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Calculus 3 : Calculus 3

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6 Diagnostic Tests 373 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #64 : Derivatives Of Functions

Let $f(x)=x^2-\frac{1}{1-x^2}$ . Which of the following gives the equation of the line normal to $f(x)$ when x=2 ?

Possible Answers:

$27x+111y=535$

$27x + 96y = 470$

$32x+9y=103$

$192x + 54y = 735$

$32x-9y=25$

Correct answer:

$27x + 96y = 470$

Explanation:

We are asked to find the normal line. This means we need to find the line that is perpendicular to the tangent line at x=2 . In order to find the tangent line, we will need to evaluate the derivative of f(x) at x=2 .

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

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

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

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

$f'(2)=4-4(\frac{1}{9})=\frac{32}{9}$

The slope of the tangent line at x=2 is $\frac{32}{9}$ . Because the tangent line and the normal line are perpendicular, the product of their slopes must equal .

(slope of tangent)(slope of normal) =

$\left ( \frac{32}{9} \right )\cdot \left ( slope\ of\ normal \right )=-1$

$slope\ of\ normal=-\frac{9}{32}$

We now have the slope of the normal line. Once we find a point through which it passes, we will have enough information to derive its equation.

Since the normal line passes through the function at x=2 , it will pass through the point $\left ( 2,f(2) \right )$ . Be careful to use the original equation for , not its derivative.

$f(2)=2^2-(1-4)^{-1}=4-(-\frac{1}{3})=\frac{13}{3}$

The normal line has a slope of $-\frac{9}{32}$ and passes through the piont $\left ( 2,\frac{13}{3} \right )$ . We can now use point-slope form to find the equation of the normal line.

$y-\frac{13}{3}=-\frac{9}{32}(x-2)$

Multiply both sides by 32(3)=96 .

$96y-416=-27(x-2)$

$27x + 96y = 470$

The answer is $27x + 96y = 470$ .

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Example Question #1 : Interpretations And Properties Of Definite Integrals

If $g(x)=\int_{0}^{x^2}f(t)dt$ , then which of the following is equal to $g'(x)$ ?

Possible Answers:

$g(f(x^{2}))$

$2xf(x^{2})$

$f(x^{2})-f(0)$

$f(2x)$

$f(x^{2})$

Correct answer:

$2xf(x^{2})$

Explanation:

According to the Fundamental Theorem of Calculus, if we take the derivative of the integral of a function, the result is the original function. This is because differentiation and integration are inverse operations.

For example, if $h(x)=\int_{a}^{x}f(u)du$ , where is a constant, then $h'(x)=f(x)$ .

We will apply the same principle to this problem. Because the integral is evaluated from 0 to $x^{2}$ , we must apply the chain rule.

$g'(x)=\frac{d}{dx}\int_{0}^{x^{2}}f(t)dt=f(x^{2})\cdot \frac{d}{dx}(x^{2})$

$=2xf(x^{2})$

The answer is $2xf(x^{2})$ .

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Example Question #11 : Calculus 3

What is the first derivative of the function $\large h(x)=x^{\frac{1}{x}}$ ?

Possible Answers:

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

$h'(x)=\ln (\frac{1}{x})\cdot x^{\frac{1}{x}-1}$

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

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

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

Correct answer:

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

Explanation:

First, let y=h(x) .

$y=x^{\frac{1}{x}}$

We will take the natural logarithm of both sides in order to simplify the exponential expression on the right.

$\ln y=\ln x^{\frac{1}{x}}$

Next, apply the property of logarithms which states that, in general, $\log x^a=a\log x$ , where is a constant.

$\ln y = \frac{1}{x}\ln x$

We can differentiate both sides with respect to .

$\frac{d}{dx}\[ln y]=\frac{d}{dx}[\frac{1}{x}\ln x]$

We will need to apply the Chain Rule on the left side and the Product Rule on the right side.

$\frac{d}{dy}[\ln y]\cdot \frac{dy}{dx}=\frac{1}{x}\cdot \frac{d}{dx}[\ln x]+\ln x\cdot \frac{d}{dx}[\frac{1}{x}]$

$\frac{1}{y}\cdot \frac{dy}{dx}=\frac{1}{x}\cdot \frac{1}{x} + \ln x\cdot \frac{-1}{x^{2}}$

Because we are looking for the derivative, we must solve for $\frac{\mathrm{d}y }{\mathrm{d} x}$ .

$\frac{dy}{dx}=y\cdot \frac{1}{x^{2}}(1-\ln x)$

However, we want our answer to be in terms of only. We now substitute $x^{\frac{1}{x}}$ in place of .

$\frac{dy}{dx}=\frac{x^{\frac{1}{x}}}{x^{2}}(1-\ln x)$

Since we let y=h(x) , we can replace $\frac{\mathrm{d} y}{\mathrm{d} x}$ with h'(x) .

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

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Example Question #6 : Fundamental Theorem Of Calculus

Let and be inverse functions, and let

$\\f(3)=4 \\g(3)=-5 \\g(4)=3 \\f'(3)=-2 \\f'(4)=4 \\g'(3)=-1$ .

What is the value of g'(4) ?

Possible Answers:

$-\frac{1}{2}$

$\frac{1}{3}$

Correct answer:

$-\frac{1}{2}$

Explanation:

Since and are inverse functions, g(f(x))=f(g(x))=x . We can differentiate both sides of the equation g(f(x)) with respect to to obtain the following:

$g'(f(x))\cdot f'(x)=1$

We are asked to find g'(4) , which means that we will need to find such that f(x)=4 . The given information tells us that f(3)=4 , which means that x=3 . Thus, we will substitute 3 into the equation.

$g'(f(3))\cdot f'(3)=1$

The given information tells us that f'(3)=-2 .

The equation then becomes $g'(4)\cdot (-2)=1$ .

We can now solve for g'(4) .

$g'(4)=-\frac{1}{2}$ .

The answer is $-\frac{1}{2}$ .

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Example Question #12 : Numerical Approximations To Definite Integrals

Consider the curve given by the parametric equations below:

$x(t)=\ln(t)$

y(t)=3^t-2^t

What is the equation of the line normal to the curve when t=1 ?

Possible Answers:

$y=\frac{-1}{\ln \frac{27}{4}}x-1$

$y=\frac{-1}{\ln \frac{27}{4}}x+1$

$y=\frac{-1}{\ln \frac{27}{4}}(x-1)$

$y-1={\ln \frac{27}{4}}(x+1)$

$y=\frac{1}{\ln \frac{4}{27}}x+1$

Correct answer:

$y=\frac{-1}{\ln \frac{27}{4}}x+1$

Explanation:

In order to find the equation of the normal line, we will need the slope of the line and a point through which it passes. If we substitute t=1 into our parametric equations, we can easily obtain the point on the curve.

x(1)=0, y(1)=1

The normal line is perpendicular to the tangent line. Thus, we should first find the slope of the tangent line.

To find the value of the tangent slope when t=1 , we will use the following formula:

$\frac{dy}{dx}=\frac{y'(t)}{x'(t)}$

$y'(t)=3^t \ln 3 -2^t \ln 2$

$y'(1)=3\ln3-2\ln 2=\ln \frac{27}{4}$

$x'(t)=\frac{1}{t}$

x'(1)=1

$\frac{dy}{dx}=\frac{y'(t)}{x'(t)}=\ln(\frac{27}{4})$

Because the normal line is perpendicular to the tangent line, the slope of the normal line is the negative reciprocal of the slope of the tangent line. Thus,

slope of normal = $\frac{-1}{\ln(\frac{27}{4})}$ .

We now have the point and slope of the normal line, so we can use point-slope form.

$y-1=\frac{-1}{\ln \frac{27}{4}}(x-0)$

The answer is $y=\frac{-1}{\ln \frac{27}{4}}x+1$ .

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Example Question #1 : Angle Between Vectors

Find the angle between these two vectors, u=(3,4) , and v=(9,-2) .

Possible Answers:

$50^\circ$

$66^\circ$

$65.5^\circ$

$65.7^\circ$

$65.66^\circ$

Correct answer:

$65.66^\circ$

Explanation:

Lets remember the formula for finding the angle between two vectors.

$\cos(\theta)=\frac{(u\cdot v)}{\left \| u\right \|\left \| v\right \|}$

$\cos(\theta)=\frac{((3\cdot 9)+(4\cdot (-2)))}{\sqrt{3^2+4^2}\sqrt{9^2+(-2)^2}}$

$\cos(\theta)=\frac{19}{\sqrt{25}\sqrt{85}}$

$\cos(\theta)=\frac{19}{5\sqrt{85}}$

$\theta=\cos^{-1}\Big(\frac{19}{5\sqrt{85}}\Big)\approx65.66^\circ$

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Example Question #1 : Angle Between Vectors

Calculate the angle between u=(1,2) , v=(3,4) .

Possible Answers:

$10^\circ$

$10.4^\circ$

$11^\circ$

$10.5^\circ$

$10.3^\circ$

Correct answer:

$10.3^\circ$

Explanation:

Lets recall the equation for finding the angle between vectors.

$\cos(\theta)=\frac{(u\cdot v)}{\left \| u\right \|\left \| v\right \|}$

$\cos(\theta)=\frac{((1\cdot 3)+(2\cdot 4))}{\sqrt{1^2+2^2}\sqrt{3^2+4^2}}$

$\cos(\theta)=\frac{11}{\sqrt{5}\sqrt{25}}$

$\cos(\theta)=\frac{11}{5\sqrt{5}}$

$\theta=\cos^{-1}\Big(\frac{11}{5\sqrt{5}}\Big)\approx 10.3^\circ$

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Example Question #2 : Angle Between Vectors

What is the angle between the vectors $\vec{a}= \hat{i}-3\hat{j} +7\hat{k}$ and $\vec{b}= -2\hat{i}+\hat{j} +4\hat{k}$ ?

Possible Answers:

$\theta\approx9.1^{\circ}$

$\theta\approx78.1^{\circ}$

$\theta\approx37.2^{\circ}$

$\theta\approx49.2^{\circ}$

Correct answer:

$\theta\approx49.2^{\circ}$

Explanation:

To find the angle between vectors, we must use the dot product formula

$a\cdot b=\left | a\right |\left | b\right |cos\theta$

where $a\cdot b$ is the dot product of the vectors and , respectively.

$\left | a\right |$ and $\left | b\right |$ are the magnitudes of vectors and , respectively.

$cos\theta$ is the angle between the two vectors.

Let vector be represented as $\vec{a}= a_{1}\hat{i}+a_{2}\hat{j} +a_{3}\hat{k}$ and vector be represented as $\vec{b}= b_{1}\hat{i}+b_{2}\hat{j} +b_{3}\hat{k}$ .

The dot product of the vectors and is $a\cdot b=a_{1}( b_{1})+a_{2}( b_{2})+a_{3}( b_{3})$ .

The magnitude of vector is $\left | a\right |=\sqrt{(a_{1})^{2}+(a_{2})^{2}+(a_{3})^{2}}$ and vector is $\left | b\right |=\sqrt{(b_{1})^{2}+(b_{2})^{2}+(b_{3})^{2}}$ .

Rearranging the dot product formula to solve for $\theta$ gives us $\theta=cos^{-1}\left ( \frac{a\cdot b}{\left | a\right |\left | b\right |}\right )$

For this problem,

$a\cdot b=a_{1}( b_{1})+a_{2}( b_{2})+a_{3}( b_{3})=1(-2)+(-3)(1)+(7)(4)$

$a\cdot b=-2-3+28=23$

$\left | a\right |=\sqrt{(a_{1})^{2}+(a_{2})^{2}+(a_{3})^{2}}=\sqrt{(1)^{2}+(-3)^{2}+(7)^{2}}$

$\left | a\right |=\sqrt{1+9+49}=\sqrt{59}$

$\left | b\right |=\sqrt{(b_{1})^{2}+(b_{2})^{2}+(b_{3})^{2}}=\sqrt{(-2)^{2}+(1)^{2}+(4)^{2}}$

$\left | b\right |=\sqrt{4+1+16}=\sqrt{21}$

$\theta=cos^{-1}\left ( \frac{a\cdot b}{\left | a\right |\left | b\right |}\right )=cos^{-1}\left ( \frac{23}{\sqrt{59}\sqrt{21}}\right )=cos^{-1}\left ( \frac{23}{\sqrt{1239}}\right )$

$\theta\approx49.2^{\circ}$

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Example Question #2 : Angle Between Vectors

What is the angle between the vectors $\vec{a}=2 \hat{i}-1\hat{j} +7\hat{k}$ and $\vec{b}= \hat{i}+2\hat{j}$ ?

Possible Answers:

$\theta\approx8.1^{\circ}$

$\theta\approx90^{\circ}$

$\theta\approx45^{\circ}$

$\theta\approx0^{\circ}$

Correct answer:

$\theta\approx90^{\circ}$

Explanation:

To find the angle between vectors, we must use the dot product formula

$a\cdot b=\left | a\right |\left | b\right |cos\theta$

where $a\cdot b$ is the dot product of the vectors and , respectively.

$\left | a\right |$ and $\left | b\right |$ are the magnitudes of vectors and , respectively.

$cos\theta$ is the angle between the two vectors.

Let vector be represented as $\vec{a}= a_{1}\hat{i}+a_{2}\hat{j} +a_{3}\hat{k}$ and vector be represented as $\vec{b}= b_{1}\hat{i}+b_{2}\hat{j} +b_{3}\hat{k}$ .

The dot product of the vectors and is $a\cdot b=a_{1}( b_{1})+a_{2}( b_{2})+a_{3}( b_{3})$ .

The magnitude of vector is $\left | a\right |=\sqrt{(a_{1})^{2}+(a_{2})^{2}+(a_{3})^{2}}$ and vector is $\left | b\right |=\sqrt{(b_{1})^{2}+(b_{2})^{2}+(b_{3})^{2}}$ .

Rearranging the dot product formula to solve for $\theta$ gives us $\theta=cos^{-1}\left ( \frac{a\cdot b}{\left | a\right |\left | b\right |}\right )$

For this problem,

$a\cdot b=a_{1}( b_{1})+a_{2}( b_{2})+a_{3}( b_{3})=2(1)+(-1)(2)+(7)(0)$

$a\cdot b=2-2+0=0$

$\left | a\right |=\sqrt{(a_{1})^{2}+(a_{2})^{2}+(a_{3})^{2}}=\sqrt{(2)^{2}+(1)^{2}+(7)^{2}}$

$\left | a\right |=\sqrt{4+1+49}=\sqrt{54}$

$\left | b\right |=\sqrt{(b_{1})^{2}+(b_{2})^{2}+(b_{3})^{2}}=\sqrt{(1)^{2}+(2)^{2}+(0)^{2}}$

$\left | b\right |=\sqrt{1+4}=\sqrt{5}$

$\theta=cos^{-1}\left ( \frac{a\cdot b}{\left | a\right |\left | b\right |}\right )=cos^{-1}\left ( \frac{0}{\sqrt{54}\sqrt{5}}\right )=cos^{-1}\left (0\right )$

$\theta\approx90^{\circ}$

The vectors are perpendicular

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Example Question #4 : Angle Between Vectors

What is the angle between the vectors $\vec{a}=-2 \hat{i}-3\hat{j} +\hat{k}$ and $\vec{b}= \hat{i}+2\hat{j}+5\hat{k}$ ?

Possible Answers:

$\theta\approx120.8^{\circ}$

$\theta\approx98.42^{\circ}$

$\theta\approx81.58^{\circ}$

$\theta\approx8.37^{\circ}$

Correct answer:

$\theta\approx98.42^{\circ}$

Explanation:

To find the angle between vectors, we must use the dot product formula

$a\cdot b=\left | a\right |\left | b\right |cos\theta$

where $a\cdot b$ is the dot product of the vectors and , respectively.

$\left | a\right |$ and $\left | b\right |$ are the magnitudes of vectors and , respectively.

$cos\theta$ is the angle between the two vectors.

Let vector be represented as $\vec{a}= a_{1}\hat{i}+a_{2}\hat{j} +a_{3}\hat{k}$ and vector be represented as $\vec{b}= b_{1}\hat{i}+b_{2}\hat{j} +b_{3}\hat{k}$ .

The dot product of the vectors and is $a\cdot b=a_{1}( b_{1})+a_{2}( b_{2})+a_{3}( b_{3})$ .

The magnitude of vector is $\left | a\right |=\sqrt{(a_{1})^{2}+(a_{2})^{2}+(a_{3})^{2}}$ and vector is $\left | b\right |=\sqrt{(b_{1})^{2}+(b_{2})^{2}+(b_{3})^{2}}$ .

Rearranging the dot product formula to solve for $\theta$ gives us $\theta=cos^{-1}\left ( \frac{a\cdot b}{\left | a\right |\left | b\right |}\right )$

For this problem,

$a\cdot b=a_{1}( b_{1})+a_{2}( b_{2})+a_{3}( b_{3})=-2(1)+(-3)(2)+(1)(5)$

$a\cdot b=-2-6+5=-3$

$\left | a\right |=\sqrt{(a_{1})^{2}+(a_{2})^{2}+(a_{3})^{2}}=\sqrt{(-2)^{2}+(-3)^{2}+(1)^{2}}$

$\left | a\right |=\sqrt{4+9+1}=\sqrt{14}$

$\left | b\right |=\sqrt{(b_{1})^{2}+(b_{2})^{2}+(b_{3})^{2}}=\sqrt{(1)^{2}+(2)^{2}+(5)^{2}}$

$\left | b\right |=\sqrt{1+4+25}=\sqrt{30}$

$\theta=cos^{-1}\left ( \frac{a\cdot b}{\left | a\right |\left | b\right |}\right )=cos^{-1}\left ( \frac{-3}{\sqrt{14}\sqrt{30}}\right )=cos^{-1}\left (\frac{-3}{\sqrt{420}}\right )$

$\theta\approx98.42^{\circ}$

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Calculus 3 : Calculus 3

Study concepts, example questions & explanations for Calculus 3

All Calculus 3 Resources

Example Questions

Example Question #64 : Derivatives Of Functions

Example Question #1 : Interpretations And Properties Of Definite Integrals

Example Question #11 : Calculus 3

Example Question #6 : Fundamental Theorem Of Calculus

Example Question #12 : Numerical Approximations To Definite Integrals

Example Question #1 : Angle Between Vectors

Example Question #1 : Angle Between Vectors

Example Question #2 : Angle Between Vectors

Example Question #2 : Angle Between Vectors

Example Question #4 : Angle Between Vectors

All Calculus 3 Resources

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