## University Calculus: Early Transcendentals (3rd Edition)

$\frac{dy}{dt} = \frac{-3}{2}$
Given that $\frac{dx}{dt} = -2$ and $x^2+y^2=25$ on differentiating with respect to time $\frac{dx^2}{dt} + \frac{d y^2}{dt} = \frac{d25}{dt}$ $2x\frac{dx}{dt} + 2y\frac{dy}{dt} =0$ when x=3, y=-4 then $-8\frac{dy}{dt} - 2\times (-2)\times3=0$ $\frac{dy}{dt} = \frac {-3}{2}$ Thus, the answer is $\frac{dy}{dt} = \frac{-3}{2}$