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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 #37 : Divergence

Find $div \vec{F}$ where F is given by

$\vec{F}=\left \langle x^2\cos(xy), yz^3, xy\right \rangle$

Possible Answers:

$2x\cos(xy)-x^2y\sin(xy)+z^3$

$\left \langle 2x\cos(xy), -x^2y\sin(xy), z^3 \right \rangle$

$2x\cos(xy)-x^2y\sin(xy)+z^3+xy$

$2x\cos(xy)+x^2y\sin(xy)+z^3$

Correct answer:

$2x\cos(xy)-x^2y\sin(xy)+z^3$

Explanation:

The divergence of a vector field is given by

$div \vec{F}=\bigtriangledown \cdot \vec{F}$

where $\bigtriangledown = \left \langle f_x, f_y, f_z\right \rangle$

In taking the dot product of the gradient and the vector field, we get the sum of the respective partial derivatives of F.

The partial derivatives are

$f_x=2x\cos(xy)-x^2y\sin(xy)$

f_y=z^3

f_z=0

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Example Question #31 : Divergence

Find $div \vec{F}$ where F is given by

$\vec{F}=\left \langle \ln(xy), x^2y^2z, \ln(z)\right \rangle$

Possible Answers:

$-\frac{y}{x}+2x^2yz-\frac{1}{z}$

$\frac{1}{x}+2x^2yz+\frac{1}{z}$

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

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

Correct answer:

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

Explanation:

The divergence of a vector field is given by

$div \vec{F}=\bigtriangledown \cdot \vec{F}$

where $\bigtriangledown = \left \langle f_x, f_y, f_z\right \rangle$

In taking the dot product of the gradient and the vector field, we get the sum of the respective partial derivatives of F.

The partial derivatives are

$f_x=\frac{y}{x}$ , f_y=2x^2yz , $f_z=\frac{1}{z}$

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Example Question #31 : Divergence

Find $div \vec{F}$ where F is given by

$\vec{F}=\left \langle \sec(5x), \ln(z), xyz\right \rangle$

Possible Answers:

$-5\sec(5x)\tan(5x)+\frac{1}{z}+xy$

$5\sec(5x)\tan(5x)+\frac{1}{z}+xyz$

$\sec(5x)\tan(5x)+\frac{1}{z}+xy$

$5\sec(5x)\tan(5x)+\frac{1}{z}+xy$

Correct answer:

$5\sec(5x)\tan(5x)+\frac{1}{z}+xy$

Explanation:

The divergence of a vector field is given by

$div \vec{F}=\bigtriangledown \cdot \vec{F}$

where $\bigtriangledown = \left \langle f_x, f_y, f_z\right \rangle$

In taking the dot product of the gradient and the vector field, we get the sum of the respective partial derivatives of F.

The partial derivatives are

$f_x=5\sec(5x)\tan(5x)$

$f_y=\frac{1}{z}$

f_z=xy

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Example Question #40 : Divergence

Find the divergence of the vector $\left \langle 2x^2y^3,y^5,\cos(z)\right \rangle$

Possible Answers:

$4xy^3+5y^4-\sin(z)$

$xy^3+5y^4-\sin(z)$

$4xy^3+7y^4+\sin(z)$

$4xy^2+7y^4-\sin(z)$

Correct answer:

$4xy^3+5y^4-\sin(z)$

Explanation:

To find the divergence of a vector $a=\left \langle A,B,C\right \rangle$ , we use the definition

$divergence=\frac{\partial A}{\partial x}+\frac{\partial B}{\partial y}+\frac{\partial C}{\partial z}$

Using the vector from the problem statement, we get

$divergence=\frac{\partial }{\partial x}(2x^2y^3)+\frac{\partial }{\partial y}(y^5)+\frac{\partial }{\partial z}(\cos(z))=4xy^3+5y^4-\sin(z)$

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Example Question #41 : Divergence

Find $div \vec{F}$ of the vector field:

$\vec{F}=\left \langle 3z, 4x, 5y\right \rangle$

Possible Answers:

3z+4x+5y

Correct answer:

Explanation:

The divergence of a vector field is given by

$div \vec{F}=\bigtriangledown \cdot \vec{F}$

where $\bigtriangledown = \left \langle f_x, f_y, f_z\right \rangle$

In taking the dot product, we are left with the sum of the respective partial derivatives of the vector function. To find the given partial derivative of the function, we must treat the other variable(s) as constants.

The partial derivatives are

f_x=0 , f_y=0 , f_z=0

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Example Question #42 : Divergence

Find $div \vec{F}$ of the vector field:

$\vec{F}=\left \langle 3xy, 5y\sin(z), 2\right \rangle$

Possible Answers:

$3y+5\sin(z)+2$

$3y+5\sin(z)$

3x+5y

$3y-5\sin(z)$

Correct answer:

$3y+5\sin(z)$

Explanation:

The divergence of a vector field is given by

$div \vec{F}=\bigtriangledown \cdot \vec{F}$

where $\bigtriangledown = \left \langle f_x, f_y, f_z\right \rangle$

The partial derivatives are

f_x=3y , $f_y=5\sin(z)$ , f_z=0

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Example Question #43 : Divergence

Find $div \vec{F}$ , where F is the following vector field:

$\vec{F}=\left \langle e^{xyz}, xy, \sec(z)\right \rangle$

Possible Answers:

$yze^{xyz}+x+\sec(z)\tan(z)$

$yze^{xyz}+xy+\sec(z)\tan(z)$

$yze^{xyz}+x-\sec(z)\tan(z)$

Correct answer:

$yze^{xyz}+x+\sec(z)\tan(z)$

Explanation:

The divergence of a vector field is given by

$div \vec{F}=\bigtriangledown \cdot \vec{F}$

where $\bigtriangledown =\left \langle f_x, f_y, f_z\right \rangle$

In taking the dot product, we get the sum of the respective partial derivatives of the vector field. To find the given partial derivative of the function, we must treat the other variable(s) as constants.

The partial derivatives are

$f_x=yze^{xyz}$

f_y=x

$f_z=\sec(z)\tan(z)$

Report an Error

Example Question #44 : Divergence

Find $div \vec{F}$ , where F is the following vector field:

$\vec{F}=\left \langle x\tan(z^2), 3yz, 8e^{xz}\right \rangle$

Possible Answers:

$\tan(z^2)+3y+8xe^{xz}$

$\tan(z^2)+3z+8e^{xz}$

$\sec^2(z^2)+3z+8xe^{xz}$

$\tan(z^2)+3z+8xe^{xz}$

Correct answer:

$\tan(z^2)+3z+8xe^{xz}$

Explanation:

The divergence of a vector field is given by

$div \vec{F}=\bigtriangledown \cdot \vec{F}$

where $\bigtriangledown =\left \langle f_x, f_y, f_z\right \rangle$

The partial derivatives are

$f_x=\tan(z^2)$

f_y=3z

$f_z=8xe^{xz}$

Report an Error

Example Question #51 : Line Integrals

Find the divergence of the following vector field:

$\vec{F}=\left \langle xy,yz, \cos(z)\right \rangle$

Possible Answers:

$y+z+\sin(z)$

$y-\sin(z)$

$y+z-\sin(z)$

$z-\sin(z)$

Correct answer:

$y+z-\sin(z)$

Explanation:

The divergence of a vector field is given by

$div\vec{F}=\bigtriangledown \cdot \vec{F}$

where $\bigtriangledown = \left \langle f_x, f_y, f_z\right \rangle$

In taking the dot product, we end up with the sum of the respective partial derivatives of the vector field. To find the given partial derivative of the function, we must treat the other variable(s) as constants.

The partial derivatives are

f_x=y , f_y=z . $f_z=-\sin(z)$

Report an Error

Example Question #45 : Divergence

Find the divergence of the vector $\left \langle 2x^2,5y,z^3\right \rangle$

Possible Answers:

4x-5+3z^2

4x^3+5+3z^2

4x+5+3z^2

4x+5+z^2

Correct answer:

4x+5+3z^2

Explanation:

To find the divergence of the vector $v=\left \langle P,Q,R\right \rangle$ , we use the following formula

$\bigtriangledown v=\frac{\partial P}{\partial x}+\frac{\partial Q}{\partial y}+\frac{\partial R}{\partial z}$

Applying to the vector from the problem statement, we get

$\frac{\partial }{\partial x}(2x^2)+\frac{\partial }{\partial y}(5y)+\frac{\partial }{\partial z}(z^3)=4x+5+3z^2$

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Calculus 3 : Calculus 3

Study concepts, example questions & explanations for Calculus 3

All Calculus 3 Resources

Example Questions

Example Question #37 : Divergence

Example Question #31 : Divergence

Example Question #31 : Divergence

Example Question #40 : Divergence

Example Question #41 : Divergence

Example Question #42 : Divergence

Example Question #43 : Divergence

Example Question #44 : Divergence

Example Question #51 : Line Integrals

Example Question #45 : Divergence

All Calculus 3 Resources

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