Confused on computing arc length

Question 1:

Compute the arc length of $r = 1 + \cos(\theta),\; 0 \leq \theta \leq 2 \pi$


$\int_{0}^{2 \pi} \sqrt{(1+\cos(\theta)^2 + (-\sin (\theta))^2}\, d \theta = 8 $

Question 2:

Determine the length of $y = \ln(\sec(x))$ between $0 \leq x \leq \pi / 4$


$\sqrt{1+\left(\frac{dy}{dx}\right)^2} = \sqrt{1+\tan^2(x)} = \sec(x)$

$L = \int_{0}^{\pi/4} \sec(x)\,dx = \ln(\sqrt{2} + 1)$

Why is there two ways to do it, yet yielding different answers? The first way got the derivative and took the magnitude of both of it, whereas the second way used the formula

$\displaystyle\int_{C} \sqrt{1+\left(\frac{dy}{dx}\right)^2}\, dx$

I tried using the above formula for the first one but got a different answer.


2 Answers 2


You are using two different formulas, which apply to two different situations:

  • In the first situation, you are computing the length of a curve wich is expressed in polar coordinate, $r$ and $\theta$. The formula for that is$$\int_a^b\sqrt{r^2(\theta)+\left(\frac{dr}{d\theta}\right)^2}\,\mathrm d\theta.$$
  • In the second situation, you are computing the length of the graph of a function $f$. The formula for that is$$\int_a^b\sqrt{1+\bigl(f'(t)\bigr)^2}\,\mathrm dt.$$

Since you have two different (although similar) situations, it is natural that you apply two different (although similar) formulas.


The applicability of formula depends upon being able to express the parametric curve in the form required by the formula.

  • In second case, since parameter curve is given in cartesian coordinates, corresponding formula for arc length, $ \displaystyle \int_{C} \sqrt{1+\left(\frac{dy}{dx}\right)^2}\, dx $ is used
  • In first case, since the curve is expressing in polar coordinates, an equivalent formulation in polar coordinates, $ \displaystyle \int_C {\sqrt {\left(r^{2}+{\frac {dr}{d\theta }}\right)^{2}}}d\theta $ is used

If parametric curves in Question 1 and Question 2 are expressed in cartesian and polar coordinates respectively, then the alternate formulas would apply as well


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.