Question

How do you graph the lemniscate `r^2=36cos2theta`?

Answer 1

Hello,

• You have to graph two curves of polar equations
  `r = 6 sqrt(cos(2 theta))` and `r = -6 sqrt(cos(2 theta))`

• It's enough to graph the first one : the second curve is the O-symmetric of the first one. The range is
  `{theta in RR | cos(2 theta) >= 0} = bigcup_(k in ZZ) [-pi/4 + k pi , pi/4 + kpi]`

• Reduce the range of `r` :
  1) `r(theta + pi) = r(theta)` : you can study `r` only on `[-pi/4 , pi/4]`.
  2) `r(-theta) = r(theta)` : you can study `r` on `[0,pi/4]` but you have to complete the curve with `Ox`-axis.

• Calculate derivative of `r` :
  `r'(theta) = 6 (-2 sin(2 theta))/(2 sqrt(cos(2 theta))) = - 6 sin(2 theta)/sqrt(cos(2 theta))`

• Study the sign of `r'` on `[0, pi/2]` : because `2theta in [0, pi]`, `sin(2 theta) >=0`, so `r'(theta) <= 0` and `r` is a decreasing function on `[0, pi/2]`.

• Find particular tangents. Remember that :
  1) if `M(theta) \ne O`, the tangent at `M(theta)` is directed by `r'(theta) vec u_(theta) + r(theta) vec v_(theta)`, where `(vec u_(theta) , vec v_(theta))` is the polar basis.
  2) if `M(theta)=O`, the tangent at `O` is directed by `vec u_(theta)`.

Here, you have :
1) tangent at `M(0) = (6,0)` : directed by `0 vec u_0 + 6 vec v_0`, so directed by `vec v_0 = (0,1)`.
2) tangent at `M(pi/4) = O` : directed by `vec u_(pi/4)`.
3) there exists a horizontal tangent : `y = r sin(theta)`, so `y' = r' sin(theta) + r cos(theta)` and so
`y' = 0 iff (r')/r = - 1/tan(theta) iff tan(2 theta)tan(theta) =1`.
But, `tan(2 theta) = (2 tan(theta))/(1-tan^2(theta))`, so you have to solve
`2 tan^2(theta) = 1-tan^2(theta) iff tan(theta) =1/sqrt(3)`
Finally, `theta = arctan(1/sqrt(3)) = pi/6`.

Graph.

1) On `[0,pi/4]` :

2) On `[-pi/4 , pi/4]` (with `Ox`-symmetry) :

3) Finally the all curve with `O`-symmetry :