Newton’s method, question 6

Topic: Newton’s method
Difficulty: Standard
Suggested time: 5 min
Marks: 3
Theory: UBC Math 100 §10.1Key results

After a dose, the concentration of a drug in the blood is $C(t) = 5t\,e^{-0.4t}$ mg/L, where $t$ is in hours. The drug takes effect once the concentration reaches 3 mg/L. Starting from $t_{0} = 1$ , use one step of Newton’s method on $f(t) = C(t) - 3$ to estimate the time at which this first happens.

Your answer

An exact expression, or a decimal to four decimal places.

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Theory UBC Math 100 §10.1Key results

Worked solution

Strategy

Rewrite “C reaches 3” as a zero of f(t) = C(t) − 3, differentiate with the product rule, and take one step from t₀ = 1. Leave e^0.4 as it is: the answer is calculator-ready.

Working

$f(t) = 5t\,e^{-0.4t} - 3$
Move the target to the left side, so the time wanted is a zero of f.
$f'(t) = 5e^{-0.4t} + 5t\cdot\left(-0.4e^{-0.4t}\right) = e^{-0.4t}(5 - 2t)$
Product rule, with the chain rule on the exponential. The constant 3 differentiates to 0.
$f(1) = 5e^{-0.4} - 3, \qquad f'(1) = 3e^{-0.4}$
Value and slope at t₀ = 1.
$t_{1} = 1 - \frac{5e^{-0.4} - 3}{3e^{-0.4}} = 1 - \frac{5}{3} + e^{0.4} = e^{0.4} - \frac{2}{3}$
Split the fraction: 3 divided by 3e^−0.4 is e^0.4.
$t_{1} = e^{0.4} - \tfrac{2}{3} \approx 0.825\ \text{h}$
About 50 minutes. The true crossing time is 0.839 h, so one step lands within a minute of it.

Answer

t_{1} = e^{0.4} - \frac{2}{3} \approx 0.825\ \text{h}

Check

C(0.825) ≈ 2.97 mg/L, just short of 3, and C is still rising there, so the true time is only a little later.

Where this goes wrong

Applying Newton’s method to C(t) itself instead of C(t) − 3, which gives t₁ = −2/3, a time before the dose. Newton’s method finds zeros, so the target value has to be moved to the left side first.