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,

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