AP Calculus BC - Curl | Practice Hub

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

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

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

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 conservative because the curl is 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.

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

Find the curl of the vector field:

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

$\left \langle 2y, 0, y-\sec(y)\tan(y)\right \rangle$

$\left \langle 2y, 0, y+\sec(y)\tan(y)\right \rangle$

$\left \langle y, 0, y-\sec(y)\tan(y)\right \rangle$

$\left \langle 2y, 0, y-\sec(y)\tan(y)\right \rangle$

Explanation

The curl of the vector field is given by

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

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.

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

Find the curl of the following vector field:

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

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

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

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

$\vec{0}$

Explanation

The curl of the vector field is given by

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

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

For the function give the curl of the gradient.

$curl(\vec{\triangledown f})=0$

$curl(\vec{\triangledown f})=4x$

$curl(\vec{\triangledown f})=\sqrt{3}$

$curl(\vec{\triangledown f})=1$

$curl(\vec{\triangledown f})=\sqrt{3}x$

Explanation

Solution 1)

This probably was deceptively easy and could have been very quickly solved without doing any calculations. This problem involves first the basic definition of a conservative vector field, and a useful theorem on conservative vector fields.

1) A vector field $\vec{F}$ is conservative if there exists a scalar function such that $\vec{F}$ is its' gradient.

2) If a vector field is conservative, its' curl must be zero.

In other words, the curl of the gradient is always zero for any scalar function.

In this problem we were given a scalar function . If we now compute the gradient, we obtain a vector field we will call $\vec{F}=\vec{\triangledown} f$ (the gradient is our vector field). Automatically we know it fits the definition of a conservative vector field because we know there is a scalar function which has $\vec{\triangledown}f$ as its' gradient. That function is .

Now we know that since the gradient is a conservative vector field, and therefore the curl must be equal to zero.

Solution 2)

Just for fun, let's see if it works by doing the actual calculation.

$\vec{\triangledown}f=\frac{\partial f}{\partial x}\hat{i}+\frac{\partial f}{\partial y}\hat{j}+\frac{\partial f}{\partial z}\hat{k}$

$\vec{\triangledown}f=\hat{i}+z\hat{j}+y\hat{k}$

$curl(\vec{\triangledown f}) = \begin{vmatrix} \hat{i}&\hat{j} &\hat{k} \ \frac{\partial }{\partial x}&\frac{\partial}{\partial y} &\frac{\partial}{\partial z} \ 1: : &:z &y \end{vmatrix}$

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

$=(1-1)\hat{i}-(0-0)\hat{j}+(0-0)\hat{k}$

Determine the curl of the following vector field:

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

$-e^z\hat{i}$

$\vec{0}$

$e^z\hat{i}$

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

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.

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

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

$y\mathbf{k}$

$\mathbf{0}$

$y\mathbf{i}$

$x\mathbf{i}$

$z\mathbf{k}$

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 .

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

$abla \times ( abla\times f)$

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

$abla\cdot( abla f)$

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

$abla \times ( abla f)$

Explanation

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

$abla \times ( abla\times f)$

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

For the other solutions:

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

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

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

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

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

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

The vector field is not conservative because the curl is not zero.

The vector field is not conservative because the curl is zero.

The vector field is conservative because the curl is not zero.

The vector field is conservative because the curl is zero.

Explanation

A vector field is conservative if its curl is equal to zero (the zero vector).

The curl of the vector field is given by

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

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

The curl is not equal to the zero vector, so the vector field is not conservative.

Find the curl of the vector field:

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

$\left \langle 0, 0, \cos(x)-2y\right \rangle$

$\left \langle 0, 0, \cos(x)+2y\right \rangle$

$\left \langle 0, 0, -\cos(x)+2y\right \rangle$

$\left \langle 0, \cos(x)-2y, 0\right \rangle$

Explanation

The curl of the vector field is given by

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

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

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

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

$\vec{F}=\left \langle x, y^4e^y, 3z^2\right \rangle$

The vector field is conservative because the curl is zero.

The vector field is not conservative because the curl is zero.

The vector field is conservative because the curl is not zero.

The vector field is not conservative because the curl is not zero.

Explanation

A vector field is conservative if its curl is equal to zero (the zero vector).

The curl of the vector field is given by

$\begin{vmatrix} \hat{i}& \hat{j}& \hat{k}\ \frac{\partial }{\partial x}&\frac{\partial }{\partial y} & \frac{\partial }{\partial z}\ x& y^4e^y\cos(y)& 3z^2 \end{vmatrix}$

$\frac{\partial }{\partial y}3z^2\hat{i}+\frac{\partial }{\partial x}y^4e^y\hat{k}+\frac{\partial }{\partial z}x\hat{j}-(\frac{\partial }{\partial y}x\hat{k}+\frac{\partial }{\partial z}y^4e^y\hat{i}+\frac{\partial }{\partial x}3z^2\hat{j})= \vec{0}$

All of the partial derivatives returned zero values for the unit vectors, so the curl indeed is equal to zero. The vector field is conservative.

Curl

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