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
4 answers
sorry, got to know the shape of the container.
mr Steve, thus how i was given the question, jst like that without the shape of the container. Pliz hlp
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
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
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
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