anonymous
  • anonymous
Find the orthogonal trajectories for the family of curves. x = Cy^4
Mathematics
jamiebookeater
  • jamiebookeater
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abb0t
  • abb0t
Find \[\frac{ ∂y }{ ∂x }\] remember two curves are orthogonal if the slopes of their tangent lines at their intersection points are negative reciprocals.
anonymous
  • anonymous
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.
abb0t
  • abb0t
Well, they found the p. derivative, but it must be in terms of x.

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anonymous
  • anonymous
how is the derivative of that equal to what they gave.
anonymous
  • anonymous
@zepdrix can you help?
zepdrix
  • 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.
anonymous
  • anonymous
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.
zepdrix
  • 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',\[\large y'=\frac{3x^2}{2y}\color{royalblue}{C} \quad = \quad \frac{3x^2}{2y}\color{royalblue}{\frac{y^2}{x^3}}\]
zepdrix
  • zepdrix
Oh you got this figured out already? :D hmm
zepdrix
  • 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
zepdrix
  • zepdrix
Or is this related to the example? Maybe I misunderstood lol
anonymous
  • anonymous
got it... thank you... are you asking me how to solve the problem now? or if what we did solved the problem?
zepdrix
  • 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 -_-
anonymous
  • anonymous
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
anonymous
  • anonymous
I just didnt get how to find the y' first, but I do now... thank you.
zepdrix
  • zepdrix
oo interesting :o

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