v=340*e^-2t sin wt where f=4Hz so w= 8pi 1) find the rate of change of v with respect to t at t=15ms 2) sketch the waveform of v for t=0 to 1s
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Originally Posted by bobbym v=340*e^-2t sin wt where f=4Hz so w= 8pi 1) find the rate of change of v with respect to t at t=15ms 2) sketch the waveform of v for t=0 to 1s $\displaystyle v=340e^{-2t}\sin(\omega t)$ $\displaystyle \frac{dv}{dt}=340[-2e^{-2t}\sin(\omega t)+e^{e^{-2t}}\omega\cos(\omega t)]=340e^{-2t}[\omega \cos(\omega t)-2\sin(t)]$ Just evalute from here Remember that 15ms =0.015s Here is a graph
did u use the chain rule to do the 1st part if so what are the steps for it please
when i differentiate v=340*e^-2t sin wt where f=4Hz so w= 8pi with repect to t do i used the chain rule if so what are the steps to differentiating it
Originally Posted by bobbym did u use the chain rule to do the 1st part if so what are the steps for it please $\displaystyle v=340e^{-2t}\sin(\omega t)$ is a product so the product rule was used.
what are the steps to doing the product rule on this equation please
Originally Posted by bobbym what are the steps to doing the product rule on this equation please Are you familiar with the product rule? Have you used it before?
i have used it a few times and get the basics of it but have not done 1 this complicated
Originally Posted by bobbym i have used it a few times and get the basics of it but have not done 1 this complicated h(t) = f(t) g(t) => h'(t) = f(t) g'(t) + f'(t) h(t). In your case f(t) = 340 e^{-2t} and h(t) = sin (wt). Get f'(t) and g'(t). Substitute f(t), f'(t), g(t) and g'(t) into the rule.
what wud that look like with my figures in plz
Originally Posted by bobbym what wud that look like with my figures in plz Go back and read post #2.
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