AP Calculus BC Flashcards: Second Derivatives Of Parametric Equations

What this deck covers

This deck focuses on Second Derivatives Of Parametric Equations, giving you a quick way to review the definitions, rules, and examples that matter most for AP Calculus BC.

How to use these flashcards

Work through these flashcards in short sessions. Try to answer each prompt before flipping the card, then revisit any cards you miss until the explanation feels automatic.

Flashcard 1: What is $\frac{dy}{dx}$ for $y(t) = e^t$ and $x(t) = t^2$?

Answer: $\frac{e^t}{2t}$. Calculate: $\frac{dy}{dt} = e^t$ and $\frac{dx}{dt} = 2t$, so $\frac{dy}{dx} = \frac{e^t}{2t}$.

Flashcard 2: Identify $\frac{dx}{dt}$ if $x(t) = e^t$.

Answer: $e^t$. The derivative of $e^t$ with respect to $t$ is $e^t$.

Flashcard 3: Identify $\frac{dy}{dx}$ for $y(t) = t^2 + 3t$ and $x(t) = 2t$.

Answer: $\frac{2t + 3}{2}$. Calculate: $\frac{dy}{dt} = 2t+3$ and $\frac{dx}{dt} = 2$, so $\frac{dy}{dx} = \frac{2t+3}{2}$.

Flashcard 4: State the second derivative formula in terms of $t$ for $x(t)$ and $y(t)$.

Answer: $\frac{d^2y}{dx^2} = \frac{d}{dt}(\frac{dy}{dx}) \cdot \frac{1}{\frac{dx}{dt}}$. Alternative form of the second derivative formula using the reciprocal of $\frac{dx}{dt}$.

Flashcard 5: What is $\frac{dy}{dx}$ for $y(t) = t^2$ and $x(t) = t^2$?

Answer: $1$. Since both functions are identical, $\frac{dy}{dx} = \frac{2t}{2t} = 1$.

Flashcard 6: Find $\frac{dx}{dt}$ for $x(t) = 2t^2 + 3t$.

Answer: $4t + 3$. Apply power rule: derivative of $2t^2$ is $4t$, derivative of $3t$ is $3$.

Flashcard 7: Calculate $\frac{dy}{dx}$ when $y(t) = t^2 + 1$ and $x(t) = 3t$.

Answer: $\frac{2t}{3}$. Calculate: $\frac{dy}{dt} = 2t$ and $\frac{dx}{dt} = 3$, so $\frac{dy}{dx} = \frac{2t}{3}$.

Flashcard 8: State the formula for the second derivative in parametric form.

Answer: $\frac{d^2y}{dx^2} = \frac{\frac{d}{dt}(\frac{dy}{dx})}{\frac{dx}{dt}}$. Apply the chain rule: differentiate $\frac{dy}{dx}$ with respect to $t$, then divide by $\frac{dx}{dt}$.

Flashcard 9: What is $\frac{dy}{dt}$ if $y(t) = \ln(t)$?

Answer: $\frac{1}{t}$. The derivative of $\ln(t)$ with respect to $t$ is $\frac{1}{t}$.

Flashcard 10: What is $\frac{dy}{dx}$ for $y(t) = e^t$ and $x(t) = t^2$?

Answer: $\frac{e^t}{2t}$. Calculate: $\frac{dy}{dt} = e^t$ and $\frac{dx}{dt} = 2t$, so $\frac{dy}{dx} = \frac{e^t}{2t}$.

Flashcard 11: What is $\frac{dy}{dx}$ for $y(t) = t^2$ and $x(t) = t^2$?

Answer: $1$. Since both functions are identical, $\frac{dy}{dx} = \frac{2t}{2t} = 1$.

Flashcard 12: What is the expression for $\frac{dy}{dx}$ if $y(t) = t^3$ and $x(t) = t$?

Answer: $3t^2$. Calculate: $\frac{dy}{dt} = 3t^2$ and $\frac{dx}{dt} = 1$, so $\frac{dy}{dx} = 3t^2$.

Flashcard 13: What is the derivative $\frac{dy}{dt}$ for $y(t) = t^3 - 4t$?

Answer: $3t^2 - 4$. Apply power rule: derivative of $t^3$ is $3t^2$, derivative of $-4t$ is $-4$.

Flashcard 14: Find $\frac{dx}{dt}$ if $x(t) = t^3 - 4t^2$.

Answer: $3t^2 - 8t$. Apply power rule: derivative of $t^3$ is $3t^2$, derivative of $-4t^2$ is $-8t$.

Flashcard 15: What is the expression for $\frac{dy}{dx}$ if $y(t) = t^3$ and $x(t) = t$?

Answer: $3t^2$. Calculate: $\frac{dy}{dt} = 3t^2$ and $\frac{dx}{dt} = 1$, so $\frac{dy}{dx} = 3t^2$.

Flashcard 16: Find $\frac{dx}{dt}$ for $x(t) = 2t^2 + 3t$.

Answer: $4t + 3$. Apply power rule: derivative of $2t^2$ is $4t$, derivative of $3t$ is $3$.

Flashcard 17: What is $\frac{dy}{dx}$ for $y(t) = \tan(t)$ and $x(t) = t$?

Answer: $\sec^2(t)$. Calculate: $\frac{dy}{dt} = \sec^2(t)$ and $\frac{dx}{dt} = 1$, so $\frac{dy}{dx} = \sec^2(t)$.

Flashcard 18: State the formula for the second derivative in parametric form.

Answer: $\frac{d^2y}{dx^2} = \frac{\frac{d}{dt}(\frac{dy}{dx})}{\frac{dx}{dt}}$. Apply the chain rule: differentiate $\frac{dy}{dx}$ with respect to $t$, then divide by $\frac{dx}{dt}$.

Flashcard 19: Identify the formula for $\frac{dy}{dx}$ in terms of $x(t)$ and $y(t)$.

Answer: $\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}}$. Use the chain rule to express the slope in terms of the parameter $t$.

Flashcard 20: Find the expression for $\frac{dy}{dx}$ given $y(t) = \sin(t)$ and $x(t) = \cos(t)$.

Answer: $-\frac{\cos(t)}{\sin(t)}$. Calculate: $\frac{dy}{dt} = \cos(t)$ and $\frac{dx}{dt} = -\sin(t)$, so $\frac{dy}{dx} = -\cot(t)$.

Flashcard 21: Calculate $\frac{dy}{dx}$ for $y(t) = 5t - t^2$ and $x(t) = t^3$.

Answer: $\frac{5 - 2t}{3t^2}$. Calculate: $\frac{dy}{dt} = 5-2t$ and $\frac{dx}{dt} = 3t^2$, so $\frac{dy}{dx} = \frac{5-2t}{3t^2}$.

Flashcard 22: Find $\frac{dx}{dt}$ if $x(t) = t^3 - 4t^2$.

Answer: $3t^2 - 8t$. Apply power rule: derivative of $t^3$ is $3t^2$, derivative of $-4t^2$ is $-8t$.

Flashcard 23: Calculate $\frac{dy}{dx}$ for $y(t) = 5t - t^2$ and $x(t) = t^3$.

Answer: $\frac{5 - 2t}{3t^2}$. Calculate: $\frac{dy}{dt} = 5-2t$ and $\frac{dx}{dt} = 3t^2$, so $\frac{dy}{dx} = \frac{5-2t}{3t^2}$.

Flashcard 24: What is the derivative $\frac{dy}{dt}$ for $y(t) = t^2 - 5t + 6$?

Answer: $2t - 5$. Apply power rule: derivative of $t^2$ is $2t$, derivative of $-5t$ is $-5$.

Flashcard 25: What is $\frac{dy}{dx}$ for $y(t) = \tan(t)$ and $x(t) = t$?

Answer: $\sec^2(t)$. Calculate: $\frac{dy}{dt} = \sec^2(t)$ and $\frac{dx}{dt} = 1$, so $\frac{dy}{dx} = \sec^2(t)$.

Flashcard 26: Identify $\frac{dy}{dx}$ if $y(t) = e^t$ and $x(t) = e^{-t}$.

Answer: $-e^{2t}$. Calculate: $\frac{dy}{dt} = e^t$ and $\frac{dx}{dt} = -e^{-t}$, so $\frac{dy}{dx} = -e^{2t}$.

Flashcard 27: What is $\frac{dx}{dt}$ for $x(t) = \tan(t)$?

Answer: $\sec^2(t)$. The derivative of $\tan(t)$ with respect to $t$ is $\sec^2(t)$.

Flashcard 28: Identify $\frac{dx}{dt}$ if $x(t) = e^t$.

Answer: $e^t$. The derivative of $e^t$ with respect to $t$ is $e^t$.

Flashcard 29: State the expression for $\frac{dy}{dx}$ if $y(t) = \cos(t)$ and $x(t) = \sin(t)$.

Answer: $-\cot(t)$. Same calculation as earlier: $\frac{dy}{dx} = -\cot(t)$.

Flashcard 30: What is the derivative $\frac{dy}{dt}$ for $y(t) = t^3 - 4t$?

Answer: $3t^2 - 4$. Apply power rule: derivative of $t^3$ is $3t^2$, derivative of $-4t$ is $-4$.

Flashcard 31: Identify $\frac{dy}{dx}$ if $y(t) = e^t$ and $x(t) = e^{-t}$.

Answer: $-e^{2t}$. Calculate: $\frac{dy}{dt} = e^t$ and $\frac{dx}{dt} = -e^{-t}$, so $\frac{dy}{dx} = -e^{2t}$

Flashcard 32: Identify the formula for $\frac{dy}{dx}$ in terms of $x(t)$ and $y(t)$.

Answer: $\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}}$. Use the chain rule to express the slope in terms of the parameter $t$.

Flashcard 33: Identify $\frac{dy}{dx}$ for $y(t) = t^2 + 3t$ and $x(t) = 2t$.

Answer: $\frac{2t + 3}{2}$. Calculate: $\frac{dy}{dt} = 2t+3$ and $\frac{dx}{dt} = 2$, so $\frac{dy}{dx} = \frac{2t+3}{2}$.

Flashcard 34: What is the derivative $\frac{dy}{dt}$ for $y(t) = t^2 - 5t + 6$?

Answer: $2t - 5$. Apply power rule: derivative of $t^2$ is $2t$, derivative of $-5t$ is $-5$.

Flashcard 35: Calculate $\frac{dy}{dx}$ when $y(t) = t^2 + 1$ and $x(t) = 3t$.

Answer: $\frac{2t}{3}$. Calculate: $\frac{dy}{dt} = 2t$ and $\frac{dx}{dt} = 3$, so $\frac{dy}{dx} = \frac{2t}{3}$.

Flashcard 36: What is $\frac{dx}{dt}$ for $x(t) = \tan(t)$?

Answer: $\sec^2(t)$. The derivative of $\tan(t)$ with respect to $t$ is $\sec^2(t)$.

Flashcard 37: What is $\frac{dy}{dt}$ if $y(t) = \ln(t)$?

Answer: $\frac{1}{t}$. The derivative of $\ln(t)$ with respect to $t$ is $\frac{1}{t}$.

Flashcard 38: State the second derivative formula in terms of $t$ for $x(t)$ and $y(t)$.

Answer: $\frac{d^2y}{dx^2} = \frac{d}{dt}(\frac{dy}{dx}) \cdot \frac{1}{\frac{dx}{dt}}$. Alternative form of the second derivative formula using the reciprocal of $\frac{dx}{dt}$.

Flashcard 39: Find the expression for $\frac{dy}{dx}$ given $y(t) = \sin(t)$ and $x(t) = \cos(t)$.

Answer: $-\frac{\cos(t)}{\sin(t)}$. Calculate: $\frac{dy}{dt} = \cos(t)$ and $\frac{dx}{dt} = -\sin(t)$, so $\frac{dy}{dx} = -\cot(t)$.