Question

If $\sqrt{x^2+y^2}=e^t$ where $t={\sin }^{-1}\left(\frac{y}{\sqrt{x^2+y^2}}\right)$ then $\frac{dy}{dx}$ is equal to

• A. $\frac{x-y}{x+y}$
• B. $\frac{x+y}{x-y}$
• C. $\frac{y-x}{y+x}$
• D. $\frac{x-y}{2x+y}$

Answer 1

Answer: (B)

$\frac{x+y}{x-y}$

$t={\sin }^{-1}\left(\frac{y}{\sqrt{x^2+y^2}}\right)$, differentiate on both sides.

$\frac{dt}{dx}=\frac{1}{\sqrt{1-{\left(\frac{y}{\sqrt{x^2+y^2}}\right)}^2}}\frac{d\left(\frac{y}{\sqrt{x^2+y^2}}\right)}{dx}(\because \frac{d\left({\sin }^{-1}x\right)}{dx}=\frac{1}{\sqrt{1-x^2}})$

$=\left(\frac{\sqrt{x^2+y^2}}{x}\right)\left[\frac{\left(\sqrt{x^2+y^2}\right)\frac{dy}{dx}-\left(\frac{2x+xy\frac{dy}{dx}}{x\sqrt{x^2+y^2}}\right)y}{(x^2+y^2)}\right]$

$=\left(\frac{\sqrt{x^2+y^2}}{x}\right)\left[\frac{(x^2+y^2)\frac{dy}{dx}-(xy+y^2\frac{dy}{dx})}{(x^2+y^2)\sqrt{x^2+y^2}}\right]$

$\therefore \frac{dt}{dx}=\frac{x^2\frac{dy}{dx}-xy}{x(x^2+y^2)}=\frac{x\frac{dy}{dx}-y}{x^2+y^2}$

Given that $\sqrt{x^2+y^2}=e^t$

$x^2+y^2=e^{2t}$ differentiate on both sides

$2xx+2y\frac{dy}{dx}=e^{2t}\left(2\right)\frac{dt}{dx}$

$x+y\frac{dy}{dx}=(x^2+y^2)\left(\frac{x\frac{dy}{dx}-y}{x^2+y^2}\right)$

$x+y=(x-y)\frac{dy}{dx}$

$\therefore \frac{dy}{dx}=\frac{x+y}{x-y}$