Water runs into a conical tank with radius 5 and altitude 10 at a constant rate of 2 cu.ft. per min. How fast is the water rising in the tank at any instant?

You want dh/dt, given dV/dt.

so write an equation relating V to h.
V= 1/3 Areatop*h, in this case the area of the top of the water is PI*radius^2 where the radius is 1/2 h.

To find the equation relating the volume (V) of the water in the tank to the height (h), we can use the formula for the volume of a cone:

V = (1/3) * A * h

Where V is the volume, A is the base area of the cone, and h is the height.

In this case, the base area of the cone is a circle with radius 5. However, the radius is not fixed in the context of the problem; it varies with the height of the water. We can express the radius in terms of the height using the given information that the altitude (the distance from the vertex to the base) is 10 units.

Since the altitude of the cone is equivalent to the height of the water, and the radius of the cone is half the altitude, we can write:

radius = (1/2) * h

Substituting this into the formula for the base area A, we get:

A = π * (radius^2) = π * (1/2 * h)^2 = π * (1/4) * h^2 = (π/4) * h^2

Now, we can substitute the expression for A into the equation for the volume V:

V = (1/3) * A * h = (1/3) * (π/4) * h^2 * h = (π/12) * h^3

Now, we want to find dh/dt, the rate at which the height h is changing with respect to time. We are given dV/dt, the rate at which the volume V is changing with respect to time, which is 2 cubic feet per minute.

To find dh/dt, we can differentiate the equation for volume V with respect to time t:

dV/dt = (π/12) * (h^3)

Now we can solve for dh/dt by isolating it:

dh/dt = (dV/dt) / ((π/12) * (h^2))

Finally, we can plug in the given values and solve for dh/dt. Assuming the water is rising, we can take the positive value of dh/dt.

Given: dV/dt = 2 cubic ft/min

We need to know the value of h at the instant we want to find dh/dt. Let's say we want to find it when h = h0. Plugging these values into the equation, we get:

dh/dt = (2) / ((π/12) * (h0^2))

So, the rate at which the water is rising in the tank at any instant when the height of the water is h0 is (2) / ((π/12) * (h0^2)) cubic feet per minute.