Answer
(a)$ − \frac{60}{7}$ units per second
(b) falling
Work Step by Step
$ \frac{x y^3}{1+y^2}=\frac{8}{5}$
Taking derivative with respect to t
$\frac{d(\frac{xy^3}{1+y^2})}{dt}= \frac{d(\frac{8}{5})}{dt}$
$\frac{d(\frac{xy^3}{1+y^2})}{dx} \frac{dx}{dt} +\frac{d(\frac{xy^3}{1+y^2})}{dy } \frac{dy}{dt}= 0$
$ \frac{(1+y^2)\frac{d(xy^3)}{dx}-xy^3 \frac{d(i+y^2)}{dx}}{(1+y^2)^2 }\frac{dx}{dt} +
\frac{(1+y^2)\frac{d(xy^3)}{dy}-xy^3 \frac{d(i+y^2)}{dy}}{(1+y^2)^2}\frac{dy}{dt} =0$
$ \frac{(1+y^2) y^3\frac{d(x)}{dx}-xy^3 \frac{d(i+y^2)}{dx}}{(1+y^2)^2 }\frac{dx}{dt} +
\frac{(1+y^2)x\frac{d(y^3)}{dy}-xy^3 \frac{d(i+y^2)}{dy}}{(1+y^2)^2}\frac{dy}{dt} =0$
$ \frac{(1+y^2) y^3\frac{d(x)}{dx}-xy^3 \frac{d(i+y^2)}{dx}}{(1+y^2 )^2 }\frac{dx}{dt} +
\frac{(1+y^2)x\frac{d(y^3)}{dy}-xy^3 \frac{d(i+y^2)}{dy}}{(1+y^2)^2}\frac{dy}{dt} =0$
$ \frac{(1+y^2) y^3}{(1+y^2)^2 }\frac{dx}{dt} + \frac{(1+y^2)x(3y^2)-xy^3 (2y)}
{(1+y^2)^2}\frac{dy}{dt} =0$
$ \frac{(1+y^2) y^3}{(1+y^2)^2 }\frac{dx}{dt} + \frac{(1+y^2)x(3y^2)-xy^3 (2y)}
{(1+y^2)^2}\frac{dy}{dt} =0$
$ \frac{(1+y^2) y^3}{(1+y^2)^2 }\frac{dx}{dt} +\frac{ 3xy^2 (1+y^2)-2xy^4 }
{(1+y^2)^2}\frac{dy}{dt} =0$
$ \frac{(1+y^2) y^3}{(1+y^2)^2 }\frac{dx}{dt} +\frac{ 3xy^2 +xy^4 }
{(1+y^2)^2}\frac{dy}{dt} =0$ .................... eq (1)
Put $\frac{dx}{dt}=6 units/s $, $x=1,y=2$ in equation (1)
(a) $\frac{dy}{dt}=?$
$ \frac{(1+2^2) 2^3}{(1+2^2)^2 }\times 6 +\frac{ 3(1)2^2 +(1)2^4 }
{(1+2^2)^2}\frac{dy}{dt} =0$
$\frac{240}{25}+\frac{28}{25} \frac{dy}{dt}=0$
$\frac{dy}{dt}=-\frac{240}{25} \times \frac{25}{28}=-\frac{240}{28}=-\frac{60}{7}$
(b)Since
$\frac{dy}{dt}<0$ $\Longrightarrow$ particle is falling at that instant
