We pour water into a conical container of a radius of 10 cm with a height of 36cm at a rate of 70 cm^3/s. At what velocity does the water level increase when it reaches 12 cm?

dh/dt=?

To find the velocity at which the water level is increasing, we need to find the derivative of the height with respect to time (dh/dt).

Let's denote the radius of the conical container as "r" (which is 10 cm in this case). The volume of a cone can be expressed as V = (1/3) * π * r^2 * h, where V is the volume, r is the radius, and h is the height.

We are given the rate of change of volume (dV/dt) as 70 cm^3/s. Since the volume of the cone is given by V = (1/3) * π * r^2 * h, we can find the derivative of the volume with respect to time (dV/dt) by differentiating the equation:

dV/dt = (1/3) * π * r^2 * (dh/dt)

Now, let's solve for (dh/dt), which represents the velocity at which the height is increasing:

(dh/dt) = (3 * dV/dt) / (π * r^2)

Substituting the given values, we have:

(dh/dt) = (3 * 70) / (π * 10^2)

Simplifying further:

(dh/dt) = 210 / (100 * π)

Now we can calculate the value:

(dh/dt) ≈ 2.12 cm/s

Therefore, the water level is increasing at a velocity of approximately 2.12 cm/s when it reaches 12 cm.