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yuki

  • 5 years ago

x = cos(2t) and y = sin(2t) For this parametric eqn. I would like to find the second derivative, but I don't seem to be doing it right. This is what I tried so far. dx/dt = -2sin(2t) dy/dt = 2cos(2t) thus dy/dx = -cot(2t). for the second derivative, d/dx(dy/dx) = -csc(2t)cot(2t) so d^2/dx^2 [y] = sec^2(x)

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  1. Yuki
    • 5 years ago
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    woops, the last sentence should have been csc^2(x)

  2. anonymous
    • 5 years ago
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    You should understand that\[\frac{d^2y}{dx^2}=\frac{d}{dx}\frac{dy}{dx}=\frac{\frac{d}{dt}(\frac{dy}{dx})}{\frac{dx}{dt}}\]

  3. anonymous
    • 5 years ago
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    \[\frac{dy}{dx}=-\frac{\cos 2t}{\sin 2t}=-\cot 2t\]

  4. Yuki
    • 5 years ago
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    right, so I found dx/dt and dy/dt

  5. Yuki
    • 5 years ago
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    yep

  6. anonymous
    • 5 years ago
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    \[\frac{d}{dt}(-\cot 2t)=2\csc ^2 x\]

  7. anonymous
    • 5 years ago
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    ^^ that's your numerator.

  8. anonymous
    • 5 years ago
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    \[\frac{dx}{dt}=-2\sin 2t\]

  9. anonymous
    • 5 years ago
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    2x in csc^2

  10. Yuki
    • 5 years ago
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    woops, my derivative of cot was completely wrong. let me try it again

  11. anonymous
    • 5 years ago
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    Man I hate this editor.

  12. anonymous
    • 5 years ago
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    Guys who programme this site - stop mucking around with frigging medals and shields and spend your time building something user-friendly!

  13. anonymous
    • 5 years ago
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    \[\frac{d^2y}{dx^2}=\frac{2 \csc^2 2t}{-2 \sin 2t}=-\csc^3 2t\]

  14. anonymous
    • 5 years ago
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    yuki, are you typing a thesis?

  15. Yuki
    • 5 years ago
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    right? I thought that was the answer too. Okay maybe my problem was not set up right. The actual problem is. The motion of a particle with x =cos(2t) and y=sin(2t). What is the magnitude of the acceleration at any time t ? I thought the acceleration had to do with f".

  16. anonymous
    • 5 years ago
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    The acceleration vector is the double time derivative of each coordinate. The acceleration is the magnitude of that vector.

  17. anonymous
    • 5 years ago
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    \[a=(\frac{d^2x}{dt^2},\frac{d^2y}{dt^2})\]

  18. anonymous
    • 5 years ago
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    \[|a|=\sqrt{x''(t)^2+y''(t)^2}\]

  19. anonymous
    • 5 years ago
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    \[a=(-4\cos 2t, -4\sin 2t)=-4r\]where r is your position vector.

  20. anonymous
    • 5 years ago
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    The acceleration points in a direction opposite to your position vector.

  21. Yuki
    • 5 years ago
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    Do we learn this in Calculus? It sounds more like multivariable. Anyways, I did get the answer and it was huge help. Thanks ! :)

  22. anonymous
    • 5 years ago
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    \[|a|=4\]

  23. anonymous
    • 5 years ago
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    This is calculus.

  24. anonymous
    • 5 years ago
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    This is calculus of parametric equations. You need to parametrize a path since in the x-y plane, a path can form loops and cut into itself, which isn't a function. If you parametrize each of the coordinates, you set it up so that each coordinate is a function of some parameter, with that parameter allowing you to plot your x-y points, but at the same time, the x and y are *functions* of t, so you may perform calculus.

  25. anonymous
    • 5 years ago
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    Hope it helped, yuki :)

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