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FoolForMath

  • 2 years ago

Fool's problem of the day, \(1.\)Find the locus of the intersection of the perpendicular tangents drawn to the curve \(4y^3=27x^2 \). Good luck!

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  1. FoolForMath
    • 2 years ago
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    This is somewhat difficult, took me some time to realize the solution.

  2. rebeccaskell94
    • 2 years ago
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    Where does one start to find the answer to this question ? you can msg me so you don't give it away if you want, I just wanna know HOW to solve it.

  3. rebeccaskell94
    • 2 years ago
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    Cheater. ;P

  4. experimentX
    • 2 years ago
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    y=3 ???

  5. FoolForMath
    • 2 years ago
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    No.

  6. pre-algebra
    • 2 years ago
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    Darn I need to read up on English definitions.ъ :(

  7. ZhangYan
    • 2 years ago
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    man, this is hard questions!!!

  8. inkyvoyd
    • 2 years ago
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    @FoolForMath , what do you mean by perp tangent/

  9. inkyvoyd
    • 2 years ago
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    @FoolForMath ?

  10. asnaseer
    • 2 years ago
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    is it x=0?

  11. FoolForMath
    • 2 years ago
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    Sorry, that's not the right answer.

  12. ladiesman123
    • 2 years ago
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    teehee :)

  13. blockcolder
    • 2 years ago
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    When you say perpendicular tangents, do you mean two tangents lines of the curve that are perpendicular to each other?

  14. cinar
    • 2 years ago
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    http://www.wolframalpha.com/input/?i=plot+4y^3%3D27x^2

  15. FoolForMath
    • 2 years ago
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    Yes, @blockcolder

  16. eliassaab
    • 2 years ago
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    If you try to do the easy problem for y = x^2, you find that the locus is the line y=-1/4.

  17. blockcolder
    • 2 years ago
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    Since the graph is symmetrical w.r.t. the y-axis, I'll consider first x>0. By implicit differentiation, we have: \[4y^3=27x^2\\ 12y^2y'=54x\\ y'=\frac{54x}{12y^2}=\frac{9x}{2y^2}\] Am I on the right track?

  18. FoolForMath
    • 2 years ago
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    Why do you Google Cinar? I am well aware of the answer.

  19. FoolForMath
    • 2 years ago
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    I am removing the spoiler.

  20. FoolForMath
    • 2 years ago
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    @blockcolder: looks good.

  21. dumbcow
    • 2 years ago
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    the set of points that represent the intercepts of the perpendicular tangents match the original function. all the intercepts lie on the same curve it was a little bit of work so i won't post my solution here...the steps were 1) find dy/dx, then the perpendicular slope is -dx/dy 2) solve for line equations for each slope 3) set them equal to find intercept points

  22. asnaseer
    • 2 years ago
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    ok - I misunderstood the question. I took it to mean the intersection of the lines that are perpendicular to the tangents to the given curve.

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