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Calculus 3 : Divergence

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Example Questions

Example Question #1651 : Calculus 3

Compute $div \vec{F}$ for

$\vec{F}=yz\ln(x)\vec{i}+2xyz\vec{j}+z\vec{k}$

Possible Answers:

$div\vec{F}=-\frac{yz}{x}+2xz$

$div\vec{F}=\frac{yz}{x}$

$div\vec{F}=2xz+1$

$div\vec{F}=0$

$div\vec{F}=\frac{yz}{x}+2xz+1$

Correct answer:

$div\vec{F}=\frac{yz}{x}+2xz+1$

Explanation:

In order to find the divergence, we need to remember the formula.

Divergence Formula:

$div\vec{F}=\frac{\partial P}{\partial x}+\frac{\partial Q}{\partial y}+\frac{\partial R}{\partial z}$ , where , , and correspond to the components of a given vector field $\vec{F}=P\vec{i}+Q\vec{j}+R\vec{k}$ .

Now lets apply this to our situation.

$div\vec{F}=\frac{\partial}{\partial x}\Big(yz\ln(x)\Big)+\frac{\partial}{\partial y}\Big(2xyz\Big)+\frac{\partial}{\partial z}\Big(z\Big)$

$div\vec{F}=\frac{yz}{x}+2xz+1$

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Example Question #1 : Line Integrals

Compute $div\vec{F}$ for

$\vec{F}=x^2z\ln(y)\vec{i}+x2^{y^2}z\vec{j}+x^x\vec{k}$

Possible Answers:

$div\vec{F}=-2xz\ln(y)-2\log(2)xyz2^{y^2}$

$div\vec{F}=2xz\ln(y)$

$div\vec{F}=2xz\ln(y)+2\log(2)xyz2^{y^2}$

$div\vec{F}=2\log(2)xyz2^{y^2}$

$div\vec{F}=0$

Correct answer:

$div\vec{F}=2xz\ln(y)+2\log(2)xyz2^{y^2}$

Explanation:

In order to find the divergence, we need to remember the formula.

Divergence Formula:

Now lets apply this to our situation.

$div\vec{F}=\frac{\partial}{\partial x}\Big(x^2z\ln(y)\Big)+\frac{\partial}{\partial y}\Big(x2^{y^2}z\Big)+\frac{\partial}{\partial z}\Big(x^x\Big)$

$div\vec{F}=2xz\ln(y)+2\log(2)xyz2^{y^2}+0$

$div\vec{F}=2xz\ln(y)+2\log(2)xyz2^{y^2}$

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Example Question #1652 : Calculus 3

Compute $div\vec{F}$ for

$\vec{F}=xy\vec{i}+xz\vec{j}+xyz\vec{k}$

Possible Answers:

$div \vec{F}=y$

$div \vec{F}=1$

$div \vec{F}=(1+x)$

$div \vec{F}=y(1+x)$

$div \vec{F}=x$

Correct answer:

$div \vec{F}=y(1+x)$

Explanation:

In order to find the divergence, we need to remember the formula.

Divergence Formula:

Now lets apply this to our situation.

$div\vec{F}=\frac{\partial}{\partial x}\Big(xy\Big)+\frac{\partial}{\partial y}\Big(xz\Big)+\frac{\partial}{\partial z}\Big(xyz\Big)$

$div \vec{F}=y+0+xy$

$div \vec{F}=y+xy=y(1+x)$

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Example Question #1 : Divergence

Find $div\vec{F}$ , where $\vec{F}=xyz\vec{i}+x^2z^{-4}\vec{j}+y^{-10}z^{3}\vec{k}$

Possible Answers:

$div\vec{F}=yz+\frac{3z^2}{y^{10}}$

$div\vec{F}=yz+\frac{3z^2}{y^{-10}}$

$div\vec{F}=yz+z^4+\frac{3z^2}{y^{10}}$

$div\vec{F}=\frac{3z^2}{y^{10}}$

$div\vec{F}=xyz+\frac{3z^2}{y^{10}}$

Correct answer:

$div\vec{F}=yz+\frac{3z^2}{y^{10}}$

Explanation:

In order to find the divergence, we need to remember the formula.

Divergence Formula:

Now lets apply this to our situation.

$div\vec{F}=\frac{\partial}{\partial x}\Big(xyz\Big)+\frac{\partial}{\partial y}\Big(x^2z^{-4}\Big)+\frac{\partial}{\partial z}\Big(y^{-10}z^3\Big)$

$div\vec{F}={yz}+0+3y^{-10}z^2=yz+\frac{3z^2}{y^{10}}$

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Example Question #1 : Divergence

Compute $div\vec{F}$ , where $\vec{F}=10x^2\ln(yz)\vec{i}+xyz\vec{j}+xz\ln(y)\vec{k}$

Possible Answers:

$div\vec{F}=20x\ln(yz)-xz-x\ln(y)$

$div\vec{F}=20x\ln(yz)-xz+x\ln(y)$

$div\vec{F}=0$

$div\vec{F}=20x\ln(yz)+xz+x\ln(y)$

$div\vec{F}=20x\ln(z)+yz+x\ln(y)$

Correct answer:

$div\vec{F}=20x\ln(yz)+xz+x\ln(y)$

Explanation:

All we need to do is calculate the partial derivatives and add them together.

$div\vec{F}=\frac{\partial}{\partial x}(10x^2\ln(yz))+\frac{\partial}{\partial y}(xyz)+\frac{\partial}{\partial z}(xz\ln(y))$

$div\vec{F}=20x\ln(yz)+xz+x\ln(y)$

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Example Question #1 : Divergence

Given the vector field

$\vec{F}=yz\vec{i}+xz\vec{j}+xy\vec{k}$

find the divergence of the vector field:

$div \vec{F}$ .

Possible Answers:

$div \vec{F}=xy+yz+xz$

$div \vec{F}=x+y$

$div \vec{F}=x+y+z$

$div \vec{F}=0$

$div \vec{F}=x+z$

Correct answer:

$div \vec{F}=0$

Explanation:

Given a vector field

$\vec{F}=P\vec{i}+Q\vec{j}+R\vec{k}$

we find its divergence by taking the dot product with the gradient operator:

$div\vec{F}=\frac{\partial P}{\partial x}+\frac{\partial Q}{\partial y}+\frac{\partial R}{\partial z}$

We know that P=yz, Q=xz, R=xy , so we have

$div\vec{F}=\frac{\partial P}{\partial x}+\frac{\partial Q}{\partial y}+\frac{\partial R}{\partial z}=0+0+0=0$

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Example Question #1 : Divergence

Suppose that $\underset{F}{\rightarrow}$ $=2x^2yz^2+y^3e^{2x}+\arctan z$ . Calculate the divergence.

Possible Answers:

$4xyz^2+2y^3e^{2x} + 2x^2z^2+3y^2e^{2x}+4x^2yz+\frac{1}{1+z^2}$

$4xyz+2x^2y^2z^2 +4x^2yz+6xy^2z+\frac{1}{1+z^2}$

$4xz^2+2y^2e^{2x} + 2x^2z+3y^2e^{2x}+4x^2yz+\frac{1}{1+z^2}$

$4x+3y^2+\frac{1}{1+z^2}$

Correct answer:

$4xyz^2+2y^3e^{2x} + 2x^2z^2+3y^2e^{2x}+4x^2yz+\frac{1}{1+z^2}$

Explanation:

We know,

$div F(x,y,z)=\frac{\partial{F}}{\partial{x}}+\frac{\partial{F}}{\partial{y}}+\frac{\partial{F}}{\partial{z}}$

Use this to obtain the correct answer

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Example Question #2 : Divergence

Given that

$\vec{F}=<2x,y,2z>$

calculate $div(\vec{F})$

Possible Answers:

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

Correct answer:

Explanation:

$div(\vec{F})=\frac{\partial{F}}{\partial{x}}+\frac{\partial{F}}{\partial{y}}+\frac{\partial{F}}{\partial{z}}$

using this formula we have

$div(\vec{F})=2+1+2=5$

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Example Question #9 : Divergence

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

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

Possible Answers:

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

$2+3x+2z\cos(z)$

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

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

Correct answer:

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

Explanation:

The divergence of a vector is given by

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

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

So, we take the partial derivative of each component of our vector with respect to x, y, and z respectively and add them together:

f_x(2x)=2

f_y(3xy)=3x

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

The derivatives were found using the following rules:

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

$\frac{\mathrm{d} }{\mathrm{d} x} \cos(x)=-\sin(x)$

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Example Question #10 : Divergence

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

$\left \langle 12y\cos(z), y^2e^x, z^2\right \rangle$

Possible Answers:

2ye^x+2z

2ye^x+z^2

$12\cos(z)+2ye^x+2z$

Correct answer:

2ye^x+2z

Explanation:

The divergence of a curve 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 of the gradient and the curve, we end up summing the respective partial derivatives (for example, the x coordinate's partial derivative with respect to x is found).

The partial derivatives are:

f_x=0

f_y=2ye^x

f_z=2z

The following rules were used to find the derivatives:

$\frac{\mathrm{d} }{\mathrm{d} x} ax=a$ , $\frac{\mathrm{d} }{\mathrm{d} x} x^n=nx^{n-1}$

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Calculus 3 : Divergence

Study concepts, example questions & explanations for Calculus 3

All Calculus 3 Resources

Example Questions

Example Question #1651 : Calculus 3

Example Question #1 : Line Integrals

Example Question #1652 : Calculus 3

Example Question #1 : Divergence

Example Question #1 : Divergence

Example Question #1 : Divergence

Example Question #1 : Divergence

Example Question #2 : Divergence

Example Question #9 : Divergence

Example Question #10 : Divergence

All Calculus 3 Resources

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