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. $\dfrac{-3}{16t^{7/2}}$ (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 2

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 3

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 4

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 5

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 6

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 7

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 8

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 9

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 10

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 11

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 $\frac{d^2y}{dx^2} = \frac{3}{-2} = -\frac{3}{2}$.

Question 12

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 13

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 14

1. $9t^2$
2. $3t^2$
3. $6t^3$
4. $9t^3$ (correct answer)
5. $6t^2$

Explanation: Finding the second derivative of a parametric curve requires $\frac{d^2y}{dx^2} = \frac{\frac{d}{dt}\left(\frac{dy/dt}{dx/dt}\right)}{dx/dt}$. We have $\frac{dy/dt} = 3t^2$ and $\frac{dx/dt} = \frac{1}{t}$, giving $\frac{dy/dx} = \frac{3t^2}{1/t} = 3t^3$. Differentiating: $\frac{d}{dt}(3t^3) = 9t^2$. Therefore, $\frac{d^2y}{dx^2} = \frac{9t^2}{1/t} = 9t^2 \cdot t = 9t^3$. Choice A shows $9t^2$, which is the derivative of \frac{dy/dx} but forgets to divide by \frac{dx/dt}. Remember that the parametric second derivative formula requires dividing by \frac{dx/dt} at the end to convert from $t$-differentiation to $x$-differentiation.

Question 15

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

Explanation: This problem involves finding the second derivative for a logarithmic-polynomial parametric curve. Given $x(t) = \ln t$ and $y(t) = t^2$ with $t > 0$, we have $\frac{dx}{dt} = \frac{1}{t}$ and $\frac{dy}{dt} = 2t$. The first derivative is $\frac{dy}{dx} = \frac{2t}{1/t} = 2t^2$. To find the second derivative: $\frac{d}{dt}(2t^2) = 4t$, so $\frac{d^2y}{dx^2} = \frac{4t}{1/t} = 4t^2$. Choice B ($\frac{2}{t^2}$) incorrectly inverts the relationship between $t$ and the derivative. The key insight for logarithmic parametric curves is that $\frac{dx}{dt} = \frac{1}{t}$ often leads to powers of $t$ in the second derivative.

Question 16

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

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 = e^t and dy/dt = e^t + t e^t = e^t (t+1), so \frac{dy}{dx} = t+1. Differentiating this with respect to t gives 1. Dividing by dx/dt yields 1 / e^t. A tempting distractor is choice C, (t+1)/e^t, which is the first derivative dy/dx, but we need the second. 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 17

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

Explanation: The skill here is finding the second derivative of parametric equations. The formula for d²y/dx² is the derivative with respect to t of dy/dx divided by dx/dt. Compute dy/dt = -1/t² and dx/dt = 2t, so dy/dx = -1/(2t³). Differentiate dy/dx to get 3/(2t⁴). Divide by 2t to obtain 3/(4t⁵). A tempting distractor is -3/(4t⁵), which keeps a negative sign from the first derivative without accounting for the positive from differentiation. A transferable strategy for second derivatives is to track signs meticulously through each step to determine concavity accurately.

Question 18

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

Explanation: The skill here is finding the second derivative of parametric equations. The formula for d²y/dx² is the derivative with respect to t of dy/dx, divided by dx/dt. First, compute dy/dx = -e^{-2t}. Then, differentiate this to get 2e^{-2t}. Finally, divide by dx/dt = e^t to obtain 2/e^{3t}. A tempting distractor like -2/e^{3t} fails because it incorrectly retains a negative sign from the first derivative without accounting for the double negative in differentiation. When computing second derivatives for parametric equations, always ensure you apply the chain rule correctly by dividing the derivative of the first derivative by dx/dt.

Question 19

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

Explanation: This problem tests the skill of finding the second derivative of parametric equations. The formula for $d^2 y / dx^2$ is $(x' y'' - y' x'') / (x')^3$, where primes denote derivatives with respect to t. For $x(t) = e^t$ and $y(t) = t e^t$, compute $x' = e^t$, $x'' = e^t$, $y' = e^t + t e^t = e^t(t + 1)$, and $y'' = e^t(t + 1) + e^t = e^t(t + 2)$. Substituting gives $[e^t \cdot e^t(t + 2) - e^t(t + 1) \cdot e^t] / (e^t)^3 = [e^{2t}(t + 2) - e^{2t}(t + 1)] / e^{3t} = e^{2t} / e^{3t} = 1/e^t$. A tempting distractor like $e^t$ fails because it might result from incorrectly inverting the denominator. A transferable strategy for second derivatives is to use the chain rule carefully when dealing with exponential functions.

Question 20

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

Explanation: This problem involves finding the second derivative for exponential parametric equations. Using the parametric second derivative formula, we start with $\frac{dx}{dt} = e^t$ and $\frac{dy}{dt} = e^t + te^t = e^t(1+t)$. The first derivative is $\frac{dy}{dx} = \frac{e^t(1+t)}{e^t} =$1 + t$. Differentiating with respect to$t$gives$\frac{d}{dt}(1+t) = 1$. Finally,$\frac{d^2y}{dx^2} = \frac{1}{e^t}$. Choice A incorrectly includes an extra$t$term, likely from misapplying the product rule. Remember that when the first derivative simplifies nicely (here to$1+t$), the second derivative calculation becomes much simpler.