Answer
\[(a)\;\;\frac{dx}{dt}=-\frac{\sqrt 5}{4}\]
\[(b)\;\;\frac{dy}{dt}=\frac{4}{\sqrt 5}\]
Work Step by Step
$4x^2+9y^2=36$ ___(1)
Since $x$ and $y$ are function of $t$
Differentiating (1) with respect to $t$
$8x\frac{dx}{dt}+18y\frac{dy}{dt}=0$
$4x\frac{dx}{dt}+9y\frac{dy}{dt}=0$ ___(2)
$(a)\;\;$ Given:- $\frac{dy}{dt}=\frac{1}{3}\;,\;x=2\;,\;y=\frac{2}{3}\sqrt 5$
Using given data in (2)
$4(2)\frac{dx}{dt}+9\left(\frac{2}{3}\sqrt 5\right)\left(\frac{1}{3}\right)=0$
$8\frac{dx}{dt}+2\sqrt 5=0$
$\frac{dx}{dt}=-\frac{\sqrt 5}{4}$
Hence $\frac{dx}{dt}=-\frac{\sqrt 5}{4}$.
$(b)\;\;$ Given:- $\Large\frac{dx}{dt}$=$3$ $\;,\;$ $x=-2\;,\;$ $y=\frac{2}{3}\sqrt 5$
Using given data in (2)
$4(-2)3+9\left(\frac{2}{3}\sqrt 5\right)\frac{dy}{dt}=0$
$-24+6\sqrt 5\large\frac{dy}{dt}$=0
$\Large\frac{dy}{dt}=\frac{24}{6\sqrt 5}=\frac{4}{\sqrt 5}$
Hence $\Large\frac{dy}{dt}=\frac{4}{\sqrt 5}$.
