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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 #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 #131 : Parametric, Polar, And Vector

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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Example Question #132 : Parametric, Polar, And Vector

Find $\frac{dy}{dx}$ for the following set of parametric equations for t = 2 .

$y = 5t^{2} + 3t$

$x = \ln t$

Possible Answers:

Does not Exist

$\frac{1}{46}$

$\frac{1}{23}$

Correct answer:

Explanation:

Finding $\frac{dy}{dx}$ of a parametric equation can be given by this formula:

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

So we must find $\frac{dy}{dt}$ and $\frac{dx}{dt}$ for when t = 2 .

$\frac{dy}{dt} = 10t + 3$ and $\frac{dx}{dt} = \frac{1}{t}$ and so

$\frac{dy}{dx} = 10t^{2} + 3t$ .

When you plug in t = 2 you get your answer .

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Example Question #133 : Parametric, Polar, And Vector

Find the derivative of the following parametric equation

$\mathbf{r}(t)=t^{3}\mathbf{i}+3ln(t)\mathbf{j}+5sin(t)\mathbf{k}$

Possible Answers:

Does not exist

$\mathbf{r'}(t)=3t^{2}\mathbf{i}+\frac{3}{t}\mathbf{j}+5cos(t)\mathbf{k}$

$\mathbf{r'}(t)=3t^{2}+\frac{3}{t}+5cos(t)$

$\mathbf{r'}(t)=2t^{2}\mathbf{i}+3t\mathbf{j}+5cos(t)\mathbf{k}$

$\mathbf{r'}(t)=t^{3}\mathbf{i}+3ln(t)\mathbf{j}+5sin(t)\mathbf{k}$

Correct answer:

$\mathbf{r'}(t)=3t^{2}\mathbf{i}+\frac{3}{t}\mathbf{j}+5cos(t)\mathbf{k}$

Explanation:

This parametric equation is described as the sum of three vectors. To find the derivative of a parametric equation, you must find the derivative of each vector, or if

$\mathbf{r}(t)=x(t)\mathbf{i}+y(t)\mathbf{j}+x(t)\mathbf{k}$ then $\mathbf{r'}(t)=x'(t)\mathbf{i}+y'(t)\mathbf{j}+x'(t)\mathbf{k}$

The derivative of the first vector is found using the power rule,

$\frac{d}{dt}(t^{c})=ct^{c-1}$ where is a constant.

The derivative of the second vector is found using the natural logarithm rule,

$\frac{d}{dt}[ln(t)]=\frac{1}{t}$ .

The derivative of the third vector is found using one of the trigonmetric rules,

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

In this case:

$\mathbf{r'}(t)=\frac{d}{dt}(t^{3})\mathbf{i}+3\frac{d}{dt}[ln(t)]\mathbf{j}+5\frac{d}{dt}[sin(t)]\mathbf{k}$

$=3t^{2}\mathbf{i}+\frac{3}{t}\mathbf{j}+5cos(t)\mathbf{k}$

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Example Question #134 : Parametric, Polar, And Vector

Find the derivative of the following parametric equation

$\mathbf{r}(t)=t\mathbf{i}+3e^{t}\mathbf{j}+sin(\pi t)\mathbf{k}$

Possible Answers:

$\mathbf{r'}(t)=t\mathbf{i}+3e^{t}\mathbf{j}+sin(\pi t)\mathbf{k}$

$\mathbf{r'}(t)=\mathbf{i}+3e^{t}\mathbf{j}+\pi cos(\pi t)\mathbf{k}$

Does not exist

$\mathbf{r'}(t)=1+3e^{t}+\pi cos(\pi t)$

$\mathbf{r'}(t)=t\mathbf{i}+9e^{t}\mathbf{j}+\pi sin(\pi t)\mathbf{k}$

Correct answer:

$\mathbf{r'}(t)=\mathbf{i}+3e^{t}\mathbf{j}+\pi cos(\pi t)\mathbf{k}$

Explanation:

This parametric equation is described as the sum of three vectors. To find the derivative of a parametric equation, you must find the derivative of each vector, or if

$\mathbf{r}(t)=x(t)\mathbf{i}+y(t)\mathbf{j}+x(t)\mathbf{k}$ , then $\mathbf{r'}(t)=x'(t)\mathbf{i}+y'(t)\mathbf{j}+x'(t)\mathbf{k}$

The derivative of the first vector is found using the power rule,

$\frac{d}{dt}(t)=1$ .

The derivative of the second vector is found using the exponential rule,

$\frac{d}{dt}(e^{t})=e^{t}$ .

The derivative of the third vector is found using one of the trigonmetric rules,

$\frac{d}{dt}[sin(at)]=acos(at)$ , where is a constant.

In this case:

$\mathbf{r'}(t)=\frac{d}{dt}(t)\mathbf{i}+3\frac{d}{dt}(e^{t})\mathbf{j}+\frac{d}{dt}[sin(\pi t)]\mathbf{k}$

$=\mathbf{i}+3e^{t}\mathbf{j}+\pi cos(\pi t)\mathbf{k}$

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Example Question #135 : Parametric, Polar, And Vector

Find the derivative of the following parametric equation

$\mathbf{r}(t)=2cos(5t)\mathbf{i}+5sin(7t)\mathbf{j}$

Possible Answers:

$\mathbf{r'}(t)=-10sin(5t)\mathbf{i}+35cos(7t)\mathbf{j}$

$\mathbf{r'}(t)=2cos(5t)\mathbf{i}+5sin(7t)\mathbf{j}$

$\mathbf{r'}(t)=-10cos(5t)\mathbf{i}+35sin(7t)\mathbf{j}$

Does not exist

$\mathbf{r'}(t)=-10sin(5t)+35cos(7t)$

Correct answer:

$\mathbf{r'}(t)=-10sin(5t)\mathbf{i}+35cos(7t)\mathbf{j}$

Explanation:

This parametric equation is described as the sum of three vectors. To find the derivative of a parametric equation, you must find the derivative of each vector, or if

$\mathbf{r}(t)=x(t)\mathbf{i}+y(t)\mathbf{j}+x(t)\mathbf{k}$ , then $\mathbf{r'}(t)=x'(t)\mathbf{i}+y'(t)\mathbf{j}+x'(t)\mathbf{k}$

The derivative of the first and second vectors are found using the following trigonometric rules,

$\frac{d}{dt}[cos(at)]=-asin(at)$ and $\frac{d}{dt}[sin(bt)]=bcos(bt)$ ,

where and are constants.

In this case:

$\mathbf{r'}(t)=2\frac{d}{dt}[cos(5t)]\mathbf{i}+5\frac{d}{dt}[sin(7t)]\mathbf{j}$

$=[2*(-sin(5t))*5]\mathbf{i}+[5*cos(7t)*7]\mathbf{j}$

$=-10sin(5t)\mathbf{i}+35cos(7t)\mathbf{j}$

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Example Question #136 : Parametric, Polar, And Vector

Find $\frac{dy}{dx}$ when x=sin(7t) and y=cos(7t) .

Possible Answers:

$\frac{dy}{dx}=-7tan(7t)$

$\frac{dy}{dx}=-7sin(7t)\mathbf{i}+7cos(7t)\mathbf{j}$

$\frac{dy}{dx}=-7sin(7t)$

$\frac{dy}{dx}=-tan(7t)$

$\frac{dy}{dx}=7cos(7t)$

Correct answer:

$\frac{dy}{dx}=-tan(7t)$

Explanation:

If x=x(t) and y=y(t) , then we can use the chain rule to define $\frac{dy}{dx}$ as

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

We then use the following trigonometric rules,

$\frac{d}{dt}[cos(at)]=-asin(at)$ and $\frac{d}{dt}[sin(bt)]=bcos(bt)$ ,

where and are constants.

In this case:

$\frac{dy}{dt}=\frac{d}{dt}(cos(7t))=-7sin(7t)$ ,

and

$\frac{dx}{dt}=\frac{d}{dt}(sin(7t))=7cos(7t)$ ,

therefore

$\frac{dy}{dx}=\frac{dy}{dt}/\frac{dx}{dt}=\frac{-7sin(7t)}{7cos(7t)}=-tan(7t)$ .

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Example Question #1 : Parametric Calculations

Calculate the length of the curve drawn out by the vector function ${\mathbf{v}(t)} = \left \langle t,\sqrt{2}}e^t,\frac{1}{2}e^{2t} \right \rangle,$ from 0<t<1 .

Possible Answers:

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

ln(2) -1

e+1

None of the other answers.

e^2 +e^4

Correct answer:

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

Explanation:

The formula for arc length of a parametric curve in space is $\int_{a}^{b} \sqrt{(\frac{d\mathbf{v}_1}{dt})^2 +(\frac{d\mathbf{v}_2}{dt})^2 +(\frac{d\mathbf{v}_3}{dt})^2} dt$ for a <t<b .

Taking derivatives of each of the vector function components and substituting the values into this formula gives

$\int_{0}^{1} \sqrt{1 + 2e^{2t} + e^{4t}} dt$

We need to recognize that underneath the square root we have a perfect square, and we can write it as

$\int_{0}^{1} \sqrt{(e^{2t} + 1)^2} dt$

Solving this we get

$\int_{0}^{1} e^{2t}+1 dt = [\frac{1}{2}e^{2t} +t]_0^1$ $= \bigg(\frac{1}{2}e^2+1\bigg)-\bigg(\frac{1}{2}\bigg) = \frac{1}{2}(e^2+1)$

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Calculus 2 : Calculus II

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

Example Question #121 : Parametric

Example Question #125 : Parametric, Polar, And Vector

Example Question #126 : Parametric, Polar, And Vector

Example Question #131 : Parametric, Polar, And Vector

Example Question #132 : Parametric, Polar, And Vector

Example Question #133 : Parametric, Polar, And Vector

Example Question #134 : Parametric, Polar, And Vector

Example Question #135 : Parametric, Polar, And Vector

Example Question #136 : Parametric, Polar, And Vector

Example Question #1 : Parametric Calculations

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