# AP Calculus AB — Applications of Calculus Principles

This is our AP Calculus AB unit test on the Applications of Calculus Principles. These questions cover parametric equations, motion, related rates, volumes of revolution, and volumes of known cross sections. These problems require you to use previous knowledge of derivatives and integrals to solve a real-life problem. These types of problems are heavily tested on the AP exam to make sure the student can apply calculus to real phenomenon.

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 Question 1
Eliminate the parameter for the following. What do these parametric equations represent in the $x-y$ plane?

$x = \sin(t) \quad$ $y = \cos(t)$

 A $\text{A}$ $\text{circle}$ $\text{of}$ $\text{radius}$ $1.$ B $\text{A}$ $\text{sinusoidal}$ $\text{function}$. C $\text{An}$ $\text{ellipse.}$ D $\text{A}$ $\text{parabola.}$
Question 1 Explanation:
The correct answer is (A). Recall the trigonometric identity below.

$\sin^2⁡(a) + \cos^2⁡(a) = 1$

Therefore, we can eliminate the parameter to arrive at the equation below:

$\sin^2⁡(t) + \cos^2⁡(t)$ $= x^2 + y^2 = 1$

The equation $x^2 + y^2 = 1$ is the equation of a circle with radius equal to $1$.
 Question 2
Identify the second derivative,$\dfrac{dy^2}{d^2x}$ for the parametric equations below.

$x = \sin(3t) \quad$ $y = 4\cos(3t)$

 A $$\dfrac{dy^2}{d^2x} = −\dfrac{16x}{y}$$ B $$\dfrac{dy^2}{d^2x} = −192y^{−2}$$ C $$\dfrac{dy^2}{d^2x} = −4y^{−3}$$ D $$\dfrac{dy^2}{d^2x} = −256y^{−3}$$
Question 2 Explanation:
The correct answer is (D). It is crucial to recall the following formula.

$\dfrac{dy^2}{d^2x}$ $= \dfrac{\frac{d\left(\frac{dy}{dx}\right)}{dt}}{\frac{dx}{dt}}$

It is also important to recall the following formula.

$\dfrac{dy}{dx}$ $= \dfrac{\frac{dy}{dt}}{\frac{dx}{dt}}$

We can find the second derivative by finding each of the separate parts of the formula.

$\dfrac{dx}{dt} = 3\cos(3t)$

$\dfrac{dy}{dt} = −12\sin(3t)$

Thus, we can find $\frac{dy}{dx}$ by substituting these into the formula above.

$\dfrac{dy}{dx} = \dfrac{−12\sin(3t)}{3\cos(3t)}$ $= −4\tan(3t)$

Then, we can find the derivative of $\frac{dy}{dx}$ with respect to $t$.

$\dfrac{d\left(\frac{dy}{dx}\right)}{dt}$ $= −12\sec^2(3t)$

And then finally, we can find the second derivative by dividing the derivative of $\frac{dy}{dx}$ with respect to $t$ by the derivative of $x$ with respect to $t$.

$\dfrac{dy^2}{d^2x} = −\dfrac{12\sec^2(3t)}{3\cos(3t)}$ $= −4 \cdot \dfrac{1}{\cos^3(3t)}$

Now, we can identify that $\cos⁡(3t) = \frac{y}{4}$. Therefore, we can substitute this into the equation above.

$\dfrac{dy^2}{d^2x} = −\dfrac{4}{\left(\frac{y}{4}\right)^3}$ $= −\dfrac{256}{y^3} = −256y^{−3}$

This can also be determined by eliminating the parameter and taking the derivative implicitly.

$x^2 + \left(\dfrac{y}{4}\right)^2$ $= x^2 + \dfrac{y^2}{16} = 1$

$2x + \dfrac{y}{8} \cdot \dfrac{dy}{dx} = 0$

$\dfrac{dy}{dx} = −\dfrac{16x}{y}$

$\dfrac{dy^2}{d^2x}$ $= −\dfrac{y(16) − (16x)\left(\frac{dy}{dx}\right)}{y^2}$ $= \dfrac{−16y^2 − 256x^2}{y^3}$ $= −256y^{−3}$
 Question 3
Identify where there exist horizontal tangents for the parametric equations below.

$x = \ln(t^2) − 6t^3 \qquad$ $y = 4t^2 − 5\ln(t)$

 A $\text{There}$ $\text{are}$ $\text{no}$ $\text{horizontal}$ $\text{tangents.}$ B $t = ± \dfrac{1}{3}$ C $t = ± \sqrt{\dfrac{5}{8}}$ D $t = 0$
Question 3 Explanation:
The correct answer is (C). Recall the formula below.

$\dfrac{dy}{dx}$ $= \dfrac{\frac{dy}{dt}}{\frac{dx}{dt}}$

We know that horizontal tangents exist where $\frac{dy}{dx} = 0$. This implies that $\frac{dy}{dt} = 0$ results in a horizontal tangent.

$\dfrac{dy}{dt} = 8t − \dfrac{5}{t}$ $= \dfrac{8t − 5}{t}$

$\dfrac{dy}{dt}$ $= \dfrac{8t − 5}{t} = 0$

$t = ± \sqrt{\dfrac{5}{8}}$

It is important to recognize that $t ≠ 0$ and that $\frac{dx}{dt} ≠ 0$ for these horizontal tangents to exist.
 Question 4
Identify the velocity and acceleration functions of the particle moving according to the position function below.

$s(t) = 4t^3 + 3\ln(t)$

 A $v(t) = 24t − \frac{3}{t^2}$ $a(t) = 12t^2 + \frac{3}{t}$ B $v(t) = 12t^2 + \frac{3}{t}$ $a(t) = 24t − \frac{3}{t^2}$ C $v(t) = 12t^2 + \frac{3}{t}$ $a(t) = 24t + \frac{3}{t^2}$ D $v(t) = 12t^2 + \frac{1}{t}$ $a(t) = 24t − \frac{1}{t^2}$
Question 4 Explanation:
The correct answer is (B). The velocity and acceleration functions would be of the form below.

$s'(t) = v(t)$ $= 12t^2 + \dfrac{3}{t}$

$s''(t) = v'(t)$ $= a(t) = 24t − \dfrac{3}{t^2}$
 Question 5
Let $s(t)$ represent the position of a particle with respect to time $t$. Find the distance traveled by the particle from $t = 0$ to $t = 4$

$s(t) = t^2 − 2t − 3$

 A $\dfrac{34}{3}$ B $−\dfrac{20}{3}$ C $9$ D $10$
Question 5 Explanation:
The correct answer is (D). Recall that the formula for distance is as follows.

Distance $= \displaystyle\int_a^b |v(t)|dt$

Do not confuse distance with displacement, as these measures are very different.
We must first find the velocity function.

$v(t) = s'(t)$ $= 2t − 2$

Then, we can integrate this function with the limits $t = 0$ to $t = 4$.

$\displaystyle\int_0^4 |2t − 2|dt$

However, it is important to recognize that this integral must be split into two different integrals.
The function $|2t − 2|$ is split into the piece-wise function below.

$|2t − 2|$ $= \begin{cases} –2t + 2, & \text{t ≤ 1} \\ ~~~2t – 2, & \text{t > 1} \end{cases}$

Therefore, the integral must be rewritten in the form below.

$\displaystyle\int_0^4 |2t − 2|dt$ $= \displaystyle\int_0^1 −2t + 2dt + \displaystyle\int_1^4 2t − 2dt$

$= [−t^2 + 2t]_0^1 + [t^2 − 2t]_1^4$

$= [(−1 + 2) − 0]$ $+~[(4)^2 − 2(4) − (1^2 − 2(1))]$

$= 1 + 9 = 10$

Therefore, the distance traveled is 10.
 Question 6
Determine the displacement of the particle with position function $s(t)$ from $t = 2$ to $t = 5$

$s(t) = −\sin\left(\dfrac{π}{2}t\right)$

 A $−1$ B $3$ C $1$ D $0$
Question 6 Explanation:
The correct answer is (A). Recall the formula for displacement.

Displacement $= \displaystyle\int_a^b v(t)dt$

Therefore, we must first find the velocity function.

$s'(t) = v(t)$ $= −\dfrac{π}{2}\cos\left(\dfrac{π}{2}t\right)$

Then, we can integrate this function with the integral below.

$\displaystyle\int_2^5 −\dfrac{π}{2}\cos\left(\dfrac{π}{2}t\right)dt$

$= −\sin\left(\dfrac{π}{2}t\right) |_2^5$ $= −(1) − ((0)) = −1$

Therefore, the displacement of the particle is $−1$.
 Question 7
Identify the speed of the particle at the given time. Do not worry about units.

$s(t) = −4t^2, \quad$ $t = 4$

 A $64$ B $32$ C $−64$ D $−32$
Question 7 Explanation:
The correct answer is (B). Recall that the speed, $s$, of a function is as follows.

$s = |v(t)|$

Thus, we must identify the velocity function,

$s'(t) = v(t) = −8t$

The speed is then as follows.

$|v(4)| = |−8(4)|$ $= |−32| = 32$
 Question 8
Identify the position function for the following. It is known that $s(0) = 1$

$v(t) = \sqrt{\tan^2(t) + 1}$

 A $s(t) = \sec(t) \tan(t) + 1$ B $s(t) = \ln|\sec(x) + \tan(x)|$ C $s(t) = \ln|\sec(x) + \tan(x)| + 1$ D $s(t) = \sec(t) \tan(t)$
Question 8 Explanation:
The correct answer is (C). It is important to recall the trigonometric identity below.

$\sec^2(a) = \tan^2(a) + 1$

Thus, the velocity function can be rewritten in the form below.

$v(t) = \sec(t)$

To find the position function, we must integrate the velocity function.

$s(t) = \displaystyle\int v(t)dt$ $= \displaystyle\int \sec(t)dt$ $= \ln|\sec(x) + \tan(x)| + C$

Now, we are given that $s(0) = 1$. We must then solve for $C$.

$\ln(\sec(0) + \tan(0)) + C$ $= \ln(1) + C$ $= 0 + C = C = 1$

Therefore, $s(t) = \ln⁡|\sec⁡(x) + \tan⁡(x) | + 1$.
 Question 9
Sand is pouring out of a bag at a constant rate of $4 \text{ cm}^3 \text{ s}^{-1}$, which forms a conical pile on the ground beneath the bag. The cone has a radius that is equal to its height. Identify the rate of change of the radius of the cone when its radius is $2 \text{ cm}$.

 A $\dfrac{1}{π} \text{ cm s}^{−1}$ B $\dfrac{1}{3π} \text{ cm s}^{−1}$ C $4 \text{ cm s}^{−1}$ D $\text{Not}$ $\text{enough}$ $\text{information}$ $\text{given.}$
Question 9 Explanation:
The correct answer is (A). First, we need a relation, or some relationship between the different pieces of given information.
The volume of a cone is of the form below.

$V = \dfrac{1}{3}πr^2h$

This will be our relation for this problem. It is also important to recognize that the rate of the sand exiting the bag is equivalent to the rate of change of the volume of the cone. Each grain of sand that adds to the pile adds volume to the cone.

$\dfrac{dv}{dt} = 4$

Now that we have all the different necessary pieces, we can begin the problem.
First, we take the derivative of the relation, which can be seen below. Be sure to take the derivative implicitly. Don’t forget to recognize that the formula is a product of two variables, being $h$ and $r$, meaning that product rule is necessary for derivation.

$\dfrac{dv}{dt}$ $= \dfrac{2}{3}πrh \cdot \dfrac{dr}{dt} + \dfrac{1}{3}πr^2 \cdot \dfrac{dh}{dt}$

As $h = r$, $\frac{dr}{dt} = \frac{dh}{dt}$, which means the volume differential can be simplified to the form below.

$\dfrac{dv}{dt}$ $= \dfrac{dr}{dt}\left(\dfrac{2}{3}r^2π + \dfrac{1}{3}πr^2\right)$

Now we must plug in all the known values.

$\dfrac{dv}{dt}$ $= 4 = \dfrac{dr}{dt}\left(\dfrac{2}{3}(2)^2π + \dfrac{1}{3}π(2)^2\right)$ $= \dfrac{dr}{dt}(4π)$

The rate of change of the radius and height of the cone is then of the form below.

$\dfrac{dr}{dt} = \dfrac{1}{π}$ cm $s^{−1}$
 Question 10
A $12$-foot ladder slides down a wall at a rate of $2$ feet per second. At what rate is the foot of the ladder sliding away from the base of the wall when the ladder is $4$ feet off the ground?

 A $\sqrt{128}$ $\text{ ft.}$ $\text{per}$ $\text{sec}$ B $\dfrac{\sqrt{128}}{2}$ $\text{ ft.}$ $\text{per}$ $\text{sec}$ C $\dfrac{8}{\sqrt{128}}$ $\text{ ft.}$ $\text{per}$ $\text{sec}$ D $\text{Not}$ $\text{enough}$ $\text{information}$ $\text{given.}$
Question 10 Explanation:
The correct answer is (C). One of the most critical steps of most related rates problems is constructing an accurate picture or model of the situation being described. It is clear from the picture above that the relationship needed to solve the problem is related to triangle properties. We know from Pythagorean’s Theorem that $a^2 + b^2 = c^2$. Thus, for this problem, the relationship is as follows, where $h$ represents the height the ladder is above the ground on the wall and $b$ represents the distance the feet of the ladder are away from the wall.

$h^2 + b^2 = 12^2$

We can then take the derivative of this relationship.

$2h \cdot \dfrac{dh}{dt} + 2b \cdot \dfrac{db}{dt} = 0$

As we know $h = 4$, and $\frac{dh}{dt} = 2$, we can solve for $b$ using Pythagorean theorem.

$12^2 − (4)^2 = b^2$

$\sqrt{128} = b$

Then, we can solve for $x$, which is $\frac{db}{dt}$, or the rate the ladder slides along the ground, by plugging in the known values into the derived equation.

$2(4)(2) + 2(\sqrt{128})(x) = 0$

$x = −\dfrac{8}{\sqrt{128}}$

Now, it is important to recognize that positive and negative signs have little meaning in related rate problems. It is impossible for a length or height to have a negative value, and similarly, a negative or positive rate simply means the rate is in an opposite direction. Therefore, the rate the foot of the ladder is sliding away from the wall is $\frac{8}{\sqrt{128}}$ ft. per sec. Be sure to include units.
 Question 11
Find the volume of the solid that is the result of the revolution of the curve $y = x^2$ around the $x$-axis on the domain $0 ≤ x ≤ 3$.

 A $\dfrac{243}{5} \text{ units}^3$ B $\dfrac{9\sqrt{3}}{5}π \text{ units}^3$ C $9π \text{ units}^3$ D $\dfrac{243π}{5} \text{ units}^3$
Question 11 Explanation:
The correct answer is (D). We shall use disk method to solve the following problem. Recall that the formula for the volume of this solid is of the form below.

$V =π \displaystyle\int_b^a (f(x))^2 dx$

Thus, we can plug in the limits and function into the formula.

$V =π \displaystyle\int_0^3 (x^2)^2 dx$ $= π\left(\dfrac{1}{5}x^5 \right) |_0^3$ $= \dfrac{243π}{5} \text{ units}^3$

We have left the label “units” as units were not given in the problem. Sometimes, a drawing of the revolved solid may be beneficial for solving problems more complicated than this one.
 Question 12
Find the volume of the solid formed by the revolution of the curve $y = x^3 − 3x − 1$ bounded by $x = 0$, $x = 2$, and $y = 1$ around the $y$-axis in cm$^3$.

 A $V = \dfrac{28}{5} \text{ cm}^3$ B $V = \dfrac{36π}{5} \text{ cm}^3$ C $V = \dfrac{28π}{5} \text{ cm}^3$ D $V = \dfrac{56π}{5} \text{ cm}^3$
Question 12 Explanation:
The correct answer is (C). We shall find this volume using shell method. Recall that the formula for shell method is as follows.

$V = 2π \displaystyle\int_a^b (r \cdot h)dx$

The picture below may help in identifying what corresponds with $r$, the radius, and $h$, the height. The shaded region represents the region that forms as a result of the revolutions of the curve. The radius is represented by $x$ and the height is represented by the difference $1 − (x^3 − 3x − 1)$, as the $y$ value corresponds with the function value. Don’t forget to add units to the final answer.

$V = 2π \displaystyle\int_0^2 x(1 − x^3 + 3x + 1)dx$

$V = 2π \left(\dfrac{28}{5}\right)$ $= \dfrac{56π}{5} \text{ cm}^3$
 Question 13
Find the volume in cubic inches of the solid formed by the revolution of the region bounded between the curves $y = x$ and $y = x^2$ around the $x$-axis.

 A $V = \dfrac{2}{15} \text{ inches}^3$ B $V = \dfrac{4π}{15} \text{ inches}^3$ C $V = \dfrac{3π}{15} \text{ inches}^3$ D $V = \dfrac{2π}{15} \text{ inches}^3$
Question 13 Explanation:
The correct answer is (D). To solve this problem, we must first identify which curve is the upper and which is the lower. This can be seen easily by a graph. It can be seen that the curves intersect at the points $(0, 0)$ and $(1, 1)$. This can also be solved for algebraically. It can be observed that for the shaded region, $y = x$ is the upper curve while $y = x^2$ is the lower curve, as the solid is being revolved around the $x$-axis. Recall that the formula for volumes between two curves is as follows, where $f(x)$ is the upper curve and $g(x)$ is the lower curve.

$V = π \displaystyle\int_a^b ((f(x))^2 − (g(x))^2)dx$

The volume is then calculated using the formula as seen below.

$V = π \displaystyle\int_0^1 ((x)^2 − (x)^2)^2)dx$ $= \dfrac{2π}{15}$

The integral must be taken with respect to $x$, as the change in $x$ represents the change in height as the integral calculates the volume formed by the bounded region. We can recognize the limits of the integral based on the fact that the integral is being taken with respect to $x$, so both the lower and upper limits will be the two values of $x$ that bound the region.
 Question 14
Find the area of the solid formed by the function $f(x)$ from $0 ≤ x ≤ 4$, bounded by the $y$-axis, and with square cross sections. The side of each cross section is equal to the height of the function above the $x$-axis.

$f(x) = − (x − 2)^2 + 4$

 A $V = \dfrac{512}{15} \text{ units}^3$ B $V = \dfrac{256}{15} \text{ units}^3$ C $V = \dfrac{32}{3} \text{ units}^3$ D $V = \dfrac{64}{3} \text{ units}^3$
Question 14 Explanation:
The correct answer is (A). Recall that the volume of a solid with known cross sections has the formula below.

$\displaystyle\int_a^b A(c)(dc)$

Where $A(c)$ is the area of the known cross section with respect to variable $c$.

Below is the graph of the function, restricted to the first quadrant. Let’s identify the volume of the solid using the function if it’s vertical cross sections were squares. Just like how sequential cross section areas were summed together using shell of disk method, the same principle is used to calculate volumes of known cross sections.

First, the formula for the area of each cross section must be identified. The area of a square is base times height or $s^2$, since all the sides are equal in length. Since the function is restricted by the $x$ and $y$ axes, the length of each square cross section will be equal to the distance between the $x$-axis and the function, which is equivalent to $y$. This means that the area of each square cross section will be $y^2$. However, since the cross sections are vertical, being perpendicular to the $x$-axis, the thickness of each cross section is not a change in $y$, but rather a change in $x$, which means the integral will need to be in terms of $x$. As the side length of the square is represented by the height of the function, the side length is equal to $−(x − 2)^2 + 4$.

$V = \displaystyle\int_0^4 (−(x − 2)^2 + 4)^2dx$ $= \dfrac{512}{15} \text{ units}^3$

The limits of the graph were obtained by observing where the graph intersects the $x$-axis. These limits could also be solved for algebraically. Don’t forget units on the final answer.
 Question 15
Identify the integral necessary to calculate the volume formed by the revolution of the curve made up of the given cross section. Do not evaluate the integral. The cross sections are bounded by the curve and the $x$-axis, where the radius of the circle is the height of the function.

$y = e^x$ $\text{with}$ $\text{semi-circular}$ $\text{cross}$ $\text{sections.}$ $−1 ≤ x ≤ 1$

 A $$V = \displaystyle\int_{−1}^1 πe^{2x}dx$$ B $$V = \displaystyle\int_{−1}^1 \dfrac{e^{2x}}{2}dx$$ C $$V = \displaystyle\int_{−1}^1 \dfrac{π}{2} \left(\dfrac{e^x}{2}\right)^2dx$$ D $$V = \displaystyle\int_{−1}^1 \dfrac{πe^{2x}}{2}dx$$
Question 15 Explanation:
The correct answer is (B). Recall that the volume of a solid with known cross sections has the formula below.

$\displaystyle\int_a^b A(c)dc$

Where $A(c)$ is the area of the known cross section with respect to variable $c$.
The area of each semi-circle is equal to the equation below, as the function height is the radius of the circle.

$A = π \cdot \dfrac{y^2}{2}$ $= \dfrac{πe^{2x}}{2}$

Then, we can integrate this function on the given bounds. The integral for the volume is then of the form below.

$V = \displaystyle\int_{−1}^1 \dfrac{πe^{2x}}{2}dx$
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