Find the exact length of the polar curve. r = e^3θ, 0 ≤ θ ≤ 2π

Solution:

Given, r = e^3θ, 0 ≤ θ ≤ 2π

We have to find the exact length of the polar curve.

The length of the polar curve is given by \(\int_{a}^{b}\sqrt{(r^{2}+(\frac{dr}{d\theta })^{2})d\theta }\)

We know, r = e^3θ

dr/dθ = 3e^3θ

Substituting the values in the formula,

= \(\int_{0}^{2\pi }\sqrt{((e^{3\theta })^{2}+(3e^{3\theta })^{2})d\theta }\)

= \(\int_{0}^{2\pi }\sqrt{((e^{3\theta })^{2}+9(e^{3\theta })^{2})d\theta }\)

Taking out common term,

= \(\int_{0}^{2\pi }\sqrt{((e^{3\theta })^{2}(1+9)d\theta }\)

= \(\int_{0}^{2\pi }e^{3\theta }\sqrt{10}\:d\theta\)

= \(\sqrt{10}\int_{0}^{2\pi }e^{3\theta }\: d\theta\)

\(\\=\sqrt{10}\left [ \frac{e^{3\theta }}{3} \right ]_{0}^{2\pi } \\ \\=\frac{\sqrt{10}}{3}\left [ e^{3\theta } \right ]_{0}^{2\pi } \\ \\=\frac{\sqrt{10}}{3}\left [ e^{3(2\pi ) } -e^{2(0)}\right] \\ \\=\frac{\sqrt{10}}{3}\left [ e^{6\pi } -e^{0}\right]\)

We know, e⁰ = 1

\(=\frac{\sqrt{10}}{3}\left [ e^{6\pi } -1\right]\)

Therefore, the exact length of the polar curve is \(\frac{\sqrt{10}}{3}\left [ e^{6\pi } -1\right]\)