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# Find the exact length of the polar curve. r = e^{5θ}, 0 ≤ θ ≤ 2π

**Solution:**

Given, r = e^{5θ}, 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^{5θ}

dr/dθ = 5e^{5θ}

Substituting the values in the formula,

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

= \(\int_{0}^{2\pi }\sqrt{((e^{5\theta })^{2}+25(e^{5\theta })^{2})d\theta }\)

Taking out common term,

= \(\int_{0}^{2\pi }\sqrt{((e^{5\theta })^{2}(1+25)d\theta }\)

= \(\int_{0}^{2\pi }e^{5\theta }\sqrt{26}\:d\theta\)

= \(\sqrt{26}\int_{0}^{2\pi }e^{5\theta }\: d\theta\)

= \(\sqrt{26}\left [ 5e^{5\theta } \right ]_{0}^{2\pi }\)

= \(\sqrt{26}\times 5\left [ e^{5\theta } \right ]_{0}^{2\pi }\)

= \(\sqrt{26}\times 5\left [ e^{5(2\pi ) } -e^{5(0)}\right]\)

= \(\sqrt{26}\times 5\left [ e^{10\pi } -e^{0}\right]\)

We know, e^{0} = 1

= \(\sqrt{26}\times 5\left [ e^{10\pi } -1\right]\)

Therefore, the exact length of the polar curve is \(\sqrt{26}\times 5\left [ e^{10\pi } -1\right]\)

## Find the exact length of the polar curve. r = e^{5θ}, 0 ≤ θ ≤ 2π

**Summary:**

The exact length of the polar curve r = e5θ, 0 ≤ θ ≤ 2π is \(\sqrt{26}\times 5\left [ e^{10\pi } -1\right]\)

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