Q69 / 160 · AP EAMCET 2012 · Mathematics
If $u=f(r)$, where $r^2=x^2+y^2$, then
$\left(\frac{\partial^2 u}{\partial x^2}+\frac{\partial^2 u}{\partial y^2}\right)$ is equal to
- A$f^{\prime \prime}(r)$
- B$f^{\prime \prime}(r)+f^{\prime}(r)$
- Correct answerC$f^{\prime \prime}(r)+\frac{1}{r} f^{\prime}(r)$
- D$f^{\prime \prime}(r)+r f^{\prime}(r)$
Solution
Given, $u=f(r)$ and $r^2=x^2+y^2$
$\begin{aligned} & \therefore \quad u_x=f^{\prime}(r) \frac{\delta r}{\delta x}=f^{\prime}(r)_r^x \\ & \Rightarrow u_{x x}=f^*(r) \times \frac{x^2}{r^2}+f^{\prime}(r) \frac{r^2-x \times x}{r^2} \\ & =f(r) \times \frac{x^2}{r^2}+f^{\prime}(r) \frac{r^2-x^2}{r^3} \\ & u_y=f^{\prime}(r) \frac{\delta r}{\delta y}=f^{\prime}(r) \frac{(2 y)}{2 \sqrt{x^2+y^2}} \\ & =f^{\prime}(r)\left(\frac{y}{r}\right) \\ & \text { and } \quad u_{y y}=f^{\prime \prime}(r) \frac{y^2}{r^2}+f^r(r) \frac{r^2-y^2}{r^3} \\ & \therefore u_{x r}+u_{y y}=f^{\prime \prime}(r)\left(\frac{x^2}{r^2}+\frac{y^2}{r^2}\right) \\ & +f^{\prime}(r)\left(\frac{r^2-x^2}{r^3}+\frac{r^2-y^2}{r^3}\right) \\ & =f^{\prime \prime}(r)\left(\frac{r^2}{r^2}\right)+f^{\prime}(r) \frac{2 r^2-\left(x^2+y^2\right)}{r^3} \\ & =f^{\prime}(r)+f^{\prime}(r) \frac{r^2}{r^3} \\ & =f^{\prime \prime}(r)+\frac{f^{\prime}(r)}{r} \\ & \end{aligned}$
Hence, the correct answer is option: 'C'.