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Calculus 3 : Divergence

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Example Questions

Example Question #1675 : Calculus 3

Find the divergence of the force field $F = (x^2+y^2)\mathbf{i}+(x^2-y^2)\mathbf{j}+(z^2)\mathbf{k}$

Possible Answers:

2(x+y+z)

2(x+y-z)

x-y-z

2(x-y+z)

Correct answer:

2(x-y+z)

Explanation:

The correct formula for divergence is $div(F) = \bigtriangledown \cdot F$ .

Remember that the result of a divergence calculation must be a scalar, not a vector.

$\bigtriangledown \cdot F = \frac{\partial}{\partial x}(x^2+y^2) +\frac{\partial}{\partial y}(x^2-y^2) +\frac{\partial}{\partial z}(z^2)$

=2x-2y+2z

=2(x-y+2) .

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Example Question #23 : Line Integrals

Evaluate the divergence of the force field $F = \sin(xy)\mathbf{i}+\cos(xz)\mathbf{j}+e^{xy}\mathbf{k}$

Possible Answers:

$y\cos(xy)\mathbf{i}-x\sin(xz)\mathbf{j}$

None of the other answers

$y\cos(xy)$

$y\cos(xy)\mathbf{i}$

$y\cos(xy)-x\sin(xz)$

Correct answer:

$y\cos(xy)$

Explanation:

The correct formula for divergence is $div(F) = \bigtriangledown \cdot F$ .

Remember that the result of a divergence calculation must be a scalar, not a vector.

$\bigtriangledown \cdot F = \frac{\partial}{\partial x}(\sin(xy)) +\frac{\partial}{\partial y}(\cos(xz)) +\frac{\partial}{\partial z}(e^{xy})$

$=y\cos(xy)+0+0$

$=y\cos(xy)$ .

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Example Question #1677 : Calculus 3

Find the divergence of the force field $F=xyz\mathbf{i}+yz\mathbf{j}+z^3\mathbf{k}$ .

Possible Answers:

$\mathbf{0}$

$3z^2\mathbf{i}+z\mathbf{j}+yz\mathbf{k}$

(y+3z+1)z

None of the other answers

$yz\mathbf{i}+z\mathbf{j}+3z^2\mathbf{k}$

Correct answer:

(y+3z+1)z

Explanation:

The correct formula for divergence is $div(F) = \bigtriangledown \cdot F$ .

Remember that the result of a divergence calculation must be a scalar, not a vector.

$\bigtriangledown \cdot F = \frac{\partial}{\partial x}(xyz) +\frac{\partial}{\partial y}(yz) +\frac{\partial}{\partial z}(z^3)$

=yz+z+3z^2

=(y+3z+1)z .

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Example Question #24 : Line Integrals

Find the divergence of the force field $F=2x\mathbf{i}+2z\mathbf{j}+e^z\mathbf{k}$ .

Possible Answers:

None of the other answers

$\mathbf{0}$

2+e^z

$2\mathbf{i}+e^z\mathbf{k}$

Correct answer:

2+e^z

Explanation:

The correct formula for divergence is $div(F) = \bigtriangledown \cdot F$ .

Remember that the result of a divergence calculation must be a scalar, not a vector.

$\bigtriangledown \cdot F = \frac{\partial}{\partial x}(2x) +\frac{\partial}{\partial y}(2z) +\frac{\partial}{\partial z}(e^z)$

=2+0+e^z

=2+e^z .

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Example Question #21 : Divergence

Find the divergence of the vector $\left \langle \sin(x),\tan(y),2yz^2\right \rangle$

Possible Answers:

$\cos(x)-\sec^2(y)+4yz$

$\sin(x)+\sec^2(y)+4yz$

$-\cos(x)+\sec(y)+4yz$

$\cos(x)+\sec^2(y)+4yz$

Correct answer:

$\cos(x)+\sec^2(y)+4yz$

Explanation:

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 $\frac{\partial }{\partial x}(\sin(x))+\frac{\partial }{\partial y}(\tan(x))+\frac{\partial }{\partial z}(2yz^2)=\cos(x)+\sec^2(y)+4yz$

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Example Question #1680 : Calculus 3

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

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

Possible Answers:

ye^x

2xz+ye^x

2xz

Correct answer:

2xz+ye^x

Explanation:

The divergence of a vector is given by

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

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

f_x=2xz

f_y=0

f_z=ye^x

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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Example Question #1681 : Calculus 3

Calculate the divergence of the following vector:

$\mathbf{F}=x^2y\mathbf{i}+3xz\mathbf{j}+z^3\mathbf{k}$

Possible Answers:

2xy+3z^2

x^2+3

2xy+3+3z

2x^2y+3x

x^2+3z^2

Correct answer:

2xy+3z^2

Explanation:

For a given vector

$\mathbf{F}=U\mathbf{i}+V\mathbf{k}+W\mathbf{k}$

The divergence is calculated by:

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

For our vector

$\triangledown\cdot \mathbf{F}=2xy+0+3z^2$

The answer is, therefore:

2xy+3z^2

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Example Question #1681 : Calculus 3

Given that F is a vector function and f is a scalar function, which of the following operations results in a vector?

Possible Answers:

$\nabla \cdot(\nabla f)$

$\nabla(\nabla \cdot f)$

$\nabla \cdot (\nabla\cdot \mathbf{F})$

$\nabla \cdot (\nabla \times \mathbf{F})$

$\nabla \times (\nabla \times \mathbf{F})$

Correct answer:

$\nabla \times (\nabla \times \mathbf{F})$

Explanation:

For all the given answers:

$\nabla \cdot (\nabla \times \mathbf{F})$ - The curl of a vector is also a vector, so the divergence of the term in parenthesis is scalar.

$\nabla \cdot (\nabla \cdot \mathbf{F})$ - The divergence of a vector is a scalar. The divergence of a scalar is undefined, so this expression is undefined.

$\nabla \cdot(\nabla f)$ - The gradient of a scalar is a vector, so the divergence of the term in parenthesis is a scalar.

$\nabla(\nabla \cdot f)$ - The divergence of a scalar doesn't exist, so this expression is undefined.

The remaining answer is:

$\nabla \times (\nabla \times \mathbf{F})$ - The curl of a vector is also a vector, so the curl of the term in parenthesis is a vector as well.

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Example Question #1682 : Calculus 3

Given that F is a vector function and f is a scalar function, which of the following expressions is valid?

Possible Answers:

$\nabla\cdot(\nabla\cdot\mathbf{F})$

$\nabla(\nabla \cdot f)$

$\nabla\cdot(\nabla\mathbf{F})$

$\nabla \cdot (\nabla \times f)$

$\nabla \cdot (\nabla f)$

Correct answer:

$\nabla \cdot (\nabla f)$

Explanation:

For each of the given answers:

$\nabla \cdot (\nabla \times f)$ - The curl of a scalar is undefined, so the term in parenthesis is invalid.

$\nabla(\nabla \cdot f)$ - The divergence of a scalar is also undefined, so the term in parenthesis is invalid.

$\nabla\cdot(\nabla\mathbf{F})$ - The gradient of a scalar is undefined as well, so the term in parenthesis is invalid.

$\nabla\cdot(\nabla\cdot\mathbf{F})$ - The divergence of a vector is a scalar. The divergence of a the term in parenthesis, which is a scalar, is undefined, so the expression is invalid.

The remaining answer must be correct:

$\nabla \cdot (\nabla f)$ - The gradient of a scalar is a vector, so the divergence of the term in parenthesis is a scalar. The expression is valid.

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Example Question #26 : Divergence

Compute the divergence of the following vector function:

$\mathbf{F}=3yz\mathbf{i}-xy\mathbf{j}+(3xz+y^3)\mathbf{k}$

Possible Answers:

-y+3z+3y^2

3z+4x

3z-x

Correct answer:

Explanation:

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

the divergence is defined by:

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

For our function:

$\frac{\partial U}{\partial x}=\frac{\partial }{\partial x}(3yz)=0$

$\frac{\partial V}{\partial y}=\frac{\partial }{\partial y}(-xy)=-x$

$\frac{\partial W}{\partial z}=\frac{\partial }{\partial z}(3xz+y^3)=3x$

Thus, the divergence of our function is:

$\nabla \cdot \mathbf{F}=-0-x+3x=2x$

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Calculus 3 : Divergence

Study concepts, example questions & explanations for Calculus 3

All Calculus 3 Resources

Example Questions

Example Question #1675 : Calculus 3

Example Question #23 : Line Integrals

Example Question #1677 : Calculus 3

Example Question #24 : Line Integrals

Example Question #21 : Divergence

Example Question #1680 : Calculus 3

Example Question #1681 : Calculus 3

Example Question #1681 : Calculus 3

Example Question #1682 : Calculus 3

Example Question #26 : Divergence

All Calculus 3 Resources

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