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anonymous
 3 years 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???
anonymous
 3 years 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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anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0did that, but how do you separate the xy term?

AccessDenied
 3 years 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)

anonymous
 3 years 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.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0This has to be implicitly differentiated

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

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

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

anonymous
 3 years 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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0and it wont let me graph that.

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

zepdrix
 3 years 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
 3 years ago
Best ResponseYou've already chosen the best response.0?? i have graph in wolfram link

anonymous
 3 years 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
 3 years 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

anonymous
 3 years 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
 3 years ago
Best ResponseYou've already chosen the best response.0yah it sure didn't :( hmm

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

anonymous
 3 years 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.

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

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

zepdrix
 3 years 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.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0im trying to get the link of the graph in here

zepdrix
 3 years 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

anonymous
 3 years 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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