- Topic
- Linear approximation
- Difficulty
- Standard
- Suggested time
- 4 min
- Marks
- 2
Linear approximation practice question 8
At a distance of metres from a small lamp, the illuminance is lux. A reader sitting 3 m from the lamp moves back to 3.1 m. Use a linear approximation to estimate the change in illuminance.
Give the change in lux, with its sign.
The question asks for a change in illuminance, not a new value. How does the slope at turn a small step in distance into a change in output?
: the slope at the start times the change in distance.
Write , so .
A fraction, or a decimal to two decimal places.
Theory UBC Math 100 §9.2CLP-1 §3.4Key results
Worked solution
Strategy
Near d = 3 the graph is close to its tangent line, so a small step in distance changes the illuminance by about the slope at 3 times the step. Moving away from the lamp makes it darker, so expect a negative answer.
Working
Power rule on 900d−2.
The rate of change at 3 m, in lux per metre.
Slope times the step of 0.1 m. The exact change is about −6.35 lux, so the estimate is close.
Answer
The answer is about negative 20 over 3 lux, a drop of about 6.67 lux.
Check
I(3) = 100 lux and I(3.1) = 900/9.61 ≈ 93.65 lux, a real drop of about 6.35 lux.
Where this goes wrong
Giving the change as positive. The light dims with distance, so the derivative is negative and the change is a drop of about 6.67 lux.
Leans on
The power rule for negative powers, from Differentiation rules
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