AP Calculus BC Flashcards: Motion Problems Parametric Vector Valued Functions

What this deck covers

This deck focuses on Motion Problems Parametric Vector Valued Functions, 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 the velocity vector $\mathbf{v}(t)$ for $x(t) = t^2$, $y(t) = t^3$?

Answer: $\langle 2t, 3t^2 \rangle$. Derivatives: $v_x = 2t$, $v_y = 3t^2$.

Flashcard 2: What is the velocity vector $\mathbf{v}(t)$ for $x(t) = t^2$, $y(t) = t^3$?

Answer: $\langle 2t, 3t^2 \rangle$. Derivatives: $v_x = 2t$, $v_y = 3t^2$.

Flashcard 3: Find the value of $\frac{dy}{dx}$ for $y(t)=t^2$, $x(t)=t+1$ at $t=1$.

Answer: $2$. At $t=1$: $\frac{2t}{1} = 2$.

Flashcard 4: Find the unit tangent vector at $t=1$ for $\mathbf{r}(t) = \langle t^2, t^3 \rangle$.

Answer: $\langle \frac{2}{\sqrt{13}}, \frac{3}{\sqrt{13}} \rangle$. Normalize $\langle 2, 3 \rangle$ with magnitude $\sqrt{13}$.

Flashcard 5: Find the position vector at $t=3$ for $\mathbf{r}(t) = \langle t^2, t^3 \rangle$.

Answer: $\langle 9, 27 \rangle$. Substitute $t=3$: $\langle 9, 27 \rangle$.

Flashcard 6: What is the horizontal component of the velocity vector for $x(t)=5t$?

Answer: $v_x(t) = 5$. Derivative of $5t$ is constant $5$.

Flashcard 7: What is the magnitude of the acceleration vector $\mathbf{a}(t) = \langle 3, 4 \rangle$?

Answer: $5$. Magnitude formula: $\sqrt{3^2 + 4^2} = \sqrt{25} = 5$.

Flashcard 8: Find $\frac{dy}{dx}$ for $y(t)=t^3$, $x(t)=t^2$ at $t=1$.

Answer: $\frac{3}{2}$. Chain rule: $\frac{ \frac{dy}{dt} }{ \frac{dx}{dt} } = \frac{3t^2}{2t} = \frac{3}{2}$

Flashcard 9: What is the horizontal distance traveled by a projectile at $t=5$ for $x(t)=10t$?

Answer: $50$. Substitute $t=5$ into horizontal position.

Flashcard 10: State the formula for speed of a particle given $v_x(t)$ and $v_y(t)$.

Answer: Speed = $\sqrt{v_x(t)^2 + v_y(t)^2}$. Pythagorean theorem for vector magnitude.

Flashcard 11: What is the formula for the curvature $\kappa(t)$ of a parametric curve?

Answer: $\kappa(t) = \frac{|x'y'' - y'x''|}{(x'^2 + y'^2)^{3/2}}$. Standard curvature formula for parametric curves.

Flashcard 12: What is the acceleration vector $\mathbf{a}(t)$ for $x(t) = 4t^2$ and $y(t) = 2t^3$?

Answer: $\mathbf{a}(t) = \langle 8, 12t \rangle$. Second derivatives: $a_x = 8$, $a_y = 12t$.

Flashcard 13: Identify the velocity vector $\mathbf{v}(t)$ given $x(t) = 3t$ and $y(t) = 4t^2$.

Answer: $\mathbf{v}(t) = \langle 3, 8t \rangle$. Take derivatives: $v_x = 3$, $v_y = 8t$.

Flashcard 14: What is the formula for total distance traveled in vector form?

Answer: $\int_a^b \| \mathbf{v}(t) \| \, dt$. Integral of speed over time interval.

Flashcard 15: Find the derivative of $x(t) = 6t^3$ with respect to $t$.

Answer: $18t^2$. Power rule: derivative of $6t^3$ is $18t^2$.

Flashcard 16: What is the formula for the binormal vector $\mathbf{B}(t)$ of a curve?

Answer: $\mathbf{B}(t) = \mathbf{T}(t) \times \mathbf{N}(t)$. Cross product of tangent and normal vectors.

Flashcard 17: State the formula for the tangent vector of a parametric curve at $t = t_0$.

Answer: $\mathbf{T}(t_0) = \frac{\mathbf{v}(t_0)}{\|\mathbf{v}(t_0)\|}$. Unit vector formula using velocity magnitude.

Flashcard 18: Find the magnitude of the velocity vector $\langle 7, 24 \rangle$.

Answer: $25$. Magnitude formula: $\sqrt{7^2 + 24^2} = \sqrt{625} = 25$.

Flashcard 19: State the formula for the position vector $\mathbf{r}(t)$ given $x(t)$ and $y(t)$.

Answer: $\mathbf{r}(t) = \langle x(t), y(t) \rangle$. Standard vector notation for position.

Flashcard 20: What is the magnitude of the acceleration vector $\mathbf{a}(t) = \langle 3, 4 \rangle$?

Answer: $5$. Magnitude formula: $\sqrt{3^2 + 4^2} = \sqrt{25} = 5$.

Flashcard 21: What is the acceleration vector $\mathbf{a}(t)$ for $x(t) = 4t^2$ and $y(t) = 2t^3$?

Answer: $\mathbf{a}(t) = \langle 8, 12t \rangle$. Second derivatives: $a_x = 8$, $a_y = 12t$.

Flashcard 22: What is the vertical component of the velocity vector for $y(t)=6t^2$?

Answer: $v_y(t) = 12t$. Derivative of $6t^2$ is $12t$.

Flashcard 23: Find the unit tangent vector at $t=1$ for $\mathbf{r}(t) = \langle t^2, t^3 \rangle$.

Answer: $\langle \frac{2}{\sqrt{13}}, \frac{3}{\sqrt{13}} \rangle$. Normalize $\langle 2, 3 \rangle$ with magnitude $\sqrt{13}$.

Flashcard 24: Find the value of $\frac{dy}{dx}$ for $y(t)=t^2$, $x(t)=t+1$ at $t=1$.

Answer: $2$. At $t=1$: $\frac{2t}{1} = 2$.

Flashcard 25: State the formula for speed of a particle given $v_x(t)$ and $v_y(t)$.

Answer: Speed = $\sqrt{v_x(t)^2 + v_y(t)^2}$. Pythagorean theorem for vector magnitude.

Flashcard 26: What is the formula for the curvature $\kappa(t)$ of a parametric curve?

Answer: $\kappa(t) = \frac{|x'y'' - y'x''|}{(x'^2 + y'^2)^{3/2}}$. Standard curvature formula for parametric curves.

Flashcard 27: What is the formula for total distance traveled in vector form?

Answer: $\int_a^b \| \mathbf{v}(t) \| \, dt$. Integral of speed over time interval.

Flashcard 28: What is the parametric equation for $y(t)$ of a projectile launched at angle $\theta$?

Answer: $y(t) = v_0 \sin(\theta) t - \frac{1}{2}gt^2$. Vertical motion with gravity acceleration $-g$.

Flashcard 29: What is the parametric equation for $x(t)$ of a projectile launched at angle $\theta$?

Answer: $x(t) = v_0 \cos(\theta) t$. Horizontal motion with constant velocity component.

Flashcard 30: Find $\frac{dy}{dx}$ for $y(t)=t^3$, $x(t)=t^2$ at $t=1$.

Answer: $\frac{3}{2}$. Chain rule: $\frac{ \frac{dy}{dt} }{ \frac{dx}{dt} } = \frac{3t^2}{2t} = \frac{3}{2}$.