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Calculus 3 : Divergence

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Example Questions

Example Question #31 : Divergence

Compute the divergence of the following vector function:

$\mathbf{F}=\cos(xy)\mathbf{i}+4y^3z^2\mathbf{j}+5xe^{2z}\mathbf{k}$ $\displaystyle \mathbf{F}=\cos(xy)\mathbf{i}+4y^3z^2\mathbf{j}+5xe^{2z}\mathbf{k}$

Possible Answers:

$-y\sin(xy)+12y^2z^2+10xe^{2z}$ $\displaystyle -y\sin(xy)+12y^2z^2+10xe^{2z}$

$-\sin(xy)+12y^2z^2+5xe^{2z}$ $\displaystyle -\sin(xy)+12y^2z^2+5xe^{2z}$

$y\sin(x)+12y^2z^3+10xe^{z}$ $\displaystyle y\sin(x)+12y^2z^3+10xe^{z}$

$-x\cos(xy)+16y^2z^2+5e^{2z}$ $\displaystyle -x\cos(xy)+16y^2z^2+5e^{2z}$

Correct answer:

$-y\sin(xy)+12y^2z^2+10xe^{2z}$ $\displaystyle -y\sin(xy)+12y^2z^2+10xe^{2z}$

Explanation:

For a vector function $\mathbf{F}=U\mathbf{i}+V\mathbf{j}+W\mathbf{k}$ $\displaystyle \mathbf{F}=U\mathbf{i}+V\mathbf{j}+W\mathbf{k}$,

the divergence is defined by:

$\nabla \cdot \mathbf{F}=\frac{\partial U}{\partial x}+\frac{\partial V}{\partial y} +\frac{\partial W}{\partial z}$ $\displaystyle \nabla \cdot \mathbf{F}=\frac{\partial U}{\partial x}+\frac{\partial V}{\partial y} +\frac{\partial W}{\partial z}$

For our function:

$\frac{\partial U}{\partial x}=\frac{\partial }{\partial x}(\cos(xy))=-y\sin(xy)$ $\displaystyle \frac{\partial U}{\partial x}=\frac{\partial }{\partial x}(\cos(xy))=-y\sin(xy)$

$\frac{\partial V}{\partial y}=\frac{\partial }{\partial y}(4y^3z^2)=12y^2z^2$ $\displaystyle \frac{\partial V}{\partial y}=\frac{\partial }{\partial y}(4y^3z^2)=12y^2z^2$

$\frac{\partial W}{\partial z}=\frac{\partial }{\partial z}(5xe^{2z})=10xe^{2z}$ $\displaystyle \frac{\partial W}{\partial z}=\frac{\partial }{\partial z}(5xe^{2z})=10xe^{2z}$

Thus, the divergence of our function is:

$\nabla \cdot \mathbf{F}=-y\sin(xy)+12y^2z^2+10xe^{2z}$ $\displaystyle \nabla \cdot \mathbf{F}=-y\sin(xy)+12y^2z^2+10xe^{2z}$

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Example Question #32 : Line Integrals

Find $div \vec{F}$ $\displaystyle div \vec{F}$, where F is given by the following:

$\vec{F}=\left \langle y\cos(z), \frac{y}{x}, z^3\cos(z)\right \rangle$ $\displaystyle \vec{F}=\left \langle y\cos(z), \frac{y}{x}, z^3\cos(z)\right \rangle$

Possible Answers:

$\frac{1}{x}+3z^2\cos(z)-z^3\sin(z)$ $\displaystyle \frac{1}{x}+3z^2\cos(z)-z^3\sin(z)$

$\left \langle 0, \frac{1}{x}, 3z^2\cos(z)-z^3\sin(z)\right \rangle$ $\displaystyle \left \langle 0, \frac{1}{x}, 3z^2\cos(z)-z^3\sin(z)\right \rangle$

$\frac{1}{x}+3z^2\cos(z)+z^3\sin(z)$ $\displaystyle \frac{1}{x}+3z^2\cos(z)+z^3\sin(z)$

$\displaystyle 0$

Correct answer:

$\frac{1}{x}+3z^2\cos(z)-z^3\sin(z)$ $\displaystyle \frac{1}{x}+3z^2\cos(z)-z^3\sin(z)$

Explanation:

The divergence of a vector function is given by

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

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

So, we must find the partial derivative of each respective component.

The partial derivatives are

f_x=0 $\displaystyle f_x=0$

$f_y=\frac{1}{x}$ $\displaystyle f_y=\frac{1}{x}$

$f_z=3z^2\cos(z)-z^3\sin(z)$ $\displaystyle f_z=3z^2\cos(z)-z^3\sin(z)$

The partial derivatives were found using the following rules:

$\frac{\mathrm{d} }{\mathrm{d} x} ax=a$ $\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} ax=a$, $\frac{\mathrm{d} }{\mathrm{d} x} f(x)g(x)=f'(x)g(x)+f(x)g'(x)$ $\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} f(x)g(x)=f'(x)g(x)+f(x)g'(x)$, $\frac{\mathrm{d} }{\mathrm{d} x} x^n=nx^{n-1}$ $\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} x^n=nx^{n-1}$, $\frac{\mathrm{d} }{\mathrm{d} x} \cos(x)=-\sin(x)$ $\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} \cos(x)=-\sin(x)$

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Example Question #1681 : Calculus 3

Find $div\vec{F}$ $\displaystyle div\vec{F}$, where F is given by the following curve:

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

Possible Answers:

$2y+3y^2-xz^2\sin(xy)$ $\displaystyle 2y+3y^2-xz^2\sin(xy)$

$\displaystyle 0$

$\left \langle 0, 3y^2, 2z\cos(xy)\right \rangle$ $\displaystyle \left \langle 0, 3y^2, 2z\cos(xy)\right \rangle$

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

Correct answer:

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

Explanation:

The divergence of a curve is given by

$div\vec{F}=\nabla \cdot \vec{F}$ $\displaystyle div\vec{F}=\nabla \cdot \vec{F}$

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

So, we must find the partial derivatives of the x, y, and z components, respectively:

f_x=0 $\displaystyle f_x=0$

$\displaystyle f_y=3y^2$

$f_z=2z\cos(xy)$ $\displaystyle f_z=2z\cos(xy)$

The partial derivatives were found using the following rules:

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Example Question #1681 : Calculus 3

Find $div \vec{F}$ $\displaystyle div \vec{F}$ of the given function:

$\vec{F}=\left \langle 4xy, y\cos(y), xy\right \rangle$ $\displaystyle \vec{F}=\left \langle 4xy, y\cos(y), xy\right \rangle$

Possible Answers:

$\cos(y)+4-\sin(y)$ $\displaystyle \cos(y)+4-\sin(y)$

$4y+\cos(y)+y\cos(y)$ $\displaystyle 4y+\cos(y)+y\cos(y)$

$\cos(y)+y(4+\sin(y))$ $\displaystyle \cos(y)+y(4+\sin(y))$

$\cos(y)+y(4-\sin(y))$ $\displaystyle \cos(y)+y(4-\sin(y))$

Correct answer:

$\cos(y)+y(4-\sin(y))$ $\displaystyle \cos(y)+y(4-\sin(y))$

Explanation:

The divergence of a vector function is given by

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

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

So, we take the respective partial derivatives of the x, y, and z-components of the vector function, and add them together (from the dot product):

f_x=4y $\displaystyle f_x=4y$

$f_y=\cos(y)-y\sin(y)$ $\displaystyle f_y=\cos(y)-y\sin(y)$

f_z=0 $\displaystyle f_z=0$

The partial derivatives were found using the following rules:

$\frac{\mathrm{d} }{\mathrm{d} x} ax=a$ $\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} ax=a$, $\frac{\mathrm{d} }{\mathrm{d} x} x^n=nx^{n-1}$ $\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} x^n=nx^{n-1}$, $\frac{\mathrm{d} }{\mathrm{d} x} f(x)g(x)=f'(x)g(x)+f(x)g'(x)$ $\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} f(x)g(x)=f'(x)g(x)+f(x)g'(x)$, $\frac{\mathrm{d} }{\mathrm{d} x} \cos(x)=-\sin(x)$ $\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} \cos(x)=-\sin(x)$

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Example Question #1681 : Calculus 3

Find $div \vec{F}$ $\displaystyle div \vec{F}$ of the vector function below:

$\vec{F}=\cos(x)\hat{i}+(xyz)\hat{j}+z^2\sin(y)\hat{k}$ $\displaystyle \vec{F}=\cos(x)\hat{i}+(xyz)\hat{j}+z^2\sin(y)\hat{k}$

Possible Answers:

$\sin(x)+xz+2z\sin(y)$ $\displaystyle \sin(x)+xz+2z\sin(y)$

$\left \langle -\sin(x),xz,2z\sin(y) \right \rangle$ $\displaystyle \left \langle -\sin(x),xz,2z\sin(y) \right \rangle$

$-\sin(x)+xz-2z\sin(y)$ $\displaystyle -\sin(x)+xz-2z\sin(y)$

$-\sin(x)+xz+2z\sin(y)$ $\displaystyle -\sin(x)+xz+2z\sin(y)$

Correct answer:

$-\sin(x)+xz+2z\sin(y)$ $\displaystyle -\sin(x)+xz+2z\sin(y)$

Explanation:

The divergence of a vector function is given by

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

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

So, in taking the dot product of the gradient and the function, we get the sum of the respective partial derivatives. To find the given partial derivative of the function, we must treat the other variable(s) as constants.

The partial derivatives are

$f_x=-\sin(x)$ $\displaystyle f_x=-\sin(x)$

f_y=xz $\displaystyle f_y=xz$

$f_z=2z\sin(y)$ $\displaystyle f_z=2z\sin(y)$

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Example Question #32 : Divergence

Find $div \vec{F}$ $\displaystyle div \vec{F}$ of the vector function below:

$\vec{F}=\left \langle xe^{\cos(y)+\sin(z)}, 2y, z\right \rangle$ $\displaystyle \vec{F}=\left \langle xe^{\cos(y)+\sin(z)}, 2y, z\right \rangle$

Possible Answers:

$e^{\cos(y)+\sin(z)}$ $\displaystyle e^{\cos(y)+\sin(z)}$

$e^{\cos(y)+\sin(z)}+2y+1$ $\displaystyle e^{\cos(y)+\sin(z)}+2y+1$

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

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

Correct answer:

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

Explanation:

The divergence of a vector function is given by

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

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

The partial derivatives are

$f_x=e^{\cos(y)+\sin(z)}$ $\displaystyle f_x=e^{\cos(y)+\sin(z)}$

f_y=2 $\displaystyle f_y=2$

f_z=1 $\displaystyle f_z=1$

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Example Question #31 : Divergence

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

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

Possible Answers:

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

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

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

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

Correct answer:

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

Explanation:

The divergence of a vector field is given by

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

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

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

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

f_y=z^3 $\displaystyle f_y=z^3$

f_z=0 $\displaystyle f_z=0$

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Example Question #31 : Divergence

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

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

Possible Answers:

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

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

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

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

Correct answer:

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

Explanation:

The divergence of a vector field is given by

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

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

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

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

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Example Question #1691 : Calculus 3

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

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

Possible Answers:

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

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

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

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

Correct answer:

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

Explanation:

The divergence of a vector field is given by

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

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

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

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

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

f_z=xy $\displaystyle f_z=xy$

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Example Question #1693 : Calculus 3

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

Possible Answers:

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

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

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

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

Correct answer:

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

Explanation:

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

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

Using the vector from the problem statement, we get

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

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Calculus 3 : Divergence

Study concepts, example questions & explanations for Calculus 3

All Calculus 3 Resources

Example Questions

Example Question #31 : Divergence

Example Question #32 : Line Integrals

Example Question #1681 : Calculus 3

Example Question #1681 : Calculus 3

Example Question #1681 : Calculus 3

Example Question #32 : Divergence

Example Question #31 : Divergence

Example Question #31 : Divergence

Example Question #1691 : Calculus 3

Example Question #1693 : Calculus 3

All Calculus 3 Resources

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