You are eating soup from a hemispherical bowl of radius 6cm at a constant rate of 40ml/min. Find the rate at which the soup level is going down in depth of the soup is 2cm.

The surface area of the soup times its speed down = 40 ml/min

By the way on ml is 1 cubic centimeter so no conversion is required
Just find surface area from pi r^2
where r is the radius of the surface when 2 cm depth is reached.

3.1831

To find the rate at which the soup level is going down, we need to use the concept of related rates. Here's how we can approach this problem step by step:

Step 1: Visualize the problem
Imagine a hemispherical bowl with a radius of 6cm, filled with soup. The soup level is going down at a constant rate of 40ml/min. We want to find the rate at which the soup level is going down when the depth of the soup is 2cm.

Step 2: Identify known and unknown quantities
Known quantities:
- Radius of the bowl (r) = 6cm
- Rate at which the soup is being consumed (ds/dt) = 40ml/min
- Depth of the soup (h) = 2cm

Unknown quantity:
- Rate at which the soup level is going down (dh/dt)

Step 3: Formulate the equation
We need to relate the known and unknown quantities using an equation. The volume of the soup in the bowl can be calculated using the formula for the volume of a hemisphere:

V = (2/3) * π * r^3

Since we are given the rate of change of the volume (ds/dt), we can differentiate both sides of the equation with respect to time (t):

dV/dt = (2/3) * π * (3r^2) * dr/dt

This equation relates the rates of change of the volume, radius, and time.

Step 4: Substitute known values
Now, let's substitute the known values into the equation:
ds/dt = 40ml/min (Given)
r = 6cm (Given)

dV/dt = (2/3) * π * (3 * 6^2) * dr/dt

Step 5: Solve for the unknown quantity
Next, we need to solve for the rate at which the soup level is going down (dh/dt). Since the soup is being consumed uniformly and the shape of the bowl is a hemisphere, the volume of the soup is directly proportional to the depth of the soup.

V = (π/6) * h^3

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

dV/dt = (π/6) * (3h^2) * dh/dt

Since we are interested in finding dh/dt, let's isolate it:

dh/dt = (6/π) * (dV/dt) * (1/h^2)

Step 6: Substitute known values and calculate
Substituting the known values for dV/dt and h:

dh/dt = (6/π) * (40ml/min) * (1/2^2)

Simplifying the equation:

dh/dt = 240/π ml/min ≈ 76.394cm^3/min

Therefore, the rate at which the soup level is going down when the depth of the soup is 2cm is approximately 76.394cm^3/min.