Divergence - AP Calculus BC

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Question

Given the vector field

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

find the divergence of the vector field:

$div \vec{F}$ .

Answer

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 , 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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Question

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

Answer

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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Question

Compute $div \vec{F}$ for

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

Answer

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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Question

Given that

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

calculate $div(\vec{F})$

Answer

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

using this formula we have

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

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Question

Compute $div\vec{F}$ for

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

Answer

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.

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

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

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

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Question

Compute $div\vec{F}$ for

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

Answer

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.

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

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

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

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Question

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

Answer

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.

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

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

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Question

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

Answer

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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Question

Compute the divergence of the vector $\left \langle xy^2,y^3z,z^3y^2\right \rangle$ .

Answer

To find the divergence of the vector

$v=\left \langle a,b,c\right \rangle$ ,

we use the formula

$div(v)=\frac{\partial a}{\partial x}+\frac{\partial b}{\partial y}+\frac{\partial c}{\partial z}$ .

Computing each partial derivative, we get

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

Adding them up gives us the correct answer.

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Question

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

Answer

To find the divergence of the vector $v=\left \langle a,b,c\right \rangle$ , we use the formula

$div(v)=\frac{\partial a}{\partial x}+\frac{\partial b}{\partial y}+\frac{\partial c}{\partial z}$ .

Computing each partial derivative, we get

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

Adding them up gives us the correct answer.

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Question

Compute the divergence of the vector.

$\left \langle yz,x^6y,z^5\right \rangle$

Answer

To find the divergence of the vector $v=\left \langle a,b,c\right \rangle$ , we use the formula

$div(v)=\frac{\partial a}{\partial x}+\frac{\partial b}{\partial y}+\frac{\partial c}{\partial z}$ .

Computing each partial derivative, we get

$\frac{\partial }{\partial x}(yz)=0, \frac{\partial }{\partial y}(x^6y)=x^6,\frac{\partial }{\partial z}(z^5)=5z^4$ .

Adding them up gives us the correct answer.

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Question

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

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

Answer

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:

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

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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Question

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

$\left \langle x^2z, z^2, ye^xz\right \rangle$

Answer

The divergence of a vector is given by

$abla \cdot \vec{F}$ , where $abla = \left \langle f_x, f_y, f_z\right \rangle$

Taking the partial respective partial derivatives of the x, y, and z components of our curve, we get

The rules used to find the derivatives are as follows:

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

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Question

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

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

Answer

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:

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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Question

Find the divergence of the vector $\left \langle 2x^3yz,3y^3,z^8\right \rangle$

Answer

The formula for the divergence of a vector $a=\left \langle A,B,C\right \rangle$ is $div(a)=\frac{\partial A}{\partial x}+\frac{\partial B}{\partial y}+\frac{\partial C}{\partial z}$ . Using the vector from the problem statement, we get $\frac{\partial }{\partial x}(2x^3yz)=6x^2yz, \frac{\partial }{\partial y}(3y^3)=9y^2,\frac{\partial }{\partial z}(z^8)=8z^7$ . Adding them up gets us the correct answer.

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Question

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

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

Answer

The divergence of a curve is given by

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

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

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

$f_z=2z\cos(xy)$

The partial derivatives were found using the following rules:

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

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Question

Find the divergence of the vector $\left \langle x^2y^2z^2,3y^5,17z\right \rangle$

Answer

To find the divergence a vector $a=\left \langle A,B,C\right \rangle$ , you use the following definition: $div(a)=\frac{\partial A}{\partial x}+\frac{\partial B}{\partial y}+\frac{\partial C}{\partial z}$ . Applying this to the vector from the problem statement, we get $\frac{\partial A}{\partial x}=2xy^2z^2, \frac{\partial B}{\partial y}=15y^4,\frac{\partial C}{\partial z}=17$ . Adding all of these up, according to the definition, will produce the correct answer.

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Question

Find the divergence of the following vector: $\left \langle yz^3,xy^5z,3xz\right \rangle$

Answer

To find the divergence a vector $a=\left \langle A,B,C\right \rangle$ , you use the following definition: $div(a)=\frac{\partial A}{\partial x}+\frac{\partial B}{\partial y}+\frac{\partial C}{\partial z}$ . Applying this to the vector from the problem statement, we get $\frac{\partial A}{\partial x}=0, \frac{\partial B}{\partial y}=5xy^4z,\frac{\partial C}{\partial z}=3x$ . Adding all of these up, according to the definition, will produce the correct answer.

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Question

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

Answer

To find the divergence a vector $a=\left \langle A,B,C\right \rangle$ , you use the following definition: $div(a)=\frac{\partial A}{\partial x}+\frac{\partial B}{\partial y}+\frac{\partial C}{\partial z}$ . Applying this to the vector from the problem statement, we get $\frac{\partial A}{\partial x}=-\sin(x), \frac{\partial B}{\partial y}=-2\cos(x),\frac{\partial C}{\partial z}=2x^2y^2z$ . Adding all of these up, according to the definition, will produce the correct answer.

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Question

Find the divergence of the vector $\left \langle 2x,6x^2y^3,z^4\right \rangle$

Answer

To find the divergence of a vector $v=\left \langle A,B,C\right \rangle$ , we apply the following definition: $div(v)=\frac{\partial A}{\partial x}+\frac{\partial B}{\partial y}+\frac{\partial C}{\partial z}$ . Applying the definition to the vector from the problem statement, we get $\frac{\partial }{\partial x}(2x)+\frac{\partial }{\partial y}(6x^2y^3)+\frac{\partial }{\partial z}(z^4)=2+18x^2y^2+4z^3$

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