UPSC 2016 Maths Optional Paper 1 Q8a — Step-by-Step Solution

10 marks · Section B

Gradient: definition, geometric meaning, computation · Vector Analysis · asked 6× in 13 yrs · Read the full method →

Question

Find $f(r)$ such that $\nabla f=\dfrac{\vec r}{r^5}$ and $f(1)=0$ .

Radial gradient $\nabla f=f'(r)\hat r$ ; match to $\vec r/r^5=r^{-4}\hat r$ giving $f'=r^{-4}$ ; integrate and fix the constant.

Since the right-hand side $\dfrac{\vec r}{r^5}$ is radial, $f$ depends on $r=|\vec r|$ only. For a function of $r$ ,

\nabla f=f'(r)\,\hat r=f'(r)\,\frac{\vec r}{r}.

Equating to $\dfrac{\vec r}{r^5}=\dfrac1{r^4}\cdot\dfrac{\vec r}{r}$ component-wise (both along $\hat r$ ):

f'(r)=\frac{1}{r^4}.

f(r)=\int r^{-4}\,dr=\frac{r^{-3}}{-3}+C=-\frac{1}{3r^3}+C.

f(1)=-\frac13+C=0\quad\Longrightarrow\quad C=\frac13.

\boxed{\;f(r)=-\frac{1}{3r^3}+\frac13=\frac13\left(1-\frac{1}{r^3}\right).\;}