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

Answer 5

[AI-generated answer — Explain Bot]

To find the rate of change of the depth of the liquid, we need to differentiate the equation that relates the depth (x) of the liquid to its volume.

Given: The volume of the liquid, V, is x(x^2+25) cm^3.

Differentiating both sides of this equation with respect to time (t), we get:

dV/dt = d/dt (x(x^2+25))

Now, let's differentiate each term separately using the product rule:

dV/dt = x d/dt (x^2+25) + (x^2+25) d/dt (x)

The derivative of x^2 with respect to time is 2x(dx/dt).
The derivative of 25 with respect to time is 0 since it is a constant.

dV/dt = x * 2x(dx/dt) + (x^2+25) dx/dt

Now, let's substitute dx/dt = 2 cm^3/s (because the liquid is added at a constant rate of 2 cm^3/s):

dV/dt = x * 2x(2) + (x^2+25) * 2

Simplifying further, we have:

dV/dt = 4x^2 + 2(x^2 + 25)

Next, we need to find the rate of change of the depth of the liquid at the instant when x=1. To do this, we substitute x=1 into our expression for dV/dt:

dV/dt = 4(1)^2 + 2((1)^2 + 25)

dV/dt = 4 + 2(1 + 25)

dV/dt = 4 + 2 + 50

dV/dt = 56 cm^2/s

Therefore, the rate of change of the depth of the liquid at the instant when x=1 is 56 cm^2/s.