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

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

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

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Answer 1 · 2018-01-13T23:20:22

Please see below.

Explanation:

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

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

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

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

enter image source here

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

Your problem states that $d = r$ $color(red)(d=r)$ . 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 $P$ $P$ on the circumference of the circle. We will consider the horizontal line that the circle is rolling on to be the $x$ $x$ -axis and the vertical line that crosses the $x$ $x$ -axis at Point $O$ $O$ to be the $y$ $y$ -axis with Point $O$ $O$ being the origin.

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

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

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

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

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

$= \frac{2 π r}{2 π} = r$ $=(2pir)/(2pi)=r$

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

$a r c P B = O B = r θ$ $arcPB=OB=rtheta$

This means the $x$ $x$ -coordinate of Point $C$ $C$ is $r θ$ $rtheta$

The $y$ $y$ -coordinate which is $C B$ $CB$ is $= r$ $=r$

Therefore, the coordinates of Point $C$ $C$ are:

$C (r θ, r)$ $C(rtheta, r)$

enter image source here

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

$x = O B - P Q$ $x=OB-PQ$

$y = C B - C Q$ $y=CB-CQ$

Let's figure out the lengths of $P Q$ $PQ$ and $C Q$ $CQ$ , and substitute them in the equations of the $x$ $x$ and $y$ $y$ 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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How do I even approach this problem? I don't know where to start.

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