Answer
a) $-2$
b) $6\sqrt 5$
Work Step by Step
$4x^2+9y^2=1$
Taking derivative with respect to t
$ \frac{d( 4x^2+9y^2))}{dt}= \frac{d(1)}{dt}$
$\frac{d (4x^2)}{dt}+\frac{d(9y^2)}{dt}=0$
$\frac{d (4x^2)}{dx} \frac{dx}{dt}+\frac{d(9y^2)}{dy} \frac{dy}{dt}=0$
$4 \frac{d (x^2)}{dx} \frac{dx}{dt}+9\frac{d(y^2)}{dy} \frac{dy}{dt}=0$
$ 4 \times2x\frac{dx}{dt}+9 \times 2 y \frac{dy}{dt}=0$
$8 x\frac{dx}{dt}+18 y \frac{dy}{dt}=0$ ........................ eq (1)
(a)Putting $ \frac{dx}{dt}=3$, $ x=\frac{1}{2\sqrt 2},y=\frac{1}{3 \sqrt 2}$ in equation (1)
$ 8\times\frac{1}{2\sqrt 2}\times 3+18\times \frac{1}{3 \sqrt 2} \frac{dy}{dt}=0$
$ 18 \times\frac{1}{3 \sqrt 2} \frac{dy}{dt}=-8 \times \frac{1}{2\sqrt 2}\times 3 $
$ \frac{dy}{dt}=- 8 \times\frac{1}{2\sqrt 2}\times 3 \times3 \sqrt 2 \times\frac{1}{18} =-2 $
(b) Putting $\frac{dy}{dt}=8$, $x=\frac{1}{3}, y=-\frac{\sqrt 5}{9}$ in equation (1)
$ 8\times \frac{1}{3} \frac{dx}{dt} - 18 \times\frac{\sqrt 5}{9}\times8=0 $
$
\frac{dx}{dt} = 18 \times \frac{\sqrt 5}{9}\times8\times 3 \times\frac{1}{8}= 6 \sqrt 5$
