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Calculus 2 : Calculus II

Study concepts, example questions & explanations for Calculus 2

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9 Diagnostic Tests 308 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #6 : Derivatives Of Parametrics

Find $\frac{dx}{dy}$ if x=cos(t) and y=sin(t) .

Possible Answers:

cot(t)

-cot(t)

-tan(t)

sec(t)

tan(x)

Correct answer:

-tan(t)

Explanation:

Write the formula to find the derivative for parametric equations.

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

$\frac{dx}{dt}= -sin(t)$

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

Substitute the knowns into the formula.

$\frac{dx}{dy}=\frac{\frac{dx}{dt}}{\frac{dy}{dt}}=\frac{-sin(t)}{cos(t)}=-tan(t)$

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Example Question #7 : Derivatives Of Parametrics

Solve for $\frac{dy}{dx}$ if x=5t+9 and y=2t+7 .

Possible Answers:

$-\frac{5}{2}$

$\frac{2}{5}$

$\frac{5}{2}$

$-\frac{2}{5}$

Correct answer:

$\frac{2}{5}$

Explanation:

We can determine that $\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}$ since the terms will cancel out in the division process.

Since x=5t+9 and y=2t+7 , we can use the Power Rule

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

$\frac{dx}{dt}=(1)5x^{1-1}=5$ and $\frac{dy}{dt}=(1)2t^{1-1}+(0)7=2$ .

Thus:

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

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Example Question #2 : Derivatives Of Parametric, Polar, And Vector Functions

Solve for $\frac{dy}{dx}$ if x=2t-5 and y=3t+8 .

Possible Answers:

$\frac{2}{3}$

$-\frac{2}{3}$

$-\frac{3}{2}$

None of the above

$\frac{3}{2}$

Correct answer:

$\frac{3}{2}$

Explanation:

We can determine that $\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}$ since the terms will cancel out in the division process.

Since x=2t-5 and y=3t+8 , we can use the Power Rule

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

$\frac{dx}{dt}=(1)2t^{1-1}-(0)5=2$ and $\frac{dy}{dt}=(1)3t^{1-1}+(0)8=3$ .

Thus:

$\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}=\frac{3}{2}}$ .

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Example Question #7 : Derivatives Of Parametrics

Solve for $\frac{dy}{dx}$ if x=9t+1 and y=4t+10 .

Possible Answers:

$-\frac{9}{4}$

$\frac{4}{9}$

$\frac{9}{4}$

$-\frac{4}{9}$

None of the above

Correct answer:

$\frac{4}{9}$

Explanation:

We can determine that $\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}$ since the terms will cancel out in the division process.

Since x=9t+1 and y=4t+10 , we can use the Power Rule

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

$\frac{dx}{dt}=(1)9t^{1-1}+(0)1=9$ and $\frac{dy}{dt}=(1)4t^{1-1}+(0)10=4$ .

Thus:

$\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}=\frac{4}{9}}$ .

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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 #13 : Derivatives Of Parametrics

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

Possible Answers:

$-\frac{7}{6}$

$-\frac{6}{7}$

None of the above

$\frac{7}{6}$

$\frac{6}{7}$

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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Calculus 2 : Calculus II

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

Example Question #6 : Derivatives Of Parametrics

Example Question #7 : Derivatives Of Parametrics

Example Question #2 : Derivatives Of Parametric, Polar, And Vector Functions

Example Question #7 : Derivatives Of Parametrics

Example Question #121 : Parametric, Polar, And Vector

Example Question #12 : Derivatives Of Parametrics

Example Question #13 : Derivatives Of Parametrics

Example Question #122 : Parametric, Polar, And Vector

Example Question #123 : Parametric, Polar, And Vector

Example Question #124 : Parametric, Polar, And Vector

All Calculus 2 Resources

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