Question 13.82: The potential function associated with a force P in space is...

(a) P_{x}=-\frac{\partial V}{\partial x}=-\frac{\partial\left[-\left(x^{2}+y^{2}+z^{2}\right)^{1 / 2}\right]}{\partial x}=x\left(x^{2}+y^{2}+z^{2}\right)^{-1 / 2}\blacktriangleleft

P_{y}=-\frac{\partial V}{\partial y}=-\frac{\partial\left[-\left(x^{2}+y^{2}+z^{2}\right)^{1 / 2}\right]}{\partial y}=y\left(x^{2}+y^{2}+z^{2}\right)^{-1 / 2}\blacktriangleleft

P_{z}=-\frac{\partial V}{\partial z}=-\frac{\partial\left[-\left(x^{2}+y^{2}+z^{2}\right)^{1 / 2}\right]}{\partial z}=z\left(x^{2}+y^{2}+z^{2}\right)^{-1 / 2}\blacktriangleleft

(b) U_{O A B D}=U_{O A}+U_{A B}+U_{B D}        

              O-A: P_{y} and P_{x} are perpendicular to O – A and do no work.      

Also, on O – A \quad x=y=0 \quad \text { and } \quad P_{z}=1   

Thus, \quad U_{O-A}=\int_{0}^{a} P_{z} d z=\int_{0}^{a} d z=a

A-B: P_{z} and P_{y} are perpendicular to A – B and do no work.

Also, on A – B \quad y=0, \quad z=a \quad \text { and } \quad P_{x}=\frac{x}{\left(x^{2}+a^{2}\right)^{1 / 2}}

Thus,

\begin{aligned}U_{A-B} & =\int_{0}^{a} \frac{x d x}{\left(x^{2}+a^{2}\right)^{1 / 2}} \\& =a(\sqrt{2}-1)\end{aligned}

B-D: P_{x} and P_{z} are perpendicular to B – D and do no work.

On  B – D, 

\begin{aligned}k & =a \\z & =a \\P_{y} & =\frac{y}{\left(y^{2}+2 a^{2}\right)^{1 / 2}}\end{aligned}

Thus,

Thus,\quad U_{O A B D}=-\Delta V_{O D}