Answer to Question #108040 in Calculus for khushi

Question

Find
dz/dt
for z = x^2 y + 4y^2 where x = cost and y = sin t using the chain rule

Answer 1

Given that $z=x^2y+4y^2$ and $x=cost \ , \ y=sint$ .

$\therefore \frac{dz}{dt}=\frac{d(x^2y+4y^2)}{dt}$

$\implies \frac{dz}{dt}=\frac{d(x^2y)}{dt} +\frac{d(4y^2)}{dt}$ $\implies \ \frac{dz}{dt}=x^2\frac{dy}{dt}+y\frac{dx^2}{dt}+4\frac{dy^2}{dt}$

Now ,using chain rule we get

$\frac{dz}{dt}=x^2\frac{dy}{dt}+y\frac{dx^2}{dx} ×\frac{dx}{dt}+4\frac{dy^2}{dy}×\frac{dy}{dt}$

Since,

$x=cost \ \implies \frac{dx}{dt}=-sint$ $and \ y=sint \ \implies \ \frac{dy}{dt}=cost$ $\therefore \ \frac{dz}{dt}=x^2cost-2xysint+8ycost.$