AP Calculus BC Quiz: Second Derivatives Of Parametric Equations

What this quiz covers

This quiz focuses on Second Derivatives Of Parametric Equations, giving you a quick way to practice the rules, question types, and explanations that matter most for AP Calculus BC.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

Question 1

1. $t < 1-\sqrt{2}$ or $t > 1+\sqrt{2}$
2. $t < -1$ or $t > 1$
3. for all real $t$
4. $1-\sqrt{2} < t < 1+\sqrt{2}$ (correct answer)

Explanation: First, find the derivatives: $\frac{dx}{dt} = e^t$ and $\frac{dy}{dt} = 3t^2-3$. Then $\frac{dy}{dx} = \frac{3t^2-3}{e^t}$. Differentiating with respect to $t$: $\frac{d}{dt}\left(\frac{dy}{dx}\right) = \frac{6t e^t - (3t^2-3)e^t}{(e^t)^2} = \frac{-3t^2+6t+3}{e^t}$. The second derivative is $\frac{d^2y}{dx^2} = \frac{(-3t^2+6t+3)/e^t}{e^t} = \frac{-3(t^2-2t-1)}{e^{2t}}$. For the curve to be concave up, we need $\frac{d^2y}{dx^2} > 0$. Since $e^{2t}$ is always positive, we need $-3(t^2-2t-1) > 0$, which simplifies to $t^2-2t-1 < 0$. The roots of $t^2-2t-1=0$ are $t = 1 \pm \sqrt{2}$. Since the parabola $z=t^2-2t-1$ opens upward, it is negative between its roots. Therefore, the curve is concave up for $1-\sqrt{2} < t < 1+\sqrt{2}$.

Question 2

1. $\frac{-3}{4e^2}$ (correct answer)
2. $\frac{-3}{2e}$
3. $\frac{-1}{4e^2}$
4. $\frac{1}{e^2}$

Explanation: First, find the derivatives with respect to $t$: $\frac{dx}{dt} = e^t$ and $\frac{dy}{dt} = \frac{1}{t+1}$. Then, $\frac{dy}{dx} = \frac{1/(t+1)}{e^t} = \frac{e^{-t}}{t+1}$. Next, differentiate $\frac{dy}{dx}$ with respect to $t$: $\frac{d}{dt}\left(\frac{dy}{dx}\right) = \frac{-e^{-t}(t+1) - e^{-t}(1)}{(t+1)^2} = \frac{-e^{-t}(t+2)}{(t+1)^2}$. Then, $\frac{d^2y}{dx^2} = \frac{-e^{-t}(t+2)/(t+1)^2}{e^t} = \frac{-(t+2)}{e^{2t}(t+1)^2}$. At $t=1$, the value is $\frac{-(1+2)}{e^{2(1)}(1+1)^2} = \frac{-3}{4e^2}$.

Question 3

1. $t < -1$ or $t > 1$
2. $-1 < t < 1$ (correct answer)
3. $t > 0$
4. $t < 0$

Explanation: First, find the derivatives: $\frac{dx}{dt} = 3t^2 - 3$ and $\frac{dy}{dt} = 2t$. Then, $\frac{dy}{dx} = \frac{2t}{3t^2-3}$. Next, differentiate $\frac{dy}{dx}$ with respect to $t$: $\frac{d}{dt}\left(\frac{dy}{dx}\right) = \frac{2(3t^2-3) - 2t(6t)}{(3t^2-3)^2} = \frac{6t^2-6-12t^2}{(3(t^2-1))^2} = \frac{-6t^2-6}{9(t^2-1)^2} = \frac{-6(t^2+1)}{9(t^2-1)^2}$. Finally, $\frac{d^2y}{dx^2} = \frac{d/dt(dy/dx)}{dx/dt} = \frac{-6(t^2+1)/[9(t^2-1)^2]}{3(t^2-1)} = \frac{-6(t^2+1)}{27(t^2-1)^3}$. For the curve to be concave up, $\frac{d^2y}{dx^2} > 0$. Since the numerator $-6(t^2+1)$ is always negative, the denominator $27(t^2-1)^3$ must be negative. This occurs when $t^2-1 < 0$, which means $t^2 < 1$, so $-1 < t < 1$.

Question 4

1. The path is always concave up.
2. The path is always concave down. (correct answer)
3. The path is concave up for $0 < t < \pi$ and concave down for $\pi < t < 2\pi$.
4. The path is concave down for $0 < t < \pi$ and concave up for $\pi < t < 2\pi$.

Explanation: First, find derivatives: $\frac{dx}{dt} = 1 - \cos(t)$ and $\frac{dy}{dt} = \sin(t)$. Then, $\frac{dy}{dx} = \frac{\sin(t)}{1-\cos(t)}$. Using trigonometric identities, this simplifies to $\cot(t/2)$. Now, $\frac{d}{dt}(\cot(t/2)) = -\frac{1}{2}\csc^2(t/2)$. The second derivative is $\frac{d^2y}{dx^2} = \frac{-\frac{1}{2}\csc^2(t/2)}{1-\cos(t)} = \frac{-1}{2\sin^2(t/2)(1-\cos t)} = \frac{-1}{2\sin^2(t/2)(2\sin^2(t/2))} = \frac{-1}{4\sin^4(t/2)}$. For $0 < t < 2\pi$, $t/2$ is in $(0, \pi)$, so $\sin(t/2) \neq 0$. Thus, $\sin^4(t/2)$ is always positive. The entire expression for $\frac{d^2y}{dx^2}$ is therefore always negative. This means the path is always concave down.

Question 5

1. $\dfrac{-3}{16t^{7/2}}$ (correct answer)
2. $\dfrac{-1}{16t^{7/2}}$
3. $\dfrac{-3}{8t^{5/2}}$
4. $\dfrac{3}{16t^{7/2}}$
5. $\dfrac{-3}{16t^{5/2}}$

Explanation: This problem tests the skill of finding second derivatives of parametric equations. The formula for the second derivative is $\frac{d^2 y}{dx^2} = \frac{ \frac{d}{dt} \left( \frac{dy/dt}{dx/dt} \right) }{ dx/dt }$. Here, $dx/dt = 2t$ and $dy/dt = \frac{1}{2} t^{-1/2}$, so $\frac{dy}{dx} = \frac{ \frac{1}{2} t^{-1/2} }{ 2t } = \frac{1}{4 t^{3/2}}$. Differentiating this with respect to t gives $\frac{d}{dt} [ \frac{1}{4} t^{-3/2} ] = \frac{1}{4} \cdot \left( -\frac{3}{2} \right) t^{-5/2} = -\frac{3}{8 t^{5/2}}$. Dividing by dx/dt yields $[-\frac{3}{8 t^{5/2}}] / 2t = -\frac{3}{16 t^{7/2}}$. A tempting distractor is choice C, $-3/(8 t^{5/2})$, which forgets the final division by dx/dt. A transferable strategy for computing parametric second derivatives is to consistently apply the formula $\frac{d}{dt}\left( \frac{dy}{dx} \right) / \frac{dx}{dt}$ and verify with specific values if possible.

Question 6

1. $\dfrac{1}{e^t}$ (correct answer)
2. $\dfrac{t+1}{e^t}$
3. $\dfrac{t}{e^t}$
4. $e^t$
5. $\dfrac{1}{t+1}$

Explanation: To find the second derivative of this parametric curve, we use $\frac{d^2y}{dx^2} = \frac{d}{dt}(\frac{dy/dt}{dx/dt}) \cdot \frac{1}{dx/dt}$. First, $\frac{dy}{dt} = e^t + te^t = e^t(1+t)$ and $\frac{dx}{dt} = e^t$, giving $\frac{dy}{dx} = \frac{e^t(1+t)}{e^t} = 1+t$. Differentiating with respect to $t$: $\frac{d}{dt}[1+t] = 1$. Finally, dividing by $\frac{dx}{dt} = e^t$ yields $\frac{d^2y}{dx^2} = \frac{1}{e^t}$. Choice B incorrectly includes $(t+1)$ in the numerator, confusing the first derivative expression with the second derivative. Remember that when $dy/dx$ simplifies to a function of $t$ alone, its derivative is straightforward before the final division by $dx/dt$.

Question 7

1. $2t^2$
2. $4t^2$ (correct answer)
3. $2t$
4. $4t$
5. $\dfrac{4}{t}$

Explanation: This problem tests the skill of finding the second derivative of parametric equations. The formula for d²y/dx² is (x' y'' - y' x'') / (x')³, where primes denote derivatives with respect to t. For x(t) = ln t and y(t) = t², compute x' = 1/t, x'' = -1/t², y' = 2t, and y'' = 2. Substituting yields [(1/t) · 2 - 2t · (-1/t²)] / (1/t)³ = (2/t + 2/t) / (1/t³) = (4/t) · t³ = 4t². A tempting distractor like 4t fails because it neglects the cubic power in the denominator. A transferable strategy for second derivatives is to handle inverse functions like logarithms by ensuring derivatives are correctly computed.

Question 8

1. $\dfrac{-2}{9t^4}$ (correct answer)
2. $\dfrac{2}{9t^4}$
3. $\dfrac{-2}{9t^2}$
4. $\dfrac{-2}{3t^2}$
5. $\dfrac{2}{3t^2}$

Explanation: This problem tests the skill of finding second derivatives of parametric equations. The formula for the second derivative is $\frac{d^2 y}{dx^2} = \frac{ \frac{d}{dt} \left( \frac{dy/dt}{dx/dt} \right) }{ dx/dt }$. Here, $dx/dt = 3t^2$ and $dy/dt = 2t$, so $\frac{dy}{dx} = 2t / 3t^2 = 2/(3t)$. Differentiating this with respect to t gives $-2/(3t^2)$. Dividing by dx/dt yields $-2/(3t^2) / 3t^2 = -2/(9t^4)$. A tempting distractor is choice C, $-2/(9t^2)$, which forgets the final division by dx/dt, resulting in a lower power in the denominator. A transferable strategy for computing parametric second derivatives is to consistently apply the formula $\frac{d}{dt}\left( \frac{dy}{dx} \right) / \frac{dx}{dt}$ and verify with specific values if possible.

Question 9

1. $\dfrac{\sec t}{\tan^2 t}$
2. $\dfrac{\sec^2 t}{\tan t}$
3. $\dfrac{1}{\sec t}$ (correct answer)
4. $\dfrac{\sec t}{\tan t}$
5. $\dfrac{1}{\tan t}$

Explanation: For this parametric curve, we need to find $\frac{d^2y}{dx^2}$ using the formula $\frac{d}{dt}(\frac{dy/dt}{dx/dt}) \cdot \frac{1}{dx/dt}$. We have $\frac{dy}{dt} = \sec t \tan t$ and $\frac{dx}{dt} = \sec^2 t$, giving $\frac{dy}{dx} = \frac{\sec t \tan t}{\sec^2 t} = \frac{\tan t}{\sec t} = \frac{\sin t}{\cos t} \cdot \cos t = \sin t$. Differentiating with respect to $t$: $\frac{d}{dt}[\sin t] = \cos t$. Finally, dividing by $\frac{dx}{dt} = \sec^2 t$ yields $\frac{d^2y}{dx^2} = \frac{\cos t}{\sec^2 t} = \cos t \cdot \cos^2 t = \cos^3 t = \frac{1}{\sec t}$. Choice A incorrectly applies the quotient rule to the original ratio instead of first simplifying $dy/dx$. Simplifying $dy/dx$ before differentiating often makes parametric second derivative calculations much cleaner.

Question 10

1. $\dfrac{2-2t}{e^{2t}}$ (correct answer)
2. $\dfrac{2-2t}{e^{t}}$
3. $\dfrac{2t}{e^{t}}$
4. $\dfrac{2}{e^{t}}$
5. $\dfrac{2-2t}{t^2}$

Explanation: This problem requires finding the second derivative of a parametric curve using $\frac{d^2y}{dx^2} = \frac{\frac{d}{dt}\left(\frac{dy/dt}{dx/dt}\right)}{dx/dt}$. We have $\frac{dy/dt} = 2t$ and $\frac{dx/dt} = e^t$, so $\frac{dy/dx} = \frac{2t}{e^t}$. Differentiating using the quotient rule: $\frac{d}{dt}\left(\frac{2t}{e^t}\right) = \frac{2e^t - 2te^t}{e^{2t}} = \frac{2e^t(1-t)}{e^{2t}} = \frac{2(1-t)}{e^t}$. Then $\frac{d^2y}{dx^2} = \frac{2(1-t)/e^t}{e^t} = \frac{2(1-t)}{e^{2t}} = \frac{2-2t}{e^{2t}}$. Choice D shows $\frac{2}{e^t}$, which omits the crucial $(1-t)$ factor from the quotient rule differentiation. When finding parametric second derivatives, carefully apply the quotient rule to \frac{dy/dx} before dividing by $\frac{dx}{dt}$.

Question 11

1. $\dfrac{2}{e^{2t}}$
2. $\dfrac{2(1-t)}{e^{2t}}$ (correct answer)
3. $\dfrac{2t}{e^{t}}$
4. $\dfrac{2(t-1)}{e^{2t}}$
5. $\dfrac{2}{e^{t}}$

Explanation: This problem requires finding d²y/dx² for x = eᵗ and y = t². Using the parametric second derivative formula d²y/dx² = [d/dt(dy/dx)]/(dx/dt), we first find dy/dx = (dy/dt)/(dx/dt) = 2t/eᵗ. Next, we differentiate dy/dx with respect to t: d/dt[2t/eᵗ] = [2·eᵗ - 2t·eᵗ]/(e²ᵗ) = 2eᵗ(1 - t)/(e²ᵗ) = 2(1 - t)/eᵗ. Finally, dividing by dx/dt = eᵗ gives d²y/dx² = [2(1 - t)/eᵗ]/eᵗ = 2(1 - t)/e²ᵗ. Choice D incorrectly has (t - 1) instead of (1 - t) in the numerator, which is a common sign error. Remember to carefully apply the quotient rule and maintain proper sign conventions throughout the calculation.

Question 12

1. $-8$ (correct answer)
2. $4\sqrt{2}$
3. $0$
4. $-4$

Explanation: First, find the derivatives with respect to $t$: $\frac{dx}{dt} = -\sin(t)$ and $\frac{dy}{dt} = 2\cos(2t)$. Then, $\frac{dy}{dx} = \frac{2\cos(2t)}{-\sin(t)}$. To find $\frac{d^2y}{dx^2}$, we compute $\frac{d}{dt}\left(\frac{dy}{dx}\right) = \frac{(-4\sin(2t))(-\sin(t)) - (2\cos(2t))(-\cos(t))}{(-\sin(t))^2} = \frac{4\sin(2t)\sin(t) + 2\cos(2t)\cos(t)}{\sin^2(t)}$. At $t=\frac{\pi}{4}$, $\frac{d}{dt}\left(\frac{dy}{dx}\right) = \frac{4\sin(\pi/2)\sin(\pi/4) + 2\cos(\pi/2)\cos(\pi/4)}{\sin^2(\pi/4)} = \frac{4(1)(\sqrt{2}/2) + 2(0)(\sqrt{2}/2)}{(\sqrt{2}/2)^2} = \frac{2\sqrt{2}}{1/2} = 4\sqrt{2}$. Also at $t=\frac{\pi}{4}$, $\frac{dx}{dt} = -\sin(\frac{\pi}{4}) = -\frac{\sqrt{2}}{2}$. Therefore, $\frac{d^2y}{dx^2} = \frac{4\sqrt{2}}{-\sqrt{2}/2} = -8$.

Question 13

1. $\frac{4}{0}$, which is undefined
2. $\frac{4}{9}$ (correct answer)
3. $\frac{4}{3}$
4. $4$

Explanation: The formula for the second derivative is $\frac{d^2y}{dx^2} = \frac{\frac{d}{dt}(\frac{dy}{dx})}{\frac{dx}{dt}}$. First, we find the derivative of $\frac{dy}{dx} = \frac{dy/dt}{dx/dt}$ with respect to $t$ using the quotient rule: $\frac{d}{dt}\left(\frac{dy}{dx}\right) = \frac{\frac{d^2y}{dt^2}\frac{dx}{dt} - \frac{dy}{dt}\frac{d^2x}{dt^2}}{(\frac{dx}{dt})^2}$. At $t=2$, this is $\frac{(4)(3) - (-1)(0)}{(3)^2} = \frac{12}{9} = \frac{4}{3}$. Now, we can find the second derivative: $\frac{d^2y}{dx^2} = \frac{4/3}{dx/dt} = \frac{4/3}{3} = \frac{4}{9}$.

Question 14

1. $-\sin(t)$/$\cos(t)$
2. $-1$ (correct answer)
3. $-2\cos(t)$/$\sin(t)$
4. $-\sin(2t)$

Explanation: First, find derivatives with respect to $t$: $\frac{dx}{dt} = 2\cos(t)$ and $\frac{dy}{dt} = -2\sin(2t)$. Then, $\frac{dy}{dx} = \frac{-2\sin(2t)}{2\cos(t)} = \frac{-2(2\sin(t)\cos(t))}{2\cos(t)} = -2\sin(t)$. Next, differentiate $\frac{dy}{dx}$ with respect to $t$: $\frac{d}{dt}(-2\sin(t)) = -2\cos(t)$. Finally, $\frac{d^2y}{dx^2} = \frac{d/dt(dy/dx)}{dx/dt} = \frac{-2\cos(t)}{2\cos(t)} = -1$.

Question 15

1. $\frac{3}{2}$
2. $\frac{3}{4t}$ (correct answer)
3. $\frac{6t}{2} = 3t$
4. $\frac{3}{2t}$

Explanation: First, find the derivatives: $\frac{dx}{dt} = 2t$ and $\frac{dy}{dt} = 3t^2$. Then, the first derivative of $y$ with respect to $x$ is $\frac{dy}{dx} = \frac{3t^2}{2t} = \frac{3}{2}t$. Next, differentiate $\frac{dy}{dx}$ with respect to $t$: $\frac{d}{dt}\left(\frac{3}{2}t\right) = \frac{3}{2}$. Finally, find the second derivative of $y$ with respect to $x$: $\frac{d^2y}{dx^2} = \frac{d/dt(dy/dx)}{dx/dt} = \frac{3/2}{2t} = \frac{3}{4t}$.

Question 16

1. $\frac{10}{3}$
2. $7$
3. $\frac{10}{9}$ (correct answer)
4. $\frac{5}{9}$

Explanation: First, find derivatives with respect to $t$: $\frac{dx}{dt} = 2t+1$ and $\frac{dy}{dt} = 4t^3+2t$. Then, $\frac{dy}{dx} = \frac{4t^3+2t}{2t+1}$. Now, differentiate $\frac{dy}{dx}$ with respect to $t$: $\frac{d}{dt}\left(\frac{dy}{dx}\right) = \frac{(12t^2+2)(2t+1) - (4t^3+2t)(2)}{(2t+1)^2}$. At $t=1$, $\frac{dx}{dt} = 3$ and $\frac{d}{dt}\left(\frac{dy}{dx}\right) = \frac{(14)(3) - (6)(2)}{3^2} = \frac{42-12}{9} = \frac{30}{9} = \frac{10}{3}$. Finally, $\frac{d^2y}{dx^2} = \frac{d/dt(dy/dx)}{dx/dt} = \frac{10/3}{3} = \frac{10}{9}$.

Question 17

1. $\frac{g''(t)}{f''(t)}$
2. $\frac{g''(t)f'(t) - g'(t)f''(t)}{(f'(t))^2}$
3. $\frac{g''(t)}{f'(t)}$
4. $\frac{g''(t)f'(t) - g'(t)f''(t)}{(f'(t))^3}$ (correct answer)

Explanation: The first derivative is $\frac{dy}{dx} = \frac{g'(t)}{f'(t)}$. To find the second derivative, we must differentiate $\frac{dy}{dx}$ with respect to $t$ and divide by $\frac{dx}{dt} = f'(t)$. Using the quotient rule, $\frac{d}{dt}\left(\frac{g'(t)}{f'(t)}\right) = \frac{g''(t)f'(t) - g'(t)f''(t)}{(f'(t))^2}$. Dividing this by $f'(t)$ gives $\frac{d^2y}{dx^2} = \frac{g''(t)f'(t) - g'(t)f''(t)}{(f'(t))^3}$.

Question 18

1. $t=0$
2. $t=4$
3. $t=2$ (correct answer)
4. There is no such value of $t$.

Explanation: First, find the derivatives with respect to $t$: $\frac{dx}{dt} = t$ and $\frac{dy}{dt} = t^3 - 4t^2$. Then, $\frac{dy}{dx} = \frac{t^3-4t^2}{t} = t^2 - 4t$ for $t \neq 0$. Next, differentiate $\frac{dy}{dx}$ with respect to $t$: $\frac{d}{dt}(t^2 - 4t) = 2t - 4$. Finally, $\frac{d^2y}{dx^2} = \frac{2t-4}{t}$. Setting this equal to zero gives $2t-4=0$, so $t=2$.

Question 19

1. $-2$ (correct answer)
2. $1$
3. $0$
4. $-1$

Explanation: First, calculate derivatives with respect to $t$: $\frac{dx}{dt} = 1 - \sin(t)$ and $\frac{dy}{dt} = 1 - \cos(t)$. Then, $\frac{dy}{dx} = \frac{1-\cos(t)}{1-\sin(t)}$. Now, differentiate $\frac{dy}{dx}$ with respect to $t$ using the quotient rule: $\frac{d}{dt}\left(\frac{dy}{dx}\right) = \frac{\sin(t)(1-\sin(t)) - (1-\cos(t))(-\cos(t))}{(1-\sin(t))^2} = \frac{\sin(t) - \sin^2(t) + \cos(t) - \cos^2(t)}{(1-\sin(t))^2} = \frac{\sin(t)+\cos(t)-1}{(1-\sin(t))^2}$. At $t=\pi$, $\frac{d}{dt}\left(\frac{dy}{dx}\right) = \frac{\sin(\pi)+\cos(\pi)-1}{(1-\sin(\pi))^2} = \frac{0-1-1}{(1-0)^2} = -2$. Also at $t=\pi$, $\frac{dx}{dt} = 1-\sin(\pi) = 1$. Therefore, $\frac{d^2y}{dx^2} = \frac{-2}{1} = -2$.

Question 20

1. $3$
2. $\frac{4}{3}$
3. $-\frac{3}{2}$ (correct answer)
4. $-\frac{2}{3}$

Explanation: The formula for the second derivative of a parametric curve is $\frac{d^2y}{dx^2} = \frac{\frac{d}{dt}(\frac{dy}{dx})}{\frac{dx}{dt}}$. We are given all the necessary values in the problem statement. Plugging them in, we get $\frac{d^2y}{dx^2} = \frac{3}{-2} = -\frac{3}{2}$.