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zonazoo
Group Title
(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
 one year ago
zonazoo Group Title
(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
 one year ago

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azolotor Group TitleBest ResponseYou've already chosen the best response.0
Multiply it out.
 one year ago

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

AccessDenied Group TitleBest ResponseYou've already chosen the best response.2
If 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)
 one year ago

zonazoo Group TitleBest ResponseYou've already chosen the best response.0
i 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.
 one year ago

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

dumbcow Group TitleBest ResponseYou've already chosen the best response.0
http://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
 one year ago

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

AccessDenied Group TitleBest ResponseYou've already chosen the best response.2
Yea, 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 \(A = \pm \sqrt{C} \)
 one year ago

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

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

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

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

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

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

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

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

zonazoo Group TitleBest ResponseYou've already chosen the best response.0
i 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.
 one year ago

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

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

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

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

zonazoo Group TitleBest ResponseYou've already chosen the best response.0
ohhh 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.
 one year ago

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

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

zonazoo Group TitleBest ResponseYou've already chosen the best response.0
sweet, it worked...
 one year ago

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

zepdrix Group TitleBest ResponseYou've already chosen the best response.0
oh cool! c:
 one year ago

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

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

zonazoo Group TitleBest ResponseYou've already chosen the best response.0
https://s3.amazonaws.com/grapher/exports/vg28uobv4e.png
 one year ago

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