Dtjytyk

There is one hockey puck with a diameter of $3$ inches. The puck is spinning around its center at a speed of $3$ counterclockwise rotations per second. At the center, the puck is traveling at a speed of $24$ inches per second at an angle of $45^circ$ to the positive $x$-axis.

(a) At time $t=0$, the center of the puck is at the origin. Find the location of the center of the puck at time $t$? (Note that time $t$ is measured in seconds.)

(b) A point on the outer edge of the puck begins at the point $left(0,frac32right)$. Find its location at time $t$?

I am not sure how to find part (a), but this is what I have for part b).

The puck has diameter of $3$ inches, a radius of $frac{3}{2}$.

I used the formula where that in order to find a point on the circle's circumference, given a circle with center $(a,b)$ and radius $r$,

$x(t) = rcos (t) + a$
$y(t) = r sin (t) + b$

the circle's center is at $(0,0)$ because of part a) statement.

Therefore $x(t) = r cos (t)$ and $y(t) = r sin (t)$. I plugged in $x(t) = 0$ and $y(t) = frac{3}{2}$ because of what was given in part b).
From there I found that $t = frac{pi}{2}$ and that $r = frac{3}{2}$,

so now the formula is $x(t) = frac{3}{2} cos (t)$ and $y(t) = frac{3}{2} sin (t)$. But now I am stuck. What do I do next for part b), and how do I solve part a)?

(a) Since the puck is traveling at an angle of $45^circ$ to the positive $x$-axis, at all times we have $x(t)=y(t)$. The distance it has travelled at time $t$ is
$$24t=sqrt{(x(t))^2+(x(t))^2} = sqrt2 x(t).$$
Therefore $x(t)=(12sqrt2) t$. and $y=x$. At time $tgeq0$, the center of the puck is at the point
$$boxed{(12sqrt2 cdot t,12sqrt2 cdot t.}$$

(b) From part (a), we know that the center of the puck is at $((12sqrt2) t,(12sqrt2) t)$ at time $t$. We can determine the location of the point by considering the displacement of the point from the center of the puck. In other words, we treat the center as stationary, and consider where the point is relative to the center. Since the puck makes 3 counterclockwise rotations per second, it spins an angle of $3cdot 2pi = 6pi$ per second. However, the angle of the point relative to the center is initially $frac{pi}{2},$ so at time $t$, the point is at an angle of $frac{pi}{2} + 6pi t.$

Finally, the radius of the puck is $1.5$, so the the displacement of the point from the center is $left(1.5cosleft(6pi t +frac{pi}{2}right),1.5sinleft(6pi t +frac{pi}{2}right)right)$. Combining this with the motion of the center of the puck, the location of the point is
begin{align*}
x(t)&=(12sqrt2) t+1.5cosleft(6pi t +frac{pi}{2}right) ,\
y(t)&=(12sqrt2) t+1.5sinleft(6pi t +frac{pi}{2}right).
end{align*}
Since $cosleft(theta+frac{pi}{2}right) = -sin theta$ and $sinleft(theta+frac{pi}{2}right)=cos theta$, the location of the point is
begin{align*}
x(t)&=(12sqrt2) t-1.5sin(6pi t),\
y(t)&=(12sqrt2) t+1.5cos(6pi t).
end{align*}

If you don't mind, I would like to work with radians instead of degrees.

First question a). The centre of the puck (let's call it $C$) is traveling at a constant speed, along a straight line which makes an angle of $frac pi4(=45^circ)$ to the positive $x$-axis. This will be the line $y=x$. Now we start at the origin and we know that the point $big(x(t), y(t)big)$ has a distance of $24t$ inches from the origin. So we solve
$$24 t= sqrt{x(t)^2+y(t)^2}=sqrt{2cdot x(t)^2}=sqrt 2 ,x(t).$$
This means $x(t)=y(t)=12sqrt 2 ,t$. So the parametrization for the centre of the puck will be $$begin{cases}x_C(t)=12sqrt 2, t\ y_C(t)= 12sqrt 2, tend{cases}$$

Now for question b). Now it is quite well known that for a point on the unit circle that spins around the origin, we have that $x=cos theta $ and $y=sin theta$, as according to this picture:

This can very easily be modified to a rotation around a point $(a,b)$ of a circle with radius $r$, like this: $x=a+rcos theta$ and $y=b+rcos theta$. Now apply this to the hockey puck. It rotates around the point $big ( x_C(t),y_C(t)big) $ and has radius $frac 32$. We also know that it makes three rotations (that's $6pi$ radians) per second and that our point (let's call it $P$) starts at $fracpi2(=90^circ)$. So we get the parametrization:
$$begin{cases} x_P(t)=&x_C(t)+frac32 cos left(6pi t+fracpi 2right) =&12sqrt2,t -frac32sin(6pi t)\
y_P(t)=&y_C(t)+frac32 sinleft(6pi t+frac pi2right )=& 12sqrt 2,t+ frac32cos (6pi t)end{cases}
$$

(a) The hockey puck is traveling along the line $y=x$, traveling 24 inches every second. Since the puck moves 24 inches when $t=1$ and the puck travels at a constant pace, we can graph:

When $t=1$, $x = y = 12sqrt{2}$. Therefore, the center of the puck at time $t$ is at:
begin{align*}
x &= 12sqrt{2}t \
y &= 12sqrt{2}t
end{align*}

(b) Let's make a parametric equation for the puck using variables:
begin{align*}
x &= a + r cos theta \
y &= b + r sin theta
end{align*}
We know that the center of the puck at time $t$ is at $12 sqrt{2} t$. We also know that the radius is $frac{3}{2}$. Filling in the variables, we get:
begin{align*}
x &= 12 sqrt{2}t + frac{3}{2} cos theta \
y &= 12 sqrt{2}t + frac{3}{2} sin theta
end{align*}
But what is $theta$? Since the puck spins $6pi$ radians every second and starts at $frac{pi}{2}$, $theta = 6pi t + frac{pi}{2}$
begin{align*}
x &= 12 sqrt{2}t + frac{3}{2} cos (6pi t + frac{pi}{2}) \
y &= 12 sqrt{2}t + frac{3}{2} sin (6pi t + frac{pi}{2})
end{align*}
Since $sin(theta+90) = cos(theta)$ and $cos(theta + 90) = -sin theta$
begin{align*}
x &= 12 sqrt{2}t - frac{3}{2} sin (6pi t) \
y &= 12 sqrt{2}t + frac{3}{2} cos (6pi t)
end{align*}