Question

Fill in the blanks to complete each of the following theorem statements:

Let $r=f\left(\theta \right)$be a differentiable function of $\theta$such that $f\text{'}\left(\theta \right)$is continuous for all $\theta \in \left[\alpha ,\beta \right]$Furthermore, assume that $r=f\left(\theta \right)$is a one-to-one function from $\left[\alpha ,\beta \right]$to the graph of the function. Then the length of the polar graph of $r=f\left(\theta \right)$on the interval $\left[\alpha ,\beta \right]$is..............

Short answer

The blank is filled by${\int }_{\alpha }^{\beta }\sqrt{{\left(f\text{'}\left(\theta \right)\right)}^2+{\left(f\left(\theta \right)\right)}^2}d\theta$

Step-by-step solution

Step 1. Given information

$r=f\left(\theta \right)$

$\alpha \le \theta \le \beta$

Step 2. Write formula of length of the arc.

As we know that when x and y are functions of the parameter θ, the arc length of the curve on the interval [α, β] is

$l={\int }_{\alpha }^{\beta }\sqrt{{\left(\frac{dx}{d\theta }\right)}^2+{\left(\frac{dy}{d\theta }\right)}^2}d\theta$

By combining this integral with the parametric equations for the curve,

$$\begin{gathered} x=r\cos \theta \\ x=f\left(\theta \right)\cos \theta \end{gathered}$$and $y=f\left(\theta \right)\sin \theta$

we have an arc length

$$\begin{gathered} l={\int }_{\alpha }^{\beta }\sqrt{{\left(\frac{d}{d\theta }\left(f(\theta \right)\cos \theta )\right)}^2+{\left(\frac{d}{d\theta }\left(f(\theta \right)\sin \theta )\right)}^2}d\theta \\ ={\int }_{\alpha }^{\beta }\sqrt{{\left(-f\left(\theta \right)\sin \theta +f\text{'}\left(\theta \right)\cos \theta \right)}^2+{\left(f\text{'}\left(\theta \right)\sin \theta +f\left(\theta \right)\cos \theta \right)}^2}d\theta \end{gathered}$$

After expanding the terms under the radical and simplifying , we get:

$l={\int }_{\alpha }^{\beta }\sqrt{{\left(f\text{'}\left(\theta \right)\right)}^2+{\left(f\left(\theta \right)\right)}^2}d\theta$