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Math Help - Differential help!

  1. #1
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    Differential help!

    y(t)=(3/2)*sin(12*pi*t)

    y't=(3/2)*cos(12*pi*t)

    Is that correct?
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  2. #2
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by No Logic Sense View Post
    y(t)=(3/2)*sin(12*pi*t)

    y't=(3/2)*cos(12*pi*t)

    Is that correct?
    Close... You forgot to apply the chain rule! You need to multiply \frac{d}{\,dt}\left[12\pi t\right]=\dots to your answer.

    --Chris
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  3. #3
    Moo
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    Hello,
    Quote Originally Posted by No Logic Sense View Post
    y(t)=(3/2)*sin(12*pi*t)

    y't=(3/2)*cos(12*pi*t)

    Is that correct?
    Nope. You forgot to use the chain rule

    [f(g(x))]'=g'(x)f'(g(x))

    Here, g(t)=12 \pi t and f(x)=\sin(x)

    Basically, you'll have to multiply by the derivative of 12 pi t, with respect to t
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  4. #4
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    Okay, let me try again:

    y't=(3/2)*cos(t)

    Is that correct?
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  5. #5
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    No wait, I can do it!

    If 12 * pi * t = u

    and sin(u) = y(t)

    du/dy = -sin(u)

    dt/dy = t

    y'(t)=(3/2) * t * -sin(12*pi*t)

    Is that now correct?
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  6. #6
    Moo
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    Quote Originally Posted by No Logic Sense View Post
    No wait, I can do it!

    If 12 * pi * t = u

    and sin(u) = y(t)

    du/dy = -sin(u)

    dt/dy = t

    y'(t)=(3/2) * t * -sin(12*pi*t)

    Is that now correct?
    Hmmm... There is a confusion... Pardon me if it is my fault


    Get back to my formula.

    g(t)=12 \pi t \implies g'(t)=12 \pi

    f(x)=\sin(x) \implies f'(x)=\cos(x)


    So the formula give :

    y'(t)=\frac 32 \cdot [f(g(t))]'=\frac 32 \cdot g'(t) \cdot f'(g(t))=\frac 32 \cdot 12 \pi \cdot \cos(12 \pi t)=\dots
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  7. #7
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    Oh yeah, you're right.

    No don't worry it wasn't your fault. It has just been ages since I did the whole diffential thing so I thought 12 * pi would dissapear, but of course it's t.

    Thanks for the answers.
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