Question

What is the arclength of `f(t) = (cos2t-sin2t,tan^2t)` on `t in [pi/12,(5pi)/12]`?

Answer 1

≈14.69 (no closed form)

Explanation:

`f(t) = (cos2t-sin2t,tan^2t)`

To find the arc length `Deltas` over a time `Deltat`, you can try to approximate it with a straight line, using the hypotenuse of the right-angled triangle with sides `Deltax(t), Deltay(t)`, which is the change in the x- and y- coordinates over the time `t`.

`Deltas ~~ sqrt(Deltax^2+Deltay^2)`

Of course, this will often be wildly inaccurate since most curves aren't exactly a straight line.

However, for a continuous function, the smaller time interval you choose, the more accurate the above approximation is (you can imagine zooming in on a function's graph: the more you zoom, the closer the function looks like a straight line)

Thus, at the limit when you shorten the time interval, you get

`ds = sqrt(dx^2+dy^2)`

To manipulate it to suit the given parametric, we can adjust the equation such that `ds` is in terms of `dt`.

`ds = sqrt((dx/dt*dt)^2+(dy/dt*dt)^2)`
`=sqrt(((dx/dt)^2+(dy/dt)^2)dt^2)`
`=sqrt((dx/dt)^2+(dy/dt)^2)dt`

Since `dx/dt=-2sin2t-2cos2t, dy/dt=2tan(t)sec^2(t)`,

`ds=sqrt((-2sin2t-2cos2t)^2+(2tan(t)sec^2(t))^2)dt`

Factorising a 2 out of the entire expression and simplifying, we have

`ds=2sqrt((cos2t+sin2t)^2+(tan(t)sec^2(t))^2)dt`

To find the total arc length from `pi/12` to `5/12pi`, we can add up tiny segments of `ds` from `t=pi/12` to `t=5/12pi`

Mathematically, this is equivalent to integrating from `t=pi/12` to `t=5/12pi`

`s=int_(pi/12)^(5/12pi)2sqrt((cos2t+sin2t)^2+(tan(t)sec^2(t))^2)dt`

Now this expression has no closed form and is approximately 14.69