Slope fields and Euler’s method, question 13

Topic: Slope fields and Euler’s method
Difficulty: Capstone
Suggested time: 10 min
Marks: 5
Theory: UBC Math 100 §12.3Key results

A ball is dropped from rest and falls with air resistance proportional to its speed. Its downward velocity $v(t)$ , in m/s after $t$ seconds, satisfies

\frac{dv}{dt} = 10 - 0.5v, \qquad v(0) = 0.

Check that $v(t) = 20 + Ce^{-t/2}$ solves the equation for every constant $C$ , and choose $C$ to match the initial condition. Then estimate $v(2)$ with two Euler steps of $\Delta t = 1$ . Give the Euler estimate minus the true value $v(2)$ : positive if Euler overestimates, negative if it underestimates.

Your answer

m/s

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

How sure are you?

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Theory UBC Math 100 §12.3Key results

Worked solution

Strategy

Verify the family by substitution, let the initial condition choose C, take two Euler steps, and compare at t = 2. Concavity predicts the sign of the difference before any arithmetic.

Working

$v' = -\tfrac{1}{2}Ce^{-t/2}, \qquad 10 - 0.5\bigl(20 + Ce^{-t/2}\bigr) = -\tfrac{1}{2}Ce^{-t/2} \quad\checkmark$
Both sides agree for every C.
$v(0) = 20 + C = 0 \implies C = -20, \qquad v(t) = 20 - 20e^{-t/2}$
The ball starts from rest and approaches 20 m/s.
$v_{1} = 0 + 1\cdot(10 - 0) = 10, \qquad v_{2} = 10 + 1\cdot(10 - 5) = 15$
Each step uses the slope at the point it starts from.
$v(2) = 20 - 20e^{-1} = 20 - \frac{20}{e} \approx 12.64$
The true value from the formula.
$v_{2} - v(2) = 15 - 20 + \frac{20}{e} = \frac{20}{e} - 5 \approx 2.36$
Positive, as concavity predicts: v″ = −0.5v′ < 0 while the ball speeds up, so Euler’s tangent steps run above the curve.

Answer

v_{2} - v(2) = \frac{20}{e} - 5 \approx 2.36 \text{ m/s}

Check

Ten Euler steps of 0.2 give about 13.03 m/s, closer to the true 12.64 m/s but still above it, so the error is positive and shrinks with the step.

Where this goes wrong

Reporting 5 − 20/e, the true value minus the estimate. The sign carries the conclusion: the solution is concave down, so Euler’s estimate is too high and the difference asked for is positive.

Leans on

Equations of the form y′ = a − by, from Exponential growth and decay

Try a similar question

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

My working

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Slope fields and Euler’s method practice question 13