Questions: Find the length of the spiral r=5 theta^2, 0 ≤ theta ≤ sqrt(5)

Solution Steps

To find the length of the spiral given by \( r = 5 \theta^2 \) for \( 0 \leq \theta \leq \sqrt{5} \), we can use the formula for the length of a curve in polar coordinates. The formula is:

\[ L = \int_{a}^{b} \sqrt{ \left( \frac{dr}{d\theta} \right)^2 + r^2 } \, d\theta \]

Here, \( r = 5 \theta^2 \) and \( \frac{dr}{d\theta} = 10 \theta \). We will integrate from \( \theta = 0 \) to \( \theta = \sqrt{5} \).

Given the spiral \( r = 5 \theta^2 \), we first compute its derivative with respect to \( \theta \): \[ \frac{dr}{d\theta} = 10 \theta \]

The formula for the length \( L \) of a curve in polar coordinates is: \[ L = \int_{a}^{b} \sqrt{ \left( \frac{dr}{d\theta} \right)^2 + r^2 } \, d\theta \] Substituting \( r = 5 \theta^2 \) and \( \frac{dr}{d\theta} = 10 \theta \), we get: \[ L = \int_{0}^{\sqrt{5}} \sqrt{(10 \theta)^2 + (5 \theta^2)^2} \, d\theta \] \[ L = \int_{0}^{\sqrt{5}} \sqrt{100 \theta^2 + 25 \theta^4} \, d\theta \] \[ L = \int_{0}^{\sqrt{5}} \sqrt{25 \theta^2 (4 + \theta^2)} \, d\theta \] \[ L = \int_{0}^{\sqrt{5}} 5 \theta \sqrt{4 + \theta^2} \, d\theta \]

To find the length, we evaluate the integral: \[ L = 5 \int_{0}^{\sqrt{5}} \theta \sqrt{4 + \theta^2} \, d\theta \] The result of this integral is: \[ L = \frac{95}{3} \]

Final Answer