HELP differentation

**mitch8726** · June 3rd 2008, 12:08 PM

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

**TheEmptySet** · June 3rd 2008, 12:36 PM

Originally Posted by mitch8726

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.

**mitch8726** · June 3rd 2008, 12:42 PM

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

**TheEmptySet** · June 3rd 2008, 01:02 PM

Originally Posted by mitch8726

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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