Slope fields and Euler’s method, question 12: UBC MATH 100 practice

Question 12 of 15Challenging3 marks · 6 minTheory UBC Math 100 §12.3Key results

After an injection, a drug is removed by an enzyme that saturates, so its concentration $C(t)$ , in mg/L after $t$ hours, satisfies

\frac{dC}{dt} = -\frac{2C}{1 + C}, \qquad C(0) = 3.

Euler’s method with $\Delta t = 0.5$ is used to estimate $C(1)$ . Without computing the steps, decide whether the estimate is above or below the true value, and why.

Below the true value: C″ = 4C/(1 + C)³ > 0, so the solution is concave up and lies above the tangent lines that Euler’s method follows.
Above the true value: C′ < 0, so the concentration is falling, and a falling curve lies below its tangent lines.
Above the true value: C″ > 0, so the solution is concave up and lies below its tangent lines.
Below the true value: C′ < 0, so the concentration is falling, and Euler steps always undershoot a falling curve.

How sure are you?

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Theory UBC Math 100 §12.3Key results

Worked solution

Strategy

Each Euler step runs along a tangent line, and a concave-up curve lies above its tangents, so the estimate runs low when the solution is concave up. Find the sign of C″ from the equation itself, with the chain rule.

Working

$f(C) = -\frac{2C}{1 + C}, \qquad f'(C) = -\frac{2(1 + C) - 2C}{(1 + C)^{2}} = -\frac{2}{(1 + C)^{2}}$
Quotient rule.
$C'' = f'(C)\,C' = \left(-\frac{2}{(1 + C)^{2}}\right)\left(-\frac{2C}{1 + C}\right) = \frac{4C}{(1 + C)^{3}} > 0$
Chain rule, since C depends on t. The concentration falls towards 0 but stays positive, so C″ > 0 throughout.
A concave-up curve lies above its tangent lines, so the first Euler step lands below the true curve. Each later step starts lower and again runs along a tangent that lies below the solution through its starting point, so the estimate of C(1) is too low.

Answer

\text{Below: } C'' = \frac{4C}{(1 + C)^{3}} > 0, \text{ so Euler underestimates.}

Check

Two Euler steps give 3, then 2.25, then about 1.56, and the true C(1) ≈ 1.62 is higher, as the concavity predicts.

Where this goes wrong

Deciding from the sign of C′. Whether the curve rises or falls says nothing about which side of its tangent lines it lies on; that is concavity, the sign of C″.