Question

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

Answer 1

`(dy)/(dx) =-\sqrt[3]{(y)/(x) }`

Explanation:

Recall that using the chain rule we can state:

`(dy)/(dt) =(dy)/(dx)*(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:

`(dy)/(dx) =(dy/dt)/(dx/dt) =(-3\,cos^2(t)*sin(t))/(3\,sin^2(t)*cos(t)) =-(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:

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