Find the exact length of the polar curve: $r = e^{2\theta}, \quad 0 \leq \theta \leq 2\pi$

Neetesh Kumar | January 2, 2025

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This is the solution to Math 1c
Assignment: 10.4 Question Number 18
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Step-by-step solution:

The length of a polar curve is given by the formula:

$L = \int_{\theta_1}^{\theta_2} \sqrt{r^2 + \left(\frac{dr}{d\theta}\right)^2} \, d\theta.$

Here, $r = e^{2\theta}$ and $\theta \in [0, 2\pi]$.

1. Compute $\frac{dr}{d\theta}$:

Differentiate $r = e^{2\theta}$ with respect to $\theta$:

$\frac{dr}{d\theta} = 2e^{2\theta}.$

2. Substitute into the Length Formula:

Using $r = e^{2\theta}$ and $\frac{dr}{d\theta} = 2e^{2\theta}$, the formula becomes:

$L = \int_{0}^{2\pi} \sqrt{\left(e^{2\theta}\right)^2 + \left(2e^{2\theta}\right)^2} \, d\theta.$

Simplify the terms inside the square root:

$L = \int_{0}^{2\pi} \sqrt{e^{4\theta} + 4e^{4\theta}} \, d\theta.$

Factor out $e^{4\theta}$:

$L = \int_{0}^{2\pi} \sqrt{5e^{4\theta}} \, d\theta.$

Simplify further:

$L = \int_{0}^{2\pi} \sqrt{5} e^{2\theta} \, d\theta.$

3. Evaluate the Integral:

Factor out $\sqrt{5}$:

$L = \sqrt{5} \int_{0}^{2\pi} e^{2\theta} \, d\theta.$

The integral of $e^{2\theta}$ is:

$\int e^{2\theta} \, d\theta = \frac{e^{2\theta}}{2}.$

Apply the bounds $\theta = 0$ to $\theta = 2\pi$:

$L = \sqrt{5} \left[ \frac{e^{2\theta}}{2} \right]_{0}^{2\pi}.$

Substitute the limits:

$L = \sqrt{5} \left( \frac{e^{4\pi}}{2} - \frac{e^{0}}{2} \right).$

Simplify:

$L = \sqrt{5} \cdot \frac{e^{4\pi} - 1}{2}.$

Final Answer:

The exact length of the polar curve is:

$L = \boxed{\frac{\sqrt{5}}{2} \left(e^{4\pi} - 1\right)}$

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