- Topic
- Linear approximation
- Difficulty
- Challenging
- Suggested time
- 6 min
- Marks
- 3
Linear approximation practice question 12
The concentration of a drug in the blood, hours after a dose, is modelled by mg/L. Find the linear approximation of at .
Both and change with . Which rules will you need to differentiate their product, and at which point is the tangent line taken?
. At the exponential is ; keep it exact.
Product rule, with the chain rule on the exponential: .
A formula in t, with exact coefficients. Use ^ for powers and brackets for every input: sin(2x), e^(x^2), log(x+1). Give exact values: 1/6, not 0.167.
Theory UBC Math 100 §9.2CLP-1 §3.4Key results
Worked solution
Strategy
The linear approximation needs C(2) and C′(2). C is a product of two functions of t, so its derivative takes the product rule, with the chain rule on the exponential.
Working
The concentration at 2 hours, about 9.7 mg/L.
Product rule; the chain rule gives the factor −¼ from the exponent.
Still rising at 2 hours, at about 2.4 mg/L per hour.
The value at 2 plus the slope times the step from 2. It can also be written 4(t + 2)/√e.
Answer
The answer is L of t equals 16 e to the minus one half, plus 4 e to the minus one half times (t minus 2); that is, 4 times (t plus 2) over root e.
Check
At t = 2.1 the line gives about 9.95 mg/L and C itself about 9.94 mg/L.
Where this goes wrong
Differentiating e−t/4 as e−t/4, without the inner derivative −¼. The slope at 2 then comes out as 24e−1/2 instead of 4e−1/2, six times too steep.
Leans on
The product rule, from Differentiation rules
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