# Water is being poured into a spherical bowl of radius 4cm at a rate of 2cm3/s. How fast is the

water level rising when it is at 2cm?

## dV/dt = 2 cm^3/s

dV/dh = surface area = pi r^2

dV/dt = dV/dh * dh/dt

so

dh/dt = dV/dt / pi r^2

so you need r, the radius of the surface 2 cm up from the bottom of the 4 cm radius sphere

I do not remember the length of a circle sector halfway from the center to the edge, look it up

## I'm getting that r is the square root of 12, so dh/dt is 1/6pi ?

## 2/12pi = 1/(6pi) agree

## thank you

## To find the rate at which the water level is rising, we need to use the concept of related rates.

Let's denote:

- V as the volume of water in the bowl at any time t (in cm^3).

- r as the radius of the water surface in the bowl at any time t (in cm).

- h as the height of the water in the bowl at any time t (in cm).

Given that the bowl is spherical and has a radius of 4 cm, we can use the volume formula for a sphere to relate the volume V and the radius r:

V = (4/3) * π * r^3

To find the rate at which the water level is rising, we need to find dh/dt, the derivative of the height of the water with respect to time.

Now, we know that the rate at which the water is being poured into the bowl is 2 cm^3/s, which means dV/dt = 2 cm^3/s.

We are given that at a certain time t, the water level is at 2 cm, so h = 2 cm. We need to find dh/dt.

To find the relationship between h and r, we can use similar triangles. Since the bowl is spherical, the relationship between h and r is given by:

h / (4 - r) = 2 / 4

h = 2(4 - r)

Now we can differentiate both sides of the equation with respect to time t:

dh / dt = 2 * dr / dt

To solve for dh/dt, we need to find dr/dt, which is the rate at which the radius is changing.

We can find the relationship between V and r by substituting the formula for V into the equation:

V = (4/3) * π * r^3

2 = (4/3) * π * r^3

Now, differentiate both sides of the equation with respect to t:

0 = 4πr^2 * dr/dt

Simplifying the equation, we get:

dr/dt = 0 / (4πr^2)

dr/dt = 0

Since dr/dt equals 0, this means that the rate at which the radius is changing is 0.

Substituting dr/dt = 0 into the equation dh / dt = 2 * dr / dt, we find:

dh / dt = 2 * 0

dh / dt = 0

Therefore, the water level is not rising when it is at 2 cm.