If u=f(r),where r^2=x^2+y^2,then left(frac{pa | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) Given $u=f(r)$ and $r^2=x^2+y^2$. First,find the partial derivatives with respect to $x$: $\frac{\partial r}{\partial x} = \frac{x}{r}$ and $\frac

Vedclass · Accepted Answer

$f^{\prime \prime}(r)+\frac{1}{r} f^{\prime}(r)$

Vedclass · Answer

(C) Given $u=f(r)$ and $r^2=x^2+y^2$. First,find the partial derivatives with respect to $x$: $\frac{\partial r}{\partial x} = \frac{x}{r}$ and $\frac{\partial r}{\partial y} = \frac{y}{r}$. Using the chain rule,$u_x = f^{\prime}(r) \frac{\partial r}{\partial x} = f^{\prime}(r) \frac{x}{r}$. Then,$u_{xx} = \frac{\partial}{\partial x} \left( f^{\prime}(r) \frac{x}{r} \right) = f^{\prime \prime}(r) \left( \frac{x}{r} \right)^2 + f^{\prime}(r) \left( \frac{r - x \frac{\partial r}{\partial x}}{r^2} \right) = f^{\prime \prime}(r) \frac{x^2}{r^2} + f^{\prime}(r) \frac{r^2 - x^2}{r^3}$. Similarly,$u_{yy} = f^{\prime \prime}(r) \frac{y^2}{r^2} + f^{\prime}(r) \frac{r^2 - y^2}{r^3}$. Adding these two: $u_{xx} + u_{yy} = f^{\prime \prime}(r) \left( \frac{x^2+y^2}{r^2} \right) + f^{\prime}(r) \left( \frac{2r^2 - (x^2+y^2)}{r^3} \right)$. Since $x^2+y^2 = r^2$,we get: $u_{xx} + u_{yy} = f^{\prime \prime}(r) \left( \frac{r^2}{r^2} \right) + f^{\prime}(r) \left( \frac{2r^2 - r^2}{r^3} \right) = f^{\prime \prime}(r) + f^{\prime}(r) \frac{r^2}{r^3} = f^{\prime \prime}(r) + \frac{1}{r} f^{\prime}(r)$.

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

Similar Questions

The binding energy per nucleon versus mass number curve for nuclei is shown in the figure. $W, X, Y$ and $Z$ are four nuclei indicated on the curve. The process that would release energy is

Which of the following aqueous solutions will exhibit the highest boiling point?

In an $n-$type silicon,which of the following statements is true?

If the area bounded by $y = ax^2$ and $x = ay^2$ for $a > 0$ is $1$,then $a =$

The instantaneous displacement of a simple pendulum oscillator is given by $x = A \cos \left( \omega t + \frac{\pi}{4} \right)$. Its speed will be maximum at time:

Vedclass Test Series

Exam Paper Generator

Online Exam Module