# AP Calculus AB FRQ Solutions 2022

Updated: January 23rd, 2023

## Question 1 (a)

Since we're given a rate function, the expression that computes the total number of vehicles that arrive at the toll is: ## Question 1 (b)

In general, the average value of a function is computed using the equation The average value of the rate function is going to be: Since this is a Calculator Allowed Problem, I was able to compute the answer as Therefore, the average rate at which vehicles arrive at the toll plaza is 375.52 vehicles per hour.

## Question 1(c)

To figure out when a rate function is increasing or decreasing, we first have to take a derivative. When we plug in t = 1 into our derivative function, A'(1) is going to be positive.

Because A'(1)>0, this implies that the rate function A(t) is increasing.

## Question 1(d)

Since we are looking for the greatest number of vehicles at the toll plaza, my first instinct is to take a derivative of N(t). By the Fundamental Theorem of Calculus, this derivative will basically cancel out with integral. So, THe next step is to set the derivative of N(t) to 0 to find critical points, or values of x that give potential extremas (max/mins) Solving for the equation above gives Using a calculator to compute t gives t = 1.47. When plotting the function: The function also as a x-intercept at 3.598.

Therefore, our two critical points are 1.47 and 3.598. Doing a "number line analysis," we can see that t = 3.598 has a maximum value. Plugging in 3.598 into our function N(t) as shown to the left, our calculator tells us the maximum value (or the greatest number of vehicles in line a the toll plaza).

## Question 2

2. Let f and g be the functions defined by f(x) = ln(x + 3) and g(x) = x + 2x³. The graphs of f and g, shown in the figure above, intersect at x = -2 and x = B, where B > 0.

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

The area is simply the integral from x = -2 and x = B of the difference between the bigger function and the smaller function (i.e., f(x) - g(x)). Note: I found B = 0.78 (the rightmost intersection) using the "intercept function" of my graphing calculator

Since calculators are allowed in this portion of the test, we used "Math-9" on a TI-84 to calculate this integral. The area between the curves is 3.6.

(b) For -2 ≤ x ≤ B, let h(x) be the vertical distance between the graphs of f and g. Is h increasing or decreasing at x = -0.5? Give a reason for your answer.

Visually, we can see that as the "vertical" line moves along the x-axis towards x = B, the length of the line gets smaller at x = -0.5. So, we expect the answer to be decreasing.

Using a more rigourous method, notice that h(x) = f(x) - g(x) (or the vertical distance at some point x can be found by the difference between f(x) and g(x). To find whether h(x) is increasing or decreasing, we must take a derivative. The last step is to plug in x = -0.5 into the derivative h'(x) and observe if the value is positive or negative.

Remember: If h'(x)>0, then h(x) is increasing. If h'(x)<0, then h(x) is decreasing. I ended up not bothering to conpute the final answer for h'(-0.5), but instead noticed that it's negative. Therefore, the vertical distance h(x) is decreasing at x = -0.5.

(c) The region enclosed by the graphs of f and g is the base of a solid. Cross sections of the solid taken perpendicular to the x-axis are squares. Find the volume of the solid.

The formula to find the volume of some shape with a known cross section is the integral of the area function A(x). Since the region between f and g is the base of the solid and the cross sections are squares. One of the sides of the square is the vertical distance between f and g (i.e., s = f(x) - g(x) = h(x)). So therefore, the area function of a square is A(s) = s^2, where s denotes the length of any side. In the problem's case, the area must be h(x)^2, and the volume can be computed using this expression: Using the "Math-9" technique to compute the final answer, we get 5.34 to be the volume of the solid.

(d) A vertical line in the xy-plane travels from left to right along the base of the solid described in part (c). The vertical line is moving at a constant rate of 7 units per second. Find the rate of change of the area of the cross section above the vertical line with respect to time when the vertical line is at position x = -0.5.

The area function cross section from Part (c) can be described using this expression: Because the vertical line is moving at a rate of 7 units per second, it's true that: Taking the derivative of A(x), using the power rule, chain rule, and other derivative rules. Also note that the expression is derive implicitly because we want to take a derivative in terms of time "t". Looks a little crazy doesn't it? Next, we are looking for the rate of change at the specific position of x = -0.5, so we can plug in x = -0.5 and dx/dt = 7 Luckily, this is a "calculator allowed" question, so our answer is -9.3 square units per second.

## Question 3

3. Let f be a differentiable function with f(4) = 3. On the interval 0 < x < 7, the graph of f', the derivative of f, consists of a semicircle and two line segments, as shown in the figure above.

(a) Find f(0) and f(5).

A general rule to get f(x) from f'(x) is to take the integral of f'(x). By the Fundamental Theorem of Calculus, To find f(x), we must take advantage of the fact that f(4) = 3. The a in the equation above can be 4, and the "C" can be f(4) = 3. Therefore, to find f(0), just plug in x = 0 into the above equation. The first operation is a property of integrals. To flip the orders of integration, add a negative sign in front. The remaining integral is the area of f' from 0 to 4, which is equivalent to the area of a semi-circle of radius 2. Also since the area of the semi-circle is beneath the x-axis, the integral is negative (i.e., -2pi). After cancelling the negatives, the final answer turns to be 2pi + 3.

The value of f(5) follows a similar logic. We are going to use this equation again. Instead, we're plugging in x = 5. The integral is equivalent to the area of the triangle formed beneath the line of f' between x = 4 and x = 5. The integral is simply 1/2. So, our final answer is 3.5.

(b) Find the x-coordinates of all points of inflection of the graph of f for 0 < x < 7. Justify your answer.

A point of inflection occurs when concavity changes. We really only have the graph of f'(x), so we're looking for a point where the slope of f' changes from positive to negative or negative to positive.

For example at x = 2, the slope of f' changes from negative to positive. x = 2 is a point of inflection.

Also, note the "corner" at x = 6. Though the second derivative does not exist at this point, the slope of f' changes from positive to negative. Therefore, x = 6 is also an inflection point.

(c) Let g be the function defined by g(x) = f(x) - x. On what intervals, if any, is g decreasing for 0 < x < 7? Show the analysis that leads to your answer.

Finding decreasing intervals involves a derivative. Taking the derivative of g(x) gives g'(x) = f'(x) - 1.

Set g'(x) = 0 to find critical points gives

f'(x) - 1 = 0

f'(x) = 1

So, looking back at the given graph of f', we are looking for x-values where f'(x) = 1. Observe that this happens at x = 5 and x = 7. Doing a number line analysis, similar to: This indicates that g' is negative from x = 0 to x = 5. This implies that g(x) is decreasing on the interval (0,5).

(d) For the function g defined in part (c), find the absolute minimum value on the interval 0 < x < 7. Justify your answer.

Absolute minimums occur on critical points or endpoints. From part (c), we know that the critical points of g(x) are at x = 5 and x = 7. So the "candidates" for absolute minimum values are x = 0, 5, and 7.

Also from part (c), we know that g(x) is decreasing on (0,5). We can deduce that x = 0 is not going to have a absolute minimum. A similar argument can be made for x = 7 since it's increasing around that point. Therefore, a minimum value does not exist at x = 7.

So, we know that x = 5 has a minimum value by process of elimination. Also, the number line analysis shows a local minimum at x = 5 since the function is decreasing to the left of x = 5 and increasing to the right of it.

To find the minimum value, we must plug in x =5 into g(x) = f(x) - x. The minimum value of g(x) is -1.

## Question 4

4. An ice sculpture melts in such a way that it can be modeled as a cone that maintains a conical shape as it decreases in size. The radius of the base of the cone is given by a twice-differentiable function r, where r(t) is measured in centimeters and t is measured in days. The table above gives selected values of r'(t), the rate of change of the radius, over the time interval 0 < t < 12.

(a) Approximate r"(8.5) using the average rate of change of r' over the interval 7 < t < 10. Show the computations that lead to your answer, and indicate units of measure.

The average rate of change formula in this context will have the form: This might be unusual to see a derivative function on the right side of the equation, but remember we are looking for the "average rate of change" of a "rate of change function."

In this case, a = 7 and b = 10 because those values are the endpoints of the interested interval. The rest is to evaluate the expression: Our final answer is 0.2 cm/day^2.

(b) Is there a time t, 0 < t < 3, for which r'(t) = -6 ? Justify your answer.

The answer is YES due to the Intermediate Value Theorem. We must also check that the r'(x) is continuous, which it is because r is twice-differentiable, and that -6 is in between -6.1 and -5 (it is!).

(c) Use a right Riemann sum with the four subintervals indicated in the table to approximate the value of the integral from 0 to 12 of r'(t)dt.

In this question, we are using R_4 to approximate the integral in question. The Right Riemann sums uses the most right r'(t) values from the table. The first rectangle has the area of (3-0)*(-5) = -15. The second has area (7-3)*(-4.4) = -17.6, third area is (10-7)*(-3.8) = -11.4, and fourth area is (12-10)(-3.5) = -7. Add these areas up gives the right Riemann sum with four subintervals of -51. (d) The height of the cone decreases at a rate of 2 centimeters per day. At time t = 3 days, the radius is 100 centimeters and the height is 50 centimeters. Find the rate of change of the volume of the cone with respect to time, in cubic centimeters per day, at time t = 3 days. (The volume V of a cone with radius r and height h is V = (1/3)(pi)(r^2)(h).)

Nice, a related rates question.

The volume of the cone is given as: Now, we want this volume equation to be of only one variable on the right hand side (e.g., either only has "r" or "h" as a variable.) So, let's dissect the question. "The height of the cone decreases at a rate of 2 centimeters per day" means that: "At time t = 3 days, the radius is 100 centimeters and the height is 50 centimeters" means:  Like in the picture above right, if we know the radius and height, we can connect the two variables with the Pythagorean theorem. We next find "c". This tells us that we can substitute a value involving "c" and "h" into "r^2":  So, the volume equation from before becomes Note that the second line simplifies the equation by distributing (1/3)(pi)(h).

Since we have an equation of one-variable, we may take a derivative in terms of "t". Remember to multiply each term by "dh/dt" by the Chain Rule. Recall that dh/dt = -2 and h = 50, so the rate of change becomes: The final answer, or rate of change of the volume at time t = 3 days, is around -10,471.976 cubic centimeters per day.

## Question 5

5. Consider the differential equation dy/dx = (1/2)sin(pi/2x)sqrt(y+7). Let y = f(x) be the particular solution to the differential equation with the initial condition f(1) = 2. The function f is defined for all real numbers.

(a) A portion of the slope field for the differential equation is given below. Sketch the solution curve through the point (1, 2). For questions like this, start with the point (1,2). Then, follow the "flow" of the slope field to get the particular solution. It should look something similar to this: (b) Write an equation for the line tangent to the solution curve in part (a) at the point (1, 2). Use the equation to approximate f(0.8).

I always start off the tangent line with the slope-intercept equation: We need the slope and the y-intercept to complete the equation. Firstly, note that the slope of the tangent line can be computed by f'(1), or plugging x = 1 into the derivative. So, the slope of the tangent line equals 3/2. To find the y-intercept "b", we'll plug (1, 2) into the tangent line equation.

Therefore, the equation of the tangent line is y = (3/2)x + (1/2). The last thing to do is to plug x = 0.8 into the tangent line, and that is our final answer.  Using the tangent line at (1,2), the approximate value for f(0.8) is (17/10) = 1.7.

(c) It is known that f''(x) > 0 for -1 < x < 1. Is the approximation found in part (b) an overestimate or an underestimate for f(0.8) ? Give a reason for your answer.

f"(0) > 0 means that f(x) is concave up between x = -1 and x = 1. In these scenarios, the tangent line is always going to be underestimating the true value of f(0.8). Here's an example: (d) Use separation of variables to find y = f(x), the particular solution to the differential equation dy/dx = (1/2)sin(pi/2x)sqrt(y+7) with the initial condition f(1) = 2.

First, we must use "separate the variables" by moving all the "y" functions to the left and all the "x" functions to the right. Then, integrate with respect of each variable After a bit of u-substitution, the solution is going to be of the form Since f(1) = 2, we can plug in x = 1 and y = 2 to get the value of C.  ## Question 6

6. Particle P moves along the x-axis such that, for time t > 0, its position is given by x_P(t) = 6 - 4e^(-t). Particle Q moves along the y-axis such that, for time t > 0, its velocity is given by v_Q(t) = 1/(t^2). At time t = 1, the position of particle Q is y_Q(1) = 2.

(a) Find v_P(t), the velocity of particle P at time t.

We know the position function of Particle P to be x_P(t) = 6 - 4e^(-t). The particle's velocity will simply be the derivative of x_P(t). Our answer is v_P(t) = 4e^t.

(b) Find a_Q(t), the acceleration of particle Q at time t. Find all times t, for t > 0, when the speed of particle Q is decreasing. Justify your answer.

Similar to part (a), we need to take the derivative of v_Q(t) = 1/(t^2) (the velocity function of Q). The acceleration function of Particle Q is a(t) = -2/(t^3).

The second portion of this question asks when the speed of the particle is decreasing. This happens when the velocity's sign (positive or negative) is opposite of the acceleration's sign. For t > 0, notice that v_Q(t) is always positive and and a_Q(t) is always negative. Therefore, for all positive times (AKA t > 0), the speed of Q is decreasing.

(c) Find y_Q(t), the position of particle Q at time t.

This question will use the velocity function of Q again, but we're going to integrate it with the fact that y_Q(1) = 2. Firstly, the integral evaluate to Next, we plug in t = 1 and y_Q = 2 to get "C".  Our specific solution is y_Q(t) = -1/t + 3.

(d) As t -> inf, which particle will eventually be farther from the origin? Give a reason for your answer.

Let's just take the limit as x approaches infinity of x_P(t) (that is given) and y_Q(t) (that we found in part c).

The limit of x_P(t) is in yellow, while the limit of y_Q(t) is in white.  We can see that the limit of x_P(t) is greater than the limit of y_Q(t). So therefore, Particle P will be furthest away from the origin.