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

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

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Answer 1 · 2017-09-22T15:59:27

See the answer below:
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Answer 2 · 2017-09-22T17:34:37

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 $4 \sqrt{2}$ $4sqrt(2)$ .

Explanation:

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

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

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

This leads us to conclude that the total distance traveled (arc length) is $4 \sqrt{2}$ $4sqrt(2)$ .

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

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How do you find the distance travelled from t=0 to t=2pit=2πt=2pi by an object whose motion is x=cos^2t, y=sin^2tx=cos2t,y=sin2tx=cos^2t, y=sin^2t?

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