# 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? Sep 22, 2017

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 $4 \sqrt{2}$.

#### Explanation:

Since $x = {\cos}^{2} \left(t\right)$ and $y = {\sin}^{2} \left(t\right)$, it follows that $x + y = {\cos}^{2} \left(t\right) + {\sin}^{2} \left(t\right) = 1$ for all values of $t$. Therefore, the motion is always on the straight line with $x y$-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 ${\cos}^{2} \left(t\right) \setminus \ge q 0$ and ${\sin}^{2} \left(t\right) \setminus \ge q 0$ for all $t$, this motion is always in the 1st quadrant of the plane where $x \setminus \ge q 0$ and $y \setminus \ge q 0$.

Now think about, for $0 \setminus \le q t \setminus \le q 2 \pi$, how the values of ${\cos}^{2} \left(t\right)$ oscillate from 1 to 0 to 1 to 0 and back to 1 again, while the values of ${\sin}^{2} \left(t\right)$ oscillate from 0 to 1 to 0 to 1 and back to 0 again. In other words, the motion traverses the line segment from $\left(1 , 0\right)$ to $\left(0 , 1\right)$ 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 $4 \sqrt{2}$.