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Arc length

Let’s start first with the formula of the arc length and then we will see a simple exercise.

Arc length formula

\displaystyle L = \int_{a}^{b}\sqrt{1 + \left[ f'(x)\right]^{2}} \ dx

Now that we have the powerful formula to calculate the arc length, let’s go with an example.

Example of arc length

Calculate the arc length of y=x^{2} from x=0 to x=1.

arc-length-example-1

The derivative of f(x)=x^{2} is f'(x) = 2x, so let’s substitute the derivative and the limits in the arc length formula.

\displaystyle L = \int_{0}^{1}\sqrt{1 + \left[ 2x \right]^{2}} \ dx

\displaystyle L = \int_{0}^{1}\sqrt{1 + 4x^{2}} \ dx

In order to solve this exercise we will use trigonometry, we will help us with the following triangle:

arc-length-trigonometryWith which we propose the following equations to be able to leave the integral in trigonometric terms:

\tan \theta = 2x

\sec \theta = \sqrt{1 + 4x^{2}}

We derive \tan \theta to obtain:

\sec^{2}\theta \ d \theta = 2 \ dx

Now we clear the 2 to leave the dx alone:

\cfrac{\sec^{2} \theta}{2} \ d \theta = dx

Once we have the previous equations, substitute in the integral of the arc length of our exercise:

\displaystyle L = \int_{0}^{1} \sqrt{1 + 4x^{2}} \ dx \Rightarrow

\displaystyle L = \int_{0}^{1} \sec \theta \ \cfrac{\sec^{2} \theta}{2} \ d\theta

We will multiply the secants to obtain a secant cube and by properties of the integrals we can extract the 2 from the denominator:

\displaystyle L = \cfrac{1}{2} \int_{0}^{1}\sec^{3} \theta \ d \theta

Consulting the integral of secant cubed, we will obtain the following:

= \left. \cfrac{1}{4}\sec \theta \tan \theta + \cfrac{1}{4} \ln \left| \sec \theta + \tan \theta \right| \right]_{0}^{1}

Now, substituting back the values of x, remember that \tan \theta = 2x and \sec \theta = \sqrt{1+ 4x^{2}}, we get:

= \cfrac{1}{4} \left[ 2x \sqrt{1 + 4x^{2}} + \ln \left| \sqrt{1 + 4x^{2}} + 2x\right|\right]_{0}^{1}

Evaluating the limits we will obtain:

= \cfrac{1}{4} \left[\left( 2\sqrt{5} + \ln\left| 2 + \sqrt{5}\right|\right) - \left(0 \right) \right]

Finally we obtain a result of:

L = \cfrac{2\sqrt{5} + \ln\left| 2 + \sqrt{5}\right|}{4}

So our arc length is:

L = \cfrac{2\sqrt{5} + \ln\left| 2 + \sqrt{5}\right|}{4}\approx 1.48 \ \text{u}

The \text{u} means units.

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