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Clairaut's Equation in Optics

  1. Jul 26, 2008 #1
    I need help solving this problem. It's a class assignment for an end of the term grade and I'm stuck. PLEASE HELP.
     

    Attached Files:

  2. jcsd
  3. Jul 26, 2008 #2
    which part are you stuck on? This is a pretty cool problem, actually.
     
  4. Jul 26, 2008 #3
    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.
     
  5. Jul 26, 2008 #4
    Ahh ok. When you first posted, I solved it by using vectors for a parameterized curve c(t) and its vector derivative c'(t) and skipped all the trig stuff, but I will take a look at it again.
     
  6. Jul 26, 2008 #5
    Alright, I don't know how long you're going to be online, but lets work through it.

    Ok, so I think their results for part c are wrong. Here is a diagram showing auxilary lines I drew, and on the diagram you can see tan(phi) = x/y (or equivalently, tan(pi/2-phi)=y/x).
    http://img70.imageshack.us/img70/5766/clairut1wg5.png

    Then taking the derivative dy/dx=tan(pi/2-phi), but here is where they get it wrong: they say dy/dx=tan(pi/2-theta), but it should be dy/dx=tan(pi/2-2*theta) since phi = 2*theta.
     
  7. Jul 26, 2008 #6
    Thanks for taking time out to explain it to me man. I really REALLY appreciate it.
     
  8. Jul 26, 2008 #7
    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.
     
  9. Jul 26, 2008 #8
    ahh ok scratch the 2nd part of what i said, its incorrect. The way it is written on the sheet is correct, have a look at the diagram modified a little bit. You can see dx/dy = tan(theta) therefore dy/dx = tan(pi/2-theta).

    http://img237.imageshack.us/img237/9184/clairut2qx8.png

    Does this make any more sense now?
     
    Last edited: Jul 26, 2008
  10. Jul 26, 2008 #9
    Yes, I can understand that.
     
  11. Jul 26, 2008 #10
    Ok on to part d. So now we have basically 2 equations - one from part c, and the new identity they give you in part d. Altogether:

    [tex]tan(\frac{\pi}{2}-\theta)=\frac{dy}{dx}[/tex]

    [tex]tan(\frac{\pi}{2}-\theta)=1/tan(\theta)[/tex]

    Can you use this to show what they want you to show?
     
  12. Jul 26, 2008 #11
    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?
     
  13. Jul 27, 2008 #12
    In part d they basically want you to show that dx/dy = tan(theta). So, you just flip each side of the equation, thats all.

    In these parts, you are basically just manipulating equations to get the result they want.

    Post again if other parts are confusing.
     
  14. Jul 27, 2008 #13
    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?
     
  15. Jul 27, 2008 #14
    Yeah, you basically have the equations from part c and d, and then the new identity they give you in part e, and you just substitute and solve stuff until you get the result they want you to show.
     
  16. Jul 27, 2008 #15
    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)

    .25w^(-1/2)*(dw/dy)^2 + y*w^(-1/2)*(dw/dy) = w^(1/2)

    .25(dw/dy)^2 +y(dw/dy) =w <--- I'm assuming this is the correct answer for H

    Now when solving Part I, I can treat (dw/dy) as if it were a variable, say v, and complete the square,

    (dw/dy)^2 + 4*y*(dw/dy) + 4y^2 = 4w+4y^2

    Now assuming that z = w+ y^2, that would mean that d/dy = dw/dy + 2y correct?

    So that would mean dz/dy = dw/dy +2y

    (dw/dy + 2y)^2 = 4(w + y^2)

    (dz/dy)^2 = 4z

    dz/dy = +,- 2z

    dz/z = +,- 2*dy

    ln(z) = +,- 2y + C

    z=e^+,-2y + C

    z=Ke^+,-2y

    x^2 + y^2 = Ke^+,-2y

    x= +,- (Ke^+,-2y) - y^2)^(1/2)

    Basically I'm asking if what I did makes sense/ is correct.
     
  17. Jul 27, 2008 #16
    One way to check your answer is to plug it back into the original DE.
     
  18. Jul 29, 2008 #17
    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: +,- (Ke^+,-2y) - y^2)^(1/2)

    The answer that maple gave is attached in an image.

    Lines 2 and 3, I believe represent the answers. I just don't know how to show that mine is equal to that.
     

    Attached Files:

  19. Jul 30, 2008 #18
    ^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 dx/dy listed above. Did I differentiate wrong or what? This is the only part that I have left. I know that the curve is a parabolic, somewhat of an ellipse in the way of concavity, I just need to prove these two are equal.
     
  20. Jul 30, 2008 #19
    Thats what I get.

    What happens when you plug x = sqrt((y+K)^2 - y^2) into the equation (dx/dy) = (-y+(x^2+y^2)^(1/2))/x and then simplify?
     
  21. Jul 30, 2008 #20
    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.
     
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