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Calculus 2 : Derivatives of Parametrics

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Example Questions

Example Question #121 : Parametric, Polar, And Vector

What is $\frac{dy}{dx}$ when x=2t+13 and y=4t+17 ?

Possible Answers:

None of the above

$-\frac{1}{2}$

$\frac{1}{2}$

Correct answer:

Explanation:

We can first recognize that

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

since cancels out when we divide.

Then, given x=2t+13 and y=4t+17 and using the Power Rule

$\frac{d}{dx}x^{n}=nx^{n-1}$ for all $n\neq0$ ,

we can determine that

$\frac{dy}{dt}=(1)4t^{1-1}+(0)17=4$ and $\frac{dx}{dt}=(1)2t^{1-1}+(0)13=2$ .

Therefore,

$\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}=\frac{4}{2}=2$ .

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Example Question #12 : Derivatives Of Parametrics

What is $\frac{dy}{dx}$ when x=-t+12 and $y=\frac{8}{t}$ ?

Possible Answers:

None of the above

$-\frac{t^{2}}{8}$

$\frac{8}{t^{2}}$

$-\frac{8}{t^{2}}$

$\frac{t^{2}}{8}$

Correct answer:

$\frac{8}{t^{2}}$

Explanation:

We can first recognize that

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

since cancels out when we divide.

Then, given x=-t+12 and $y=\frac{8}{t}$ or $y=8t^{-1}$ and using the Power Rule

$\frac{d}{dx}x^{n}=nx^{n-1}$ for all $n\neq0$ ,

we can determine that

$\frac{dy}{dt}=(-1)8t^{-1-1}=-\frac{8}{t^{2}}$ and $\frac{dx}{dt}=(1)(-1)t^{1-1}+(0)12=-1$ .

Therefore,

$\frac{dy}{dx}=\frac{-\frac{8}{t^{2}}}{-1}=\frac{8}{t^{2}}$ .

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Example Question #121 : Parametric

What is $\frac{dy}{dx}$ when x=-6t+10 and y=7t-2 ?

Possible Answers:

$\frac{6}{7}$

$-\frac{7}{6}$

None of the above

$-\frac{6}{7}$

$\frac{7}{6}$

Correct answer:

$-\frac{7}{6}$

Explanation:

We can first recognize that

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

since cancels out when we divide.

Then, given x=-6t+10 and y=7t-2 and using the Power Rule

$\frac{d}{dx}x^{n}=nx^{n-1}$ for all $n\neq0$ ,

we can determine that

$\frac{dy}{dt}=(1)7t^{1-1}-(0)2=7$ and $\frac{dx}{dt}=(1)(-6)t^{1-1}+(0)10=-6$ .

Therefore,

$\frac{dy}{dx}=\frac{7}{-6}=-\frac{7}{6}$ ..

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Example Question #122 : Parametric, Polar, And Vector

Find the derivative of the curve defined by the parametric equations.

x(t)=3sin(t)

y(t)=t^2+4t

Possible Answers:

$\frac{dy}{dx}=-\frac{2t+4}{3sin(t)cos(t)}$

$\frac{dy}{dx}=3cos(t)+2t+4$

$\frac{dy}{dx}=\frac{6t^2cos(t)+10tcos(t)-4cos(t)}{9cos^2(t)}$

$\frac{dy}{dx}=\frac{2t+4}{3cos(t)}$

Correct answer:

$\frac{dy}{dx}=\frac{2t+4}{3cos(t)}$

Explanation:

The first derivative of a parametrically defined curve is found by computing

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

We need to find the derivatives of y(t) and x(t) separately, and then find the quotient of the derivatives.

You will need to know that

$\frac{d}{dt}sin(t)=cos(t)$ and that $\frac{d}{dt}t^n=nt^{n-1}$ .

$x'(t)=\frac{dx}{dt}=3cos(t)$

$y'(t)=\frac{dy}{dt}=2t+4$

Thus,

$\frac{dy}{dx}=\frac{2t+4}{3cos(t)}$

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Example Question #123 : Parametric, Polar, And Vector

Find the derivative of the following parametric function:

x=t^3+2t^2+4t, y=2t

Possible Answers:

$\frac{2}{3t^2+4t+4}$

$\frac{3t^2+4t+4}{2}$

3t^2+4t+4

Correct answer:

$\frac{2}{3t^2+4t+4}$

Explanation:

The derivative of a parametric equation is given by the following equation:

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

The derivative of the equation for is

$\frac{dy}{dt}=2$

and the derivative of the equation for is

$\frac{dx}{dt}=3t^2+4t+4$

The derivatives were found using the following rule:

$\frac{d}{dx}(x^n)=nx^{n-1}$

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Example Question #124 : Parametric, Polar, And Vector

Find the derivative of the following parametric function:

x=3t+5 , $y=te^t+\cos^2(t)$

Possible Answers:

$\frac{e^t-2\cos(t)\sin(t)}{3}$

$\frac{e^t+te^t+2\cos(t)\sin(t)}{3}$

$\frac{e^t+te^t-2\cos(t)\sin(t)}{3}$

$\frac{2\cos(t)\sin(t)}{3}$

Correct answer:

$\frac{e^t+te^t-2\cos(t)\sin(t)}{3}$

Explanation:

The derivative of a parametric function is given by

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

So, we must find the derivative of the functions with respect to t:

$y'=e^t+te^t-2\cos(t)\sin(t)$ , x'=3

The derivatives were found using the following rules:

$\frac{d}{dx}(x^n)=nx^{n-1}$ , $\frac{d}{dx}(e^x)=e^x$ $\frac{d}{dx}\cos(t)=-\sin(t)$ , $\frac{d}{dx}f(g(x))=f'(g(x))\cdot g'(x)$

Simply divide the derivatives to get your answer.

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Example Question #121 : Parametric

What is $\frac{dy}{dx}$

if x=10t-3 and y=2t+9 ?

Possible Answers:

$-\frac{1}{5}$

None of the above

$\frac{1}{5}$

Correct answer:

$\frac{1}{5}$

Explanation:

We can first recognize that

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

since cancels out when we divide.

Then, given x=10t-3 and y=2t+9 and using the Power Rule

$\frac{d}{dx}x^{n}=nx^{n-1}$ for all $n\neq0$ , we can determine that

$\frac{dx}{dt}=(1)10t^{1-1}-(0)3=10$ and

$\frac{dy}{dt}=(1)2t^{1-1}+(0)9=2$ .

Therefore,

$\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}=\frac{2}{10}=\frac{1}{5}$ .

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Example Question #125 : Parametric, Polar, And Vector

What is $\frac{dy}{dx}$

if x=3t+2 and y=9t-5 ?

Possible Answers:

$\frac{1}{3}$

None of the above

$-\frac{1}{3}$

Correct answer:

Explanation:

We can first recognize that

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

since cancels out when we divide.

Then, given x=3t+2 and y=9t-5 and using the Power Rule

$\frac{d}{dx}x^{n}=nx^{n-1}$ for all $n\neq0$ , we can determine that

$\frac{dx}{dt}=(1)3t^{1-1}+(0)2=3$ and

$\frac{dy}{dt}=(1)9t^{1-1}-(0)5=9$ .

Therefore,

$\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}=\frac{9}{3}=3$ .

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Example Question #126 : Parametric, Polar, And Vector

What is $\frac{dy}{dx}$

if x=5t-1 and y=t+5 ?

Possible Answers:

$-\frac{1}{5}$

None of the above

$\frac{1}{5}$

Correct answer:

$\frac{1}{5}$

Explanation:

We can first recognize that

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

since cancels out when we divide.

Then, given x=5t-1 and y=t+5 and using the Power Rule

$\frac{d}{dx}x^{n}=nx^{n-1}$ for all $n\neq0$ , we can determine that

$\frac{dx}{dt}=(1)5t^{1-1}-(0)1=5$ and

$\frac{dy}{dt}=(1)t^{1-1}+(0)5=1$ .

Therefore,

$\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}=\frac{1}{5}$ .

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Example Question #11 : Derivatives Of Parametrics

Find the derivative of the following parametric function at t=0 :

$x=t^2\cos(t)+3t$ ,

y=te^t

Possible Answers:

$\frac{1}{3}$

$-\frac{1}{3}$

$\infty$

Correct answer:

$\frac{1}{3}$

Explanation:

The derivative of a parametric function is given by the following:

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

So, we must find the derivative of each function at the t given:

y'=e^t+te^t , $x'=2t\cos(t)-t^2\sin(t)+3$

The derivatives were found using the following rules:

$\frac{d}{dx}(e^x)=e^x$ , $\frac{d}{dx}(x^n)=nx^{n-1}$ ,

$\frac{d}{dx}\sin(x)=\cos(x)$ ,

$\frac{d}{dx}f(x)g(x)=f'(x)g(x)+f(x)g'(x)$

Next, plug in the given t=0 into each derivative function:

y'(0)=1 , x'(0)=3

Finally, divide $\frac{y'}{x'}$ to get a final answer of $\frac{1}{3}$ .

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Calculus 2 : Derivatives of Parametrics

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

Example Question #121 : Parametric, Polar, And Vector

Example Question #12 : Derivatives Of Parametrics

Example Question #121 : Parametric

Example Question #122 : Parametric, Polar, And Vector

Example Question #123 : Parametric, Polar, And Vector

Example Question #124 : Parametric, Polar, And Vector

Example Question #121 : Parametric

Example Question #125 : Parametric, Polar, And Vector

Example Question #126 : Parametric, Polar, And Vector

Example Question #11 : Derivatives Of Parametrics

All Calculus 2 Resources

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