# Envelope of family of curves

1. Jan 30, 2008

### gtfitzpatrick

1. The problem statement, all variables and given/known data

Find envelope of the family of curves x^2cosΘ + y^2sinΘ = a^2 where Θ is the parameter

2. Relevant equations

3. The attempt at a solution
I tried differentiating and putting it = to 0 but this is coming up very messy, is there something i'm not seeing here?thanks
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Jan 30, 2008

### EnumaElish

3. Jan 30, 2008

### gtfitzpatrick

Thanks for getting back to me but i'm not sure i follow. Do i solve 1 for x say then sub it back in and solve 2?

4. Jan 30, 2008

### EnumaElish

That would be one way. Or you can try solving either equation for theta then substitute it into the other equation.

Last edited: Jan 30, 2008
5. Feb 4, 2008

### gtfitzpatrick

I differentiated eq1 with respect to theta and got

-x^2sin theta + y^2cos theta

which is eq 2 but i just cant figure out how to isolate The theta in such a way that i can sub it back into eq 1?

6. Feb 4, 2008

### foxjwill

Try doing a system of equations:

\begin{align} x^2 \cos\theta + y^2 \sin\theta &= a^2\\ y^2 \cos\theta - x^2 \sin\theta &= 0 \end{align}

You can solve it using linear combinations with your "variables" being $$\cos\theta$$ and $$\sin\theta$$.

7. Feb 4, 2008

### gtfitzpatrick

Thanks, for getting back to me but i'm not sure i know what you mean by linear combinations?

8. Feb 4, 2008

### foxjwill

Did you ever learn how to solve systems of linear combinations? For example, let's say we have the system

\begin{align} x + 2y &= 6\\ 3x + 7y &= 0 \end{align}

We can solve this by multiplying (1) by -3 then adding the two equations together resulting in:

$$-6y + 7y = 6 + 0$$

which you can then solve for y. Once you have y, you can plug it into one of the original equations and solve for x.

For your problem, you can use an analogous method by solving, instead of for x and y as above, for $$\sin\theta$$ and $$\cos\theta.$$

9. Feb 6, 2008

### gtfitzpatrick

i've solved the 2 eqs simultaneously by multiplying eq 1 by x^2 and eq 1 by y^2
which gives me

cos theta=a^2x^2/x^4=y^4

and throwing that back in for

sin theta = a^2y^2/y^4+x^4

Im not sure is this the answer? or where do i go from there?
thanks

10. Feb 8, 2008

### gtfitzpatrick

since theta is the parameter i think i should be trying to eliminate it from the equations

tan theta = y^2/x^2

theta = tan ^-1 y^2/x^2

which i sub back in to give

x^2cos (tan^-1 y^2/x^2) + y^2sin (tan^-1 y^2/x^2) = a^2

i'm going mad trying to work this out!!!

11. Feb 8, 2008

### EnumaElish

What identity can you use to tie sin theta and cos theta together?

Suppose I thought there is the identity "sqrt[sin theta] * sqrt[cos theta] = 2 for any theta."

Then I can write sqrt[a^2x^2/(x^4+y^4)] * sqrt[a^2y^2/(x^4+y^4)] - 2 = 0, which would be the envelope.

12. Feb 9, 2008

### gtfitzpatrick

thanks for being so patiant with me but i dont understand where
"sqrt[sin theta] * sqrt[cos theta] = 2 comes from

13. Feb 9, 2008

### HallsofIvy

Staff Emeritus
There is no such identity. Enuma Elish said "suppose I thought that" because he was trying to not to do the problem for you! Do you know of any real identity you could use instead?

14. Feb 9, 2008

### gtfitzpatrick

cos^2theta +sin^theta = 1?

15. Feb 9, 2008

### gtfitzpatrick

thanks for that got it now, i hope