Question

assume x and y are both differentiable functions of t.

Answer 1

First Substitute x=-1 into the equation 4$x^{2}$ + 3$y^{3}$ = 28, and you get y=2.

Take the derivative of the equation 4$x^{2}$ + 3$y^{3}$ = 28, using implicit differentiation:

8x + 9$y^{2}$$\frac{dy}{dx}$ = 0

Rearrange this in terms of $\frac{dy}{dx}$:

$\frac{dy}{dx} = - \frac{8x}{9y^{2} }$

Keep in mind that $\frac{dy}{dx}$ is equal to $\frac{ \frac{dy}{dt} }{ \frac{dx}{dt} }$

Substitute and you will get the following:

$\frac{ \frac{dy}{dt} }{ \frac{dx}{dt} } = - \frac{8x}{9y^{2} }$

Put in the values for x, y, and $\frac{dy}{dt}$ and solve:

$\frac{ \frac{dy}{dt} }{ \frac{dx}{dt} } = - \frac{8x}{9y^{2} }$

$\frac{8}{ \frac{dx}{dt} } = \frac{8}{36}$

$\frac{dx}{dt} = 36$