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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 #1 : Curl

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

Possible Answers:

$\nabla \times (\nabla f)$

$\nabla\cdot(\nabla f)$

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

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

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

Correct answer:

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

Explanation:

The cross product of a scalar function is undefined. The expression in the parenthesis of:

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

is the cross product of a scalar function, therefore the entire expression is undefined.

For the other solutions:

$\nabla\cdot(\nabla\times\mathbf{F})$ - The cross product of a vector is also a vector, and the divergence of a vector is defined. This expression is a scalar.

$\nabla\cdot(\nabla f)$ - The gradient of a scalar is a vector, and the divergence of a vector is defined. This expression is also a scalar.

$\nabla(\nabla\cdot\mathbf{F})$ - The divergence of a vector is scalar, and the gradient of a scalar is defined. This expression is a vector.

$\nabla \times (\nabla f)$ - The gradient of a scalar is a vector, and the curl of a vector is defined. This expression is a vector.

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Example Question #9 : Curl

Compute the curl of the following vector function:

$\mathbf{F}=-y^2\mathbf{i}+3y\mathbf{j}$

Possible Answers:

$-3\mathbf{k}$

$-2y\mathbf{i}$

$3\mathbf{i}-2y\mathbf{k}$

$-2y\mathbf{i}+3\mathbf{k}$

$2y\mathbf{k}$

Correct answer:

$2y\mathbf{k}$

Explanation:

For a vector function $\mathbf{F}=U\mathbf{i}+V\mathbf{j}+W\mathbf{k}$ , the curl is given by:

$\nabla \times \mathbf{F}=\textup{det}\begin{vmatrix} \mathbf{i}&\mathbf{j} &\mathbf{k} \\ \frac{\partial}{\partial x}&\frac{\partial}{\partial y} &\frac{\partial}{\partial z} \\ U& V& W \end{vmatrix}$

For this function, we calculate the curl as:

$\nabla \times \mathbf{F}=(0-0)\mathbf{i}-(0-0)\mathbf{j}+(0+2y)\mathbf{k}=2y\mathbf{k}$

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Example Question #10 : Curl

Compute the curl of the following vector function:

$\mathbf{F}=(y^2+z^2)\mathbf{i}+y\mathbf{j}+z\mathbf{k}$

Possible Answers:

$2y\mathbf{i}+\mathbf{j}$

$(2y-1)\mathbf{i}-\mathbf{k}$

$(1-2z)\mathbf{i}+\mathbf{j}$

$2z\mathbf{j}-2y\mathbf{k}$

$-2z\mathbf{i}-\mathbf{k}$

Correct answer:

$2z\mathbf{j}-2y\mathbf{k}$

Explanation:

For a vector function $\mathbf{F}=U\mathbf{i}+V\mathbf{j}+W\mathbf{k}$ , the curl is given by:

For this function, we calculate the curl as:

$\nabla \times \mathbf{F} =\textup{det}\begin{vmatrix} \mathbf{i}&\mathbf{j} &\mathbf{k} \\ \frac{\partial}{\partial x}&\frac{\partial}{\partial y} &\frac{\partial}{\partial z} \\ (y^2+z^2)& y& z \end{vmatrix}$

$\nabla\times \mathbf{F}=(0-0)\mathbf{i}-(0-2z)\mathbf{j}+(0-2y)\mathbf{k}= 2z\mathbf{j}-2y\mathbf{k}$

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Example Question #1731 : Calculus 3

Find the curl of the vector function:

$\vec{F}=\left \langle 2xy, xyz, z^2\right \rangle$

Possible Answers:

$\left \langle 0, 0, 0\right \rangle$

$\left \langle -xy, 0, yz\right \rangle$

$\left \langle -xy, 0, yz-2x\right \rangle$

$\left \langle xy, 0, yz+2x\right \rangle$

Correct answer:

$\left \langle -xy, 0, yz-2x\right \rangle$

Explanation:

The curl of the function is given by

$\bigtriangledown X \vec{F}$

First, we must write the determinant in order to take the cross product:

$\begin{vmatrix} \hat{i}&\hat{j} &\hat{k} \\ \frac{\partial }{\partial x}&\frac{\partial }{\partial y} &\frac{\partial }{\partial z} \\ 2xy& xyz&z^2 \end{vmatrix}$

where i, j, and k are the unit vectors corresponding to the x, y, and z direction respectively.

$\\ \frac{\partial }{\partial y}z^2\hat{i}+\frac{\partial }{\partial x} xyz\hat{k}+\frac{\partial }{\partial z}2xy\hat{j}-(\frac{\partial }{\partial y}2xy\hat{k}+\frac{\partial }{\partial z}xyz\hat{i}+\frac{\partial }{\partial x}z^2\hat{j})\\ \\=0+yz\hat{k}+0-2x\hat{k}-xy\hat{i}\\ \\=\left \langle -xy, 0, yz-2x\right \rangle$

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 #1732 : Calculus 3

Find the curl of the vector function:

$\vec{F}=e^x\hat{i}+x^2y\hat{j}+x\hat{k}$

Possible Answers:

$\left \langle 0, 1, 2xy\right \rangle$

$\left \langle 0, -1, -2xy\right \rangle$

$\left \langle 0, -1, 2xy\right \rangle$

$\left \langle y, -1, 2xy\right \rangle$

Correct answer:

$\left \langle 0, -1, 2xy\right \rangle$

Explanation:

The curl of the function is given by

$\bigtriangledown X \vec{F}$

First, we must write the determinant in order to take the cross product:

$\begin{vmatrix} \hat{i}&\hat{j} &\hat{k} \\ \frac{\partial }{\partial x}&\frac{\partial }{\partial y} &\frac{\partial }{\partial z} \\ e^x& x^2y&x \end{vmatrix}$

where i, j, and k are the unit vectors corresponding to the x, y, and z direction respectively.

$\\ \frac{\partial }{\partial y}x\hat{i}+\frac{\partial }{\partial x} x^2y\hat{k}+\frac{\partial }{\partial z}e^x\hat{j}-(\frac{\partial }{\partial y}e^x\hat{k}+\frac{\partial }{\partial z}x^2y\hat{i}+\frac{\partial }{\partial x}x\hat{j})\\ \\=0+2xy\hat{k}+0-0-0-1\hat{j}\\ \\=\left \langle 0, -1, 2xy\right \rangle$

The derivatives were found using the following rules:

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

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Example Question #1733 : Calculus 3

Find the curl of the vector function:

$\vec{F}=\left \langle x^2,yz,\sin(z)\right \rangle$

Possible Answers:

$\left \langle 0, 0, -y\right \rangle$

$\left \langle y, 0, 0\right \rangle$

$\left \langle 0,-y, 0\right \rangle$

$\left \langle -y, 0, 0\right \rangle$

Correct answer:

$\left \langle -y, 0, 0\right \rangle$

Explanation:

The curl of a vector function $\vec{F}=\left \langle P, Q, R\right \rangle$ is given by

$\begin{vmatrix} \hat{i}&\hat{j} &\hat{k} \\ \frac{\partial }{\partial x}&\frac{\partial }{\partial y} &\frac{\partial }{\partial z} \\ x^2& yz& \sin(z) \end{vmatrix}$

where i, j, and k are the unit vectors corresponding to the x, y, and z direction respectively.

Next, we take the cross product. One can do this by multiplying across from the top left to the lower right, and continuing downward, and then subtracting the terms multiplied from top right to the bottom left. To find the given partial derivative of the function, we must treat the other variable(s) as constants:

$\frac{\partial }{\partial y}\sin(z)\hat{i}+\frac{\partial }{\partial x}yz\hat{k}+\frac{\partial }{\partial z}x^2\hat{j}-(\frac{\partial }{\partial y}x^2\hat{k}+\frac{\partial }{\partial z}yz\hat{i}+\frac{\partial }{\partial y}\sin(z)\hat{j})=-y\hat{i}$

The partial derivatives were found using the following rule:

$\frac{\mathrm{d} }{\mathrm{d} x} ax=a$

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Example Question #1734 : Calculus 3

Determine if the vector field is conservative or not, and why:

$\left \langle xz, y^2z, z\right \rangle$

Possible Answers:

The vector field is conservative because the curl is not equal to $\vec{0}$ .

The vector field is not conservative because the curl does not equal to $\vec{0}$ .

The vector field is not conservative because the curl is equal to $\vec{0}$ .

The vector field is conservative because the curl is equal to $\vec{0}$ .

Correct answer:

The vector field is not conservative because the curl does not equal to $\vec{0}$ .

Explanation:

The curl of the function is given by the cross product of the gradient and the vector function. If a vector function is conservative if the curl equals zero.

First, we can write the determinant in order to take the cross product of the two vectors:

$\begin{vmatrix} \hat{i}&\hat{j} &\hat{k} \\ \frac{\partial }{\partial x}&\frac{\partial }{\partial y} &\frac{\partial }{\partial z} \\ xz&y^2z &z \end{vmatrix}$

where i, j, and k are the unit vectors corresponding to the x, y, and z direction respectively.

$\frac{\partial }{\partial y}z\hat{i}+\frac{\partial }{\partial x}y^2z\hat{k}+\frac{\partial }{\partial z}xz\hat{j}-(\frac{\partial }{\partial y}xz\hat{k}+\frac{\partial }{\partial z}y^2z\hat{i}+\frac{\partial }{\partial x}z\hat{j})=x\hat{j}-y^2\hat{i}\neq\vec{0}$

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Example Question #1735 : Calculus 3

Find the curl of the following vector field, in vector form:

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

Possible Answers:

$\left \langle 0, 0, \cos(x)+\sin(y)\right \rangle$

$\left \langle 0, 0, \cos(y)+\sin(x)\right \rangle$

$\left \langle 0, 0, \cos(x)-\sin(y)\right \rangle$

$\left \langle 0, 0, \cos(x)+\sin(x)\right \rangle$

Correct answer:

$\left \langle 0, 0, \cos(x)+\sin(y)\right \rangle$

Explanation:

The curl of the vector field is given by:

$\begin{vmatrix} \hat{i}& \hat{j}& \hat{k}\\ \frac{\partial }{\partial x}&\frac{\partial }{\partial y} & \frac{\partial }{\partial z}\\ \cos(y)&\sin(x) &\cos(z) \end{vmatrix}$

where i, j, and k are the unit vectors corresponding to the x, y, and z direction respectively.

$\frac{\partial }{\partial y}\cos(z)\hat{i}+\frac{\partial }{\partial x}\sin(x)\hat{k}+\frac{\partial }{\partial z}\cos(y)\hat{j}-(\frac{\partial }{\partial y}\cos(y)\hat{k}+\frac{\partial }{\partial z}\sin(x)\hat{i}+\frac{\partial }{\partial x}\cos(z)\hat{j})=0+\cos(x)\hat{k}+0+\sin(y)\hat{k}+0+0=\left \langle 0, 0, \cos(x)+\sin(y)\right \rangle$

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Example Question #16 : Curl

Find the curl of the following vector field, in vector form:

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

Possible Answers:

$\left \langle -x, 0, z-x\sin(xy)\right \rangle$

$\left \langle -x, 0, z+y\sin(xy)\right \rangle$

$\left \langle -x, 0, z+x\sin(xy)\right \rangle$

$\left \langle 0, -x, z+x\sin(xy)\right \rangle$

Correct answer:

$\left \langle -x, 0, z+x\sin(xy)\right \rangle$

Explanation:

The curl of the vector field is given by:

$\begin{vmatrix} \hat{i}& \hat{j}& \hat{k}\\ \frac{\partial }{\partial x}&\frac{\partial }{\partial y} & \frac{\partial }{\partial z}\\ \cos(xy)&xz &z^2 \end{vmatrix}$

where i, j, and k are the unit vectors corresponding to the x, y, and z direction respectively.

$\frac{\partial }{\partial y}z^2\hat{i}+\frac{\partial }{\partial x}xz\hat{k}+\frac{\partial }{\partial z}\cos(xy)\hat{j}-(\frac{\partial }{\partial y}\cos(xy)\hat{k}+\frac{\partial }{\partial z}xz\hat{i}+\frac{\partial }{\partial x}z^2\hat{j})=0=z\hat{k}+0+x\sin(xy)\hat{k}-x\hat{i}+0=\left \langle -x, 0, z+x\sin(xy) \right \rangle$

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Example Question #17 : Curl

Find the curl of the following vector field:

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

Possible Answers:

$\left \langle y, 0, -x\right \rangle$

$\vec{0}$

$\left \langle y, 0, x\right \rangle$

$\left \langle -y, 0, -x\right \rangle$

Correct answer:

$\left \langle -y, 0, -x\right \rangle$

Explanation:

The curl of the vector field is given by

$\begin{vmatrix} \hat{i}& \hat{j}& \hat{k}\\ \frac{\partial }{\partial x}&\frac{\partial }{\partial y} & \frac{\partial }{\partial z}\\ xy& yz& \cos(z) \end{vmatrix}$

$\frac{\partial }{\partial y}\cos(z)\hat{i}+\frac{\partial }{\partial x}yz\hat{k}+\frac{\partial }{\partial z}xy\hat{j}-(\frac{\partial }{\partial y}xy\hat{k}+\frac{\partial }{\partial z}yz\hat{i}+\frac{\partial }{\partial x}\cos(z)\hat{j})=-x\hat{k}-y\hat{i}=\left \langle -y, 0, -x\right \rangle$

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Calculus 3 : Calculus 3

Study concepts, example questions & explanations for Calculus 3

All Calculus 3 Resources

Example Questions

Example Question #1 : Curl

Example Question #9 : Curl

Example Question #10 : Curl

Example Question #1731 : Calculus 3

$\left \langle -xy, 0, yz-2x\right \rangle$

$\left \langle -xy, 0, yz-2x\right \rangle$

Example Question #1732 : Calculus 3

Example Question #1733 : Calculus 3

Example Question #1734 : Calculus 3

Example Question #1735 : Calculus 3

Example Question #16 : Curl

Example Question #17 : Curl

All Calculus 3 Resources

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