from a handpicked tutor in LIVE 1-to-1 classes

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

**Solution:**

Given, r = e^{2θ}, 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^{2θ}

dr/dθ = 2e^{2θ}

Substituting the values in the formula,

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

= \(\int_{0}^{2\pi }\sqrt{((e^{2\theta })^{2}+4(e^{2\theta })^{2})d\theta }\)

Taking out common term,

= \(\int_{0}^{2\pi }\sqrt{((e^{2\theta })^{2}(1+4)d\theta }\)

= \(\int_{0}^{2\pi }e^{2\theta }\sqrt{5}\:d\theta\)

= \(\sqrt{5}\int_{0}^{2\pi }e^{2\theta }\: d\theta\)

= \(\\\frac{\sqrt{5}}{2}\left [ e^{2\theta } \right ]_{0}^{2\pi } \\ \\\frac{\sqrt{5}}{2}\left [ e^{2(2\pi ) } -e^{2(0)}\right] \\ \\\frac{\sqrt{5}}{2}\left [ e^{4\pi } -e^{0}\right] \\ \\\frac{\sqrt{5}}{2}\left [ e^{4\pi } -1\right]\)

Therefore, the length of the polar curve is \(\frac{\sqrt{5}}{2}\left [ e^{4\pi } -1\right]\)

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

**Summary:**

The exact length of the polar curve r = e^{2θ}, 0 ≤ θ ≤ 2π is \(\frac{\sqrt{5}}{2}\left [ e^{4\pi } -1\right]\).

visual curriculum