# Search results

1. ### Clairaut's Equation in Optics

No, this makes sense. I actually came up with this stuff like an hour or two ago, but like you said, it looked so simple, I thought that there was no way that it could be right and that I was missing a crucial step and making dumb mistakes like I usually do. Thank you so much for your help. You...
2. ### Clairaut's Equation in Optics

Looks like a parabola, but I also want to say I've seen something else like this such as cosine.
3. ### Clairaut's Equation in Optics

y = (x^2+K^2)/(2K)

y= 2x-2
5. ### Clairaut's Equation in Optics

So for part L, how would I find y = f(x). Same strategies?
6. ### Clairaut's Equation in Optics

dx/dy = k/((2ky+y^2)^(1/2)) k/((2ky+y^2)^(1/2)) - (-y+((sqrt((y+K)^2 - y^2))^2 +y^2)/sqrt((y+K)^2 - y^2) k/((2ky+y^2)^(1/2)) - (-y+(y+K)^2+y^2-y^2)^(1/2) / sqrt((y+K)^2 - y^2) k/((2ky+y^2)^(1/2)) - (-y+((y+K)^2+y^2-y^2)^(1/2) / sqrt((y+K)^2 - y^2) k/((2ky+y^2)^(1/2)) - (-y+(y+K)) /...
7. ### Clairaut's Equation in Optics

f= y^3 +2y^2 +y
8. ### Clairaut's Equation in Optics

Damn, alright. Even though I know thats right, you're saying to seperate the variables to get the equation to say dy = e^x * dx and then integrate.
9. ### Clairaut's Equation in Optics

I'd differentiate y= e^x which is also e^x thereby showing that e^x = e^x
10. ### Clairaut's Equation in Optics

So you're saying I should come up with an answer of y=((1-k^2)/2k) ?
11. ### Clairaut's Equation in Optics

I have no idea what to do with it. Plugging it back in just proves k^2 + 2ky = "same thing". So I'm lost, and on the last damned step.
12. ### Clairaut's Equation in Optics

And that didn't work either... :frown:
13. ### Clairaut's Equation in Optics

Looking at it again, I'm going to try plugging it back into the equation in H.
14. ### Clairaut's Equation in Optics

And I realize now that that won't work since dx/dy = x' with respect to y...
15. ### Clairaut's Equation in Optics

I take it since I've used every equation besides Clairaut's, I need to plug dx/dy into that to see what curve y is.
16. ### Clairaut's Equation in Optics

If my math is correct I get (dx/dy) = k/(2Ky + K^2)^(1/2). When I differentiated x I came up with .5((y+K)^2-y^2)^(-1/2)*(2(y+K))-2y which equates to: 2K/2((y+K)^2-y^2)^(1/2)) Please tell me I did that right.
17. ### Clairaut's Equation in Optics

^I've come to realize that the above is wrong. It should be x^2 = +,-(y+K)^2 - y^2 I'm still having trouble proving that that answer is the same as (dx/dy) = (-y+(x^2+y^2)^(1/2))/x I differentiated the first equation, and came up with (K/((y+K)^2-y^2)) which doesn't relate to the...
18. ### Clairaut's Equation in Optics

Well, I solved/checked it another way and came up with the same answer. What I'm having problems with now is understanding the answer maple gave. My professor gave us a worksheet that demonstrated what the answer was since he was having problems generating the slope fields. My answer was...
19. ### Clairaut's Equation in Optics

If you come back on and get a chance, you think you could help me with I, more so in see if the answer I obtained is correct. For Part H I solved it as follows: x*(dx/dy)^2 + 2y*(dx/dy) = x w=x^2 .5w^(-1/2)*dw = dx w^(1/2)*(.5w^(-1/2)*(dw/dy))^2 + 2y*.5w^(-1/2)*dw/dx = w^(1/2)...
20. ### Clairaut's Equation in Optics

Alright. Yeah, over analyzing a lot. So part E would be the same thing then? Just manipulate the ODE until I'm able to come up with the identity?
21. ### Clairaut's Equation in Optics

So you're saying that dy/dx = cot(theta). I'm sorry but other than that I don't understand where you are going. I can tell I'm over analyzing already. When asking to "derive the relationship" does that simple translate to, state what this means and not literally differentiate?
22. ### Clairaut's Equation in Optics

Yes, I can understand that.
23. ### Clairaut's Equation in Optics

I understand what you're saying, but my professor never pointed that out. I don't know if it was a typo or if that is how the problem is supposed to read.
24. ### Clairaut's Equation in Optics

Thanks for taking time out to explain it to me man. I really REALLY appreciate it.
25. ### Clairaut's Equation in Optics

As bad as it may sound, I'm having problems understanding part C through E. I've tinkered around with the other portions of the problem, but for some reason, I can't seem to understand those parts. I think I'm over analyzing more than I need to.