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 vectorsandare orthogonal. Find all real values of.

Possible Answers:

Correct answer:

Explanation:

Example Question #2 :Calculus 3

The vector-valued functionparemeterizes a curve, where.

Find a tangent vector toat 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 #3 :Calculus 3

Letf(x)=\ln (1-x^2). Which of the following is equal tof'(x)?

Possible Answers:

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

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

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

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

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

Correct answer:

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

Explanation:

Notice thatf(x)可以表示为一个复合函数,即function within a function. If we letg(x)=1-x^2andh(x)=\ln (x), thenf(x)=h(g(x)). In order to differentiatef(x), we will need to apply the Chain Rule, as shown below:

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

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

Then, we need to findg'(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 isf'(x)=\frac{2x}{x^2-1}.

Example Question #1 :Numerical Approximations To Definite Integrals

Evaluate:

Possible Answers:

cannot be determined

Correct answer:

Explanation:

First, we can write out the first few terms of the sequence, whereranges 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, whereis the common ratio between the terms. Becausein 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 #3 :Calculus 3

Findif.

Possible Answers:

Correct answer:

Explanation:

We will have to find the first derivative ofwith respect tousing implicit differentiation. Then, we can find, which is the second derivative ofwith 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 differentiatewith respect to. This will require us to employ the Quotient Rule.

We will replacewith.

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

The answer is.

Example Question #1 :Calculus 3

Which of the following represents the graph of the polar functionin Cartestian coordinates?

Possible Answers:

Correct answer:

Explanation:

First, mulitply both sides by r.

Then, use the identitiesand.

The answer is.

Example Question #1 :Calculus 3

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

Possible Answers:

undefined

1-e^{2}

e^{2}-e

e^{2}-1

e+1

Correct answer:

e^{2}-e

Explanation:

We can use the substitution technique to evaluate this integral.

Let.

We will differentiatewith respect to.

, which means that.

We can solve forin terms of, which gives us.

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

We will now substitutein for the, and we will substitutefor.

\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 ise^{2}-e.

Example Question #3 :Calculus 3

Evaluate the following limit:

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

Possible Answers:

\infty

\frac{1}{4}

Does not exist.

\frac{1}{2}

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}

Asbecomes 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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