Question

Observe the following statements:
$I$: If $x=r\cos \theta ,y=r\sin \theta$, then $\frac{{\partial }^2\theta }{\partial x^2}+\frac{{\partial }^2\theta }{\partial y^2}=0$
$II$: If $x=r\cos \theta ,y=r\sin \theta ,then(\frac{\partial \theta }{\partial x}{)}^2+(\frac{\partial \theta }{\partial y}{)}^2=0$

Which of the above statements is correct?

• A. Only I
• B. Only II
• C. Both I and II
• D. Neither I nor II

Answer 1

Answer: (A)

Only I

Given : $x=r\cos \theta y=r\sin \theta$
$\frac{x}{r}=\cos \theta \frac{y}{r}=\sin \theta$
$\cos \theta =\frac{x}{r}\sin \theta =\frac{y}{r}$
$\theta ={\cos }^{-1}\left(\frac{x}{r}\right)\theta ={\sin }^{-1}\left(\frac{y}{r}\right)$

Differentiate partially w.r.t ${}^{\prime }{x}^{\prime }$ and ${}^{\prime }{y}^{\prime }$

$\frac{\partial \theta }{\partial x}=-\frac{1}{\sqrt{1-{\left(\frac{x}{r}\right)}^2}}\times \frac{1}{r}\frac{\partial \theta }{\partial y}=-\frac{1}{\sqrt{1-{\left(\frac{y}{r}\right)}^2}}\times \frac{1}{r}$
$=\frac{-\sqrt{r^2}}{\sqrt{r^2-x^2}}\times \frac{1}{r}=\frac{-\sqrt{r^2}}{\sqrt{r^2-y^2}}\times \frac{1}{r}$
$=\frac{-1}{\sqrt{r^2-x^2}}=\frac{1}{\sqrt{r^2-y^2}}$
Differentiate the equation below partially again with ${}^{\prime }{x}^{\prime }$ and ${}^{\prime }{y}^{\prime }$

$\frac{\partial \theta }{\partial x}=\frac{-1}{\sqrt{r^2-x^2}}\frac{\partial \theta }{\partial y}=\frac{1}{\sqrt{r^2-y^2}}$
$\frac{{\partial }^2\theta }{{\partial x}^2}=\frac{-1}{2\sqrt{r^2-x^2}}\times -2x\frac{{\partial }^2\theta }{{\partial y}^2}=\frac{1}{\sqrt{r^2-y^2}}\times -2y$
$=\frac{1}{\sqrt{r^2-x^2}}=\frac{-1}{\sqrt{r^2-y^2}}$
So the given statements are :

$I:\frac{{\partial }^2\theta }{{\partial x}^2}+\frac{{\partial }^2\theta }{{\partial y}^2}=0$
$\Rightarrow \frac{1}{\sqrt{r^2-x^2}}-\frac{1}{\sqrt{r^2-y^2}}$
Substitute 'x' and 'y' with $r\cos \theta$ and $r\sin \theta$

$\Rightarrow \frac{1}{\sqrt{r^2-r^2{\cos }^2\theta }}-\frac{1}{\sqrt{r^2-r^2{\sin }^2\theta }}$
$=\frac{1}{\sqrt{r^2}(1-{\cos }^2\theta )}-\frac{1}{\sqrt{r^2}(1-{\sin }^2\theta )}$
$=\frac{1}{r\sqrt{{\sin }^2\theta }}-\frac{1}{r\sqrt{{\cos }^2\theta }}$
$=\frac{1}{r\sin \theta }-\frac{1}{r\cos \theta }=0$.