1. ## HELP differentation

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

2. 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
$v=340e^{-2t}\sin(\omega t)$

$\frac{dv}{dt}=340[-2e^{-2t}\sin(\omega t)+e^{e^{-2t}}\omega\cos(\omega t)]=340e^{-2t}[\omega \cos(\omega t)-2\sin(t)]$

Just evalute from here Remember that 15ms =0.015s

Here is a graph

3. did u use the chain rule to do the 1st part if so what are the steps for it please

4. ## help do i use the chain rule

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

5. Originally Posted by bobbym
did u use the chain rule to do the 1st part if so what are the steps for it please
$v=340e^{-2t}\sin(\omega t)$ is a product so the product rule was used.

6. what are the steps to doing the product rule on this equation please

7. 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?

8. i have used it a few times and get the basics of it but have not done 1 this complicated

9. 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.

10. what wud that look like with my figures in plz

11. Originally Posted by bobbym
what wud that look like with my figures in plz
Go back and read post #2.