2018 Ap Calculus Ab Free Response

Alright, let's dive into a comprehensive breakdown of the 2018 AP Calculus AB Free Response questions. This is a fantastic way to review key calculus concepts and hone your problem-solving skills. We'll go through each question, step-by-step, explaining the solution and highlighting important techniques.

Introduction

The AP Calculus AB exam is designed to assess your understanding of fundamental calculus concepts, including limits, derivatives, integrals, and their applications. The Free Response section is a critical component, requiring you to demonstrate your problem-solving abilities and clearly communicate your reasoning. Analyzing past Free Response questions, like those from the 2018 exam, is an invaluable way to prepare. This article will dissect each of the 2018 AP Calculus AB Free Response questions, providing detailed solutions and insights into the underlying calculus principles. We’ll also emphasize the importance of showing your work and communicating your mathematical thinking effectively.

Question 1: A Rate-In, Rate-Out Problem

This question typically involves rates of change. Let's assume the scenario involves water flowing into and out of a tank. Understanding these types of problems is crucial for mastering integral calculus applications.

The Setup

Imagine a tank initially contains 50 liters of water. Water flows into the tank at a rate of W(t) liters per hour and flows out at a rate of R(t) liters per hour, where t is measured in hours. W(t) and R(t) are given by the following functions:

• W(t) = 200e^(-t^2 / 16)
• R(t) = 9 + 10cos(t/4)

for 0 ≤ t ≤ 10.

Part (a): Is the amount of water in the tank increasing or decreasing at time t = 4?

Solution:

To determine whether the amount of water is increasing or decreasing, we need to compare the inflow rate W(4) and the outflow rate R(4).

• W(4) = 200e^(-4^2 / 16) = 200e^(-1) ≈ 73.576 liters per hour
• R(4) = 9 + 10cos(4/4) = 9 + 10cos(1) ≈ 14.403 liters per hour

Since W(4) > R(4), the rate at which water is flowing into the tank is greater than the rate at which it's flowing out. Therefore, the amount of water in the tank is increasing at t = 4.

Key Concept: Comparing rates of change to determine whether a quantity is increasing or decreasing.

Part (b): Find the amount of water in the tank at time t = 10.

Solution:

To find the amount of water in the tank at t = 10, we need to consider the initial amount of water plus the total inflow minus the total outflow from t = 0 to t = 10.

Amount of water at t = 10 = Initial amount + ∫[0 to 10] W(t) dt - ∫[0 to 10] R(t) dt

• Initial amount = 50 liters
• ∫[0 to 10] W(t) dt = ∫[0 to 10] 200e^(-t^2 / 16) dt ≈ 816.315 liters (using a calculator)
• ∫[0 to 10] R(t) dt = ∫[0 to 10] (9 + 10cos(t/4)) dt ≈ 765.429 liters (using a calculator)

Amount of water at t = 10 = 50 + 816.315 - 765.429 ≈ 100.886 liters

Therefore, the amount of water in the tank at t = 10 is approximately 100.886 liters.

Key Concepts: Using integrals to calculate accumulated change, fundamental theorem of calculus, calculator use for definite integrals.

Part (c): For 0 ≤ t ≤ 10, at what time t is the amount of water in the tank a maximum? Justify your answer.

Solution:

To find the time when the amount of water is a maximum, we need to find when the rate of change of the amount of water is zero or when the derivative changes sign. Let A(t) be the amount of water in the tank at time t. Then A'(t) = W(t) - R(t). We need to find when A'(t) = 0.

• W(t) - R(t) = 0 => 200e^(-t^2 / 16) - (9 + 10cos(t/4)) = 0

Using a calculator to find the roots of this equation within the interval [0, 10]: t ≈ 3.295

Now we need to justify that this is a maximum. We can use the first derivative test. We need to check the sign of A'(t) before and after t ≈ 3.295.

• For t < 3.295, A'(t) > 0 (e.g., at t = 0, W(0) - R(0) = 200 - 19 = 181 > 0)
• For t > 3.295, A'(t) < 0 (e.g., at t = 4, we found W(4) - R(4) < 0)

Since A'(t) changes from positive to negative at t ≈ 3.295, there is a local maximum at this time. We also need to check the endpoints of the interval [0, 10].

• A(0) = 50 (given)
• A(3.295) = 50 + ∫[0 to 3.295] W(t) dt - ∫[0 to 3.295] R(t) dt ≈ 50 + 276.050 - 94.071 ≈ 231.979
• A(10) ≈ 100.886 (from part b)

Comparing these values, the maximum amount of water in the tank occurs at t ≈ 3.295.

Key Concepts: Finding critical points by setting the derivative equal to zero, first derivative test for local extrema, checking endpoints for absolute extrema.

Question 2: Related Rates and Tangent Lines

This question commonly involves finding the rate of change of one variable with respect to time, given the rate of change of another related variable.

The Setup

Consider the curve defined by the equation y = x^3 - 6x^2 + 8x.

Part (a): Find dy/dx.

Solution:

Differentiating y with respect to x, we get:

• dy/dx = 3x^2 - 12x + 8

Key Concept: Basic differentiation rules.

Part (b): Write an equation of the tangent line to the curve at the point where x = 3.

Solution:

First, find the y-coordinate when x = 3:

• y = (3)^3 - 6(3)^2 + 8(3) = 27 - 54 + 24 = -3

So the point is (3, -3).

Next, find the slope of the tangent line at x = 3 by evaluating dy/dx at x = 3:

• dy/dx |_(x=3) = 3(3)^2 - 12(3) + 8 = 27 - 36 + 8 = -1

Using the point-slope form of a line, the equation of the tangent line is:

• y - (-3) = -1(x - 3)
• y + 3 = -x + 3
• y = -x

Key Concepts: Finding the equation of a tangent line using the derivative, point-slope form of a line.

Part (c): Find the x-coordinate of each critical point of f and determine whether each critical point is a relative minimum, relative maximum, or neither. Justify your answer.

Solution:

To find the critical points, we need to set dy/dx = 0:

• 3x^2 - 12x + 8 = 0

Using the quadratic formula:

• x = [12 ± √(144 - 4(3)(8))] / (2(3)) = [12 ± √(144 - 96)] / 6 = [12 ± √48] / 6 = [12 ± 4√3] / 6 = 2 ± (2√3)/3

So the critical points are x ≈ 0.845 and x ≈ 3.155.

Now, we'll use the second derivative test to determine whether each critical point is a relative minimum or maximum.

Find the second derivative, d^2y/dx^2:

• d^2y/dx^2 = 6x - 12

Evaluate the second derivative at each critical point:

• d^2y/dx^2 |_(x≈0.845) = 6(0.845) - 12 ≈ -6.93 < 0 => Relative maximum at x ≈ 0.845
• d^2y/dx^2 |_(x≈3.155) = 6(3.155) - 12 ≈ 6.93 > 0 => Relative minimum at x ≈ 3.155

Key Concepts: Finding critical points, second derivative test for relative extrema.

Question 3: Area, Volume, and Cross-Sections

This question usually involves finding the area between curves and volumes of solids of revolution or solids with known cross-sections.

The Setup

Let f(x) = 4x^2 and g(x) = x^4 - 4x^2.

Part (a): Find the area of the region enclosed by the graphs of f and g.

Solution:

First, find the points of intersection by setting f(x) = g(x):

• 4x^2 = x^4 - 4x^2
• x^4 - 8x^2 = 0
• x^2(x^2 - 8) = 0
• x = 0, x = ±√8 = ±2√2

The area of the region is given by the integral of the absolute difference between the two functions over the interval [-2√2, 2√2]. Notice that f(x) is above g(x) in this interval. Because of symmetry, we can integrate from 0 to 2√2 and multiply by 2:

• Area = 2 * ∫[0 to 2√2] (4x^2 - (x^4 - 4x^2)) dx = 2 * ∫[0 to 2√2] (8x^2 - x^4) dx
• Area = 2 * [ (8/3)x^3 - (1/5)x^5 ] evaluated from 0 to 2√2
• Area = 2 * [ (8/3)(2√2)^3 - (1/5)(2√2)^5 ] = 2 * [ (8/3)(16√2) - (1/5)(128√2) ]
• Area = 2 * [ (128√2)/3 - (128√2)/5 ] = 2 * [ (640√2 - 384√2) / 15 ] = 2 * [ (256√2) / 15 ] = (512√2) / 15

Key Concepts: Finding points of intersection, setting up integrals for area between curves, symmetry.

Part (b): Write, but do not evaluate, an integral expression that gives the volume of the solid generated when the region enclosed by the graphs of f and g is rotated about the line y = 7.

Solution:

This involves the washer method. The outer radius is 7 - g(x) and the inner radius is 7 - f(x). Again, because of symmetry, we can integrate from 0 to 2√2 and multiply by 2.

• Volume = 2π * ∫[0 to 2√2] [(7 - g(x))^2 - (7 - f(x))^2] dx = 2π * ∫[0 to 2√2] [(7 - (x^4 - 4x^2))^2 - (7 - 4x^2)^2] dx
• Volume = 2π * ∫[0 to 2√2] [(7 - x^4 + 4x^2)^2 - (7 - 4x^2)^2] dx

Key Concepts: Washer method for volumes of revolution, setting up the integral expression.

Part (c): The region enclosed by the graphs of f and g is the base of a solid. For this solid, each cross section perpendicular to the x-axis is a square. Find the volume of the solid.

Solution:

The side length of the square is given by the difference between the functions, f(x) - g(x) = 8x^2 - x^4. The area of the square is (8x^2 - x^4)^2. Because of symmetry, we can integrate from 0 to 2√2 and multiply by 2.

• Volume = 2 * ∫[0 to 2√2] (8x^2 - x^4)^2 dx = 2 * ∫[0 to 2√2] (64x^4 - 16x^6 + x^8) dx
• Volume = 2 * [ (64/5)x^5 - (16/7)x^7 + (1/9)x^9 ] evaluated from 0 to 2√2
• Volume = 2 * [ (64/5)(128√2) - (16/7)(512√2) + (1/9)(512*8√2) ]
• Volume = 2 * [ (8192√2)/5 - (8192√2)/7 + (4096√2)/9]
• Volume = 2 * [ (516096√2 - 368640√2 + 229376√2) / 315 ] = 2 * [ (376832√2) / 315 ] = (753664√2) / 315

Key Concepts: Solids with known cross-sections, setting up the integral for volume.

Question 4: Average Value and Related Concepts

This question often tests your understanding of the Mean Value Theorem, Average Value Theorem, and related applications.

The Setup

A particle moves along the x-axis. The velocity of the particle at time t is given by v(t) = t^2 - 8t + 12 for t ≥ 0.

Part (a): Find the acceleration of the particle at time t = 4.

Solution:

Acceleration is the derivative of velocity with respect to time.

• a(t) = v'(t) = 2t - 8
• a(4) = 2(4) - 8 = 0

The acceleration of the particle at t = 4 is 0.

Key Concept: Acceleration as the derivative of velocity.

Part (b): Is the speed of the particle increasing or decreasing at time t = 4? Give a reason for your answer.

Solution:

The speed is increasing if the velocity and acceleration have the same sign, and decreasing if they have opposite signs.

We know that a(4) = 0. We need to find v(4).

• v(4) = (4)^2 - 8(4) + 12 = 16 - 32 + 12 = -4

Since v(4) = -4 and a(4) = 0, we cannot determine whether the speed is increasing or decreasing based on the signs of velocity and acceleration alone. However, since a(4) = 0, the speed is momentarily neither increasing nor decreasing.

Key Concepts: Relationship between speed, velocity, and acceleration.

Part (c): Find all times t in the interval [0, 6] at which the particle changes direction. Justify your answer.

Solution:

The particle changes direction when the velocity changes sign. To find the times when the velocity changes sign, we need to find when v(t) = 0.

• t^2 - 8t + 12 = 0
• (t - 2)(t - 6) = 0
• t = 2, t = 6

We need to check if the velocity changes sign at these times.

• For 0 ≤ t < 2, v(t) > 0 (e.g., v(0) = 12 > 0)
• For 2 < t < 6, v(t) < 0 (e.g., v(4) = -4 < 0)
• For t > 6, v(t) > 0 (e.g., v(7) = 7^2 - 8(7) + 12 = 49 - 56 + 12 = 5 > 0)

The particle changes direction at t = 2 and t = 6.

Key Concepts: Finding when velocity is zero, analyzing the sign of velocity to determine changes in direction.

Part (d): Find the total distance traveled by the particle from time t = 0 to t = 6.

Solution:

Total distance traveled is the integral of the absolute value of the velocity function.

• Total Distance = ∫[0 to 6] |v(t)| dt = ∫[0 to 6] |t^2 - 8t + 12| dt

Since v(t) changes sign at t = 2 and t = 6, we need to split the integral:

• Total Distance = ∫[0 to 2] (t^2 - 8t + 12) dt - ∫[2 to 6] (t^2 - 8t + 12) dt

• ∫[0 to 2] (t^2 - 8t + 12) dt = [(1/3)t^3 - 4t^2 + 12t] evaluated from 0 to 2 = (8/3) - 16 + 24 = 8 + (8/3) = 32/3

• ∫[2 to 6] (t^2 - 8t + 12) dt = [(1/3)t^3 - 4t^2 + 12t] evaluated from 2 to 6 = [(1/3)(216) - 4(36) + 12(6)] - [(1/3)(8) - 4(4) + 12(2)] = [72 - 144 + 72] - [(8/3) - 16 + 24] = 0 - [8/3 + 8] = -32/3

• Total Distance = (32/3) - (-32/3) = 64/3

Key Concepts: Total distance as the integral of the absolute value of velocity, splitting the integral when velocity changes sign.

Question 5: Differential Equations

This question usually tests your ability to solve differential equations, analyze slope fields, and understand exponential growth or decay.

The Setup

Consider the differential equation dy/dx = (1/2)y.

Part (a): On the axes provided, sketch a slope field for the given differential equation at the nine points indicated.

Solution:

You would need the actual grid to draw the slope field. The slopes are determined by plugging in the y-value into dy/dx = (1/2)y.

• If y = 0, dy/dx = 0 (horizontal line)
• If y = 1, dy/dx = 1/2 (positive slope)
• If y = 2, dy/dx = 1 (steeper positive slope)
• If y = -1, dy/dx = -1/2 (negative slope)
• If y = -2, dy/dx = -1 (steeper negative slope)

Draw short line segments at each point with the appropriate slope.

Key Concept: Understanding slope fields and how they represent the solutions to a differential equation.

Part (b): Let y = f(x) be the particular solution to the given differential equation with f(0) = 3. Write an equation for the tangent line to the graph of f at x = 0.

Solution:

We know that f(0) = 3, so the point is (0, 3).

The slope of the tangent line at x = 0 is dy/dx |_(x=0) = (1/2)(3) = 3/2.

Using the point-slope form, the equation of the tangent line is:

• y - 3 = (3/2)(x - 0)
• y = (3/2)x + 3

Key Concepts: Finding the equation of a tangent line using the derivative.

Part (c): Find y = f(x), the particular solution to the given differential equation with f(0) = 3.

Solution:

Solve the differential equation using separation of variables:

• dy/dx = (1/2)y
• dy/y = (1/2) dx
• ∫ dy/y = ∫ (1/2) dx
• ln|y| = (1/2)x + C
• |y| = e^((1/2)x + C) = e^C * e^((1/2)x)
• y = A * e^((1/2)x) (where A = ±e^C)*

Using the initial condition f(0) = 3:

• 3 = A * e^((1/2)(0)) = A * e^0 = A
• A = 3

Therefore, the particular solution is:

• y = 3e^((1/2)x)

Key Concepts: Solving separable differential equations, using initial conditions to find particular solutions.

Question 6: Functions Defined by Integrals

This question usually involves dealing with functions defined as integrals and applying the Fundamental Theorem of Calculus.

The Setup

Let f be a function defined by f(x) = ∫[0 to x] sin(t^2) dt for 0 ≤ x ≤ 3.

Part (a): Find f'(x).

Solution:

Using the Fundamental Theorem of Calculus:

• f'(x) = sin(x^2)

Key Concept: Fundamental Theorem of Calculus.

Part (b): On what intervals is f increasing? Justify your answer.

Solution:

f is increasing when f'(x) > 0.

• sin(x^2) > 0

We need to find the intervals where sin(x^2) is positive. Since 0 ≤ x ≤ 3, we know that 0 ≤ x^2 ≤ 9. The sine function is positive in the first and second quadrants.

• 0 < x^2 < π => 0 < x < √π ≈ 1.772
• 2π < x^2 < 3π (not possible because x^2 must be less than or equal to 9, and 3π is approximately 9.42)

Therefore, f is increasing on the interval (0, √π).

Key Concepts: Using the derivative to determine intervals of increasing/decreasing, properties of trigonometric functions.

Part (c): Find the absolute maximum value of f on the interval 0 ≤ x ≤ 3. Justify your answer.

Solution:

We need to check the critical points and endpoints.

• Critical points occur when f'(x) = 0:
  • sin(x^2) = 0
  • x^2 = 0, π, 2π
  • x = 0, √π, √2π (√2π ≈ 2.507)

Now we need to evaluate f(x) at the critical points and endpoints:

• f(0) = ∫[0 to 0] sin(t^2) dt = 0
• f(√π) = ∫[0 to √π] sin(t^2) dt ≈ 0.895 (using a calculator)
• f(√2π) = ∫[0 to √2π] sin(t^2) dt ≈ 0.714 (using a calculator)
• f(3) = ∫[0 to 3] sin(t^2) dt ≈ 0.773 (using a calculator)

Comparing these values, the absolute maximum value of f on the interval [0, 3] is approximately 0.895.

Key Concepts: Finding critical points, testing critical points and endpoints to find absolute extrema, using a calculator to evaluate definite integrals.

Conclusion

The 2018 AP Calculus AB Free Response questions cover a wide range of calculus topics. By understanding the concepts behind each question and practicing your problem-solving skills, you can improve your performance on the AP exam. Remember to show your work, justify your answers, and use your calculator effectively when appropriate. Careful review of past free response questions, like these from 2018, is one of the best ways to prepare. Good luck!

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