An area minimum constrained by the trace

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Before you start

Areas in polar coordinates. Read the relevant method, then decide which parts you can solve without hints.

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A family of polar curves has $r=k+\frac{2}{k}\cos\theta$ , where $k>0,\quad k^2\geq2$ .

Part a

Show that the enclosed area is π[k² + $c^{2}$ /(2k²)], with c the stated numerator.

Part b

Find $k^{2}$ at the minimum permitted area.

Part c

Give the minimum area as a coefficient of π.

Model solution and review criteria

Part a

For $k^{2}$ ≥ 2, r ≥ 0 and the curve traces one closed region. Expand $r^{2}$ and integrate over 0 ≤ θ ≤ 2π. The cosine term integrates to 0 and cos² to π, giving π[k² + 4/(2k²)].

Check the nonnegative-radius condition.
Expand and integrate with the factor ½.

Part b

Writing t = $k^{2}$ ≥ 2, area/π = t + 4/(2t). Its derivative is 1 − 4/(2t²) ≥ $\frac{1}{2}$ on this domain, so the minimum is at t = 2.

Use the domain constraint.
Reject the unconstrained stationary point outside it.

Part c

At $k^{2}$ = 2, area/π = 2 + $\frac{2}{2}$ = 3.

Substitute the permitted boundary value.

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