Theta is Also a Sinusoid

Calculus Level 4

In the x y xy plane, the coordinates of a moving point are as specified below:

x = cos ( sin ( t ) ) y = sin ( sin ( t ) ) x = \cos(\sin(t))\hspace{1cm}y=\sin(\sin(t))

If t t denotes time, determine the average speed of the moving point from t = π 2 t = -\frac{\pi}{2} to t = π 2 t = \frac{\pi}{2}

Details and Assumptions: All angles are in radians.

1 2 {\frac{1}{2}} π 2 {\frac{\pi}{2}} 2 π {\frac{2}{\pi}} 0 0

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2 solutions

Let v a v_a be the average speed and s s be the travel space and the time is t = π 2 π 2 = π t = \frac{\pi}{2} - \frac{-\pi}{2} = \pi . Then, v a = travel space t i m e = s t v_a = \frac{\text{travel space}}{time} = \frac{s}{t} . Since sin ( t ) [ 1 , 1 ] , cos(sin(t)) \sin(t) \in [- 1, 1], \space \text{ cos(sin(t))} always will be positive, v a = s t = d s t = ( ( d x ) 2 + ( d y ) 2 ) 1 2 t = v_a = \frac{s}{t} = \frac{\int ds}{t} = \frac{\int ((dx)^2 + (dy)^2)^{\frac{1}{2}}}{t} = = π / 2 π / 2 ( cos 2 ( t ) ( sin 2 ( sin ( t ) ) + cos 2 ( sin ( t ) ) ) 1 2 π d t = π / 2 π / 2 cos ( t ) π d t = = \int_{-\pi/2}^{\pi/2} \frac{\left(\cos^2(t) \cdot (\sin^2(\sin(t)) + \cos^2(\sin(t))\right)^{\frac{1}{2}}}{\pi} dt = \int_{-\pi/2}^{\pi/2} \frac{|\cos(t)|}{\pi} dt = = ( 1 π ) ( sin ( t ) ) π / 2 π / 2 = 2 π = (\frac{1}{\pi})\cdot \left(\sin(t) \right)_{-\pi/2}^{\pi/2} = \frac{2}{\pi}

Steven Chase
Sep 10, 2016

One can certainly use calculus to solve, but it's also apparent that in the interval t = π 2 t = -\frac{\pi}{2} to t = π 2 t = \frac{\pi}{2} , s i n ( t ) sin(t) goes from -1 radian to 1 radian. Therefore, we have 2 radians per π \pi units of time.

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