String PatternsIntroduction Polygons and Stars Cyclic groups Snooker on a circular table Introduction
Notice below that there is a circle which the ball never enters.
This is true for any θ, if we draw all chords subtending angle 2θ rather than just the ones traced by the ball
(remember if θ is a rational multiple of 180, you'll get finitely many lines)
You can see this in the activity below, which draws not all chords, but 200 chords to give you the idea.
Considering the diagrams below, we can see that the radius of this excluded circle is
r = R cos θ
We can make more interesting curves formed by an infinite set of tangents.
How about a plank of wood of length 'L' starting vertical and falling downwards:
At each point of the black curve there is a position of the plank that is a tangent.
How can we find the equation of the black curve?
Well, when the angle of the plank is θ, the equation of the plank is
y = -x tanθ + Lsinθ - Equation 1
Also, for each θ there is exactly one point on the black curve.
So we can write the (x, y) positions of the curve as functions of θ x = x(θ) and y = y(θ).
We want to find the functions 'x' and 'y' in terms of θ
To do this you need to know from calculus that the gradient of a curve is dy/dx and that
|
dy dx |
= |
dy/dθ dx/dθ |
From equation 1, the gradient is also -tanθ
| - tanθ | = |
dy/dθ dx/dθ |
Differentitate Equation 1 with respect to θ:
|
dy dθ |
= | - tanθ |
dx dθ |
- x sec2θ + Lcosθ |
(using the product rule)
Substitute for tanθ and simplify: x sec2θ = Lcosθ
So x = Lcos3θ
And y = -xtanθ + Lcosθ = Lsin3θ
Eliminating θ, x2/3 + y2/3 = L2/3
This curve is called the 'envelope' of the set of red lines.
In general, given a set of lines we define their envelope as a curve which, for each of its points, has one of the lines as a tangent.
We can use exactly the same method as above for any set of lines.
A line is of the form ax + by = c.
A family of lines is of the form a(t)x + b(t)y = c(t) - Equation 1
As the value of 't' changes we get an infinite family of lines.
In the above example, 't' was the angle θ, but that need not be so in general.
As above, since each point of the envelope curve has one of these lines as a tangent, we can label each point of the envelope as (x(t), y(t)).
As 't' changes these points trace out the envelope, one value of 't' giving one point on the curve.
If you find that hard to imagine, here is an example from before.
You can see how each value of 't' gives a point (x, y) on the curve that depends on 't'.
The gradient of the envelope curve at (x, y) is dy/dx and
|
dy dx |
= |
dy/dt dx/dt |
The tangent line from the above family has equation a(t)x + b(t)y = c(t) - Equation 1
Rearranging, and comparing to y = mx + c, you will see that the gradient is also
| - |
a(t) b(t) |
So,
| - |
a(t) b(t) |
= |
dy/dt dx/dt |
- Equation 2 |
Now differentiate Equation 1 with respect to 't':
|
da dt |
x + a |
dx dt |
+ |
db dt |
y + b |
dy dt |
= |
dc dt |
Cross multiply equation 2 and substitute, and the 2nd and 4th terms will disappear:
|
da dt |
x + |
db dt |
y | = |
dc dt |
In summary,
To find the envelope curve of a family of curves a(t)x + b(t)y = c(t), solve the equation: |
|
da dt |
x + |
db dt |
y | = |
dc dt |
What a nice result! We just differentitate a(t), b(t) and c(t) in the equation of the lines with respect to 't'.
Let's apply this to another nail board!
|
|
|
|
These sorts of shapes are made by joining lines of string as shown below, stepping one nail at a time:
To use our result we need to find the family of lines in terms of 't'.
From the diagram below, you can see that the equations are:
| y = - |
1 - t t |
x + 1 - t | for 0 = t = 1 - Equation 1 |
Simplifying, this is:
| ty = (t - 1)x + t - t2 |
Differentiate with respect to 't':
| y = x + 1 - 2t |
And substitute into Equation 1:
| x + 1 - 2t = - |
1 - t t |
x + 1 - t |
| tx + t - 2t2 = tx - x + t - t2 |
| So x = t2 and y = x + 1 - 2t = t2 + 1 - 2t = (1 - t)2 |
Our envelope is given by the equations x = t2, y = (1 - t)2
Eliminating 't', this gives vx + vy = 1