(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

- katieb

I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!

At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your **free** account and access **expert** answers to this

and **thousands** of other questions

- anonymous

@zepdrix

- anonymous

Multiply it out.

- anonymous

did that, but how do you separate the xy term?

Looking for something else?

Not the answer you are looking for? Search for more explanations.

## More answers

- AccessDenied

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)

- anonymous

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.

- anonymous

This has to be implicitly differentiated

- dumbcow

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

- dumbcow

oh sorry , i plugged in 4 for C in wolfram example

- AccessDenied

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 \(A = \pm \sqrt{C} \)

- zepdrix

You figure this one out yet zon? :D

- anonymous

I am putting it in that graphing calculator now to see if it worked.

- anonymous

well i guess i did something wrong, i ended up getting ln y - ln x = C

- anonymous

and it wont let me graph that.

- zepdrix

\[\huge \ln\left(\frac{y}{x}\right)=C \qquad \rightarrow \qquad e^{\ln\left(\frac{y}{x}\right)}=e^C\] \[\large \frac{y}{x}=e^C \qquad \rightarrow \qquad \color{royalblue}{y=e^cx}\]

- zepdrix

Try graphing that blue one maybe :o hmm

- zepdrix

Where are these natural logs coming from? :D I don't understand what you did XD

- dumbcow

?? i have graph in wolfram link

- anonymous

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.

- zepdrix

oh the next step is to take the integral? Hmm yah i guess I'm coming up with that answer also then :o

- anonymous

when i graphed that blue equation it was a straight line.. but didnt look to be orthogonal to the originals

- zepdrix

yah it sure didn't :( hmm

- dumbcow

??? integral ... i thought the point was to find dy/dx

- anonymous

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.

- anonymous

back where it was y' = y/x, it needed to be -1/y' = y/x

- anonymous

so now i get 1/2x^2 + 1/2y^2 = C

- anonymous

sweet, it worked...

- zepdrix

Multiply both sides by 2, get rid of the halves. Just absorb the 2 into the C. Ooo you have a circle :O interesting.

- zepdrix

oh cool! c:

- anonymous

im trying to get the link of the graph in here

- zepdrix

Upper right corner, theres a blue button, it's a like...a square with an arrow in it

- anonymous

https://s3.amazonaws.com/grapher/exports/vg28uobv4e.png

- anonymous

anyways, there is the picture... so it worked out... that was a tough one... thanks for all the help.

Looking for something else?

Not the answer you are looking for? Search for more explanations.