Question

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

Answer 1

The arc length `L = (3pi sqrt2)/4`

Explanation:

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

`L = int_a^b sqrt(1+(dy/dx)^2) dx`

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

`y = sqrt(18-x^2)`

`dy/dx = 1/(2sqrt(18-x^2)) * d/dx(18-x^2)`

`dy/dx = 1/(2sqrt(18-x^2)) * (-2x) = (-x)/sqrt(18-x^2)`

Plugging in what we have here...

`L = int_0^3 sqrt(1+(dy/dx)^2) dx`

`L = int_0^3 sqrt(1+((-x)/sqrt(18-x^2))^2) dx`

`L = int_0^3 sqrt(1+(x^2)/(18-x^2)) dx`

`L = int_0^3 sqrt(((18-x^2)+x^2)/(18-x^2)) dx`

`L = int_0^3 sqrt(18/(18-x^2)) dx`

Using the formula `intdx/sqrt(a^2-x^2) = arcsin(x/a) + C`

`L = sqrt18 int_0^3 dx/sqrt(18-x^2) = [sqrt18arcsin(x/sqrt18)]_0^3`

`= sqrt18 arcsin(3/sqrt18) - sqrt18 arcsin(0/sqrt18)`

`= sqrt18 arcsin(1/sqrt2) - sqrt18 arcsin(0)`

`= sqrt18 * (pi/4)`

`L = (3pi sqrt2)/4`

Final Answer

Answer 2

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 `sqrt18 = 3sqrt2`

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

`y = sqrt(18-(3)^2) = sqrt(18-9) = sqrt9 = 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 `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 `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`

Final Answer