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inkyvoyd Group Title

Definition of acceleration (calculus) "The position of an object is described by the parametric equations x=lnt and y=5t^2. What is the acceleration of the object in m/sec2 when t=2?"

  • one year ago
  • one year ago

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  1. inkyvoyd Group Title
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    I have gotten this problem wrong for calculating \(\sqrt{\frac{d^2y}{dt^2}+\frac{d^2x}{dt^2}}\) instead of \(\frac{d^2y}{dx^2}\) What am I doing wrong?

    • one year ago
  2. inkyvoyd Group Title
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    @hartnn , can you help me please?

    • one year ago
  3. inkyvoyd Group Title
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    I just need to know what I should be calculating for acceleration, and why. thanks!

    • one year ago
  4. hartnn Group Title
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    i would have calculated d^2y/dx^2 because i didn't know the sqrt formula :P

    • one year ago
  5. inkyvoyd Group Title
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    But, d^2y/dx^2 is the movement of y with respect to x - it doesn't say the particle is accelerating does it?

    • one year ago
  6. hartnn Group Title
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    if x is position, dy/dx gives velocity and d2y/dx2 gives the acceleration. so, it does.

    • one year ago
  7. inkyvoyd Group Title
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    no, dx/dt gives velocity - velocity is the derivative of distance with respect to time right?

    • one year ago
  8. hartnn Group Title
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    oh, yes, sorry....

    • one year ago
  9. inkyvoyd Group Title
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    I mean, the units don't add up when you do d^2y/dx^2 if you have y and x in parametric terms of t >.<

    • one year ago
  10. inkyvoyd Group Title
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    I've tried to email my teacher about it like 3 times but I don't think she's read what I've said - she just keeps repeating that the acceleration is d^2y/dx^2

    • one year ago
  11. hartnn Group Title
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    i am sure , just realised , that d2y/dx2 is NOT acceleration

    • one year ago
  12. inkyvoyd Group Title
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    Okay - how should I prove to my teacher that d^2y/dx^2 is NOT acceleration?

    • one year ago
  13. hartnn Group Title
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    units.....x and y are distances..... d2y/dx2 is dimensionless. it doesn't have the unit of acceleration.

    • one year ago
  14. inkyvoyd Group Title
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    Okay - any other ways? I tried emailing her saying units don't add up but I'm not sure if she read. >.<

    • one year ago
  15. hartnn Group Title
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    d2y/dx2 gives change in dy/dx w.r.t x which is in no way, acceleration

    • one year ago
  16. inkyvoyd Group Title
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    because the change in y with respect to x isn't velocity right?

    • one year ago
  17. hartnn Group Title
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    yeah, it isn't velocity , its just the slope

    • one year ago
  18. inkyvoyd Group Title
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    So then my teacher argues that d^2y/dx^2=\(\frac{\frac{d}{dt}(\frac{dx}{dt})}{\frac{dx}{dt}}\) Which unfortunately has a d/t in it >.<

    • one year ago
  19. hartnn Group Title
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    that equality doesn't hold!

    • one year ago
  20. inkyvoyd Group Title
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    it does, but it has nothing to do with the definition of acceleration in two dimensions

    • one year ago
  21. inkyvoyd Group Title
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    yet I'm afraid my teacher will attempt to end the discussion and get back to doing other stuff (it's an online course) by simply telling me to ignore units.

    • one year ago
  22. hartnn Group Title
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    you know the right thing,thats enough

    • one year ago
  23. inkyvoyd Group Title
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    my grade is hurt though >.<

    • one year ago
  24. inkyvoyd Group Title
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    Ah well I'll think of something - thank you very much!

    • one year ago
  25. hartnn Group Title
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    hmm...welcome ^_^

    • one year ago
  26. amistre64 Group Title
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    spose we equate t in terms of x x=lnt ; e^x = t sub in e^x into the t for y y = 5e^(x^2) y' = 10x e^(x^2) y'' = 20x^2 e^(x^2) when t=2, x=ln2 y'' = 20(ln2)^2 e^(ln2^2) = 15.536 or so ....if my idea is correct

    • one year ago
  27. hartnn Group Title
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    product rule for y'' ?

    • one year ago
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is replying to Can someone tell me what button the professor is hitting...

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