Questions: Find the length of the spiral r=5 θ^2, 0 ≤ θ ≤ √32. The length of the spiral is . (Type an exact answer, using π as needed.)

Solution Steps

To find the length of the spiral given by \( r = 5 \theta^2 \) from \( \theta = 0 \) to \( \theta = \sqrt{32} \), we can use the formula for the length of a curve in polar coordinates: \[ L = \int_{a}^{b} \sqrt{ \left( \frac{dr}{d\theta} \right)^2 + r^2 } \, d\theta \] Here, \( r = 5 \theta^2 \) and \( \frac{dr}{d\theta} = 10 \theta \). We will integrate from \( \theta = 0 \) to \( \theta = \sqrt{32} \).

The spiral is defined by the equation \( r = 5 \theta^2 \). To find the length of the spiral, we first compute the derivative of \( r \) with respect to \( \theta \): \[ \frac{dr}{d\theta} = 10 \theta \]

The formula for the length \( L \) of a curve in polar coordinates is given by: \[ L = \int_{a}^{b} \sqrt{ \left( \frac{dr}{d\theta} \right)^2 + r^2 } \, d\theta \] Substituting \( r \) and \( \frac{dr}{d\theta} \) into the integral, we have: \[ L = \int_{0}^{\sqrt{32}} \sqrt{(10 \theta)^2 + (5 \theta^2)^2} \, d\theta \]

The integrand simplifies to: \[ \sqrt{100 \theta^2 + 25 \theta^4} = \sqrt{25 \theta^2 (4 + \theta^2)} = 5 \theta \sqrt{4 + \theta^2} \] Thus, the length integral becomes: \[ L = \int_{0}^{\sqrt{32}} 5 \theta \sqrt{4 + \theta^2} \, d\theta \]

After evaluating the integral, we find: \[ L = \frac{1040}{3} \]

Final Answer