- Topic
- Related rates
- Difficulty
- Standard
- Suggested time
- 5 min
- Marks
- 3
Related rates practice question 9
A spherical balloon is inflated at a constant 100 cm³/s. How fast is its radius increasing at the instant the radius is 5 cm?
You know how fast the volume grows and want how fast the radius grows. What ties the two rates together at every instant?
Differentiate the volume formula with respect to time. The chain rule turns into .
You are given and want when .
An exact expression, or a decimal to two places.
Theory UBC Math 100 §5CLP-1 §3.2Key results
Worked solution
Strategy
Both the volume and the radius change with time. Differentiate the volume formula with respect to time to relate the two rates, then solve for the one you want.
Working
The relationship holds at every instant, so it can be differentiated in time.
Differentiate. The chain rule supplies the dr/dt factor.
Rearrange for the unknown rate before putting numbers in.
Substitute the given instant. The surface area 4πr² = 100π cancels the inflation rate neatly.
Answer
The answer is 1/π cm/s, approximately 0.318 cm/s.
Check
Adding 1 cm³, a hundredth of a second of inflation, to the 523.6 cm³ balloon takes r from 5 to about 5.00318 cm, a rate of 0.318 cm/s.
Where this goes wrong
Assuming a constant volume rate means a constant radius rate. It does not: as the balloon grows, the same 100 cm³/s spreads over a larger surface, so the radius grows more slowly.
Leans on
Chain rule in time, from The chain rule and logarithms
Saved in this browser only.