Calculus 3 : Calculus 3

Study concepts, example questions & explanations for Calculus 3

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Example Questions

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Example Question #1 : Calculus 3

Suppose the vectors and are orthogonal. Find all real values of .

Possible Answers:

Correct answer:

Explanation:

Example Question #2 : Calculus 3

The vector-valued function paremeterizes a curve , where .

 

Find a tangent vector to at the point .

Possible Answers:

Correct answer:

Explanation:

Example Question #2 : Calculus 3

Let .

Find .

Possible Answers:

Correct answer:

Explanation:

Example Question #4 : Calculus 3

Let .

Find its linear approximation at .

Possible Answers:

Hint: Use Taylor's formula.


Correct answer:

Explanation:

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}.

Example Question #11 : Integrals

Evaluate:

Possible Answers:

cannot be determined

Correct answer:

Explanation:

First, we can write out the first few terms of the sequence , where  ranges from 1 to 3.

Notice that each term , is found by multiplying the previous term by . Therefore, this sequence is a geometric sequence with a common ratio of . We can find the sum of the terms in an infinite geometric sequence, provided that , where  is the common ratio between the terms. Because  in this problem,  is indeed less than 1. Therefore, we can use the following formula to find the sum, , of an infinite geometric series.

The answer is .

 

Example Question #171 : Functions, Graphs, And Limits

Find  if .

Possible Answers:

Correct answer:

Explanation:

We will have to find the first derivative of  with respect to  using implicit differentiation. Then, we can find , which is the second derivative of  with respect to .

We will apply the chain rule on the left side.

We now solve for the first derivative with respect to .

In order to get the second derivative, we will differentiate with respect to . This will require us to employ the Quotient Rule.

We will replace  with .

But, from the original equation, . Also, if we solve for , we obtain .

The answer is .

Example Question #1 : Interpretations And Properties Of Definite Integrals

Which of the following represents the graph of the polar function  in Cartestian coordinates?

Possible Answers:

Correct answer:

Explanation:

First, mulitply both sides by r. 

Then, use the identities  and .

The answer is .

Example Question #1 : Calculus 3

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

Possible Answers:

1-e^{2}

e^{2}-1

undefined

e+1

e^{2}-e

Correct answer:

e^{2}-e

Explanation:

We can use the substitution technique to evaluate this integral.

Let .

We will differentiate  with respect to .

, which means that .

We can solve for  in terms of , which gives us .

We will also need to change the bounds of the integral. When , , and when , .

We will now substitute  in for the , and we will substitute  for .

\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.

Example Question #1 : Calculus 3

Evaluate the following limit:

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

Possible Answers:

\frac{1}{2}

Does not exist.

\infty

\frac{1}{4}

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 .

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

 

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

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