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Calculus 3 : Calculus 3

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6 Diagnostic Tests 373 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

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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Example Question #31 : Divergence

Compute the divergence of the following vector function:

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

Possible Answers:

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

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

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

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

Correct answer:

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

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:

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

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

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

Thus, the divergence of our function is:

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

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Example Question #32 : Divergence

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

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

Possible Answers:

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

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

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

Correct answer:

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

Explanation:

The divergence of a vector function is given by

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

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

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

The partial derivatives are

f_x=0

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

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

The partial derivatives were found using the following rules:

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

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Example Question #33 : Divergence

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

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

Possible Answers:

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

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

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

Correct answer:

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

Explanation:

The divergence of a curve is given by

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

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

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

f_x=0

f_y=3y^2

$f_z=2z\cos(xy)$

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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Example Question #34 : Divergence

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

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

Possible Answers:

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

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

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

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

Correct answer:

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

Explanation:

The divergence of a vector function 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 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

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

f_z=0

The partial derivatives were found using the following rules:

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

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Example Question #35 : Divergence

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

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

Possible Answers:

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

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

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

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

Correct answer:

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

Explanation:

The divergence of a vector function is given by

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

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

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

$f_x=-\sin(x)$

f_y=xz

$f_z=2z\sin(y)$

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Example Question #36 : Divergence

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

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

Possible Answers:

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

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

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

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

Correct answer:

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

Explanation:

The divergence of a vector function is given by

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

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

The partial derivatives are

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

f_y=2

f_z=1

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Calculus 3 : Calculus 3

Study concepts, example questions & explanations for Calculus 3

All Calculus 3 Resources

Example Questions

Example Question #1681 : Calculus 3

Example Question #1681 : Calculus 3

Example Question #1682 : Calculus 3

Example Question #26 : Divergence

Example Question #31 : Divergence

Example Question #32 : Divergence

Example Question #33 : Divergence

Example Question #34 : Divergence

Example Question #35 : Divergence

Example Question #36 : Divergence

All Calculus 3 Resources

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