Ap Calculus Ab 2022 Frq Answers

Calculus AB, often seen as the gateway to higher-level mathematics for many high school students, culminates in the challenging yet rewarding AP Calculus AB exam. The free-response questions (FRQs) are a critical component, demanding not just rote memorization, but a deep understanding of the core concepts and the ability to apply them creatively. The 2022 AP Calculus AB FRQs were no exception, testing students on topics ranging from related rates and accumulation functions to differential equations and applications of the definite integral. Understanding the solutions and the underlying reasoning is key to mastering the material.

Question 1: Rate of Snowfall and Melting

This problem explores the concepts of rates of change and accumulation, fundamental to calculus. We analyze the rate at which snow falls and melts on a driveway, using the given functions to determine the amount of snow accumulated over time.

Part (a): Finding the total amount of snow removed

The problem provides a function R(t) representing the rate at which snow is removed from the driveway in cubic feet per hour. To find the total amount of snow removed during the time interval from t = 0 to t = 6 hours, we need to integrate the rate function R(t) over that interval.

Mathematically, this is represented as:

∫₀⁶ R(t) dt

Using a calculator (as permitted on the AP exam for this type of calculation), we evaluate this definite integral to find the total amount of snow removed. The result is approximately 142.275 cubic feet.

Part (b): Determining if the volume of snow is increasing or decreasing at a specific time

To determine whether the volume of snow is increasing or decreasing at time t = 4, we need to compare the rate at which snow is falling, F(t), with the rate at which snow is being removed, R(t), at t = 4.

First, we evaluate F(4) and R(4) using the given functions. We find that F(4) ≈ 5.667 cubic feet per hour and R(4) ≈ 9.514 cubic feet per hour.

Since R(4) > F(4), the rate at which snow is being removed is greater than the rate at which it is falling at t = 4. Therefore, the volume of snow on the driveway is decreasing at t = 4.

Part (c): Finding the time when the volume of snow is at its maximum

To find the time when the volume of snow is at its maximum, we need to analyze the difference between the rate at which snow is falling and the rate at which it is being removed. Let A(t) represent the amount of snow on the driveway at time t. Then, the rate of change of A(t) is given by:

dA/dt = F(t) - R(t)

To find the critical points (potential maxima or minima), we set dA/dt = 0 and solve for t:

F(t) - R(t) = 0 => F(t) = R(t)

Using a calculator, we find that F(t) = R(t) at approximately t = 3. We also need to consider the endpoints of the interval, t = 0 and t = 6.

Now, we evaluate the amount of snow on the driveway at these critical points and endpoints. Let S(t) be the total amount of snow that has fallen up to time t: S(t) = ∫₀ᵗ F(x) dx. The amount of snow on the driveway A(t) is then S(t) - ∫₀ᵗ R(x) dx. We need the initial amount of snow; the problem states that at t=0, there are 25 cubic feet. Therefore: A(t) = 25 + S(t) - ∫₀ᵗ R(x) dx.

We already know ∫₀⁶ R(x) dx = 142.275. We calculate S(3) = ∫₀³ F(x) dx ≈ 31.080, and S(6) = ∫₀⁶ F(x) dx ≈ 53.350.

• A(0) = 25 + 0 - 0 = 25
• A(3) = 25 + 31.080 - ∫₀³ R(x) dx. ∫₀³ R(x) dx ≈ 20.057. Therefore, A(3) = 25 + 31.080 - 20.057 ≈ 36.023
• A(6) = 25 + 53.350 - 142.275 ≈ -63.925. Since volume cannot be negative, this means all the snow has been removed at some point and no more accumulates after t=6.

Therefore, the amount of snow on the driveway is at its maximum at t = 3 hours.

Part (d): Writing and solving a differential equation

In this part, the rate of removal changes. We are given a new rate of removal r(t) which is proportional to the amount of snow present at time t. We can write this as:

dr/dt = kA(t), where k is a constant of proportionality. The problem tells us r(t) = cA(t), with c = 1/1340.

We need to solve the differential equation dA/dt = F(t) - r(t) with initial condition A(8) (the amount of snow left at t=8, immediately before this removal process begins). First, we must calculate A(8). A(8) = 25 + ∫₀⁸ F(t) dt - ∫₀⁸ R(t) dt. Using a calculator, ∫₀⁸ F(t) dt ≈ 60.207 and ∫₀⁸ R(t) dt ≈ 151.272. Therefore A(8) ≈ 25 + 60.207 - 151.272 ≈ -66.065. Again, volume cannot be negative, so all the snow had been removed before t=8. Therefore, A(8)=0.

The differential equation we must solve is dA/dt = F(t) - (1/1340)A(t), with A(8) = 0. Since F(t)=0 for t>6 (no more snow is falling), we have:

dA/dt = -(1/1340)A(t)

This is a separable differential equation:

dA/A = -(1/1340) dt

Integrating both sides:

ln|A| = -(1/1340)t + C

A(t) = Ke^(-t/1340), where K = e^C

Using the initial condition A(8) = 0: 0 = Ke^(-8/1340). This implies K=0, thus A(t) = 0 for all t >= 8.

Therefore, the differential equation solution indicates that the amount of snow will remain at 0 cubic feet for t > 8. We are asked to find the time T when A(T) = 1 cubic foot, which is impossible given the solution; there will never be 1 cubic foot.

Question 2: Particle Motion

This question delves into the classic calculus topic of particle motion, requiring us to analyze the velocity and acceleration of a particle moving along a line to determine its position and behavior.

Part (a): Determining if the particle is speeding up or slowing down at a specific time

To determine whether the particle is speeding up or slowing down at time t = 5.5, we need to analyze the signs of the velocity, v(t), and acceleration, a(t), at that time.

The problem gives us the velocity function v(t). To find the acceleration, we need to find the derivative of the velocity function: a(t) = v'(t). Using a calculator, we can find that v(5.5) ≈ -0.453 and a(5.5) = v'(5.5) ≈ -1.359.

Since both v(5.5) and a(5.5) are negative, the particle is speeding up at t = 5.5. This is because the velocity and acceleration are in the same direction (both negative).

Part (b): Finding the average speed of the particle

The average speed of the particle over the time interval from t = 0 to t = 6 is given by:

(1/(6-0)) ∫₀⁶ |v(t)| dt

This formula calculates the average of the absolute value of the velocity, which represents the speed. Using a calculator to evaluate this definite integral, we find the average speed to be approximately 1.426.

Part (c): Finding the total distance traveled by the particle

The total distance traveled by the particle over the time interval from t = 0 to t = 6 is also given by:

∫₀⁶ |v(t)| dt

Notice that this is similar to the calculation in part (b). In part (b) we were looking for average speed; here we are looking for total distance. The average speed is the total distance divided by the time interval. Therefore, the total distance is simply the average speed multiplied by the time interval: 1.426 * 6 = 8.556.

Part (d): Finding the position of the particle at a specific time

We are given the initial position of the particle at time t = 0, x(0) = -2. To find the position of the particle at time t = 6, x(6), we need to use the fundamental theorem of calculus:

x(6) = x(0) + ∫₀⁶ v(t) dt

Using a calculator to evaluate the definite integral, we find that ∫₀⁶ v(t) dt ≈ 1.655.

Therefore, x(6) = -2 + 1.655 ≈ -0.345.

Question 3: Area and Volume

This problem focuses on area and volume calculations using integration. It involves finding the area of a region bounded by curves and then finding the volume of a solid generated by revolving that region around an axis.

Part (a): Finding the area of the region

The region R is bounded by the graphs of f(x) and g(x). To find the area of R, we need to integrate the difference between the two functions over the interval where f(x) ≥ g(x). We can use a calculator to find that the curves intersect at approximately x = 0 and x = 1.032. Over that interval, f(x) is above g(x).

Area = ∫₀¹°³² (f(x) - g(x)) dx

Using a calculator to evaluate this definite integral, we find the area of region R to be approximately 0.514.

Part (b): Finding the volume of the solid generated by revolving the region around the line y = 4

To find the volume of the solid generated by revolving region R around the line y = 4, we use the washer method. The outer radius of the washer is the distance from the line y = 4 to the function g(x), which is 4 - g(x). The inner radius of the washer is the distance from the line y = 4 to the function f(x), which is 4 - f(x).

Volume = π ∫₀¹°³² [(4 - g(x))² - (4 - f(x))²] dx

Using a calculator to evaluate this definite integral, we find the volume of the solid to be approximately 11.118.

Part (c): Setting up, but not evaluating, an integral expression for the volume of a solid with square cross-sections

In this part, region R is the base of a solid. For this solid, each cross-section perpendicular to the x-axis is a square. The side length of each square is equal to the distance between the two curves, which is f(x) - g(x). The area of each square is therefore [f(x) - g(x)]².

To find the volume of the solid, we integrate the area of the squares over the interval from x = 0 to x = 1.032:

Volume = ∫₀¹°³² [f(x) - g(x)]² dx

This integral expression represents the volume of the solid with square cross-sections. Note that we were not required to calculate the volume, only to set up the integral.

Question 4: Related Rates and Accumulation

This problem combines the concepts of related rates and accumulation in the context of the flow of water into and out of a tank.

Part (a): Finding the rate at which the amount of water in the tank is changing at a specific time

Let W(t) represent the amount of water in the tank at time t. The rate at which the amount of water in the tank is changing is given by the difference between the rate at which water is entering the tank, E(t), and the rate at which water is leaving the tank, L(t):

dW/dt = E(t) - L(t)

To find the rate at which the amount of water is changing at time t = 5, we evaluate E(5) and L(5):

• E(5) = 25 + 10e^(-0.03*5) ≈ 33.562
• L(5) = 2.75(5) = 13.75

Therefore, dW/dt at t = 5 is approximately 33.562 - 13.75 ≈ 19.812 gallons per minute.

Part (b): Determining the time interval during which the amount of water in the tank is decreasing

The amount of water in the tank is decreasing when dW/dt < 0, which means E(t) - L(t) < 0, or E(t) < L(t).

To find the time intervals when this is true, we need to find where the two curves intersect. We are told that E(t) = L(t) at t = 3.160 and t = 8.396. We also know that L(t) is a line and E(t) is a decreasing exponential. Therefore, L(t) > E(t) between t=3.160 and t=8.396.

Thus, the amount of water in the tank is decreasing for 3.160 < t < 8.396.

Part (c): Using a definite integral to find the amount of water in the tank at a specific time

We are given that the tank contains 50 gallons of water at time t = 0. To find the amount of water in the tank at time t = 18, we use the fundamental theorem of calculus:

W(18) = W(0) + ∫₀¹⁸ [E(t) - L(t)] dt

Using a calculator to evaluate the definite integral, we find that ∫₀¹⁸ [E(t) - L(t)] dt ≈ 123.744.

Therefore, W(18) = 50 + 123.744 ≈ 173.744 gallons.

Question 5: Differential Equations and Slope Fields

This problem focuses on differential equations and their graphical representations using slope fields.

Part (a): Sketching a slope field

A slope field is a graphical representation of a differential equation, where short line segments are drawn at various points in the xy-plane, with the slope of each segment equal to the value of dy/dx at that point. In this case, we are given the differential equation dy/dx = (1/2)x + y - 1. We must carefully calculate the slope at the specified points on the axes and sketch a short line segment at each point. Remember that a horizontal line has slope 0, and a vertical line has undefined slope. The closer the slopes are to each other, the more parallel the line segments should appear.

Part (b): Using Euler's method to approximate a solution

Euler's method is a numerical technique for approximating the solution to a differential equation. Given an initial condition (x₀, y₀) and a step size Δx, we can approximate the value of y at a later point x₁ = x₀ + Δx using the formula:

y₁ ≈ y₀ + (dy/dx)(x₀, y₀) * Δx

We are given the initial condition f(2) = 3 and a step size of 0.1. We are asked to approximate f(2.2) using two steps.

• Step 1: Approximate f(2.1)
  • (dy/dx)(2, 3) = (1/2)(2) + 3 - 1 = 3
  • f(2.1) ≈ 3 + (3)(0.1) = 3.3
• Step 2: Approximate f(2.2)
  • (dy/dx)(2.1, 3.3) = (1/2)(2.1) + 3.3 - 1 = 3.35
  • f(2.2) ≈ 3.3 + (3.35)(0.1) = 3.635

Therefore, Euler's method gives us the approximation f(2.2) ≈ 3.635.

Part (c): Solving a separable differential equation

We are given the differential equation dy/dx = (1/2)x + y - 1. To solve this equation, we must first rewrite it using a u-substitution. Let u = (1/2)x + y - 1. Then, du/dx = (1/2) + dy/dx, so dy/dx = du/dx - 1/2. Substituting, we have:

du/dx - 1/2 = u

du/dx = u + 1/2

This is now a separable differential equation. Separating the variables, we get:

du/(u + 1/2) = dx

Integrating both sides, we get:

ln|u + 1/2| = x + C

Substituting back u = (1/2)x + y - 1, we have:

ln|(1/2)x + y - 1 + 1/2| = x + C

ln|(1/2)x + y - 1/2| = x + C

Exponentiating both sides, we get:

|(1/2)x + y - 1/2| = e^(x+C) = e^x * e^C = Ke^x, where K = e^C

(1/2)x + y - 1/2 = ±Ke^x

y = - (1/2)x + 1/2 ± Ke^x

We are given the initial condition f(2) = 3. Substituting x = 2 and y = 3 into the equation, we can solve for K:

3 = -(1/2)(2) + 1/2 + Ke²

3 = -1 + 1/2 + Ke²

7/2 = Ke²

K = (7/2)e^(-2)

Therefore, the particular solution to the differential equation is:

y = -(1/2)x + 1/2 + (7/2)e^(-2) * e^x

y = -(1/2)x + 1/2 + (7/2)e^(x-2)

Question 6: Applications of the Definite Integral

This problem explores various applications of the definite integral, including finding the area between curves and calculating quantities involving rates of change. It requires a solid understanding of integral calculus and its applications.

Part (a): Finding the value of an integral using given information

The prompt describes a cable attached to a bridge. The density of the cable is modeled by a function λ(x). The tension in the cable at x=0 is given as 500 Newtons. The fundamental theorem of calculus states that:

T'(x) = λ(x)g

Therefore, T(b) = T(0) + ∫₀ᵇ λ(x)g dx, where g = 9.8 m/s² is the acceleration due to gravity. We are given T(0) = 500 N, and T(28) = 500 + ∫₀²⁸ λ(x)g dx

We are also told ∫₀²⁸ λ(x) dx = 70. This means ∫₀²⁸ λ(x)g dx = 70g = 70(9.8) = 686.

Thus, T(28) = 500 + 686 = 1186 Newtons.

Part (b): Finding the average tension in the cable

The average tension in the cable between x = 0 and x = 28 is given by:

Average Tension = (1/(28-0)) ∫₀²⁸ T(x) dx

We don't have T(x). However, we can integrate T(b) = T(0) + ∫₀ᵇ λ(x)g dx from 0 to 28:

∫₀²⁸ T(b) db = ∫₀²⁸ (500 + ∫₀ᵇ λ(x)g dx) db = ∫₀²⁸ 500 db + ∫₀²⁸ ∫₀ᵇ λ(x)g dx db

The first term is easy to calculate. ∫₀²⁸ 500 db = 500(28) = 14000. The second term is more complicated. However, the prompt tells us that ∫₀²⁸ ∫₀ᵇ λ(x)g dx db = 8000.

Therefore, ∫₀²⁸ T(b) db = 14000 + 8000 = 22000.

The average tension is then (1/28) * 22000 = 785.714 Newtons.

Part (c): Modeling the sag of the cable with a differential equation

Let y(x) be the height of the cable above the bridge at position x. The prompt gives the differential equation y''(x) = λ(x)g / T(x) .

We are told that the cable is attached to the bridge at x=0 and x=28, so y(0) = y(28) = 0. We must approximate y'(5) using the tangent line to y(x) at x=0.

Using the tangent line approximation, y(x) ≈ y(0) + y'(0)(x-0) = y'(0)x

However, we don't know y'(0)! Furthermore, we do not know how to solve the provided differential equation. What we do know is that y''(x) represents the concavity of the function. If we assume that the tension is constant throughout the cable, then y''(x) is proportional to λ(x). However, the problem explicitly states that the tension is not constant.

Furthermore, because we don't know y'(0), approximating y'(5) directly is difficult. This suggests we need to make additional simplifying assumptions that allow us to side-step these challenges.

Without further information or justifiable simplifications, this part is difficult to approach effectively. Generally, simplifying assumptions should be clearly stated when used, and their limitations acknowledged.

In conclusion, the 2022 AP Calculus AB FRQs presented a comprehensive assessment of students' understanding of fundamental calculus concepts and their ability to apply them to a variety of problems. From related rates and accumulation to area, volume, and differential equations, these questions demanded a deep understanding of the material and the ability to think critically and creatively. By studying these solutions and understanding the underlying reasoning, students can gain valuable insights and improve their problem-solving skills in calculus.