A snowball of a radius r cm  is melting at a rate of 800 π cm^3 / s .  Find the rate of change of its radius  when the radius is 20 cm . Express the rate of change of its radius as −a/b cm/s where a is __and b is __

To solve this problem, we need to use the concept of related rates in calculus. Specifically, we need to relate the rate at which the volume of the snowball is changing to the rate at which its radius is changing.

The volume \( V \) of a sphere with radius \( r \) is given by:
\[ V = \frac{4}{3} \pi r^3 \]

We are given that the volume is decreasing at a rate of \( 800 \pi \) cm³/s. This means:
\[ \frac{dV}{dt} = -800 \pi \, \text{cm}^3/\text{s} \]

We need to find the rate at which the radius \( r \) is changing when \( r = 20 \) cm. Let's denote this rate by \( \frac{dr}{dt} \).

First, we differentiate the volume \( V \) with respect to time \( t \):
\[ \frac{dV}{dt} = \frac{d}{dt} \left( \frac{4}{3} \pi r^3 \right) \]

Using the chain rule:
\[ \frac{dV}{dt} = 4 \pi r^2 \frac{dr}{dt} \]

We can now plug in the given values:
\[ -800 \pi = 4 \pi (20)^2 \frac{dr}{dt} \]

Solving for \( \frac{dr}{dt} \):
\[ -800 \pi = 4 \pi \cdot 400 \frac{dr}{dt} \]
\[ -800 \pi = 1600 \pi \frac{dr}{dt} \]
\[ \frac{dr}{dt} = \frac{-800 \pi}{1600 \pi} \]
\[ \frac{dr}{dt} = -\frac{1}{2} \, \text{cm/s} \]

So, the rate of change of the radius when the radius is 20 cm is \( -\frac{1}{2} \) cm/s. This can be expressed as \( -\frac{a}{b} \) where \( a = 1 \) and \( b = 2 \).

Thus, \( a \) is \( 1 \) and \( b \) is \( 2 \).

Would you like more details or have any questions? Here are some related questions to explore further:

1. How does the rate of change of the radius vary as the radius decreases?
2. What is the volume of the snowball when the radius is 20 cm?
3. How long will it take for the snowball to completely melt?
4. What is the relationship between the rate of change of volume and the surface area of the sphere?
5. Can the rate of change of radius be positive? Why or why not?
6. How does the rate of change of radius compare to the rate of change of volume?
7. What would happen to the rate of change of the radius if the melting rate were doubled?
8. How would you set up and solve the related rates problem if the melting rate were given in cubic inches per second?

**Tip:** When dealing with related rates problems, always start by identifying the known rates and the rates you need to find, then relate them using the appropriate geometric formulas and differentiation.

Learn how to solve a related rates problem involving a melting snowball with a given rate of volume change. Find the rate of change of its radius when the radius is 20 cm. Discover the step-by-step calculus solution to express the rate of radius change as -1/2 cm/s.

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