Question

If y^3 = 25 x^2, determine \frac{dx}{dt} when x = 5 and \frac{dy}{dt} = -2

I'm new to this topic and i'm confused since its y^3 instead of just y, thanks in advance

Answer 1

Okay this is slightly harder, but the same concepts apply.
\[y^3=25x^2\]Differentiating both sides yields: \[3y^2\frac{dy}{dx}=50x\]Divide both sides by 3y^2 to give us our dy/dx:
\[\frac{dy}{dx}=\frac{50x}{3y^2}\]
We can find our answer using this equation:\[\frac{dx}{dt} = \frac{dx}{dy} * \frac{dy}{dt} \]We know \[\frac{dx}{dy} = \frac{1}{\frac{dy}{dx}} = \frac{3y^2}{50x} \]\[\frac{dx}{dy} (5) = \frac{3y^2}{250}\]and \[\frac{dy}{dt} = -2 \]
Hence:
\[\frac{dx}{dt} = \frac{-6y^2}{250} \]

Answer 2

@tom982 I have to reply in webwork and it doesn't accept the answer.. or is there more steps to it?

Answer 3

if x = 5, then you can plug that into y^3 = 25x^2 and solve for y

once you know y, you can plug that into \[\Large \frac{dx}{dt} = \frac{-6y^2}{250}\]
don't forget to reduce or simplify fully as much as possible

Answer 4

Thanks Jim. I missed a pretty crucial step!

Answer 5

ohhh i see, first solve for y then plug in, thanks guys @tom982 @jim_thompson5910

Answer 6

np