i have answered the question of the time to reach the greatest voltage rate of change with the equation v=100sin (200pit+pi/4). My answer is t=-0.00125. However i have to evaluate t from the following expression, d2v/dt2 = - (4x10^6 pi/2)sin (200 pit + pi/4) = 0. How do i do this?

Question

rational · Answer

take the integral

rational · Answer

$\large   \frac{d^2v}{dt^2} = -(4*10^6\pi/2) \sin (200 \pi t + \pi/4)$

rational · Answer

$\large   \frac{dv}{dt} = \int   -(4*10^6\pi/2) \sin (200 \pi t + \pi/4) dt$

anonymous · Answer

i dont understand how i re-write it because my next part was dv/dt is max when cos(200pit + pi/4) is 1 or -1 so t =-1/800 ot 1/800 which gives t= -0.00125 or 0.00125

ganeshie8 · Answer

yes, take the integral first

ganeshie8 · Answer

$\large   \frac{dv}{dt} = \int   -(4*10^6\pi/2) \sin (200 \pi t + \pi/4) dt$

$\large   ~~~~ =   -(4*10^6\pi/2)\int \sin (200 \pi t + \pi/4) dt$

$\large   ~~~~ =   [4*10^6\pi/(2*200\pi)] \cos (200 \pi t + \pi/4) $

$\large   ~~~~ =   10^4 \cos (200 \pi t + \pi/4) $

ganeshie8 · Answer

this reaches max-value, when cos term becomes 1

ganeshie8 · Answer

so u need to solve :
$\cos (200 \pi t + \pi/4) = 1$

for $t$

ganeshie8 · Answer

if that makes any sense...

anonymous · Answer

struggling with this sorry, how do i make cos (200pit+pi/4) show for t

ganeshie8 · Answer

take $\cos^{-1}$ both sides

ganeshie8 · Answer

$\cos (200 \pi t + \pi/4) = 1$

take $\cos^{-1}$ both sides :

$\cos^{-1}\cos (200 \pi t + \pi/4) = \cos^{-1}(1)$

$200 \pi t + \pi/4 = 0$


solve $t$

anonymous · Answer

so is the answer t = -0.00125 or 0.00125seconds?

anonymous · Answer

thanks for all your help

ganeshie8 · Answer

yes, becoz cos(x) = cos(-x)

u need to take both positive and negative values

ganeshie8 · Answer

u wlc :)