 2(6)(1) + 2(8)(dy/dt) = 0. 12 + 16(dy/dt) = 0. 16(dy/dt) = -12. dy/dt = -12/16 = -3/4.

  The top of the ladder is sliding down the wall at a rate of 3/4 feet per second. The negative sign indicates that y is decreasing.

Question 4: Particle Motion

A particle moves along the x-axis with velocity given by v(t) = t sin(t^2) for t ≥ 0. The particle is at position x = 2 when t = 0.

(a) Find the acceleration of the particle at time t.

(b) Find the position of the particle at time t.

(c) Find the total distance traveled by the particle from t = 0 to t = 3.

• Solution:

  (a) Acceleration is the derivative of velocity: a(t) = v'(t) = d/dt (t sin(t^2)). Use the product rule: a(t) = sin(t^2) + t * cos(t^2) * 2t = sin(t^2) + 2t^2 cos(t^2).

  (b) Position is the integral of velocity: x(t) = ∫v(t) dt = ∫t sin(t^2) dt. Use u-substitution: u = t^2, du = 2t dt, so t dt = (1/2) du.

  x(t) = ∫(1/2) sin(u) du = -(1/2) cos(u) + C = -(1/2) cos(t^2) + C.
  
  Use the initial condition x(0) = 2: 2 = -(1/2) cos(0) + C.  2 = -(1/2) + C.  C = 5/2.
  
  Therefore, x(t) = -(1/2) cos(t^2) + 5/2.
  

  (c) Total distance traveled is ∫(from 0 to 3) |v(t)| dt = ∫(from 0 to 3) |t sin(t^2)| dt. Since t ≥ 0, we only need to consider when sin(t^2) changes sign. sin(t^2) = 0 when t^2 = nπ, where n is an integer. So, t = √(nπ). In the interval [0, 3], we need to find the values of n for which √(nπ) ≤ 3. nπ ≤ 9. n ≤ 9/π ≈ 2.86. So, n = 0, 1, 2. This means v(t) changes sign at t = 0, √π, and √(2π).

  We need to evaluate the integral ∫(from 0 to 3) |t sin(t^2)| dt. You'll need to use your calculator to evaluate this definite integral. The result is approximately 2.341. *Make sure your calculator is in radian mode.*
  

Question 5: Theorems and Justifications

Let f be a differentiable function such that f(2) = 5 and f'(x) ≥ 3 for all x in the interval [2, 6].

(a) Use the Mean Value Theorem to show that f(6) ≥ 17.

(b) Find the minimum possible value for f(6). Justify your answer.

(c) Is it possible that f(x) = 3x + cos(πx) for all x in the interval [2, 6]? Explain your reasoning.

• Solution:

  (a) The Mean Value Theorem states that there exists a c in (2, 6) such that f'(c) = (f(6) - f(2)) / (6 - 2). We are given that f'(x) ≥ 3 for all x in [2, 6], so f'(c) ≥ 3.

  Therefore, (f(6) - f(2)) / (6 - 2) ≥ 3.  (f(6) - 5) / 4 ≥ 3.  f(6) - 5 ≥ 12.  f(6) ≥ 17.
  

  (b) The minimum possible value for f(6) occurs when f'(x) = 3 for all x in [2, 6]. In this case, f(x) is a linear function with slope 3. Using the point-slope form, f(x) - 5 = 3(x - 2). f(x) = 3x - 6 + 5 = 3x - 1.

  So, f(6) = 3(6) - 1 = 18 - 1 = 17.  The minimum possible value for f(6) is 17.
  
  *Justification:*  Since f'(x) ≥ 3, the smallest possible value for f'(x) is 3.  If f'(x) is always 3, then f(x) is a linear function with slope 3 passing through the point (2, 5).  This gives us the minimum possible value for f(6).
  

  (c) Let g(x) = 3x + cos(πx). Then g'(x) = 3 - πsin(πx). We need to check if g'(x) ≥ 3 for all x in [2, 6].

  Since -1 ≤ sin(πx) ≤ 1, then -π ≤ πsin(πx) ≤ π. Therefore, 3 - π ≤ 3 - πsin(πx) ≤ 3 + π.  Approximately, -0.14 ≤ g'(x) ≤ 6.14.
  
  However,  since π > 0,  3 - πsin(πx) could be less than 3 for some x.  For example, if sin(πx) = 1, then g'(x) = 3 - π < 3.
  
  Therefore, it is *not* possible that f(x) = 3x + cos(πx) for all x in the interval [2, 6] because its derivative is not always greater than or equal to 3.
  

Question 6: Accumulation Functions

Let f be a function defined by f(x) = ∫(from 0 to x) (t^3 - 6t^2 + 8t) dt for 0 ≤ x ≤ 3.

(a) Find the x-coordinate of each critical point of f in the interval (0, 3).

(b) Identify whether each critical point found in part (a) is a local minimum, a local maximum, or neither. Justify your answer.

(c) Find the absolute maximum value of f on the interval [0, 3].

• Solution:

  (a) By the Fundamental Theorem of Calculus, f'(x) = x^3 - 6x^2 + 8x. Critical points occur where f'(x) = 0 or f'(x) is undefined. Since f'(x) is a polynomial, it is defined for all x. So, we only need to find where f'(x) = 0.

  x^3 - 6x^2 + 8x = 0.  x(x^2 - 6x + 8) = 0.  x(x - 2)(x - 4) = 0.  So, x = 0, x = 2, and x = 4.  However, we are only interested in the interval (0, 3), so the critical point is x = 2.
  

  (b) To determine if x = 2 is a local minimum or maximum, we can use the second derivative test or the first derivative test.

  *First Derivative Test:*  We analyze the sign of f'(x) around x = 2.
  *   For 0 < x < 2, choose x = 1: f'(1) = 1(1 - 2)(1 - 4) = 1(-1)(-3) = 3 > 0.  So, f(x) is increasing.
  *   For 2 < x < 3, choose x = 2.5: f'(2.5) = 2.5(2.5 - 2)(2.5 - 4) = 2.5(0.5)(-1.5) < 0.  So, f(x) is decreasing.
  
  Since f'(x) changes from positive to negative at x = 2, f(x) has a local maximum at x = 2.
  

  (c) The absolute maximum value of f on [0, 3] must occur at either an endpoint (x = 0 or x = 3) or at a critical point (x = 2). We need to evaluate f(0), f(2), and f(3).

  *   f(0) = ∫(from 0 to 0) (t^3 - 6t^2 + 8t) dt = 0.
  *   f(2) = ∫(from 0 to 2) (t^3 - 6t^2 + 8t) dt = [(1/4)t^4 - 2t^3 + 4t^2] |_(from 0 to 2) = (4 - 16 + 16) - 0 = 4.
  *   f(3) = ∫(from 0 to 3) (t^3 - 6t^2 + 8t) dt = [(1/4)t^4 - 2t^3 + 4t^2] |_(from 0 to 3) = (81/4 - 54 + 36) - 0 = 81/4 - 18 = (81 - 72)/4 = 9/4 = 2.25.
  
  The absolute maximum value of f on the interval [0, 3] is 4, which occurs at x = 2.
  

By working through this practice exam and thoroughly understanding each solution, you'll significantly improve your preparedness and confidence for the actual AP Calculus AB exam. Good luck!