What is the Parametric Expression for the Lemniscate of Bernoulli?

[jamesbob]

The lemniscate of Bernoulli is the curve that is the locus of points the product of whose distances from two fixed centres (called the foci) a distance of 2c apart is the cosntant $$[c^2$$. If the foci have Cartesian coordinates $$(\pmc, 0)$$ the Cartesian equation of the lemniscate is

$$([x-c]^2 + y^2)([x + c]^2 + y^2) = c^4$$​

or

$$(x^2 + y^2)^2 = 2c^2(x^2 - y^2).$$​

a) Show that the lemniscate of Bernoulli may be expressed parametrically by

$$x(t) = \sqrt{2c}\frac{\cost}{1 + \sin^2t}$$, $$y(t) = \sqrt{2c}\frac{costsint}{1 + sin^2t}$$​

where $$t \epsilon[-\pi, \pi).$$ For t out of this interval the curve repeats on itself.)

 

[benorin]

Should read "If the foci have Cartesian coordinates $$(\pm c, 0)$$..."

 

[Fermat]

and x(t) is missing a cos(t) on the numerator

 

[benorin]

that explains a lot

 

[benorin]

I'll fix the question

Modified quote:

The lemniscate of Bernoulli is the curve that is the locus of points the product of whose distances from two fixed centres (called the foci) a distance of 2c apart is the cosntant $$[c^2$$. If the foci have Cartesian coordinates $$(\pm c, 0)$$ the Cartesian equation of the lemniscate is

$$([x-c]^2 + y^2)([x + c]^2 + y^2) = c^4$$​

or

$$(x^2 + y^2)^2 = 2c^2(x^2 - y^2).$$​

a) Show that the lemniscate of Bernoulli may be expressed parametrically by

$$x(t) = \sqrt{2}c \frac{\cos t}{1 + \sin^2t}$$, $$y(t) = \sqrt{2}c\frac{\cos t \sin t}{1 + sin^2t}$$​

where $$t \in[-\pi, \pi).$$ For t out of this interval the curve repeats on itself.)

note that the c is no longer under the square root

 

[benorin]

a) Just plug-in x=x(t) and y=y(t) into either of the given equations and show that it reduces to an identity and note that x(t) and y(t) are periodic functions, having as a funamental period $$t \in [-\pi, \pi).$$. BTW, use \in rather than \epsilon for "element of".

 

[jamesbob]

Thanks very much for fixing my post and for your help! So i just plug

$$x = \sqrt{2c}\frac{cost}{1 + \sin^2t} and y = \sqrt{2c}\frac{costsint}{1 + sin^2t}$$ into the 2 equations?

 

[Fermat]

That's right, as benorin posted. Also remember about the rt(2c). It should be rt(2)*c.

BTW, if you'ld like a challenge, try and find the slope of the curve at the origin

 

[0rthodontist]

Just substituting in doesn't prove it though. It shows that any point of the form (x(t), y(t)) is on the lemniscate--it doesn't show that every point on the lemniscate is of the form (x(t), y(t)).

I'm not sure what the simplest way to show the latter is. One way I can think of is expanding out your second formula for the lemniscate, which will give you a quadratic equation in x^2. Solve that equation for x^2 and from that find how many solutions there are for a given y (which is the number of values of x that are possible for a given y). Then show that for a given y(t), the parametric equation yields the same number of x(t) as there are solutions x for that y.

 

[jamesbob]

Ok I'm still stuck on this and get the feeling I am doing something silly and overcomplicated. I tried fitting those x and y values to the first equation and got:

$$([\sqrt{2}c\frac{cost}{1 + sin^2t} - c]^2 + \sqrt{2}c\frac{costsint}{1 + sin^2t}) \times ([\sqrt{2}c\frac{cost}{1 + sin^2t} + c]^2 + \sqrt{2}c\frac{costsint}{1 + sin^2t})$$

So thismultiplies to:

$$(2c^2\frac{\cos^2t}{1 + \sin^2t} - c^2 + 2c^2\frac{\cos^2t\sin^2t}{(1 + sin^2t)^2}) \times (2c^2\frac{\cos^2t}{(1 + \sin^2t)^2} + c^2 + 2c^2\frac{\cos^2t\sin^2t}{(1 + sin^2t)^2})$$

I have to run, ill continue posting my calculations later, tho I am sure its wrong already.

 

[jamesbob]

Argh why is my latex code not working. There should be cos's and sin's on the top line of the first part

 

[benorin]

If you use \cos or \sin, either leave a space after them and before the argument or use {} around the arguement, for example: \cos t = $$\cos t$$, but \cost = $$\cost$$

 

[benorin]

Sometimes I get just a blank space where an unrecognized function is like \log is OK, but \Log is blank.

 

[benorin]

And, by the way, using the second form of the equation given is far easier...

 

[benorin]

Try the easier equation...

Let us substitute

$$x(t) = \sqrt{2}c \frac{\cos t}{1 + \sin^2t}$$

$$y(t) = \sqrt{2}c\frac{\cos t \sin t}{1 + sin^2t}$$

into

$$(x^2 + y^2)^2 = 2c^2(x^2 - y^2)$$

to get


$$\left[\left( \sqrt{2}c \frac{\cos t}{1 + \sin^2t}\right) ^2 + \left( \sqrt{2}c\frac{\cos t \sin t}{1 + sin^2t}\right) ^2 \right] ^2 = 2c^2\left[\left( \sqrt{2}c \frac{\cos t}{1 + \sin^2t}\right) ^2 - \left( \sqrt{2}c\frac{\cos t \sin t}{1 + sin^2t}\right) ^2 \right]$$

now simplify to get something like 0=0

 

[jamesbob]

Im being really dumb here and just cannot simplify this. So far i get:

$$4c^4\frac{\cos^4 t}{(1 + \sin^2 t)^4} + 4c^4\frac{\cos^4 t\sin^2 t}{(1 + sin^2 t)^4} + 4c^4\frac{\cos^4 t\sin^4 t}{(1 + sin^2 t)^4} = 4c^4(\frac{cos^2 t}{(1 + \sin^2 t)^2} - \frac{\cos^4 t\sin^2 t}{(1 +sin^2 t)^2})$$

This somehow equates

$$\frac{\cos^4 t}{(1 + \sin^2 t)^4} + \frac{\cos^4 t\sin^2 t}{(1 + sin^2 t)^4} + \frac{\cos^4 t\sin^4 t}{(1 + sin^2 t)^4} = \frac{cos^2 t}{(1 + \sin^2 t)^2} - \frac{\cos^4 t\sin^2 t}{(1 +sin^2 t)^2} ?$$

 

[jamesbob]

ok i finally got this, il code it later when i get a chance, but thanks for your help