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Calculus 3 : Curl

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Example Questions

Example Question #1 : Curl

Calculate the curl for the following vector field.

$\vec{F}=x^3y^2\ \vec{i}+x^2y^3z^4\ \vec{j}+x^2z^2\ \vec{k}$

Possible Answers:

$curl\ \vec{F}=\Big(4x^2y^3z^3\Big)\vec{i}+\Big(2xz^2\Big)\vec{j}+\Big(-2xy^3z^4+2x^3y\Big)\vec{k}$

$curl\ \vec{F}=0$

$curl\ \vec{F}=\Big(-4x^2y^3z^3\Big)\vec{i}+\Big(-2xz^2\Big)\vec{j}+\Big(2xy^3z^4-2x^3y\Big)\vec{k}$

$curl\ \vec{F}=\Big(2x^2z^3\Big)\vec{i}+\Big(2yz^2\Big)\vec{j}+\Big(-2x^3y\Big)\vec{k}$

Correct answer:

$curl\ \vec{F}=\Big(-4x^2y^3z^3\Big)\vec{i}+\Big(-2xz^2\Big)\vec{j}+\Big(2xy^3z^4-2x^3y\Big)\vec{k}$

Explanation:

In order to calculate the curl, we need to recall the formula.

$\\curl\ \vec{F}=\triangledown \times \vec{F}=\begin{vmatrix} \vec{i} & \vec{j} & \vec{k}\\ \frac{\partial}{\partial x}& \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ P & Q & R \end{vmatrix}\\ \\ \\ =\Big(\frac{\partial R}{\partial y}-\frac{\partial Q}{\partial z}\ \Big)\vec{i}+\Big(\frac{\partial P}{\partial z}-\frac{\partial R}{\partial x}\ \Big)\vec{j}+\Big(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}\ \Big)\vec{k}$

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 out situation.

$curl\ \vec{F}=\triangledown \times \vec{F}=\begin{vmatrix} \vec{i} & \vec{j} & \vec{k}\\\\ \frac{\partial}{\partial x}& \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\\\ x^3y^2 & x^2y^3z^4 & x^2z^2 \end{vmatrix}$

$\\=\Big(\frac{\partial }{\partial y}\Big(x^2z^2\Big)-\frac{\partial}{\partial z}\Big(x^2y^3z^4\Big)\ \Big)\vec{i}+\Big(\frac{\partial}{\partial z}\Big(x^3y^2\Big)-\frac{\partial}{\partial x}\Big(x^2z^2\Big)\ \Big)\vec{j}+\Big(\frac{\partial}{\partial x}\Big(x^2y^3z^4\Big)-\frac{\partial}{\partial y}\Big(x^3y^2\Big)\ \Big)\vec{k}$

$=\Big(0-4x^2y^3z^3\Big)\vec{i}+\Big(0-2xz^2\Big)\vec{j}+\Big(2xy^3z^4-2x^3y\Big)\vec{k}$

Thus the curl is

$=\Big(-4x^2y^3z^3\Big)\vec{i}+\Big(-2xz^2\Big)\vec{j}+\Big(2xy^3z^4-2x^3y\Big)\vec{k}$

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Example Question #2 : Curl

Calculate the curl for the following vector field.

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

Possible Answers:

$curl \ \vec{F}=\Big(2yz^2-x^2y\Big)\vec{i}+\Big(x\Big)\vec{j}+\Big(2xyz\Big)\vec{k}$

$curl \ \vec{F}=\Big(-x^2y\Big)\vec{i}+\Big(-x\Big)\vec{j}+\Big(xyz\Big)\vec{k}$

$curl \ \vec{F}=\Big(2yz^2\Big)\vec{i}+\Big(z\Big)\vec{j}+\Big(2xz\Big)\vec{k}$

$curl \ \vec{F}=\Big(-2yz^2+x^2y\Big)\vec{i}+\Big(-x\Big)\vec{j}+\Big(-2xyz\Big)\vec{k}$

$curl \ \vec{F}=0$

Correct answer:

$curl \ \vec{F}=\Big(2yz^2-x^2y\Big)\vec{i}+\Big(x\Big)\vec{j}+\Big(2xyz\Big)\vec{k}$

Explanation:

In order to calculate the curl, we need to recall the formula.

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 out situation.

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

$=\Big(2yz^2-x^2y\Big)\vec{i}+\Big(x-0\Big)\vec{j}+\Big(2xyz-0\Big)\vec{k}$

Thus the curl is

$=\Big(2yz^2-x^2y\Big)\vec{i}+\Big(x\Big)\vec{j}+\Big(2xyz\Big)\vec{k}$

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Example Question #3 : Curl

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

Possible Answers:

None of the other answers

All of the other answers

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

1+xy+z^2

$y\mathbf{k}$

Correct answer:

$y\mathbf{k}$

Explanation:

Curl is probably best remembered by the determinant formula

$\bigtriangledown \times \mathbf{F}(x,y,z)$ , which is used here as follows.

$\bigtriangledown \times \mathbf{F}(x,y,z) = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ \frac{\partial}{\partial x} &\frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ 1& xy & z^2 \end{vmatrix}$

$=\begin{vmatrix} \frac{\partial}{\partial y} &\frac{\partial}{\partial z} \\ xy & z^2 \end{vmatrix}\mathbf{i}- \begin{vmatrix} \frac{\partial}{\partial x} &\frac{\partial}{\partial z} \\ 1 & z^2 \end{vmatrix} \mathbf{j}+ \begin{vmatrix} \frac{\partial}{\partial x} &\frac{\partial}{\partial y} \\ 1 & xy \end{vmatrix}\mathbf{k}$

$=(\frac{\partial}{\partial y}z^2 -\frac{\partial}{\partial z}(xy))\mathbf{i} -(\frac{\partial}{\partial x}z^2 - \frac{\partial}{\partial z}(1))\mathbf{j} + (\frac{\partial}{\partial x}(xy)-\frac{\partial}{\partial y}(1))\mathbf{k}$ $=0\mathbf{i} - 0\mathbf{j}+ (y-0)\mathbf{k}$

$=y\mathbf{k}$

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Example Question #1 : Curl

Let be any arbitrary real valued vector field. Find the div(curl F)

Possible Answers:

$\pm\frac{1}{2}$

Correct answer:

Explanation:

Take any field, the curl gives us the amount of rotation in the vector field. The purpose of the divergence is to tell us how much the vectors move in a linear motion.

When vectors are moving in circular motion only, there are no possible linear motion. Thus the divergence of the curl of any arbitrary vector field is zero.

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Example Question #1 : Curl

Evaluate the curl of the force field $F = (x^3+1)\mathbf{i}+xy\mathbf{j}+\ln(z)\mathbf{k}$ .

Possible Answers:

$\mathbf{0}$

$z\mathbf{k}$

$y\mathbf{i}$

$y\mathbf{k}$

$x\mathbf{i}$

Correct answer:

$y\mathbf{k}$

Explanation:

To evaluate the curl of a force field, we use Curl $(F)=\bigtriangledown \times F.$

$\bigtriangledown \times F = \begin{vmatrix} \mathbf{i}&\mathbf{j} &\mathbf{k} \\ \frac{\partial}{\partial x}&\frac{\partial}{\partial y} &\frac{\partial}{\partial z} \\ (x^3 +1)& xy & \ln(z) \end{vmatrix}$ . Start

$=(\frac{\partial}{\partial y} \ln(z) - \frac{\partial}{\partial z}(xy))\mathbf{i} -(\frac{\partial}{\partial x}\ln(z)-\frac{\partial}{\partial z}(x^3+1))\mathbf{j} +(\frac{\partial}{\partial x}(xy)-\frac{\partial}{\partial y}(x^3+1))\mathbf{k}$ Evaluate along the first row using cofactor expansion.

$= y\mathbf{k}$ . Evaluate partial derivatives. All terms except the 2nd to last one are .

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Example Question #1 : Curl

Calculate the curl $\triangledown \times \mathbf{F}$ of the following vector:

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

Possible Answers:

$z\mathbf{i}+x\mathbf{j}-y\mathbf{k}$

$-z\mathbf{i}+x\mathbf{j}-y\mathbf{k}$

$z\mathbf{i}+x\mathbf{j}+y\mathbf{k}$

$z\mathbf{i}-x\mathbf{j}+y\mathbf{k}$

$-z\mathbf{i}-x\mathbf{j}-y\mathbf{k}$

Correct answer:

$z\mathbf{i}+x\mathbf{j}+y\mathbf{k}$

Explanation:

The curl of a vector $\mathbf{F}=U\mathbf{i}+V\mathbf{j}+W\mathbf{k}$

is defined by the determinant of the following 3x3 matrix:

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

For the given vector, we can calculate this determinant

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

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Example Question #7 : Curl

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

Possible Answers:

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

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

$\nabla \times(\nabla f)$

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

$\nabla(\nabla\cdot f)$

Correct answer:

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

Explanation:

For each of the given expressions:

$\nabla(\nabla \cdot f)$ - The divergence of a scalar function does not exist, so this expression is undefined.

$\nabla(\nabla \cdot\mathbf{F})$ - The dot product of a vector function is a scalar, so the gradient of the term in parenthesis results in a vector.

$\nabla \cdot(\nabla \cdot\mathbf{F})$ - The divergence of a vector function is a scalar. Taking the divergence of the term in parenthesis would be taking the divergence of a scalar, which doesn't exist. This expression is undefined.

$\nabla \times(\nabla f)$ - The gradient of a scalar function is a vector. Thus, the curl of the term in parenthesis is also a vector.

The remaining answer is:

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

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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 #1 : Curl

Compute the curl of the following vector function:

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

Possible Answers:

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

$2y\mathbf{k}$

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

$-2y\mathbf{i}$

$-3\mathbf{k}$

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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Calculus 3 : Curl

Study concepts, example questions & explanations for Calculus 3

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