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 one year ago
(x + y)^2 = Cx^2
I need to get the x's on one side and the y's on the other so i can take the derivative of them... how do i do that with the (x + y)^2???
 one year ago
(x + y)^2 = Cx^2 I need to get the x's on one side and the y's on the other so i can take the derivative of them... how do i do that with the (x + y)^2???

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zonazoo
 one year ago
Best ResponseYou've already chosen the best response.0did that, but how do you separate the xy term?

AccessDenied
 one year ago
Best ResponseYou've already chosen the best response.2If we're trying to avoid implicit differentiation, I wonder if taking the square root of both sides might help. We just have to be careful that we respect the properties of that (the square of either a positive or negative is positive, so we have two cases to deal with)

zonazoo
 one year ago
Best ResponseYou've already chosen the best response.0i like that idea... i dont remember what implicit differentiation is though, so maybe that might be better... I just need to be able to take the derivative of both sides, and I just thought the x's had to be on one side and y's on the other... b/c thats all the other examples are like that.

azolotor
 one year ago
Best ResponseYou've already chosen the best response.0This has to be implicitly differentiated

dumbcow
 one year ago
Best ResponseYou've already chosen the best response.0http://www.wolframalpha.com/input/?i=%28x%2By%29^2+%3D+4x^2 you can use implicit differentiation or solve for y and get dy/dx directly either way works here

dumbcow
 one year ago
Best ResponseYou've already chosen the best response.0oh sorry , i plugged in 4 for C in wolfram example

AccessDenied
 one year ago
Best ResponseYou've already chosen the best response.2Yea, it appears to work either way: http://puu.sh/1Xdb6 A little tricky dealing with the +/ cases, so I replaced it with an "A" for some of the work. :p \(A = \pm \sqrt{C} \)

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.0You figure this one out yet zon? :D

zonazoo
 one year ago
Best ResponseYou've already chosen the best response.0I am putting it in that graphing calculator now to see if it worked.

zonazoo
 one year ago
Best ResponseYou've already chosen the best response.0well i guess i did something wrong, i ended up getting ln y  ln x = C

zonazoo
 one year ago
Best ResponseYou've already chosen the best response.0and it wont let me graph that.

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.0\[\huge \ln\left(\frac{y}{x}\right)=C \qquad \rightarrow \qquad e^{\ln\left(\frac{y}{x}\right)}=e^C\] \[\large \frac{y}{x}=e^C \qquad \rightarrow \qquad \color{royalblue}{y=e^cx}\]

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.0Try graphing that blue one maybe :o hmm

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.0Where are these natural logs coming from? :D I don't understand what you did XD

dumbcow
 one year ago
Best ResponseYou've already chosen the best response.0?? i have graph in wolfram link

zonazoo
 one year ago
Best ResponseYou've already chosen the best response.0i did the first suggestion and took the square root. (x+y)^2 = Cx^2 .... that is the problem. C = (x+y)^2/x^2 but back to the original, i took the square root. x + y = C^.5 x , then i moved all around, took the derivative... and then substituted C back in and I ended up with 1/y = 1/x.... then took the integral of that.

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.0oh the next step is to take the integral? Hmm yah i guess I'm coming up with that answer also then :o

zonazoo
 one year ago
Best ResponseYou've already chosen the best response.0when i graphed that blue equation it was a straight line.. but didnt look to be orthogonal to the originals

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.0yah it sure didn't :( hmm

dumbcow
 one year ago
Best ResponseYou've already chosen the best response.0??? integral ... i thought the point was to find dy/dx

zonazoo
 one year ago
Best ResponseYou've already chosen the best response.0ohhh my... okay i know what i did wrong... i took the y' and not the 1/y' to make it orthogonal... but there are two parts, the first is to get the derivative of the orthogonal trajectories, and then to take the integral to find the general solution.

zonazoo
 one year ago
Best ResponseYou've already chosen the best response.0back where it was y' = y/x, it needed to be 1/y' = y/x

zonazoo
 one year ago
Best ResponseYou've already chosen the best response.0so now i get 1/2x^2 + 1/2y^2 = C

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.0Multiply both sides by 2, get rid of the halves. Just absorb the 2 into the C. Ooo you have a circle :O interesting.

zonazoo
 one year ago
Best ResponseYou've already chosen the best response.0im trying to get the link of the graph in here

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.0Upper right corner, theres a blue button, it's a like...a square with an arrow in it

zonazoo
 one year ago
Best ResponseYou've already chosen the best response.0anyways, there is the picture... so it worked out... that was a tough one... thanks for all the help.
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