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AP Calculus BC : Derivatives of Parametric, Polar, and Vector Functions

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Example Questions

Example Question #174 : Derivatives

Given the parametric curve

x= 2t^2 , $y=\sin{t}$

Evaluate $\frac{dy}{dx}$ when $t = \frac{\pi}{4}$ .

Possible Answers:

$\frac{\sqrt{2}}{2\pi}$

$\frac{2\pi}{\sqrt{2}}$

$\sqrt{2}$

$\frac{2}{\pi}$

Correct answer:

$\frac{\sqrt{2}}{2\pi}$

Explanation:

To find $\frac{dy}{dx}$ , we can use the formula $\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}}$ .

$\frac{dy}{dt} = \frac{d}{dt}(\sin(t)) = \cos(t)$ .

And $\frac{dx}{dt} = \frac{d}{dt}(2t^2) = 4t$ .

Hence $\frac{dy}{dx} = \frac{\cos(t)}{4t}$ .

Plugging in $t = \frac{\pi}{4}$ , we get

$\frac{\cos(\pi/4)}{4\frac{\pi}{4}} = \frac{\sqrt{2}/2}{\pi} = \frac{\sqrt{2}}{2\pi}.$

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Example Question #11 : Derivatives Of Parametric, Polar, And Vector Functions

Find $\frac{\mathrm{d} y}{\mathrm{d} x}$ of the following function:

x=t^2e^t , y=t+3

Possible Answers:

2te^t+t^2e^t

$\frac{1}{e^t(t+2)}$

2te^t

Correct answer:

$\frac{1}{e^t(t+2)}$

Explanation:

To find the derivative of the parametric function, we must use the following:

$\frac{\mathrm{d} y}{\mathrm{d} x}= \frac{\frac{\mathrm{d} y}{\mathrm{d} t}}{\frac{\mathrm{d} x}{\mathrm{d} t}}$

So, we must take the derivative of each component with respect to t:

$\frac{\mathrm{d} x}{\mathrm{d} t}=2te^t+t^2e^t$

The derivative was found using the following rules:

$\frac{\mathrm{d} }{\mathrm{d} x} f(x)g(x)=f'(x)g(x)+f(x)g'(x)$ , $\frac{\mathrm{d} }{\mathrm{d} x} x^n=nx^{n-1}$ , $\frac{\mathrm{d} }{\mathrm{d} x} e^x=e^x$

$\frac{\mathrm{d} y}{\mathrm{d} t}=1$

This derivative was found using the following rule:

$\frac{\mathrm{d} }{\mathrm{d} x} ax=a$

Dividing the two and factoring, we get

$\frac{\mathrm{d} y}{\mathrm{d} t}=\frac{t}{2te^t+t^2e^t}=\frac{t(1)}{te^t(t+2)}=\frac{1}{e^t(t+2)}$

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Example Question #73 : Computation Of Derivatives

Find $\frac{\mathrm{d} x}{\mathrm{d} t}$ , where

x^2=t^3

Possible Answers:

$\frac{3t^2}{2}$

$\frac{3t^2}{2x}$

$\frac{3x^2}{2t}$

Correct answer:

$\frac{3t^2}{2x}$

Explanation:

To find the derivative of x with respect to t, we must differentiate both sides of the parametric equation with respect to t:

$2x \frac{\mathrm{d} x}{\mathrm{d} t}=3t^2$

The derivatives were found using the following rules:

$\frac{\mathrm{d} }{\mathrm{d} x} f(g(x))=f'(g(x))\cdot g'(x)$ , $\frac{\mathrm{d} }{\mathrm{d} x} x^n=nx^{n-1}$

Note that the chain rule was used when taking the derivative of x^2 .

Solving, we get

$\frac{\mathrm{d} x}{\mathrm{d} t}=\frac{3t^2}{2x}$

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Example Question #181 : Derivatives

Screen shot 2016 03 30 at 4.58.19 pm

A particle moves around the xy plane such that its position as a function of time is given by the parametric function:

$\begin{pmatrix} x(t)\\y(t) \end{pmatrix}=\begin{pmatrix} \sin(3t)\\\sin(2t) \end{pmatrix}$ .

What is the slope, $\frac{dy}{dx}$ , of the particle's trajectory when $t=\pi/12$ ?

Possible Answers:

$\frac{\sqrt{2}}{3}$

$\frac{\sqrt{2}}{2}$

$\frac{\sqrt{6}}{3}$

$\frac{\sqrt{6}}{2}$

$\frac{3\sqrt{2}}{2}$

Correct answer:

$\frac{\sqrt{6}}{3}$

Explanation:

Evaluate the slope as

$\frac{dy}{dx}=\frac{dy/dt}{dx/dt}$ .

We have

$\frac{dy}{dt}=2\cos(2t)$

and

$\frac{dx}{dt}=3\cos(3t)$

$\frac{dy}{dx}=\frac{2\cos(2t)}{3\cos(3t)}$

Evaluating this when $t=\pi/12$ gives

$\frac{dy}{dx}=\frac{2\cos(\frac{\pi}{6})}{3\cos(\frac{\pi}{4})}=\frac{2\left (\frac{\sqrt{3}}{2} \right )}{3\left (\frac{1}{\sqrt{2}} \right )}=\frac{\sqrt{6}}{3}$ .

Remark: This curve is one example from family of curves called Lissajous figures, which can be observed on oscilloscopes.

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AP Calculus BC : Derivatives of Parametric, Polar, and Vector Functions

Study concepts, example questions & explanations for AP Calculus BC

All AP Calculus BC Resources

Example Questions

Example Question #174 : Derivatives

Example Question #11 : Derivatives Of Parametric, Polar, And Vector Functions

Example Question #73 : Computation Of Derivatives

Example Question #181 : Derivatives

All AP Calculus BC Resources

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