How do I even approach this problem? I don't know where to start.

Let P be a point at a distance d from the center of a circle of radius r. The curve traced out by P as the circle rolls along a straight line is called a trochoid. The cycloid is the special case of a trochoid with d=r. Using the same parameter theta as for the cycloid and, assuming the line is the x-axis and theta=0 when P is at one of its lowest points, show that parametric equations of the trochoid are:
x=rtheta-dsintheta
y=r-dcostheta
Sketch the trochoid for the cases d>r and d<r.

1 Answer
Sean
Jan 13, 2018

Please see below.

Explanation:

.

Point P, which is shown as a red dot, is at a distance d from the center of a circle. Below, you can see the three different possibilities for point P.

(1), Point P is at a distance d which is smaller than the radius of the circle (color(red)(d < r)).

(2), Point P is at a distance d equal to the radius of the circle (color(red)(d=r))

(3), Point P is at a distance d larger than the radius of the circle (color(red)(d > r))

enter image source here

In either case, as the circle rolls along the horizontal line, Point P moves along the curve shown in red. That curve is called a Trochoid.

Your problem states that color(red)(d=r). This special case of a trochoid is called a Cycloid and is shown below:

enter image source here

In the picture below, you can see Point P on the circumference of the circle. We will consider the horizontal line that the circle is rolling on to be the x-axis and the vertical line that crosses the x-axis at Point O to be the y-axis with Point O being the origin.

Initially, the circle was back where Point P was resting on the origin. As the circle began to roll, Point P moved from the origin to its currently shown location, i.e, it traveled the distance of OP.

Let's consider the Point C, the center of the circle and figure out its coordinates.

Its x-coordinate is =OB and its y-coordinate is =CB. OB=arc PB (both shown in yellow) since Point P is a fixed point on the circle and was originally where O is.

Therefore, if we can figure out the length of arc PB we will have the x-coordinate of C. arc PB is facing central angle theta. The formula for the perimeter (circumference) of a circle is:

P=2pir where r is the radius of the circle. But perimeter P covers 360^@ or 2pi radians. The piece of the perimeter that would face a small angle of 1 radian would be:

=(2pir)/(2pi)=r

Then it would stand to reason that the piece of the perimeter facing an angle of theta radians would be:

arcPB=OB=rtheta

This means the x-coordinate of Point C is rtheta

The y-coordinate which is CB is =r

Therefore, the coordinates of Point C are:

C(rtheta, r)

enter image source here

Now, let's figure out the coordinates of Point P.

x=OB-PQ

y=CB-CQ

Let's figure out the lengths of PQ and CQ, and substitute them in the equations of the x and y coordinates.

In triangle Delta CPQ,:

sintheta=(PQ)/(PC)=(PQ)/r

costheta=(CQ)/(PC)=(CQ)/r

PQ=rsintheta

CQ=rcostheta

Now, we can plug these into the coordinate equations:

x=rtheta-rsintheta

y=r-rcostheta

which is exactly what the problem wanted you to prove if you replace d with r.

To show that the formulas in the problem stand for a Trochoid regardless of whether d is smaller than, equal to, or larger than r, let's look at the picture below:

enter image source here

This picture shows Point P outside the circle. Here,

OB=arcZB=r theta

CB=r

arcPD=d theta

x-coordinate of Point P=OB-PQ

y-coordinate of Point P=CB-CQ

PQ=CP sintheta=dsintheta

CQ=CPcostheta=dcostheta

x=r theta-dsintheta

y=r-dcostheta

which proves the formulas given in the problem.

Similarly, if Point P were inside the circle. i,e d < r, it can be shown that the formulas hold.

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