Question

suppose 4x^2 + 9y^2 = 36 where x and y are functions of t. a) if dy/dt= 1/3, find dx/dt when x=2 and y= (2/3)*(square root of 5) b) if dx/dt= 3 find dy/dt when x= -2 and y= (2/3)*(square root of 5)

Answer 1

\[
4x^2 + 9y^2 = 36
\]Differentiate: \[
4\left(2x\frac{dx}{dt}\right) + 9\left(2y\frac{dy}{dt}\right) = 0
\]Now win.