-
When , the position is
. When , the position is . When , the position is , when , the position is , and when , the position is . The plot of these positions is
below.
For each , we have and . Solving for in the equation gives , which can be substituted into the
second equation to get . Therefore,
every point of the form satisfies .
From part (ii), the points that the particle occupies are all on
the line with equations . We
care about the points on this line from to inclusive, so the plot is just the
line segment connecting (the
point for ) to (the point for ). The plot is below.
We have and , so for any position that the particle occupies, . Therefore, the
particle is always somewhere on the unit circle.
Indeed, every point on the unit circle has coordinates for exactly one real
number with . The graph of the path of the particle is
-
In each of the three tables below, the left column contains
values of , the middle column
contains the corresponding values of , and the right column contains the corresponding values of
. Although it is not
particularly difficult to write down exact values for each of the
trigonometric ratios in the table, we want to plot the points so the
values are all given rounded to three digits past the decimal point.
Below is a plot of the points above including a sketch of the
curve.
Every point on the
parametric curve satisfies
and for some real number
. Using the double-angle formula
for , we get , so . By the Pythagorean
identity, .
Substituting gives .
We will label the point on Circle 2 that is originally at by and we will label the centre of Circle
2 by . Since the circumferences of
the circles are the same, Circle 2 will return to its original position
after rolling exactly once around Circle 1. This means will travel exactly once around the
circle of radius centred at
the origin.
We will let represent the
angle made by the positive -axis
and the ray from the origin to .
For example, the diagram below depicts the position of the outer circle
when . The origin
is labelled by .
We wish to find the coordinates of in terms of the angle . In the diagram below, we have added to
the previous diagram a line through the origin and , a line segment connecting to , as well as a horizontal line through
. The horizontal line intersects
Circle 2 at to the right of . The line through and intersects the point of tangency of the
two circles at and it also
intersects Circle 2 at (the
points and are different). A perpendicular has
also been drawn from to intersecting at .
The line connecting the centres of tangent circles always passes
through the point of tangency, which justifies the implicit claim in the
previous paragraph that passes
through the point of tangency. It also implies that the length of is . Therefore, the coordinates of
are . Since is horizontal by construction, is parallel to the axis, which implies .
Because Circle 2 is rolling along Circle 2 without slipping, the arc
from to the axis along Circle 1 has the same length
as the arc . This means as well since the two
circles have the same radius. Since is a line, we get that .
Suppose has coordinates and has coordinates . The radius of Circle 2 is
, so . This means and . Thus,
where the last equality is by trigonometric identities. Therefore, and . A plot of this
curve is given below.
As Circle 2 rolls around the inside of Circle 1, if a line is
drawn through the point of tangency perpendicular to the mutual tangent,
then the line will pass through the centre of both circles. Such a line
contains a diameter of both circles, and since the radius of Circle 1
equals the diameter of Circle 2, the centre of Circle 1 is always on
Circle 2. In the diagram below, Circle 2 has been rotated by some
positive angle between and . The origin is labelled by
, the current point of tangency is
labelled by , the centre of
Circle 2 is labelled by , the
other point at which Circle 2 intersects the -axis is labelled by , and is labelled by .
We wish to determine the current location of the point on Circle 2
that started at . Circle 2 does
not slip as it rolls, so this point is the same distance from in the clockwise direction along both
circles.
Since is on the circumference
of Circle 2 and is the centre of
Circle 2, a well-known circle property implies that . The radius of
Circle 1 is , so provided we
measure angles in radians, the length of the arc is . The radius of Circle 2 is , so the length of arc is . These quantities are equal, so the arcs and have equal length. Therefore, the
original point of tangency is at .
By drawing similar diagrams for obtuse and reflex angles, it can be
similarly shown that the original point of tangency is always on the
diameter of Circle 1. Now note half the circumference of Circle 1 is
equal in length to the circumference of Circle 2, so when Circle 2 has
rolled exactly half way around Circle 1, the original point of tangency
is exactly where Circle 1 intersects the negative -axis. It follows by symmetry that the
original point of tangency will go from to and back to as Circle 2 rolls around Circle
1.
We will first work out, in general, a pair of parametric
equations for , the original point
of tangency. As in part (d), is
the origin, is the centre of
Circle 2, and the measure of the angle made by the positive -axis and the ray from to is . In the diagram below, a horizontal
line is drawn through
intersecting Circle 2 at to the
right of and the line defined by
, for the same reasoning that was
used in part (d), passes through the mutual point of tangency labelled
by . We label by .
As mentioned, , , and are on a line, and so . Therefore, the coordinates
of are . If we
let be , then by reasoning similar to
that which was used in part (d), the coordinates of are For now, assume that is small enough that Circle 2 has not
yet rolled far enough for to have
returned to the circumference of Circle 1. The length of arc on Circle 2 is equal to the length of
arc on Circle 1, which is equal
to since the radius of Circle 1
is (provided we use radians as
our unit of angle measure). Because is parallel to , , so (note that this angle is measured clockwise from
, so in the diagram, the angle
measures more than radians).
The radius of Circle 2 is , so the
length of arc is . The arcs and are equal, so , which can be solved for
to get , and we note that
, so is positive.
Now suppose that is such that
has already returned to Circle 1
times. Between consecutive times
that returns to Circle 1, the arc
increases in length by , the circumference of Circle 2.
Therefore, instead of the equation , we get since the arc
length from to is still , but to get equality, we need to
account for the complete
revolutions of Circle 2.
Solving this equation for
gives . However, since we will be
taking and of , we can ignore the quantity since is an integer. Therefore, the
coordinates of when makes an angle of with the positive -axis are
Before answering the given questions, we note that if , (that is, the radius of
Circle 2 is twice that of Circle 1), then the coordinates simplify to
, meaning the point always has a -coordinate of . This gives another proof of the result
in (e).
When , the
coordinates of are Using standard
trigonometric identities, one can show that and .
Substituting into the coordinates above, we get leading to the rather tidy
expression for the coordinates of . Therefore, and , so .
For , we get
For , we get
Focusing for a moment on the equations for , if we substitute , we get and , which means the point
initially at on Circle 2 has
not returned to its original position yet. This is because the
circumference of Circle 1 is not an integer multiple of the
circumference of Circle 2. Indeed, the circumference of Circle 1 is
, but the circumference of
Circle 2 is .
When Circle 2 first reaches the point where it is tangent to Circle
at , some points on Circle 2 have been
tangent to Circle 1 more than once. To be precise, , so
every point on Circle 2 has been tangent to Circle 1 at least once, but
the points on an arc of length on Circle 2 have been
tangent to Circle 1 twice. Specifically, since ,
exactly half of the points on Circle 2 have been tangent to Circle 1
twice and the rest have been tangent once.
If Circle 2 rolls around Circle 1 again, then half the points on
Circle 2 will be tangent to Circle 1 exactly twice, and the other half
will be tangent exactly once. This gives a total of three times for each
point. Indeed, , so three times the circumference of Circle 2 is an
integer multiple of the circumference of Circle 1. Therefore, returns to for the first time when the
circumference of Circle 2 wraps around the inside of the circumference
of Circle 1 exactly 2 times.
Algebraically, if we substitute , we get and , so the curve for is travelled periodically
every two times Circle 2 rolls along Circle 1.
To avoid confusion, we will refer to the circle with as Circle 3 and the circle
with as Circle 2.
The circumference of Circle 1 is an integer multiple of the
circumference of Circle 3, so point reaches after Circle 3 rolls around Circle
1 exactly once.
Suppose the centre of Circle 3 revolves around the origin at a rate
of radian per second and the
centre of Circle 2 revolves around the origin at a rate of radians per second. By assuming this,
we will have that both circles take exactly seconds for to return to its original position for
the first time.
If Circle 3 is instead rolled clockwise around Circle 1, then the
position of will be where and are as given for above. This is because,
for example, point is in the same
position after rolling radians counterclockwise
as it would be if it had rolled radians clockwise.
Thus, if Circle 3 rotated clockwise instead of counterclockwise, its
position at time is given by
where
If Circle 2 is rotated counterclockwise
(as before) at radians per
second, then at time the angle is
, and so the coordinates of are
which are the exact same coordinates as
Circle 3 at time . Of course, if
Circle 3 is rotated clockwise, then it travels the same path as if it
were rotated counterclockwise but in the opposite direction. Thus, if
both Circle 2 and Circle 3 are rotated counterclockwise, then travels the same path in opposite
directions.