Question

Determine an expression for dy dx given that x = sin3 (t)andy = cos3 (t)

Answer 1

Answer:

$\frac{dy}{dx} =-\sqrt[3]{\frac{y}{x} }$

Step-by-step explanation:

Recall that using the chain rule we can state:

$\frac{dy}{dt} =\frac{dy}{dx}*\frac{dx}{dt}$

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

$x=sin^3(t)\\dx/dt = 3 sin^2(t)* cos(t)$

And similarly:

$y=cos^3(t)\\dy/dt=-3\,cos^2(t)*sin(t)$

Therefore, dy/dx can be determined by the quotient of the expressions we just found:

$\frac{dy}{dx} =\frac{dy/dt}{dx/dt} =\frac{-3\,cos^2(t)*sin(t)}{3\,sin^2(t)*cos(t)} =-\frac{cos(t)}{sin(t)}$

now notice that we can find   $cos(t) = \sqrt[3]{y}$  from the expression for y,

and   $sin(t) = \sqrt[3]{x}$  from its expression for x.

Therefore dy/dx can be written in terms of x and y as:

$\frac{dy}{dx} =-\frac{cos(t)}{sin(t)}=-\sqrt[3]{\frac{y}{x} }$