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Calculus 3 : Curl

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

Example Questions

Example Question #11 : Curl

Find the curl of the vector function:

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

Possible Answers:

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

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

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

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

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 #12 : Curl

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 #13 : Curl

Find the curl of the vector function:

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

Possible Answers:

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

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

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

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

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 #14 : Curl

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 is equal to $\vec{0}$ .

The vector field is not conservative because the curl does not 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 #15 : Curl

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:

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

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

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

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

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 #91 : Line Integrals

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

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

Possible Answers:

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

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

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

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

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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Example Question #1741 : Calculus 3

Determine the curl of the following vector field:

$\vec{F}=6x\hat{i}+(e^z+y)\hat{j}+z\hat{k}$

Possible Answers:

$-e^z\hat{i}$

$-(e^z+y)\hat{i}$

$e^z\hat{i}$

$\vec{0}$

Correct answer:

$-e^z\hat{i}$

Explanation:

The curl of the vector field is given by

$\begin{vmatrix} \hat{i}& \hat{j}& \hat{k}\\ \frac{\partial }{\partial x}&\frac{\partial }{\partial y} & \frac{\partial }{\partial z}\\ 6x&e^z+y & z \end{vmatrix}$

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

$\frac{\partial }{\partial y}z\hat{i}+\frac{\partial }{\partial x}(e^z+j)\hat{k}+\frac{\partial }{\partial z}6x\hat{j}-(\frac{\partial }{\partial y}6x\hat{k}+\frac{\partial }{\partial z}(e^z+j)\hat{i}+\frac{\partial }{\partial x}z\hat{j})=-e^z\hat{i}$

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Example Question #1742 : Calculus 3

Determine the curl of the following vector field:

$\vec{F}=\ln(y)\hat{i}+xz^2\hat{j}+y\hat{k}$

Possible Answers:

$(1-2xz)\hat{i}+(z^2-\frac{1}{y})\hat{k}$

$(1+2xz)\hat{i}+(z^2+\frac{1}{y})\hat{k}$

$(1-xz^2)\hat{i}+(z^2-\frac{1}{y})\hat{k}$

$(1-2xz)\hat{i}+(z^2-\ln(y))\hat{k}$

Correct answer:

$(1-2xz)\hat{i}+(z^2-\frac{1}{y})\hat{k}$

Explanation:

The curl of the vector field is given by

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

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

$\frac{\partial }{\partial y}y\hat{i}+\frac{\partial }{\partial x}xz^2\hat{k}+\frac{\partial }{\partial z}\ln(y)\hat{j}-(\frac{\partial }{\partial y}\ln(y)\hat{k}+\frac{\partial }{\partial z}xz^2\hat{i}+\frac{\partial }{\partial x}y\hat{j})=(1-2xz)\hat{i}+(z^2-\frac{1}{y})\hat{k}$

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Example Question #20 : Curl

Find the curl of the following vector field:

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

Possible Answers:

$(2x-2xy)\hat{j}$

$(2x+2xy)\hat{k}$

2x-2xy

$(2x-2xy)\hat{k}$

$\vec{0}$

Correct answer:

$(2x-2xy)\hat{k}$

Explanation:

The curl of the vector field is given by

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

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

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

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Calculus 3 : Curl

Study concepts, example questions & explanations for Calculus 3

All Calculus 3 Resources

Example Questions

Example Question #11 : Curl

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

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

Example Question #12 : Curl

Example Question #13 : Curl

Example Question #14 : Curl

Example Question #15 : Curl

Example Question #91 : Line Integrals

Example Question #17 : Curl

Example Question #1741 : Calculus 3

Example Question #1742 : Calculus 3

Example Question #20 : Curl

All Calculus 3 Resources

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