# What is the arc length of f(x)=sqrt(18-x^2)  on x in [0,3]?

Jun 12, 2018

The arc length $L = \frac{3 \pi \sqrt{2}}{4}$

#### Explanation:

The formula for the arc length $L$ of a curve is:

$L = {\int}_{a}^{b} \sqrt{1 + {\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}^{2}} \mathrm{dx}$

First, let's find the derivative of our function, which I'll rename $y$ for simplicity.

$y = \sqrt{18 - {x}^{2}}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{2 \sqrt{18 - {x}^{2}}} \cdot \frac{d}{\mathrm{dx}} \left(18 - {x}^{2}\right)$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{2 \sqrt{18 - {x}^{2}}} \cdot \left(- 2 x\right) = \frac{- x}{\sqrt{18 - {x}^{2}}}$

Plugging in what we have here...

$L = {\int}_{0}^{3} \sqrt{1 + {\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}^{2}} \mathrm{dx}$

$L = {\int}_{0}^{3} \sqrt{1 + {\left(\frac{- x}{\sqrt{18 - {x}^{2}}}\right)}^{2}} \mathrm{dx}$

$L = {\int}_{0}^{3} \sqrt{1 + \frac{{x}^{2}}{18 - {x}^{2}}} \mathrm{dx}$

$L = {\int}_{0}^{3} \sqrt{\frac{\left(18 - {x}^{2}\right) + {x}^{2}}{18 - {x}^{2}}} \mathrm{dx}$

$L = {\int}_{0}^{3} \sqrt{\frac{18}{18 - {x}^{2}}} \mathrm{dx}$

Using the formula $\int \frac{\mathrm{dx}}{\sqrt{{a}^{2} - {x}^{2}}} = \arcsin \left(\frac{x}{a}\right) + C$

$L = \sqrt{18} {\int}_{0}^{3} \frac{\mathrm{dx}}{\sqrt{18 - {x}^{2}}} = {\left[\sqrt{18} \arcsin \left(\frac{x}{\sqrt{18}}\right)\right]}_{0}^{3}$

$= \sqrt{18} \arcsin \left(\frac{3}{\sqrt{18}}\right) - \sqrt{18} \arcsin \left(\frac{0}{\sqrt{18}}\right)$

$= \sqrt{18} \arcsin \left(\frac{1}{\sqrt{2}}\right) - \sqrt{18} \arcsin \left(0\right)$

$= \sqrt{18} \cdot \left(\frac{\pi}{4}\right)$

$L = \frac{3 \pi \sqrt{2}}{4}$

Jun 12, 2018

Alternatively (and I didn't realize this until after I solved it brute-force with calculus), you could just use the fact that this curve is part of a circle with a radius of $\sqrt{18} = 3 \sqrt{2}$

Since the ending x-point of the arc is at $x = 3$, the ending y-coordinate will be at:

$y = \sqrt{18 - {\left(3\right)}^{2}} = \sqrt{18 - 9} = \sqrt{9} = 3$

Since the circle is centered at the origin, and the x and y coordinates of the endpoint of the arc are equal (i.e. the line $y = x$ passes through the endpoint), the endpoint must be at a bearing of $\frac{\pi}{4}$ with respect to either the x or y axis.

Since the starting point is on the y-axis, this means that the arc covers $\frac{\pi}{4}$ radians. Using the basic algebraic formula for arc length, we can see much more simply that:

L_"circle" = ("angle") * ("radius") = (pi/4) * (3sqrt2) = (3pi sqrt2)/4