## Formula to find the arc length

\displaystyle S = \int_{\alpha}^{\beta}\sqrt{\left( \cfrac{d\rho}{d\theta}\right)^{2} + \rho^{2}} \ d \theta

## Example. Let’s calculate the arc length of a cardioid

The cardioid to which we are going to find its arc length is \rho = 2 (1 + \cos \theta), graphically it looks like this:

\rho = 2(1 + \cos \theta)

As it says in the formula, we need to calculate the derivative of \rho. To do so, let’s multiply the 2 with the parenthesis:

\rho = 2 + 2 \cos \theta

And now we proceed to derive, derivative of a constant is equal to zero and derivative of cosine is equal to -\sin:

\rho ' = 0 - 2\sin \theta

\rho ' = -2 \sin \theta

What you have to do is substitute in the integral the values of the derivative and the square of \rho:

\displaystyle S = \int_{\alpha}^{\beta}\sqrt{(-2\sin \theta)^{2} + (2(1 + \cos \theta))^{2}} \ d\theta

### And what are our limits of angles in radians of \alpha and \beta?

There are two options to place the limits in the integral. The first option is to place the limits from zero to \pi, that the result will only give us half the value of the full length, which means that you have to multiply the result by 2.

The second option is to place the limits from zero to 2\pi, which represents that you will give once and for all the return to the cardioid. With the two options you reach the same result, let’s continue:

We will put a 2 to the integral because we are going to integrate from zero to \pi and we squared to obtain:

\displaystyle S = 2\int_{0}^{\pi}\sqrt{4\sin^{2}\theta + 4(1 + 2 \cos \theta + \cos^{2} \theta)} \ d\theta

Multiply the parenthesis where the 4 is:

\displaystyle S = 2\int_{0}^{\pi}\sqrt{4\sin^{2} \theta + 4 + 8 \cos \theta + 4 \cos^{2} \theta} \ d \theta

We are going to order the term 4\sin^{2}\theta and the term 4\cos^{2}\theta and we will factor the 4:

\displaystyle S = 2\int_{0}^{\pi}\sqrt{4(\sin^{2}\theta + \cos^{2} \theta) + 4 + 8 \cos \theta} \ d \theta

Remember a very powerful trigonometric identity:

\sin^{2}\theta + \cos^{2}\theta = 1

With the trigonometric identity, let’s reduce terms:

\displaystyle S = 2\int_{0}^{\pi}\sqrt{4 + 4 + 8\cos \theta} \ d \theta

Let’s sum the 4‘s:

\displaystyle S = 2\int_{0}^{\pi}\sqrt{8 + 8\cos \theta} \ d \theta

Let’s factorize 8:

\displaystyle S = 2\int_{0}^{\pi}\sqrt{8(1 + \cos \theta)} \ d \theta

And that 8 we can represent it as 2^{2}\cdot 2:

\displaystyle S = 2 \int_{0}^{\pi}\sqrt{2^{2}\cdot 2(1 + \cos \theta)} \ d \theta

That 2^{2}\cdot 2 can be taken from the square root and then, by integral properties, remove it from the integral:

\displaystyle S = 2 \int_{0}^{\pi} \left(2\sqrt{2} \right) \sqrt{1 + \cos \theta} \ d \theta

\displaystyle S = 2\left( 2\sqrt{2}\right)\int_{0}^{\pi}\sqrt{1 + \cos \theta} \ d\theta

Multiply terms:

\displaystyle S = 4\sqrt{2}\int_{0}^{\pi}\sqrt{1 + \cos \theta} \ d \theta

### Well, and how is it integrated?

Relax, that is easily integrated using the following trigonometric identity:

\cos^{2}\left(\cfrac{\theta}{2}\right) = \cfrac{1+\cos \theta}{2}

That trigonometric identity we will apply to it a square root to obtain:

\cos \left( \cfrac{\theta}{2}\right) = \cfrac{\sqrt{1 + \cos \theta}}{\sqrt{2}}

Now to the integral to which we arrived, we are going to multiply and divide a \sqrt{2} to be equal to the trigonometric identity:

\displaystyle S = 4\sqrt{2} \int_{0}^{\pi} \sqrt{2}\cfrac{\sqrt{1 + \cos \theta}}{\sqrt{2}} \ d \theta

By properties of integrals, we will extract the \sqrt{2} from the integral that is found as a numerator:

\displaystyle S = 4 \sqrt{2}\cdot\sqrt{2}\int_{0}^{\pi}\cfrac{\sqrt{1 + \cos \theta}}{\sqrt{2}} \ d \theta

Now our integral is similar to what we did in the trigonometric identity! Just replace what is in the integral by \cos\left(\frac{\theta}{2}\right) and multiply the two square roots of 2 and the 4:

\displaystyle S = 8\int_{0}^{\pi}\cos \left( \cfrac{\theta}{2} \right) \ d \theta

To solve that integral, we will replace u to \frac{\theta}{2}, we will derive and clear:

u = \cfrac{\theta}{2}

du = \cfrac{d\theta}{2}

2 \ du = d \theta

Let’s substitute those terms in the integral:

\displaystyle S = 8 \int_{0}^{\pi} \cos (u ) (2 \ du)

Let’s apply properties of integrals to take the 2 and multiply it with the 8:

\displaystyle S = 16 \int_{0}^{\pi}\cos(u) \ du

Remember that the integral of cosine is equal to sine, so let’s integrate:

S = 16 \left. \sin(u)\right|_{0}^{\pi}

Let’s return the \frac{\theta}{2}:

S = 16 \left. \sin\left( \cfrac{\theta}{2} \right) \right|_{0}^{\pi}

Using a calculator in radians, let’s evaluate the sine:

S = 16 \left( 1 -0\right)

And finally our result of the cardioid arch length is:

S = 16 \ \text{u}

*The \text{u} of the final result means “units”, that is, the length is 16 units.*

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