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July 11, 2014

July 11, 2014

Posted by **Sam** on Wednesday, December 11, 2013 at 5:07pm.

(a) At time t=0, the center of the puck is at the origin. Where is the center of the puck at time t? (Time t is measured in seconds.)

(b) A point on the outer edge of the puck begins at the point (0, 3/2). What is its location at time t?

I tried solving this problem but I don't know where to start. This almost looks like a physics problem. How do you incorporate the "spinning"?

- Math - Precalculus - Parametric Equations/Geometry -
**Steve**, Wednesday, December 11, 2013 at 5:14pm(a) the horizontal and vertical speeds are 24/√2 = 6√2, so at time t, the center is at

(6√2 t, 6√2 t)

(b) the puck is a circle of radius 3/2, and angular velocity = 3*2π /s

The point (0,3/2) corresponds to θ=0

So, we have, relative to the puck center,

x = 3/2 cosθ

y = 3/2 sinθ

But θ = 6πt, so point P on the puck has coordinates

x = 3/2 cos(6πt) + 6√2 t

y = 3/2 sin(6πt) + 6√2 t

- Math - Precalculus - Parametric Equations/Geometry -
**Sam**, Wednesday, December 11, 2013 at 5:24pmThanks, but I was wondering how you got the horizontal and vertical speeds, sorry I'm not that good at parametric equations yet :( Also how did you get that the angular velocity is 3*2pi / s?

Thanks

- Math - Precalculus - Parametric Equations/Geometry -
**Steve**, Wednesday, December 11, 2013 at 5:29pm2pi radians per rotation. 3 rotations/sec

- precalc - correction -
**Steve**, Wednesday, December 11, 2013 at 5:32pmOops. (0,3/2) corresponds to θ=π/2, so

x = 3/2 cos(6πt+π/2) + 6√2 t

y = 3/2 sin(6πt+π/2) + 6√2 t

or

x = -3/2 sin(6πt) + 6√2 t

y = 3/2 cos(6πt) + 6√2 t

- Math - Precalculus - Parametric Equations/Geometry -
**musicnmath**, Thursday, December 12, 2013 at 5:33pmWell, I would just like to point out to steve that 24/√2 = 12√2, not 6√2.

And Samuel? Ahem.

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