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wcaprar Group Title

I just don't seem to be able to understand how to get the second derivative of a parametric equation. Can someone walk me through it?

  • 2 years ago
  • 2 years ago

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  1. wcaprar Group Title
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    Okay, my parametric equation is x=3+3t^2, y+3t^2+t^3. to find dy/dx you divide (dy/dt)/(dx/dt). Which I can do no problem.

    • 2 years ago
  2. wcaprar Group Title
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    For the first derivative I got for 1-((3t^2)/(4t)). Which the online homework problem says is correct.

    • 2 years ago
  3. wcaprar Group Title
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    To find the second derivative I have to find d/dt (dy/dx). I separated the two parts and got 0 for the one, of course, and then used the quotient rule on the -3t^2/4t part. Is this correct?

    • 2 years ago
  4. amistre64 Group Title
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    a paramentric eqution is also a vector equation: r(t) = <x(t),y(t)> right?

    • 2 years ago
  5. wcaprar Group Title
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    Yes

    • 2 years ago
  6. amistre64 Group Title
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    x=3+3t^2 , y=3t^2+t^3 x'=6t , y=6t+3t r'(t) = < 6t , 6t+3t^2 > which is what youve got for the tangent

    • 2 years ago
  7. wcaprar Group Title
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    Lol, I mixed two problems in what I explained, but yes, go on

    • 2 years ago
  8. amistre64 Group Title
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    r''(t) is just the next derivative. r''(t) = <6, 6+6t> if im reading it right

    • 2 years ago
  9. amistre64 Group Title
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    http://tutorial.math.lamar.edu/Classes/CalcII/ParaTangent.aspx might help

    • 2 years ago
  10. amistre64 Group Title
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    bout 3/4 of the way down it goes into 2nd derivative

    • 2 years ago
  11. wcaprar Group Title
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    Right, but the question is asking for it in a certain form. Here is the question:

    • 2 years ago
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  12. amistre64 Group Title
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    a tiff? really?

    • 2 years ago
  13. amistre64 Group Title
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    you got that in a jpeg?

    • 2 years ago
  14. wcaprar Group Title
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    Lol, here it is in jpg

    • 2 years ago
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  15. wcaprar Group Title
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    I just sliced a portion of the window to get the picture

    • 2 years ago
  16. amistre64 Group Title
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    \[\frac{d^2y}{dx^2}=\frac{d}{dx}(\frac{dy}{dx})=\frac{\frac{d}{dt}(\frac{dy}{dx})}{\frac{dx}{dt}}\]

    • 2 years ago
  17. amistre64 Group Title
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    soo dy/dx = 1+ t/2 stick this in the derivative thing for: 1/2 right?

    • 2 years ago
  18. amistre64 Group Title
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    \[\frac{d}{dt}(\frac{dy}{dx})\to \frac{d}{dt}(1+t/2)\]

    • 2 years ago
  19. wcaprar Group Title
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    Yeah, you're right. But what I have is a dy/dx = 1-3t^2/4t

    • 2 years ago
  20. amistre64 Group Title
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    or the right dy/dx .. lol

    • 2 years ago
  21. wcaprar Group Title
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    Sorry, I starting typing the first, which is pretty easy when I was actually going for a second problem. But this is the spot I always get confused. With that dy/dx I would lose the one and use quotient rule to solve for -3t^2/4t, wouldn't I? But when I do that it says I got the problem wrong

    • 2 years ago
  22. amistre64 Group Title
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    reduce it by t and you got -3t/4 which is simpler to derive

    • 2 years ago
  23. amistre64 Group Title
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    not reduce ..... simplify

    • 2 years ago
  24. wcaprar Group Title
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    Oh my Gosh!! I can't believe I wasn't seeing that! Thank you!

    • 2 years ago
  25. amistre64 Group Title
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    youre welcome

    • 2 years ago
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