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Calculus 3 : Calculus 3

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Example Questions

Example Question #1 : Calculus 3

Suppose the vectors ci+10j+ck and ci-2j-k are orthogonal. Find all real values of .

Possible Answers:

$4\ \textup{and}\ -5$

$2\ \textup{and}\ 10$

$-3\ \textup{and}\ 3$

$-4\ \textup{and}\5$

Correct answer:

$-4\ \textup{and}\5$

Explanation:

$-4\ \textup{and}\ 5$

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Example Question #2 : Calculus 3

The vector-valued function r(t) paremeterizes a curve , where $r(t)=\left \langle e^{2t}, 3t+1, t^{2} \right \rangle$ .

Find a tangent vector to at the point $\left ( e^{10}, 16, 25 \right )$ .

Possible Answers:

$\left \langle e^{10}, 16, 25 \right \rangle$

$\left \langle -2e^{10}, -3, -10\right \rangle$

$\left \langle e, 0, 1 \right \rangle$

$\left \langle 2e^{10}, 3, 10 \right \rangle$

$\left \langle 1,1,1 \right \rangle$

Correct answer:

$\left \langle 2e^{10}, 3, 10 \right \rangle$

Explanation:

$\left \langle 2e^{10}, 3, 10 \right \rangle$

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Example Question #2 : Calculus 3

Let $f\left ( x,y,z \right )=\sqrt{3x^{2}y+z^{3}}$ .

Find $f_{x}\left ( x,y,z \right )$ .

Possible Answers:

$\frac{-3xy}{\sqrt{3x^{2}y-z^{3}}}$

$\frac{3xy}{\sqrt{3x^{2}y+z^{3}}}$

$\sqrt{3x^{2}y+z^{3}}$

$\frac{1}{\sqrt{3x^{2}y+z^{3}}}$

3xy

Correct answer:

$\frac{3xy}{\sqrt{3x^{2}y+z^{3}}}$

Explanation:

$\frac{3xy}{\sqrt{3x^{2}y+z^{3}}}$

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Example Question #4 : Calculus 3

Let $f(x,y)=e^{3x+y+1}$ .

Find its linear approximation at $\left ( 0,0 \right )$ .

Possible Answers:

$L\left ( x,y \right )=3ex$

$L\left ( x,y \right )=ey$

$L\left ( x,y \right )=e+3ex+ey$

$L\left ( x,y \right )=e-3ex-ey$

$L\left ( x,y \right )=3ex-ey$

Hint: Use Taylor's formula.

Correct answer:

$L\left ( x,y \right )=e+3ex+ey$

Explanation:

$L\left ( x,y \right )=e+3ex+ey$

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Example Question #1 : Calculus 3

Let $f(x)=\ln (1-x^2)$ . Which of the following is equal to $f'(x)$ ?

Possible Answers:

$f'(x)=x^2\ln x-\ln x$

$f'(x)=2x\ln (1-x^2)$

$f'(x)=\frac{2x}{x^2-1}$

$f'(x)=\frac{\ln x}{1-x^2}$

$f'(x)=\frac{1}{1-x^2}$

Correct answer:

$f'(x)=\frac{2x}{x^2-1}$

Explanation:

Notice that $f(x)$ can be expressed as a composite function, i.e. a function within a function. If we let $g(x)=1-x^2$ and $h(x)=\ln (x)$ , then $f(x)=h(g(x))$ . In order to differentiate $f(x)$ , we will need to apply the Chain Rule, as shown below:

$f'(x)=h'(g(x))\cdot g'(x)$

First, we need to find $h'(x)$ , which equals $\frac{1}{x}$ .

Then, we need to find $g'(x)$ by applying the Product Rule.

$g'(x)=-2x$

$f'(x)=h'(g(x))\cdot g'(x)$ $=\frac{1}{1-x^2}\cdot (-2x)=\frac{2x}{x^2-1}$

The answer is $f'(x)=\frac{2x}{x^2-1}$ .

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Example Question #11 : Integrals

Evaluate:

$\sum_{n=1}^{\infty }4(\frac{-1}{2})^{n-1}$

Possible Answers:

$\frac{8}{3}$

$\infty$

cannot be determined

Correct answer:

$\frac{8}{3}$

Explanation:

First, we can write out the first few terms of the sequence $a_{n}=$ $4(\frac{-1}{2})^{n-1}$ , where ranges from 1 to 3.

$a_{1}=4(\frac{-1}{2})^{1-1}=4$

$a_{2}=4(\frac{-1}{2})^{2-1}=-2$

$a_{3}=4(\frac{-1}{2})^{3-1}=1$

Notice that each term $\left ( n>1 \right )$ , is found by multiplying the previous term by $-\frac{1}{2}$ . Therefore, this sequence is a geometric sequence with a common ratio of $-\frac{1}{2}$ . We can find the sum of the terms in an infinite geometric sequence, provided that |r|<1 , where is the common ratio between the terms. Because $r=-\frac{1}{2}$ in this problem, $\left | r \right |$ is indeed less than 1. Therefore, we can use the following formula to find the sum, , of an infinite geometric series.

$S=\frac{a_{1}}{1-r}=\frac{4}{1-(\frac{-1}{2})}=\frac{4}{3/2}=\frac{8}{3}$

The answer is $\frac{8}{3}$ .

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Example Question #171 : Functions, Graphs, And Limits

Find $\frac{d^{2}y}{dx^{2}}$ if y^2-x^2 = 4 .

Possible Answers:

$\frac{x^2-4}{\sqrt{x^2+4}}$

$\frac{x^2}{(x^2-4)^\frac{3}{2}}$

$\frac{4}{(x^{2}+4)^{\frac{3}{2}}}$

$\frac{x^2+4}{(x^2-4)^\frac{3}{2}}$

$\frac{x}{\sqrt{4+x^2}}$

Correct answer:

$\frac{4}{(x^{2}+4)^{\frac{3}{2}}}$

Explanation:

We will have to find the first derivative of with respect to using implicit differentiation. Then, we can find $\frac{d^{2}y}{dx^{2}}$ , which is the second derivative of with respect to .

$\frac{d}{dx}[y^2-x^2]=\frac{d}{dx}[4]$

We will apply the chain rule on the left side.

$\frac{d}{dx}[y^2]\cdot \frac{dy}{dx}-\frac{d}{dx}[x^2]=0$

$2y\cdot \frac{dy}{dx}-2x=0$

We now solve for the first derivative with respect to .

$\frac{dy}{dx}=\frac{x}{y}$

In order to get the second derivative, we will differentiate $\frac{x}{y}$ with respect to . This will require us to employ the Quotient Rule.

$\frac{d}{dx}\frac{dy}{dx}=\frac{d^{2}y}{dx^{2}}=\frac{d}{dx}[\frac{x}{y}]=\frac{y\cdot \frac{d}{dx}[x]-x\cdot \frac{d}{dx}[y]}{y^{2}} =\frac{y\cdot 1-x\cdot \frac{dy}{dx}}{y^{2}}$

We will replace $\frac{\mathrm{d}y }{\mathrm{d} x}$ with $\frac{x}{y}$ .

$\frac{y-x(\frac{x}{y})}{y^{2}}=\frac{y^{2}-x^{2}}{y^{3}}$

But, from the original equation, $y^{2}-x^{2}=4$ . Also, if we solve for , we obtain $y = (4+x^{2})^{\frac{1}{2}}$ .

$\frac{d^{2}y}{dx^{2}}=\frac{y^{2}-x^{2}}{y^{3}}=\frac{4}{(x^{2}+4)^{\frac{3}{2}}}$

The answer is $\frac{4}{(x^{2}+4)^{\frac{3}{2}}}$ .

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Example Question #1 : Calculus 3

Which of the following represents the graph of the polar function $r = 4\cos\theta$ in Cartestian coordinates?

Possible Answers:

x^2+(y-2)^2 =4

x^2 +y^2 =4

(x-2)^2 +y^2 =4

(x-2)^2 +(y-2)^2 =4

(x-1)^2 +(y-2)^2 =4

Correct answer:

(x-2)^2 +y^2 =4

Explanation:

$r = 4\cos\theta$

First, mulitply both sides by r.

$r^2 =4r\cos\theta$

Then, use the identities x^2+y^2 =r^2 and $x = r\cos\theta$ .

x^2+y^2=4x

x^2-4x+y^2=0

x^2 -4x + 4 +y^2 =4

(x-2)^2 +y^2 =4

The answer is (x-2)^2 +y^2 =4 .

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Example Question #1 : Calculus 3

$\int_{-1}^{0}e^{1-t}dt =$

Possible Answers:

$1-e^{2}$

$e^{2}-e$

$undefined$

$e^{2}-1$

$e+1$

Correct answer:

$e^{2}-e$

Explanation:

We can use the substitution technique to evaluate this integral.

Let u=1-t .

We will differentiate with respect to .

$\frac{\mathrm{d}u }{\mathrm{d} t}=-1$ , which means that ${\mathrm{d} u}=-{\mathrm{d} t}$ .

We can solve for ${\mathrm{d}t }$ in terms of ${\mathrm{d} u}$ , which gives us ${\mathrm{d}t }=-{\mathrm{d} u}$ .

We will also need to change the bounds of the integral. When t=-1 , u=1-(-1) , and when t=0 , u=1-0=1 .

We will now substitute in for the 1-t , and we will substitute $-{\mathrm{d}u }$ for ${\mathrm{d} t}$ .

$\int_{2}^{1}-e^{u}du$

$\int_{2}^{1}-e^{u}du = -e^{u}|_{2}^{1}=-e^{1}-(-e^{2})=e^{2}-e^{1}$

The answer is $e^{2}-e$ .

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Example Question #1 : Calculus 3

Evaluate the following limit:

$\lim_{x\rightarrow \infty }\frac{1-x^4}{x^2-4x^4}$

Possible Answers:

$\infty$

$\frac{1}{2}$

$\frac{1}{4}$

Does not exist.

Correct answer:

$\frac{1}{4}$

Explanation:

First, let's multiply the numerator and denominator of the fraction in the limit by $\frac{1}{x^{4}}$ .

$\lim_{x\rightarrow \infty }\frac{1-x^4}{x^2-4x^4}$ $=\lim_{x\rightarrow \infty }\frac{(\frac{1}{x^4})(1-x^4)}{\frac{1}{x^{4}}(x^2-4x^4)}$

$=\lim_{x\rightarrow \infty }\frac{\frac{1}{x^4}-1}{\frac{1}{x^2}-4}$

As becomes increasingly large the $\frac{1}{x^{4}}$ and $\frac{1}{x^{2}} ^{}$ terms will tend to zero. This leaves us with the limit of $\frac{1}{4}$ .

$\lim_{x\rightarrow \infty }\frac{-1}{-4}=\frac{1}{4}$ .

The answer is $\frac{1}{4}$ .

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Calculus 3 : Calculus 3

Study concepts, example questions & explanations for Calculus 3

All Calculus 3 Resources

Example Questions

Example Question #1 : Calculus 3

Example Question #2 : Calculus 3

Example Question #2 : Calculus 3

Example Question #4 : Calculus 3

Example Question #1 : Calculus 3

Example Question #11 : Integrals

Example Question #171 : Functions, Graphs, And Limits

Example Question #1 : Calculus 3

Example Question #1 : Calculus 3

Example Question #1 : Calculus 3

All Calculus 3 Resources

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