Question

when the depth of liquid in a container is x cm, the volume of the liquid is x(x^2+25) cm^3. Liquid is added to the container at a constant rate of 2 cm^3/s. Find the rate of change of the depth of the liquid at the instant when x=1

Answer 1

sorry, got to know the shape of the container.

Answer 2

mr Steve, thus how i was given the question, jst like that without the shape of the container. Pliz hlp

Answer 3

hmmm. well, let's see. Ah, I see they gave the formula for the volume, so we don't have to derive it from the shape of the container.

v = x(x^2+25) = x^3 + 25x^2
dv/dt = (3x^2 + 50x) dx/dt
so,
2 = (3+50) dx/dt
dx/dt = 2/53

Answer 4

A little late but here is the answer:

You know that the rate of change of the volume is dV/dt = 2 (given in the question).
You also know that V = x^3 + 25x

To find the rate of change of the depth of the liquid, we find

dx/dt = dx/dV x dV/dt

dV/dx = 3x^2 + 25, so dx/dV is 1 over that

so dx/dt = 2 / (3x^2 + 25)

Pluggin in x = 11 into the equation, we get

dx/dt = 0.00515 cm/s