Answer:
[tex]\frac{dy}{dx} =-\sqrt[3]{\frac{y}{x} }[/tex]
Step-by-step explanation:
Recall that using the chain rule we can state:
[tex]\frac{dy}{dt} =\frac{dy}{dx}*\frac{dx}{dt}[/tex]
and therefore solve for dy/dx as long as dx/dt is different from zero.
Then we find dy/dt and dx/dt,
Given that
[tex]x=sin^3(t)\\dx/dt = 3 sin^2(t)* cos(t)[/tex]
And similarly:
[tex]y=cos^3(t)\\dy/dt=-3\,cos^2(t)*sin(t)[/tex]
Therefore, dy/dx can be determined by the quotient of the expressions we just found:
[tex]\frac{dy}{dx} =\frac{dy/dt}{dx/dt} =\frac{-3\,cos^2(t)*sin(t)}{3\,sin^2(t)*cos(t)} =-\frac{cos(t)}{sin(t)}[/tex]
now notice that we can find [tex]cos(t) = \sqrt[3]{y}[/tex] from the expression for y,
and [tex]sin(t) = \sqrt[3]{x}[/tex] from its expression for x.
Therefore dy/dx can be written in terms of x and y as:
[tex]\frac{dy}{dx} =-\frac{cos(t)}{sin(t)}=-\sqrt[3]{\frac{y}{x} }[/tex]
