Q 64. - Calculus

Question

Use at least two methods to prove that $\frac{dr}{dt}=0$ when $r=\sqrt{x^2+y^2}$ $x=\alpha \cos t,\text{ and }y=\alpha \sin t$ if $\alpha$ is constant.

Step-by-Step Solution

Answer

Above relation is proved using chain rule.

$\frac{dr}{dt}=\frac{\partial r}{\partial x}\frac{dx}{dt}+\frac{\partial r}{\partial y}\frac{dy}{dt}$

Step 1: Given Information

It is given that

$r=\sqrt{x^2+y^2},x=\alpha \cos t\text{ and }y=\alpha \sin t$

Step 2: Applying Chain Rule

Using chain rule

$\frac{dr}{dt}=\frac{\partial r}{\partial x}\frac{dx}{dt}+\frac{\partial r}{\partial y}\frac{dy}{dt}$

Solving for $\frac{\partial r}{\partial x}$

$\frac{\partial r}{\partial x}=\frac{\partial }{\partial x}\left(\sqrt{x^2+y^2}\right)$

$=\frac{1}{2}{\left(x^2+y^2\right)}^{-\frac{1}{2}}\frac{\partial }{\partial x}\left(x^2+y^2\right)$

$=\frac{1}{2\sqrt{x^2+y^2}}(2x+0)$

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

Finding $\frac{\partial r}{\partial y}$

$\frac{\partial r}{\partial y}=\frac{\partial }{\partial y}\left(\sqrt{x^2+y^2}\right)$

$=\frac{1}{2}{\left(x^2+y^2\right)}^{-\frac{1}{2}}\frac{\partial }{\partial y}\left(x^2+y^2\right)$

$=\frac{1}{2\sqrt{x^2+y^2}}(0+2y)$

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

Step 3 Differentiating w.r.t t

$\frac{dx}{dt}=\frac{d}{dt}\alpha \cos t=-\alpha \sin t$

Also

$\frac{dy}{dt}=\frac{d}{dt}\alpha \sin t=\alpha \cos t$

Solving for $\frac{dz}{dt}=\frac{\partial z}{\partial x}\frac{dx}{dt}+\frac{\partial z}{\partial y}\frac{dy}{dt}$

$=\left(\frac{x}{\sqrt{x^2+y^2}}\right)(-\alpha \sin t)+\left(\frac{y}{\sqrt{x^2+y^2}}\right)\left(\alpha \cos t\right)$

$=\frac{\alpha }{\sqrt{x^2+y^2}}(-x\sin t+y\cos t)$

$=\frac{\alpha }{\sqrt{(\alpha \cos t{)}^2+(\alpha \sin t{)}^2}}(-\alpha \cos t\sin t+\alpha \sin t\cos t)$

$=\frac{\alpha }{\sqrt{{\alpha }^2\left({\cos }^{2}t+{\sin }^{2}t\right)}}·0=0$

Step 4: Solve using direct derivative

Use $x=\alpha \cos t\text{ and }y=\alpha \sin t\text{ in }r=\sqrt{x^2+y^2}$

Hence,

$r=\sqrt{x^2+y^2}$

$=\sqrt{(\alpha \cos t{)}^2+(\alpha \sin t{)}^2}$

$=\sqrt{{\alpha }^2}$

Differentiating $r=\alpha$ wrt $t$

$\frac{dr}{dt}=\frac{d}{dt}\alpha =0$