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Calculus 3 : Divergence

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Example Questions

Example Question #1702 : Calculus 3

Find the divergence of the following vector field:

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

Possible Answers:

$x\sec^2(xz)e^{\tan(xz)}+2ye^{y^2}+3z^2$

$\sec^2(xz)e^{\tan(xz)}+2ye^{y^2}+3z^2$

$z\sec^2(xz)e^{\tan(xz)}+2ye^{y^2}+3z^2$

$z\sec(xz)e^{\tan(xz)}+2ye^{y^2}+3z^2$

$-z\sec^2(xz)e^{\tan(xz)}+2ye^{y^2}+3z^2$

Correct answer:

$z\sec^2(xz)e^{\tan(xz)}+2ye^{y^2}+3z^2$

Explanation:

The divergence of the vector field is given by

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

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

Taking the dot product gives us the sum of the respective partial derivatives of the vector field. For higher order partial derivatives, we work from left to right for the given variables.

The partial derivatives are

$f_x=z\sec^2(xz)e^{\tan(xz)}$ , $f_y=2ye^{y^2}$ , f_z=3z^2

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Example Question #52 : Line Integrals

Find the divergence of the vector $\left \langle 3x,2xy^2,z^4\right \rangle$

Possible Answers:

5+xy+4z^3

3+4xy+4z^3

7xy+4z^3

3+4xy

Correct answer:

3+4xy+4z^3

Explanation:

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

$div(v)=\frac{\partial P}{\partial x}+\frac{\partial Q}{\partial y}+\frac{\partial R }{\partial z}$

Using the vector from the problem statement, we get

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

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Example Question #53 : Line Integrals

Find the divergence of the vector $\left \langle \sin(x),x^4y,xyz\right \rangle$

Possible Answers:

$\cos(x)+x^4+xy$

$\cos(x)+x^3y+xy$

$\cos(x)+4x^3y+y$

$-\cos(x)+4x^3y+xy$

Correct answer:

$\cos(x)+x^4+xy$

Explanation:

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

$div(v)=\frac{\partial P}{\partial x}+\frac{\partial Q}{\partial y}+\frac{\partial R }{\partial z}$

Using the vector from the problem statement, we get

$\frac{\partial }{\partial x}(\sin(x))+\frac{\partial }{\partial y}(x^4y)+\frac{\partial }{\partial z}(xyz)=\cos(x)+x^4+xy$

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Example Question #54 : Line Integrals

Find the divergence of the following vector field:

$\vec{F}=\left \langle y^3z, xy^2, e^{xyz}\right \rangle$

Possible Answers:

$y^3+2xy+xye^{xyz}$

$2xy+e^{xyz}$

$2xy+yze^{xyz}$

$2xy+xye^{xyz}$

Correct answer:

$2xy+xye^{xyz}$

Explanation:

The divergence of the vector field

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

where $\bigtriangledown = \left \langle \frac{\partial }{\partial x}, \frac{\partial }{\partial y}, \frac{\partial }{\partial z}\right \rangle$

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

The partial derivatives are

f_x=0 , f_y=2xy , $f_z=xye^{xyz}$

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Example Question #61 : Line Integrals

Find the divergence of the vector field:

$\vec{F}=\left \langle x^5e^y, e^{xy}, z\right \rangle$

Possible Answers:

$5x^4+xe^{xy}+1$

$5x^4e^y+ye^{xy}+1$

$5x^4e^y+xe^{xy}+1$

$5x^4e^y+xe^{xy}+z$

Correct answer:

$5x^4e^y+xe^{xy}+1$

Explanation:

The divergence of the vector field

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

where $\bigtriangledown = \left \langle \frac{\partial }{\partial x}, \frac{\partial }{\partial y}, \frac{\partial }{\partial z}\right \rangle$

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

The partial derivatives are

f_x=5x^4e^y , $f_y=xe^{xy}$ , f_z=1

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Example Question #62 : Line Integrals

Find the divergence of the following vector field:

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

Possible Answers:

x^3

$\sec^2(y)e^{\tan(y)}+3x^2y+2x$

3x^2y

Correct answer:

x^3

Explanation:

The divergence of a vector field is given by

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

where $\bigtriangledown = \left \langle \frac{\partial }{\partial x}, \frac{\partial }{\partial y} , \frac{\partial }{\partial z}\right \rangle$

When we take the dot product of the gradient with the vector field, 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=0 , f_y=x^3 , f_z=0

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Example Question #63 : Line Integrals

Determine the divergence of the vector field:

$\vec{F}=\left \langle x^4y^2, e^{\cos(y)}, x+y\right \rangle$

Possible Answers:

$4x^3y^2-\sin(y)e^{\cos(y)}$

$4x^3y^2-\sin(y)e^{\cos(y)}+x+y$

$4x^3y^2+\sin(y)e^{\cos(y)}$

$4x^3-\sin(y)e^{\cos(y)}$

Correct answer:

$4x^3y^2-\sin(y)e^{\cos(y)}$

Explanation:

The divergence of a vector field is given by

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

where $\bigtriangledown = \left \langle \frac{\partial }{\partial x} , \frac{\partial }{\partial y}, \frac{\partial }{\partial z}\right \rangle$

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

The partial derivatives are

f_x=4x^3y^2 , $f_y=-\sin(y)e^{\cos(y)}$ , f_z=0

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Example Question #51 : Divergence

Find the divergence of the vector field:

$\vec{F}=\left \langle 5\cos(y), 6x, z^2e^{3\tan(z)}\right \rangle$

Possible Answers:

$2ze^{3\tan(z)}+6z^2\sec(z)e^{3\tan(z)}$

$6ze^{3\tan(z)}+6z^2\sec^2e^{3\tan(z)}$

$2ze^{3\tan(z)}+6z\sec^2e^{3\tan(z)}$

$2ze^{3\tan(z)}+3z^2\sec^2e^{3\tan(z)}$

$2ze^{3\tan(z)}+6z^2\sec^2e^{3\tan(z)}$

Correct answer:

$2ze^{3\tan(z)}+6z^2\sec^2e^{3\tan(z)}$

Explanation:

The divergence of a vector field is given by

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

where $\bigtriangledown = \left \langle \frac{\partial }{\partial x} , \frac{\partial }{\partial y}, \frac{\partial }{\partial z}\right \rangle$

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

The partial derivatives are

f_x=0 , f_y=0 , $f_z=2ze^{3\tan(z)}+6z^2\sec^2e^{3\tan(z)}$

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Example Question #65 : Line Integrals

Find the divergence of the following vector field:

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

Possible Answers:

$y\sec(x)(1+x\tan(x))+2y$

$\sec(x)(1+x\tan(x))+3y$

$y\sec(x)(1+x\sec(x)\tan(x))+2y$

$\sec(x)(1+x\tan(x))+2y$

Correct answer:

$y\sec(x)(1+x\tan(x))+2y$

Explanation:

The divergence of a vector field is given by $\bigtriangledown \cdot \vec{F}$ , where $\bigtriangledown=\left \langle \frac{\partial }{\partial x}, \frac{\partial }{\partial y}, \frac{\partial }{\partial z}\right \rangle$

When we take the dot product of the gradient with the vector field, we are left with the sum of the respective partial derivative of the vector field.

The partial derivatives are

$f_x=y\sec(x)(1+x\tan(x))$ , f_y=2y , f_z=0

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Example Question #52 : Divergence

Find the divergence of the vector field:

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

Possible Answers:

$-\sin(xy)e^{\cos(xy)}+3z^2$

$-y\sin(xy)e^{\cos(xy)}+4z^2$

$y\sin(xy)e^{\cos(xy)}+3z^2$

$-y\sin(xy)e^{\cos(xy)}+3z^2$

Correct answer:

$-y\sin(xy)e^{\cos(xy)}+3z^2$

Explanation:

The divergence of a vector field is given by

$div \vec{F}=\bigtriangledown \cdot \vec{F}$ , where $\bigtriangledown = \left \langle \frac{\partial }{\partial x}, \frac{\partial }{\partial y}, \frac{\partial }{\partial z} \right \rangle$

When we take 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

$f_x=-y\sin(xy)e^{\cos(xy)}$ , f_y=0 , f_z=3z^2

Report an Error

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Calculus 3 : Divergence

Study concepts, example questions & explanations for Calculus 3

All Calculus 3 Resources

Example Questions

Example Question #1702 : Calculus 3

Example Question #52 : Line Integrals

Example Question #53 : Line Integrals

Example Question #54 : Line Integrals

Example Question #61 : Line Integrals

Example Question #62 : Line Integrals

Example Question #63 : Line Integrals

Example Question #51 : Divergence

Example Question #65 : Line Integrals

Example Question #52 : Divergence

All Calculus 3 Resources

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