Chapter 4, Problem 4.23P

(a)

To determine

Whether the force is conservative or not, the corresponding potential energy, and verify $F=-\nabla U$.

Answer to Problem 4.23P

The force is conservative, the corresponding potential energy is $-\frac{1}{2}\left(k{x}^2-2k{y}^2-3k{z}^2\right)$, and verified $F=-\nabla U$ for the given function.

Explanation of Solution

Write the expression for curl of a function.

    $\begin{array}{c}\nabla \times F=\left(\begin{array}{ccc}i& j& k\\ \frac{\partial }{\partial x}& \frac{\partial }{\partial y}& \frac{\partial }{\partial z}\\ F_x& F_y& F_z\end{array}\right)\\ =i\left[\frac{\partial F_z}{\partial y}-\frac{\partial F_y}{\partial z}\right]-j\left[\frac{\partial F_z}{\partial z}-\frac{\partial F_x}{\partial x}\right]+k\left[\frac{\partial F_y}{\partial x}-\frac{\partial F_x}{\partial y}\right]\end{array}$        (I)

Use $k\left(x\widehat{i}+2y\widehat{j}+3z\widehat{k}\right)$ for $F$ in the equation (I).

    $\begin{array}{c}\nabla \times F=\widehat{i}\left[\frac{\partial \left(3kz\right)}{\partial y}-\frac{\partial \left(2ky\right)}{\partial z}\right]-\widehat{j}\left[\frac{\partial \left(3kz\right)}{\partial z}-\frac{\partial \left(kx\right)}{\partial x}\right]+\widehat{k}\left[\frac{\partial \left(2ky\right)}{\partial x}-\frac{\partial \left(kx\right)}{\partial y}\right]\\ =0\end{array}$

So the force is conservative.

Potential energy of the system of masses is the negative integral of gravitational force to bring the masses from infinite distance to a distance $r$.

    $\begin{array}{c}U\left(r\right)=-{\displaystyle \underset{r_0}{\overset{r}{\int }}F\left(r^{\prime }\right)\cdot d{r}^{\prime }}\\ U-U_0=-{\displaystyle \underset{0}{\overset{x}{\int }}{F}_{x}dx}-{\displaystyle \underset{0}{\overset{y}{\int }}{F}_{y}dy}-{\displaystyle \underset{0}{\overset{z}{\int }}{F}_{z}dz}\end{array}$        (II)

At the origin the potential energy is zero. Use $k\left(x\widehat{i}+2y\widehat{j}+3z\widehat{k}\right)$, and $0$ for $U_0$ in the equation (II).

    $\begin{array}{c}U-0=-{\displaystyle \underset{0}{\overset{x}{\int }}\left(kx\right)dx}-{\displaystyle \underset{0}{\overset{y}{\int }}\left(2yk\right)dy}-{\displaystyle \underset{0}{\overset{z}{\int }}\left(3kz\right)dz}\\ U=-\left(\frac{1}{2}k{x}^2+k{y}^2+\frac{3}{2}k{z}^2\right)\\ =-\frac{1}{2}\left(k{x}^2-2k{y}^2-3k{z}^2\right)\end{array}$        (III)

Write the general form of gradient of a function $U$.

    $\nabla U=\widehat{x}\frac{\partial U}{\partial x}+\widehat{y}\frac{\partial U}{\partial y}+\widehat{z}\frac{\partial U}{\partial z}$        (IV)

Use $-\frac{1}{2}\left(k{x}^2-2k{y}^2-3k{z}^2\right)$ for $U$ in the equation (IV), and solve.

    $\begin{array}{c}\nabla U=\widehat{i}\frac{\partial }{\partial x}\left[-\frac{1}{2}\left(k{x}^2-2k{y}^2-3k{z}^2\right)\right]+\widehat{y}\frac{\partial }{\partial y}\left[-\frac{1}{2}\left(k{x}^2-2k{y}^2-3k{z}^2\right)\right]+\widehat{k}\frac{\partial }{\partial z}\left[-\frac{1}{2}\left(k{x}^2-2k{y}^2-3k{z}^2\right)\right]\\ =\frac{1}{2}k\left(2x\right)\widehat{i}+k\left(2y\right)\widehat{j}+\frac{3}{2}k\left(2z\right)\widehat{k}\\ =k\left(x\widehat{i}+2y\widehat{j}+3z\widehat{k}\right)\end{array}$        (V)

Equation (V) and given force are same. So the result $F=-\nabla U$ is verified.

Conclusion:

Therefore, the force is conservative, the corresponding potential energy is $-\frac{1}{2}\left(k{x}^2-2k{y}^2-3k{z}^2\right)$, and verified $F=-\nabla U$ for the given function.

(b)

To determine

Whether the force is conservative or not, the corresponding potential energy, and verify $F=-\nabla U$.

Answer to Problem 4.23P

The force is conservative, the corresponding potential energy is $-kxy$, and verified $F=-\nabla U$ for the given function.

Explanation of Solution

Use $k\left(y\widehat{i}+x\widehat{j}+0\widehat{k}\right)$ for $F$ in the equation (I).

    $\begin{array}{c}\nabla \times F=\widehat{i}\left[\frac{\partial \left(0\right)}{\partial y}-\frac{\partial \left(kx\right)}{\partial z}\right]-\widehat{j}\left[\frac{\partial \left(0\right)}{\partial z}-\frac{\partial \left(ky\right)}{\partial x}\right]+\widehat{k}\left[\frac{\partial \left(x\right)}{\partial x}-\frac{\partial \left(ky\right)}{\partial y}\right]\\ =\widehat{i}\left[0-0\right]-\widehat{j}\left[0-0\right]+\widehat{k}\left[k-k\right]\\ =0\end{array}$

So the force is conservative.

At the origin the potential energy is zero. Use $k\left(y\widehat{i}+x\widehat{j}+0\widehat{k}\right)$, and $0$ for $U_0$ in the equation (II).

    $\begin{array}{c}U-0=-{\displaystyle \underset{0}{\overset{x}{\int }}\left(ky\right)dx}-{\displaystyle \underset{0}{\overset{y}{\int }}\left(kx\right)dy}-{\displaystyle \underset{0}{\overset{z}{\int }}0dz}\\ U=-kxy-kyx\\ =-2kxy\\ =-kxy\end{array}$        (VI)

Since $2k$ is also a constant.

Use $-kxy$ for $U$ in the equation (IV), and solve.

    $\begin{array}{c}\nabla U=\widehat{i}\frac{\partial }{\partial x}\left(-kxy\right)+\widehat{y}\frac{\partial }{\partial y}\left(-kxy\right)+\widehat{k}\frac{\partial }{\partial z}\left(-kxy\right)\\ =k\left(y\right)\widehat{i}+k\left(x\right)\widehat{j}+\frac{3}{2}k\left(0\right)\widehat{k}\\ =k\left(y\widehat{i}+x\widehat{j}+0\widehat{k}\right)\end{array}$        (VII)

Equation (VII) and given force are same. So the result $F=-\nabla U$ is verified.

Conclusion:

Therefore, the force is conservative, the corresponding potential energy is $-kxy$, and verified $F=-\nabla U$ for the given function.

(c)

To determine

Whether the force is conservative or not, the corresponding potential energy, and verify $F=-\nabla U$.

Answer to Problem 4.23P

The force is not conservative, there is no corresponding potential energy, and so the result $F=-\nabla U$ is not verified for the given function.

Explanation of Solution

Use $k\left(-y\widehat{i}+x\widehat{j}+0\widehat{k}\right)$ for $F$ in the equation (I).

    $\begin{array}{c}\nabla \times F=\widehat{i}\left[\frac{\partial \left(0\right)}{\partial y}-\frac{\partial \left(kx\right)}{\partial z}\right]-\widehat{j}\left[\frac{\partial \left(0\right)}{\partial z}-\frac{\partial \left(-ky\right)}{\partial x}\right]+\widehat{k}\left[\frac{\partial \left(x\right)}{\partial x}-\frac{\partial \left(-ky\right)}{\partial y}\right]\\ =\widehat{i}\left[0-0\right]-\widehat{j}\left[0-0\right]+\widehat{k}\left[k-\left(-k\right)\right]\\ =2k\widehat{k}\end{array}$

So the force is not conservative. Hence there is no corresponding potential energy.

Conclusion:

Therefore, the force is not conservative, there is no corresponding potential energy, and so the result $F=-\nabla U$ is not verified for the given function.