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  1. #1
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    HELP differentation

    i have derivered the equation 340sinwt in respect of t to 340wcoswt then i need 2 do the 2nd derivative i did the and got -340w^2*sinwt when iput t=0.005 in the equation the final answer should be 340 but i get a negative number in the equation w=100Pie
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  2. #2
    Behold, the power of SARDINES!
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    Quote Originally Posted by mitch8726 View Post
    i have derivered the equation 340sinwt in respect of t to 340wcoswt then i need 2 do the 2nd derivative i did the and got -340w^2*sinwt when iput t=0.005 in the equation the final answer should be 340 but i get a negative number in the equation w=100Pie
    I think we need some more information...

    Let me see if I have this

    f(t)=340\sin(\omega t)

    f'(t)=340 \omega \cos(\omega t)

    f''(t)=-340 \omega ^{2} \sin(\omega t)

    Now we evaluate at t=0.005 and get

    f''(0.005)=-340 \omega ^{2} \sin(0.005 \omega )

    I don't understand what you want/need to do from here.

    Are you given that \omega = 100 \pi? or trying to deduce that?

    P.S. \pi is pi, pie is dessert.
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  3. #3
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    i was given that w=100pi i have worked out that the graph has a rate of change 0 at t=0.005 but have to find out the maximum which i already know is 340 i just cant get it to work
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  4. #4
    Behold, the power of SARDINES!
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    Quote Originally Posted by mitch8726 View Post
    i was given that w=100pi i have worked out that the graph has a rate of change 0 at t=0.005 but have to find out the maximum which i already know is 340 i just cant get it to work
    To find the maximum of a fundtion you find the critical values of the first derivatve by setting it equal to zero.

    0=340 \omega \cos(\omega t) \iff \cos(\omega t)=0 \iff \omega t =\cos^{-1}(0)

    \omega t =\frac{\pi}{2}+\pi k, \mbox{ where } k \in \mathbb{N} \iff t=\frac{(2k+1)\pi }{2 \omega}

    If we can use the 2nd derivative test to see if this is a max or a min so we get

    f''\left( \frac{(2k+1)\pi }{2 \omega} \right)=-340 \omega ^{2} \sin\left( \frac{(2k+1)\pi }{2} \right)=-340\omega^{2}(-1)^k=340\omega(-1)^{k+1}

    Now if k is even f'' is negative so its a max, if k is odd f'' is postive so its a min

    Plugging our critical value into the f we get

     <br />
f(\frac{(2k+1)\pi }{2 \omega})=340\sin\left( \frac{(2k+1)\pi }{2 } \right)=340(-1)^k<br />

    From our prevoius observation we know that max occurs when k is even and the min when k is odd.

    I hope this helps.

    Good luck.
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