## zonazoo 2 years ago Find the orthogonal trajectories for the family of curves. x = Cy^4

1. abb0t

Find $\frac{ ∂y }{ ∂x }$ remember two curves are orthogonal if the slopes of their tangent lines at their intersection points are negative reciprocals.

2. zonazoo

I guess I just dont understand how to get that. I am looking at the example and in the example they use the problem y^2 = Cx^3 and they rewrite it as y' = 3y/2x I don't understand how to get to that point from the example or from my problem I listed above.

3. abb0t

Well, they found the p. derivative, but it must be in terms of x.

4. zonazoo

how is the derivative of that equal to what they gave.

5. zonazoo

@zepdrix can you help?

6. zepdrix

Hmm I haven't seen this type of problem before. I think I understand what they did in the example, lemme see if I can explain it.

7. zonazoo

I am okay once I find y'.... but I don't understand how to get that in the example or in the problem I have.

8. zepdrix

$\large y^2=Cx^3 \qquad \rightarrow \qquad \color{royalblue}{C=\frac{y^2}{x^3}}$Keep that blue piece in mind, we'll need it later. Differentiating the black equation with respect to x,$\large 2yy'=3Cx^2$ Solving for y',$\large y'=\frac{3x^2}{2y}\color{royalblue}{C} \quad = \quad \frac{3x^2}{2y}\color{royalblue}{\frac{y^2}{x^3}}$

9. zepdrix

Oh you got this figured out already? :D hmm

10. zepdrix

$\large x=Cy^4 \qquad \rightarrow \qquad \color{royalblue}{C=\frac{x}{y^4}}$Derivative of black part,$\large 1=4Cy^3y'\qquad \rightarrow \qquad y'=\frac{1}{\color{royalblue}{C}4y^3}$ $\large y'=\frac{1}{4y^3}\color{royalblue}{\frac{y^4}{x}}$ $\large y'=\frac{y}{4x}$ Hmm what do we need to do from here :O

11. zepdrix

Or is this related to the example? Maybe I misunderstood lol

12. zonazoo

got it... thank you... are you asking me how to solve the problem now? or if what we did solved the problem?

13. zepdrix

I'm just not familiar with these types of problems. Wasn't sure if that was what you were looking for :D lol Grr why didn't we cover this in Diff EQ -_-

14. zonazoo

ohhh, well once we have y', we are supposed to replace y' with -1/y'... since they are orthogonal... we then solve for y' again... so we would get y' = -4x/y then we have to separate the variables... y y' = -4x and integrate y^2/2 = -2x^2 + C then move the x's over y^2/2 + 2x^2 = C and we are done... just have to graph it now

15. zonazoo

I just didnt get how to find the y' first, but I do now... thank you.

16. zepdrix

oo interesting :o