Rates of change / Differentiation (1 Viewer)

[skillz]

Hi guys. Havent posted in ages so i thought i'd add a quick q.

When the depth of liquid in a container is x cm the volume is x(x^2+36)cm ^3. Liquid is added to the container at a rate of 3cm^3/s. Find the rate of change of the depth of liquid at the instant when x=11.

I don't know how to actually set up the information given.

cheers

 

[bboyelement]

V=x(x^2+36)
dV/dt=3
and we are finding dx/dt

V = x^3+36x
dV/dx = 3x^2+36

dx/dt = dx/dV * dV/dt (so we would have to invert dV/dx)

= 1/(3x^2+36) *3

and so we sub x=11 into 3/(3x^2+36)

and so we get dx/dt = 1/133 cm/s

btw. next time post in ext 1