Newton’s method, question 11

Topic: Newton’s method
Difficulty: Challenging
Suggested time: 8 min
Marks: 4
Theory: UBC Math 100 §10.1Key results

A rectangle stands on the $x$ -axis with its top corners on the curve $y = \cos x$ , at $(-x, \cos x)$ and $(x, \cos x)$ for some $0 < x < \tfrac{\pi}{2}$ . Its area $A(x)$ is greatest where $A'(x) = 0$ , an equation with no solution in closed form. Apply one step of Newton’s method to the function $A'$ itself, starting from $x_{0} = \tfrac{\pi}{4}$ , to estimate the $x$ that gives the greatest area.

Your answer

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

How sure are you?

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Theory UBC Math 100 §10.1Key results

Worked solution

Strategy

Build the area function, then treat A′ as the function whose zero Newton’s method hunts: its slope is A″. Evaluate both at π/4, where sine and cosine are exact, and take one step.

Working

$A(x) = 2x\cos x$
Width 2x, height cos x.
$A'(x) = 2\cos x - 2x\sin x, \qquad A''(x) = -4\sin x - 2x\cos x$
The product rule, twice: differentiating 2x sin x gives 2sin x + 2x cos x.
$A'\!\left(\tfrac{\pi}{4}\right) = \sqrt{2}\left(1 - \tfrac{\pi}{4}\right), \qquad A''\!\left(\tfrac{\pi}{4}\right) = -\sqrt{2}\left(2 + \tfrac{\pi}{4}\right)$
sin(π/4) = cos(π/4) = √2/2.
$x_{1} = \frac{\pi}{4} - \frac{\sqrt{2}\left(1 - \frac{\pi}{4}\right)}{-\sqrt{2}\left(2 + \frac{\pi}{4}\right)} = \frac{\pi}{4} + \frac{4 - \pi}{8 + \pi} \approx 0.8624$
The √2 cancels; multiply top and bottom by 4.

Answer

x_{1} = \frac{\pi}{4} + \frac{4 - \pi}{8 + \pi} \approx 0.8624

Check

A′(0.8624) ≈ −0.0086, much closer to 0 than A′(π/4) ≈ 0.30, and the exact maximiser is 0.8603, just left of the estimate.

Where this goes wrong

Taking the step with the derivative of A instead of A″, or dropping one of the two sin x terms when differentiating A′. Newton’s method on A′ divides A′ by its own slope, which is A″.