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inkyvoyd

  • 2 years ago

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?"

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  1. inkyvoyd
    • 2 years ago
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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?

  2. inkyvoyd
    • 2 years ago
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    @hartnn , can you help me please?

  3. inkyvoyd
    • 2 years ago
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    I just need to know what I should be calculating for acceleration, and why. thanks!

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

  5. inkyvoyd
    • 2 years ago
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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?

  6. hartnn
    • 2 years ago
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    if x is position, dy/dx gives velocity and d2y/dx2 gives the acceleration. so, it does.

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

  8. hartnn
    • 2 years ago
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    oh, yes, sorry....

  9. inkyvoyd
    • 2 years ago
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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 >.<

  10. inkyvoyd
    • 2 years ago
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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

  11. hartnn
    • 2 years ago
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    i am sure , just realised , that d2y/dx2 is NOT acceleration

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

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

  14. inkyvoyd
    • 2 years ago
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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. >.<

  15. hartnn
    • 2 years ago
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    d2y/dx2 gives change in dy/dx w.r.t x which is in no way, acceleration

  16. inkyvoyd
    • 2 years ago
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    because the change in y with respect to x isn't velocity right?

  17. hartnn
    • 2 years ago
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    yeah, it isn't velocity , its just the slope

  18. inkyvoyd
    • 2 years ago
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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 >.<

  19. hartnn
    • 2 years ago
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    that equality doesn't hold!

  20. inkyvoyd
    • 2 years ago
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    it does, but it has nothing to do with the definition of acceleration in two dimensions

  21. inkyvoyd
    • 2 years ago
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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.

  22. hartnn
    • 2 years ago
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    you know the right thing,thats enough

  23. inkyvoyd
    • 2 years ago
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    my grade is hurt though >.<

  24. inkyvoyd
    • 2 years ago
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    Ah well I'll think of something - thank you very much!

  25. hartnn
    • 2 years ago
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    hmm...welcome ^_^

  26. amistre64
    • 2 years ago
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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

  27. hartnn
    • 2 years ago
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    product rule for y'' ?

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