Parametric Equations (College Board AP® Calculus BC): Exam Questions

A curve in the $xy$-plane is defined parametrically as

$\begin{array}{rcl}x& =& t^2+\cos \left(t\right)\\ y& =& 1+\sin \left(t\right)\end{array}$

What is the slope of the line tangent to the curve at $t=\pi$?

A curve is defined parametrically by

$\begin{array}{l}x=5\cos t\\ y=4\sin t\end{array}$

The equation of the curve in terms of $x$ and $y$ only is

A curve is defined parametrically by

$\begin{array}{rcl}x& =& t^3-3t\\ y& =& t^3-12t\end{array}$

where $0\le t\le 10$.

Find the coordinates of the point on the curve at which the tangent is horizontal.

A particle moves along a curve so that its position at time $t$ is $\left(x\left(t\right),y\left(t\right)\right)$, where

$\begin{array}{rcl}\frac{dx}{dt}& =& 24t^3-6t^2+1\\ \frac{dy}{dt}& =& 12t^2+2t-1\end{array}$

The particle is at the point $\left(3,9\right)$ at time $t=1$.

An expression for $x\left(t\right)$ in terms of $t$ is

The motion of a particle at position $\left(x\left(t\right),y\left(t\right)\right)$ at time $t$ where $t\ge 0$ is defined by

$\begin{array}{rcl}\frac{dx}{dt}& =& 1-5\sin t\\ \frac{dy}{dt}& =& 4+\cos t\end{array}$

The total distance travelled by the particle at time $t=\pi$ is

A curve is defined by the parametric equations

$\begin{array}{l}x=\frac{1}{4}{t}^4\\ y=t^3-t\end{array}$

An expression for $\frac{d^{2}y}{d{x}^2}$ is

What is the equation of the vertical tangent to the curve

$\begin{array}{rcl}x& =& t+\sin \left(2t\right)\\ y& =& -\cos \left(2t\right)\end{array}$

where $0\le t\le \frac{\pi }{2}$?

A curve is defined parametrically by

$\begin{array}{rcl}x& =& \frac{1}{2}{t}^2-\ln t\\ y& =& 2t-1\end{array}$

where $t\ge 1$.

Which of the following integrals gives the correct arc length from $t=1$ to $t=8$?

A curve defined parametrically is shown below, with parameter $t$ where $0\le t\le 2$.

Which of the following parametric equations could represent the curve above?

A particle is moving along a curve in the $xy$-plane with its position at time $t$ given by $\left(x\left(t\right),y\left(t\right)\right)$, where $t\ge 0$ and

$\begin{array}{rcl}\frac{dx}{dt}& =& e^{\sqrt{x}}\\ \frac{dy}{dt}& =& \sqrt{1+t^3}\end{array}$

The initial position of the particle is $\left(12,10\right)$.

The position of the $y$-coordinate of the particle at time $t=5$ is