Question

How do you find the arc length of the curve `y = (x^4/8) + (1/4x^2)` from [1, 2]?

Answer 1

If you don't remember the arc length formula, you can use the distance formula:

`D(x) = sqrt((Deltax)^2 + (Deltay)^2)`

`s = D(x) = sumsqrt((Deltax)^2 + (Deltay)^2/(Deltax)^2*(Deltax)^2)`

`= sumsqrt(1 + ((Deltay)/(Deltax))^2)(Deltax)`

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

This is just a "dynamic", infinitesimally-short-distance formula that accumulates over an interval of constantly increasing `x`. The general strategy is to get common denominators, perhaps complete the square, and get the square root to go away.

So, first let's take the derivative and square it.

`(dy)/(dx) = x^3/2 + x/2 = 1/2(x^3 + x)`

`((dy)/(dx))^2 = 1/4(x^3 + x)^2 = 1/4 (x^6 + 2x^4 + x^2)`

Now you can insert this as:

`= int_1^2 sqrt(1 + ((dy)/(dx))^2)dx`

`= int_1^2 sqrt(1 + (x^6 + 2x^4 + x^2)/4)dx`

Cross-multiply:
`= int_1^2 sqrt((4 + x^6 + 2x^4 + x^2)/4)dx`

Factor out `sqrt(1/4)` and reorder terms:
`= 1/2int_1^2 sqrt(x^6 + 2x^4 + x^2 + 4)dx`

Now I want to somehow get this into a perfect square under a square root if possible. This was:

`= color(blue)(1/2int_1^2 sqrt((x^3 + x)^2 + 4)dx)`

...and as it turns out, there is no answer in terms of standard mathematical functions, so we have to stop writing here.

All you can do is numerically evaluate this on your calculator or Wolfram Alpha, so if it weren't for the boundaries, this would not be that fun!

The numerical result is `color(blue)(~~2.84218)`.

graph{(y - x^4/8 - x^2/4)*sqrt(1^2 - (x-1)^2)/sqrt(0.5^2 - (x-1.5)^2) = 0 [-2.92, 5.85, -0.955, 3.428]}