Posted by Sam on Wednesday, December 11, 2013 at 5:07pm.
A hockey puck of diameter 3 inches is spinning around its center at a speed of 3 counterclockwise rotations per second. The center of the puck is traveling at a speed of 24 inches per second at an angle of 45 degrees to the positive real axis.
(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 5:24pm
Thanks, 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?
- Math - Precalculus - Parametric Equations/Geometry - Steve, Wednesday, December 11, 2013 at 5:29pm
2pi radians per rotation. 3 rotations/sec
- precalc - correction - Steve, Wednesday, December 11, 2013 at 5:32pm
Oops. (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
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 5:33pm
Well, I would just like to point out to steve that 24/√2 = 12√2, not 6√2.
And Samuel? Ahem.
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