Question

If y^3 = 16 x^2, determine \frac{dx}{dt} when x = 4 and \frac{dy}{dt} = -1 .

Answer 1

To get y, I solved for it and got y^3=256, however, after I cube the 256 and plug it into the original equation, I still don't get the right answer

Answer 2

start by writing
\[3y^2y'=32xx'\]

Answer 3

replace \(x\) by \(4\) and \(x'\) by \(-1\) and \(y\) by \(4\sqrt[3]{4}\)

Answer 4

oh that is wrong, you are solving for \(x'\) so replace \(y'\) by \(-1\)

Answer 5

ah so it should be (3(4*3(4)^(3/2))(-1)=32(4)?