Question

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

\frac{dx}{dt} =

Answer 1

\[3y^2y'=32xx'\] make the replacements \(x=4, y'=-3\) and solve for \(x'\)

Answer 2

ok

Answer 3

oh i guess you need \(y\) too but you know when \(x=4\) you have \(y^3=16\times 4^2=256\) so you can solve for \(y\) as well

Answer 4

-3/4 y^2???

Answer 5

i forgot that you need \(y\) as well

Answer 6

ok

Answer 7

you got that?

Answer 8

how would i find y?

Answer 9

from what i wrote above. replace x by 4 in the original equation and solve for y

Answer 10

ok lets see