Question

How do you find the distance travelled from t=0 to `t=2pi` by an object whose motion is `x=cos^2t, y=sin^2t`?

Answer 1

See the answer below:

Answer 2

An interesting thing about this example is that there is a non-calculus-based way of doing it as well, to get the same answer for the distance (arc length) equal to `4sqrt(2)`.

Explanation:

Since `x=cos^{2}(t)` and `y=sin^{2}(t)`, it follows that `x+y=cos^{2}(t)+sin^{2}(t)=1` for all values of `t`. Therefore, the motion is always on the straight line with `xy`-equation `x+y=1`, which is equivalent to `y=-x+1` (a straight line with a slope of `-1` and a `y`-intercept of `1`).

Also, since `cos^{2}(t)\geq 0` and `sin^{2}(t)\geq 0` for all `t`, this motion is always in the 1st quadrant of the plane where `x\geq 0` and `y\geq 0`.

Now think about, for `0\leq t\leq 2pi`, how the values of `cos^{2}(t)` oscillate from 1 to 0 to 1 to 0 and back to 1 again, while the values of `sin^{2}(t)` oscillate from 0 to 1 to 0 to 1 and back to 0 again. In other words, the motion traverses the line segment from `(1,0)` to `(0,1)` four times. By the Pythagorean Theorem (draw an appropriate right triangle), this line segment has length `sqrt(1^{2}+1^{2})=sqrt(2)`.

This leads us to conclude that the total distance traveled (arc length) is `4sqrt(2)`.