Question

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

• A. $\frac{x-y}{x+y}$
• B. $\frac{x+y}{x-y},x>0$
• C. $\frac{y-x}{y+x},x<0$
• D. $\frac{x-y}{2x+y}$

Answer 1

Answer: (B), (C)

The correct options are
B $\frac{x+y}{x-y},x>0$
C $\frac{y-x}{y+x},x<0$

Given

$\sqrt{x^2+y^2}=e^t,t={\sin }^{-1}\left(\frac{y}{\sqrt{x^2+y^2}}\right)$

$\sin t=\frac{y}{\sqrt{x^2+y^2}}$

$\therefore \cos t=\sqrt{1-{\sin }^{2}t}=\frac{x}{\sqrt{x^2+y^2}},x>0$

$=\frac{-x}{\sqrt{x^2+y^2}},x<0$

$\sin t=\frac{y}{\sqrt{x^2+y^2}}$

$e^t\sin t=y(\because e^t=\sqrt{x^2+y^2})$

$(e^t\sin t+e^t\cos t)\frac{dt}{dx}=\frac{dy}{dx}\to 1$

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

$t=\ln \left(\sqrt{x^2+y^2}\right)$

Differentiate on both sides w.r.t x

$\frac{dt}{dx}=\frac{1}{\sqrt{x^2+y^2}}.\frac{2x+2y\frac{dy}{dx}}{2\sqrt{x^2+y^2}}=\frac{x+y\frac{dy}{dx}}{(x^2+y^2)}$

For $x>0$

$1\Rightarrow (\sqrt{x^2+y^2}.\frac{y}{\sqrt{x^2+y^2}}+\sqrt{x^2+y^2}\frac{x}{\sqrt{x^2+y^2}})\frac{dt}{dx}=\frac{dt}{dx}$

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

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

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

$\therefore \frac{dy}{dx}=\frac{x^2+xy}{x^2-xy}=\frac{x+y}{x-y},x>0$

For $x<0$

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

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

$\therefore \frac{dy}{dx}=\frac{yx-x^2}{x^2+xy}=\frac{y-x}{y+x},x<0$