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Find the length of the spiral
r=theta for
0 <= theta <= 2.

Find the length of the spiral $r=\theta$ for $0 \leq \theta \leq 2$ .

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Find derivatives of trigonometric functions using the product and quotient rules

Full solution

Q. Find the length of the spiral

r=\theta

for

0 \leq \theta \leq 2

.

Calculate spiral length: Calculate the length of the spiral using the integral formula for polar coordinates: $L = \int \sqrt{r^2 + \left(\frac{dr}{d\theta}\right)^2} d\theta$ .
Find $\frac{dr}{d\theta}:$ First, find $\frac{dr}{d\theta}$ . Since $r=\theta$ , $\frac{dr}{d\theta} = \frac{d\theta}{d\theta} = 1$ .
Plug into integral formula: Now, plug $r=\theta$ and $\frac{dr}{d\theta}=1$ into the integral formula: $L = \int_{0}^{2} \sqrt{\theta^2 + 1^2} d\theta$ .
Simplify integral: Simplify the integral: $L = \int_{0}^{2} \sqrt{\theta^2 + 1} \, d\theta$ .
Use trigonometric substitution: Now we need to solve the integral. This is a bit tricky, but we can use a trigonometric substitution or look up the integral in a table.
Change limits of integral: Let's use a trigonometric substitution. Set $\theta = \tan(u)$ , then $d\theta = \sec^2(u) \, du$ .
Simplify integral with identity: When $\theta=0$ , $u=0$ , and when $\theta=2$ , $u=\arctan(2)$ . So we need to change the limits of the integral.
Complex integral solution: The integral becomes $L = \int_{0}^{\arctan(2)} \sqrt{\tan^2(u) + 1} \sec^2(u) \, du$ .
Final integral result: Simplify the integral using the identity $\tan^2(u) + 1 = \sec^2(u)$ : $L = \int_{0}^{\arctan(2)} \sec^3(u) \, du$ .
Final integral result: Simplify the integral using the identity $\tan^2(u) + 1 = \sec^2(u)$ : $L = \int_{0}^{\arctan(2)} \sec^3(u) \, du$ .This integral is still complex, but we can look it up or use a reduction formula. The result is $L = \left[\frac{1}{2} \sec(u) \tan(u) + \frac{1}{2} \ln|\sec(u) + \tan(u)|\right]_{0}^{\arctan(2)}$ .

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