Q. 12. - Calculus

Question

What is the formula for computing the arc length of a polar curve $r=f\left(\theta \right)$ where $\theta \in \left[\alpha ,\beta \right]$ What conditions on the polar function$f\left(\theta \right)$ are necessary for this formula to hold?

Step-by-Step Solution

Answer

The formula for the arc length of the polar graph in the interval $\left[\alpha ,\beta \right]$ is:

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

The required conditions for the polar function $f\left(\theta \right)$are necessary for this formula to hold: ${\int }_a^{\beta }\sqrt{{\left(f^{\text{'}}\left(\theta \right)\right)}^2+\left(f\right(\theta ){)}^2}d\theta$

Step 1: Find the formula for the arc length of the polar graph in the interval [ α , β ]  

The arc length is the distance measured along the curved line that makes up the arc.

A polar curve's arc length is defined as follows.

Let $r=f\left(\theta \right)$is any differentiable function of $\theta$ such that $f^{\text{'}}\left(\theta \right)$ is continuous for all $\theta \in [\alpha ,\beta ]$

Moreover, consider that $r=f$is a one-one function from $[\alpha ,\beta ]$ to the function graph. then in the interval $[\alpha ,\beta ]$ the arc length of the polar graph,

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

Step 2: The objective is to find What conditions on the polar function r = f ( θ )   are necessary for this formula to hold?

To determine the length of the arc for the curve $r=f\left(\theta \right)$the function $f\left(\theta \right)$must be differentiable and one to one.

The derivative $f\left(\theta \right)$ of  that is $f^{\text{'}}\left(\theta \right)$is continuous for all $\theta \in [\alpha ,\beta ]$.

As a result, for this formula to work, $f\left(\theta \right)$ must be differentiable, continuous, and one to one.

Therefore the answer is:

${\int }_a^{\beta }\sqrt{{\left(f^{\text{'}}\left(\theta \right)\right)}^2+\left(f\right(\theta ){)}^2}d\theta$