What is Clairaut's Equation and How Does it Apply to Optics?

[jnbfive]

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

 

[maze]

which part are you stuck on? This is a pretty cool problem, actually.

 

[jnbfive]

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.

 

[maze]

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.

 

[maze]

Alright, I don't know how long you're going to be online, but let's 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.

 

[jnbfive]

Thanks for taking time out to explain it to me man. I really REALLY appreciate it.

 

[jnbfive]

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.

 

[maze]

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?

 

[jnbfive]

Yes, I can understand that.

 

[maze]

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:

$$tan(\frac{\pi}{2}-\theta)=\frac{dy}{dx}$$

$$tan(\frac{\pi}{2}-\theta)=1/tan(\theta)$$

Can you use this to show what they want you to show?

 

[jnbfive]

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?

 

[maze]

In part d they basically want you to show that dx/dy = tan(theta). So, you just flip each side of the equation, that's all.

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

Post again if other parts are confusing.

 

[jnbfive]

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?

 

[maze]

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.

 

[jnbfive]

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.

 

[maze]

One way to check your answer is to plug it back into the original DE.

 

[jnbfive]

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.

 

[jnbfive]

^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.

 

[maze]

^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.

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?

 

[jnbfive]

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.

 

[maze]

Yes that's correct I believe (you can cancel the 2's btw). Now what are you going to do with it?

 

[jnbfive]

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.

 

[jnbfive]

And I realize now that that won't work since dx/dy = x' with respect to y...

 

[jnbfive]

Looking at it again, I'm going to try plugging it back into the equation in H.

 

[jnbfive]

And that didn't work either...

 

[jnbfive]

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.

 

[flexflow]

Multiply by x both sides

Then, start substitute.

 

[jnbfive]

So you're saying I should come up with an answer of y=((1-k^2)/2k) ?

 

[maze]

Ok, let's go to a simpler example and then expand back to the main problem after the concept is clearer. If I asked:
verify that y=e^x is a solution to the DE (dy/dx) = y by plugging the solution into the DE,

what would you do?

 

[jnbfive]

I'd differentiate y= e^x which is also e^x thereby showing that e^x = e^x