﻿ Euler Math Toolbox - Examples

# Envelopes of Lines

by R. Grothmann

I want to compute the famous curve, which appears as a polar set of a ladder with fixed length, leaning at a wall (see the image below).

If a is the height of the ladder at the wall, then sqrt(1-a^2) is the distance of the foot from the wall (assuming the ladder has length 1), and we get the following equation for the line.

```>eq &= y/a+x/sqrt(1-a^2)=1
```
```                         y        x
- + ------------ = 1
a             2
sqrt(1 - a )

```

We can solve this equation for y, and get the equations of a set of lines.

```>&solve(eq,y), function y(x,a) &= y with %[1]
```
```                                       2
a sqrt(1 - a ) - a x
[y = --------------------]
2
sqrt(1 - a )

2
a sqrt(1 - a ) - a x
--------------------
2
sqrt(1 - a )

```

The maximal height at each x is the curve, we are looking for (the envelope). So we try to find it, making sure to pick the correct solution.

```>&assume(x>0); sol &= solve(diff(y(x,a),a)=0,a)
```
```                              2/3    2/3
sqrt(sqrt(3) I x    + x    + 2)
[a = - -------------------------------,
sqrt(2)
2/3    2/3
sqrt(sqrt(3) I x    + x    + 2)
a = -------------------------------,
sqrt(2)
2/3    2/3
sqrt(- sqrt(3) I x    + x    + 2)
a = - ---------------------------------,
sqrt(2)
2/3    2/3
sqrt(- sqrt(3) I x    + x    + 2)                  2/3
a = ---------------------------------, a = - sqrt(1 - x   ),
sqrt(2)
2/3
a = sqrt(1 - x   )]

```
```>function f(x) &= factor(at(y(x,a),sol[6]))
```
```                                 2/3        1/3
sqrt(1 - x   ) (x - x   )
- -------------------------
1/3
x

```

There is the following parametrization of the curve.

```>&assume(cos(t)>0); &trigsimp(f(sin(t)^3))
```
```                                  3
cos (t)

```

Now we plot everything, first the ladders.

```>a=(0.1:0.05:0.9)'; plot2d("y(x,a)",a=0,b=1,c=0,d=1):
```

Then the function.

```>plot2d("f",add=1,color=2,thickness=2):
```

# Other Examples

We can use the very same code to study other functions, like the segments, such that the y-abscissa plus the x-abscissa are constant.

```>eq &= y/a+x/(1-a)=1
```
```                            y     x
- + ----- = 1
a   1 - a

```

We can solve this equation for y, and get the equations of a set of lines.

```>&assume(x>0); ...
&solve(eq,y); function y(x,a) &= y with %
```
```                                    2
a x + a  - a
------------
a - 1

```

The maximal height at each x is the curve, we are looking for (the envelope). So we try to find it, making sure to pick the correct solution.

```>&solve(diff(y(x,a),a)=0,a), function f(x) &= factor(y(x,a) with %[1])
```
```                  [a = 1 - sqrt(x), a = sqrt(x) + 1]

3/2
x    - 2 x + sqrt(x)
--------------------
sqrt(x)

```

Now we plot everything, first the ladders.

```>a=(0.1:0.05:0.9)'; plot2d("y(x,a)",a=0,b=1,c=0,d=1); ...
```

Here is another example wheret the y-abscissa times the x-abscissa is constant.

```>eq &= y/a+a*x=1
```
```                             y
- + a x = 1
a

```

We can solve this equation for y, and get the equations of a set of lines.

```>&solve(eq,y); function y(x,a)&=rhs(%[1])
```
```                                    2
a - a  x

```

The maximal height at each x is the curve, we are looking for (the envelope). The result is very simple this time.

```>&assume(x>0);  ...
&solve(diff(y(x,a),a)=0,a), function f(x)&=factor(at(y(x,a),%[1]))
```
```                                    1
[a = ---]
2 x

1
---
4 x

```

Now we can plot everything.

```>a=exp(-4:0.2:4)'; plot2d("y(x,a)",a=0,b=1,c=0,d=1); ...