Homework Help: Parametric Expression

1. Feb 25, 2006

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.)

2. Feb 25, 2006

benorin

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

3. Feb 25, 2006

Fermat

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

4. Feb 25, 2006

benorin

that explains a lot

5. Feb 25, 2006

benorin

I'll fix the question

Modified quote:

note that the c is no longer under the square root

Last edited: Feb 26, 2006
6. Feb 25, 2006

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".

7. Feb 26, 2006

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?

Last edited: Feb 26, 2006
8. Feb 26, 2006

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

9. Feb 26, 2006

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.

Last edited: Feb 26, 2006
10. Feb 27, 2006

jamesbob

Ok i'm still stuck on this and get the feeling im 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 im sure its wrong already.

Last edited: Feb 27, 2006
11. Feb 27, 2006

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

12. Feb 27, 2006

benorin

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

13. Feb 27, 2006

benorin

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

14. Feb 27, 2006

benorin

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

15. Feb 27, 2006

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

Last edited: Feb 27, 2006
16. Feb 28, 2006

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

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

Last edited: Feb 28, 2006
17. Feb 28, 2006

jamesbob

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