Orthogonal trajectories

[JJ007]

\(\displaystyle x^2+y^2=cx^3\)

How do you differentiate this?
Do you use separation of variables afterwords?

Thanks

 

[Deleted member 4993]

\(\displaystyle x^2+y^2=cx^3\)

How do you differentiate this?
Do you use separation of variables afterwords?

Thanks

Using implicit differentiation.

suppose I needed to find y' from the following function:

x[sup:1drcxtez]2[/sup:1drcxtez] + 2x + y[sup:1drcxtez]2[/sup:1drcxtez] = y[sup:1drcxtez]4[/sup:1drcxtez]

then

2x + 2 + 2y*y' = 4y[sup:1drcxtez]3[/sup:1drcxtez]*y'

4y[sup:1drcxtez]3[/sup:1drcxtez]*y' - 2y*y' = 2x + 2

y' = (x + 1)/[y*(2y[sup:1drcxtez]2[/sup:1drcxtez] - 1)]

Use the same method for your problem

 

[JJ007]

\(\displaystyle x^2+y^2=cx^3\)

How do you differentiate this?
Do you use separation of variables afterwords?

Thanks

Using implicit differentiation.

suppose I needed to find y' from the following function:

x[sup:iwwr8zuz]2[/sup:iwwr8zuz] + 2x + y[sup:iwwr8zuz]2[/sup:iwwr8zuz] = y[sup:iwwr8zuz]4[/sup:iwwr8zuz]

then

2x + 2 + 2y*y' = 4y[sup:iwwr8zuz]3[/sup:iwwr8zuz]*y'

4y[sup:iwwr8zuz]3[/sup:iwwr8zuz]*y' - 2y*y' = 2x + 2

y' = (x + 1)/[y*(2y[sup:iwwr8zuz]2[/sup:iwwr8zuz] - 1)]

Use the same method for your problem

\(\displaystyle \frac{dy}{dx}= \frac{3cx^2-2x}{-2y}\)?

\(\displaystyle c=\frac{x^2+y^2}{-x^3}\)

\(\displaystyle \frac{dy}{dx}=\frac{3(\frac{y^2}{-x^3})-2x}{-2y}\)

\(\displaystyle \frac{dy}{dx}=\frac{2y}{3(\frac{y^2}{-x^3})-2x}\)?

 

[Deleted member 4993]

\(\displaystyle \frac{dy}{dx}= \frac{3cx^2-2x}{-2y}\)? ..............Why negetive "2y"?

\(\displaystyle c=\frac{x^2+y^2}{-x^3}\)..............Why negetive x[sup:3n1lnn11]3[/sup:3n1lnn11]?

\(\displaystyle \frac{dy}{dx}=\frac{3(\frac{y^2}{-x^3})-2x}{-2y}\)

\(\displaystyle \frac{dy}{dx}=\frac{2y}{3(\frac{y^2}{-x^3})-2x}\)?