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zonazoo

  • 3 years ago

Find the orthogonal trajectories for the family of curves. x = Cy^4

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  1. abb0t
    • 3 years ago
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    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
    • 3 years ago
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    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
    • 3 years ago
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    Well, they found the p. derivative, but it must be in terms of x.

  4. zonazoo
    • 3 years ago
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    how is the derivative of that equal to what they gave.

  5. zonazoo
    • 3 years ago
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    @zepdrix can you help?

  6. zepdrix
    • 3 years ago
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    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
    • 3 years ago
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    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
    • 3 years ago
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    \[\large y^2=Cx^3 \qquad \rightarrow \qquad \color{royalblue}{C=\frac{y^2}{x^3}}\]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
    • 3 years ago
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    Oh you got this figured out already? :D hmm

  10. zepdrix
    • 3 years ago
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    \[\large x=Cy^4 \qquad \rightarrow \qquad \color{royalblue}{C=\frac{x}{y^4}}\]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
    • 3 years ago
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    Or is this related to the example? Maybe I misunderstood lol

  12. zonazoo
    • 3 years ago
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    got it... thank you... are you asking me how to solve the problem now? or if what we did solved the problem?

  13. zepdrix
    • 3 years ago
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    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
    • 3 years ago
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    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
    • 3 years ago
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    I just didnt get how to find the y' first, but I do now... thank you.

  16. zepdrix
    • 3 years ago
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    oo interesting :o

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