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Calculus 3 : Calculus 3

Study concepts, example questions & explanations for Calculus 3

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

Example Questions

Example Question #45 : Angle Between Vectors

Find the angle $\theta$ to the nearest degree between the two vectors

u=<-3,-9>

v=<2,7>

Possible Answers:

105^o

133^o

82^o

178^o

Correct answer:

178^o

Explanation:

In order to find the angle between the two vectors, we follow the formula

$cos(\theta)=\frac{u\bullet v}{||u||*||v||}$

and solve for

$\theta$

Using the vectors in the problem, we get

$cos(\theta)=\frac{<-3,-9>\bullet <2,7>}{\sqrt{(-3)^2+(-9)^2}*\sqrt{2^2+7^2}}=\frac{-6+(-63)}{\sqrt{90}*\sqrt{53}}$

Simplifying we get

$cos(\theta)=\frac{-69}{\sqrt{4770}}$

To solve for

$\theta$

we find the

$cos^{-1}$

of both sides and get

$cos^{-1}(cos(\theta))=cos^{-1}(\frac{-69}{\sqrt{4770}})$

and find that

$\theta=178^o$

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Example Question #46 : Angle Between Vectors

Find the angle $\theta$ in degrees between the two vectors

u=<5,-4>

v=<9,12>

Possible Answers:

57^o

92^o

81^o

35^o

Correct answer:

92^o

Explanation:

In order to find the angle between the two vectors, we follow the formula

$cos(\theta)=\frac{u\bullet v}{||u||*||v||}$

and solve for

$\theta$

Using the vectors in the problem, we get

$cos(\theta)=\frac{<5,-4>\bullet <9,12>}{\sqrt{5^2+(-4)^2}*\sqrt{9^2+12^2}}=\frac{45+(-48)}{\sqrt{41}*\sqrt{225}}$

Simplifying we get

$cos(\theta)=\frac{-3}{\sqrt{9225}}$

To solve for

$\theta$

we find the

$cos^{-1}$

of both sides and get

$cos^{-1}(cos(\theta))=cos^{-1}(\frac{-3}{\sqrt{9225}})$

and find that

$\theta=92^o$

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Example Question #46 : Vectors And Vector Operations

Find the angle between the vectors and , where $a=\left \langle 0,0,5\right \rangle$ and $b=\left \langle 3,0,4\right \rangle$

Note: Use the dot product formula when finding the answer

Possible Answers:

$\theta=\cos^{-1}(\frac{10}{25})$

$\theta=\cos^{-1}(\frac{2}{25})$

$\theta=\cos^{-1}(\frac{20}{24})$

$\theta=\cos^{-1}(\frac{20}{25})$

Correct answer:

$\theta=\cos^{-1}(\frac{20}{25})$

Explanation:

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

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

$a\cdot b=0+0+(5*4)=20$

$\left | a\right |=\sqrt{0^2+0^2+5^2}=5$

$\left | b\right |=\sqrt{3^2+0^2+4^2}=5$

$20=25\cos(\theta)$ , and solving for theta, we get

$\theta=\cos^{-1}(\frac{20}{25})$

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Example Question #45 : Angle Between Vectors

Determine the cosine of the angle between the following vectors:

$\left \langle 2, -1, -3\right \rangle, \left \langle 0, 1, 5\right \rangle$

Possible Answers:

Correct answer:

Explanation:

The cosine of the angle, denoted by $\theta$ , between two vectors is given by the dot product of the vectors, which is the sum of the products of the corresponding components.

For our two vectors, the dot product is given by

$2(0)+1(-1)-3(5)=-16=\cos(\theta)$

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Example Question #41 : Vectors And Vector Operations

Find the angle between $\vec{A}=2i+3j+k$ and $\vec{B}=3i+4j+12k$ in degrees

Possible Answers:

Correct answer:

Explanation:

Step 1: Calculate

$A\cdot B=\begin{pmatrix} 2\\3 \\1 \end{pmatrix}\begin{pmatrix} 3\\4 \\12 \end{pmatrix}=6+12+12=30$

Step 2: Find the respective magnitudes of A and B

$A=\sqrt{2^2+3^2+1^2}=\sqrt{14}\\B=\sqrt{3^2+4^2+12^2}=13$

Step 3:

Use the formula to find the angle between two vectors and .

Let the angle between the vectors be $\theta$ . Then

$\cos\theta=\frac{a.b}{||a||||b||}$

$\cos \theta=\frac{30}{13\sqrt{14}}$

$\theta=\cos^{-1}\left[\frac{30}{13\sqrt{14}}\right]=51.92\approx52^\circ$

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Example Question #51 : Vectors And Vector Operations

Find the angle between the two vectors. Round to the nearest degree.

u=<4,8>

v=<7,-7>

Possible Answers:

$\theta=54^o$

$\theta=134^o$

$\theta=22^o$

$\theta=108^o$

Correct answer:

$\theta=108^o$

Explanation:

In order to find the angle between the two vectors, we follow the formula

$cos(\theta)=\frac{u\bullet v}{||u||*||v||}$

and solve for

$\theta$

Using the vectors in the problem, we get

$cos(\theta)=\frac{<4,8>\bullet <7,-7>}{\sqrt{4^2+8^2}*\sqrt{7^2+(-7)^2}}=\frac{4(7)+8(-7)}{\sqrt{80}*\sqrt{98} }$

Simplifying we get

$cos(\theta)=\frac{-28}{\sqrt{7840}}$

To solve for

$\theta$

we find the

$cos^{-1}$

of both sides and get

$cos^{-1}(cos(\theta))=cos^{-1}(\frac{-28}{\sqrt{7840}})$

and find that

$\theta=108^o$

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Example Question #51 : Vectors And Vector Operations

Find the angle between the two vectors. Round to the nearest degree.

u=<9,-5>

v=<5,9>

Possible Answers:

$\theta=135^o$

$\theta=60^o$

$\theta=90^o$

$\theta=45^o$

Correct answer:

$\theta=90^o$

Explanation:

In order to find the angle between the two vectors, we follow the formula

$cos(\theta)=\frac{u\bullet v}{||u||*||v||}$

and solve for

$\theta$

Using the vectors in the problem, we get

$cos(\theta)=\frac{<9,-5>\bullet <5,9>}{\sqrt{9^2+(-5)^2}*\sqrt{5^2+9^2}}=\frac{9(5)+(-5)9}{\sqrt{106}*\sqrt{106} }$

Simplifying we get

$cos(\theta)=0$

To solve for

$\theta$

we find the

$cos^{-1}$

of both sides and get

$cos^{-1}(cos(\theta))=cos^{-1}(0)$

and find that

$\theta=90^o$

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Example Question #61 : Calculus 3

Find the angle between the two vectors. Round to the nearest degree.

u=<1,-8>

v=<0,6>

Possible Answers:

$\theta=81^o$

$\theta=71^o$

$\theta=173^o$

$\theta=116^o$

Correct answer:

$\theta=173^o$

Explanation:

In order to find the angle between the two vectors, we follow the formula

$cos(\theta)=\frac{u\bullet v}{||u||*||v||}$

and solve for

$\theta$

Using the vectors in the problem, we get

$cos(\theta)=\frac{<1,-8>\bullet <0,6>}{\sqrt{1^2+(-8)^2}*\sqrt{0^2+6^2}}=\frac{1(0)+(-8)6}{\sqrt{65}*\sqrt{36} }$

Simplifying we get

$cos(\theta)=\frac{-48}{\sqrt{2340}}$

To solve for

$\theta$

we find the

$cos^{-1}$

of both sides and get

$cos^{-1}(cos(\theta))=cos^{-1}(\frac{-48}{\sqrt{2340}})$

and find that

$\theta=173^o$

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Example Question #53 : Vectors And Vector Operations

Calculate the angle between the vectors $\mathbf a = \left \langle 1,-1,0\right \rangle$ and $\mathbf b = \left \langle 3,0,0\right \rangle$ , and express the measurement of the angle in degrees.

Possible Answers:

$30^\circ$

$60^\circ$

$135^\circ^{}$

$90^\circ$

$45^\circ$

Correct answer:

$45^\circ$

Explanation:

The angle $\Theta$ between the vectors $\mathbf a = \left \langle a_1,a_2,a_3\right \rangle$ and $\mathbf b = \left \langle b_1,b_2,b_3\right \rangle$ is given by the following equation:

$\cos\Theta=\frac{\mathbf a \cdot \mathbf b}{|\mathbf a| \ |\mathbf b|}$

where $\mathbf a \cdot \mathbf b$ represents the cross product of the vectors $\mathbf a$ and $\mathbf b$ , and $|\mathbf a|$ and $|\mathbf b|$ represent the respective magnitudes of the vectors $\mathbf a$ and $\mathbf b$ .

We are given the vectors $\mathbf a = \left \langle 1,-1,0\right \rangle$ and $\mathbf b = \left \langle 3,0,0\right \rangle$ . Calculate $\mathbf$ $|\mathbf a|$ , $|\mathbf b|$ , and $\mathbf a \cdot \mathbf b$ , and then substitute these results into the formula for the angle between these vectors, as shown:

$|\mathbf a|=\sqrt{(1)^2+(-1)^2+(0)^2}=\sqrt{2}$ ,

$|\mathbf b|=\sqrt{(3)^2+(0)^2+(0)^2}=\sqrt{9}=3$ ,

and

$\mathbf a \cdot \mathbf b = (3)(1)+(-1)(0)+(0)(0)=3$ .

Hence,

$\cos\Theta=\frac{3}{3\sqrt{2}}=\frac{1}{\sqrt{2}}=\frac{\sqrt{2}}{2}$

The principal angle $\Theta$ for which $\cos\Theta=\frac{\sqrt{2}}{2}$ is $\Theta=45^\circ$ . Hence, the angle between the vectors $\mathbf a = \left \langle 1,-1,0\right \rangle$ and $\mathbf b = \left \langle 3,0,0\right \rangle$ measures $45^\circ$ .

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Example Question #55 : Vectors And Vector Operations

Find the angle between the gradient vector $\vec{\triangledown}f(0,y,z)$ and the vector $\vec{u}=4\hat{i}-3\hat{j}+\sqrt{3}\hat{k}$ where is defined as:

$f(x,y,z)=\sqrt{x+1}+y+z$

Possible Answers:

$\theta = \cos^{-1}\left(\frac{1}{3}(\sqrt{7}+\sqrt{3}) \right )$

$\theta = \cos^{-1}\left(\frac{\sqrt{7}}{3} \right )$

$\theta = \cos^{-1}\left(\frac{\sqrt{21}}{3} \right )$

$\theta = \cos^{-1}\left(\sqrt{21} \right )$

$\theta = \cos^{-1}\left(\frac{1}{3}(\sqrt{21}-\sqrt{7}) \right )$

Correct answer:

$\theta = \cos^{-1}\left(\frac{1}{3}(\sqrt{21}-\sqrt{7}) \right )$

Explanation:

Find the angle between the gradient vector $\vec{\triangledown}f(0,y,z)$ and the vector $\vec{u}=4\hat{i}-3\hat{j}+\sqrt{3}\hat{k}$ where is defined as:

$f(x,y,z)=\sqrt{x+1}+y+z$

_____________________________________________________________

Compute the gradient by taking the partial derivative for each direction:

$\vec{\triangledown}f(x,y,z)=\frac{\partial f}{\partial x}\hat{i}+\frac{\partial f}{\partial y}\hat{j}+\frac{\partial f}{\partial z}\hat{k}$

$\vec{\triangledown}f(x,y,z)=\frac{1}{2\sqrt{x+1}}\hat{i}+\hat{j}+\hat{k}$

At x = 0 we have:

$\vec{\triangledown}f(0,y,z)=\frac{1}{2}\hat{i}+\hat{j}+\hat{k}$

____________________________________________________________

The angle $\theta$ between two vectors $\vec{v_1}$ and $\vec{v_2}$ can be found using the dot product:

$\vec{v_1}\cdot\vec{v_2}=\left \| \vec{v_1}\right \|\left \| \vec{v_2}\right \|cos(\theta )$

We wish to find the angle $\theta$ between the two vectors:

$\vec{u}=4\hat{i}-3\hat{j}+\sqrt{3}\hat{k}$

$\vec{\triangledown}f(0,y,z)=\frac{1}{2}\hat{i}+\hat{j}+\hat{k}$

Compute the dot product between $\vec{u}$ and $\vec{\triangledown}f(0,y,z)$ ,

$\vec{\triangledown}f(0,y,z)\cdot\vec{u}=\left( \frac{1}{2}\hat{i}+\hat{j}+\hat{k} \right )\cdot\left( 4\hat{i}-3\hat{j}+\sqrt{3}\hat{k}\right )=-1+\sqrt{3}$

Therefore the dot product is:

$\vec{\triangledown}f(0,y,z)\cdot\vec{u}=\sqrt{3}-1$

Compute the magnitude of $\vec{u}$

$\left \| \vec{u}\right \|=\sqrt{16+9+3}=\sqrt{28}=2\sqrt{7}$

Compute the magnitude of $\vec{\triangledown}f(0,y,z)$ ,

$\left \| \vec{\triangledown}f(0,x,y)\right \|=\sqrt{1/4 +2}=\sqrt\frac{9}{4}=\frac{3}{2}$

Now put it all together:

$\vec{\triangledown}f(0,y,z)\cdot\vec{u} =\left \| \vec{\triangledown}f(0,x,y)\right \|\left \| \vec{u}\right \|cos(\theta )$

$\sqrt{3}-1 =3\sqrt{7}\cos(\theta)$

$\theta = \cos^{-1}\left(\frac{\sqrt{21}-\sqrt{7}}{3} \right )$

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

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Calculus 3 : Calculus 3

Study concepts, example questions & explanations for Calculus 3

All Calculus 3 Resources

Example Questions

Example Question #45 : Angle Between Vectors

Example Question #46 : Angle Between Vectors

Example Question #46 : Vectors And Vector Operations

Example Question #45 : Angle Between Vectors

Example Question #41 : Vectors And Vector Operations

Example Question #51 : Vectors And Vector Operations

Example Question #51 : Vectors And Vector Operations

Example Question #61 : Calculus 3

Example Question #53 : Vectors And Vector Operations

Example Question #55 : Vectors And Vector Operations

All Calculus 3 Resources

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