- Topic
- Related rates
- Difficulty
- Introductory
- Suggested time
- 3 min
- Marks
- 2
Related rates practice question 4
For one species of fish, the mass in grams of a fish centimetres long is modelled by the relation below. A fish is 30 cm long and its length is growing at 0.2 cm per day. How fast is its mass increasing?
Give the answer in grams per day.
As the fish lengthens, its mass grows too. How does a rate in centimetres per day become a rate in grams per day?
Both and change with time. Differentiate with respect to ; the chain rule brings in .
A number, as a decimal or a fraction.
Theory UBC Math 100 §5CLP-1 §3.2Key results
Worked solution
Strategy
Differentiate the relation with respect to time while L is still a variable, then substitute the length and its growth rate at this instant.
Working
Chain rule: L depends on t.
Substitute L = 30 and dL/dt = 0.2 only now.
A check: in percentage terms the mass grows three times as fast as the length, and 2 % of 270 g is 5.4 g.
Answer
The answer is 5.4 grams per day.
Check
A day later the fish is 30.2 cm long and M = 0.01(30.2)³ ≈ 275.4 g, about 5.4 g more than the 270 g it weighs now.
Where this goes wrong
Substituting L = 30 before differentiating. M = 270 is then a constant and its derivative is 0; the relation has to be differentiated while L is still a variable.
Leans on
The chain rule, from The chain rule and logarithms
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