Question

V = 100 sin (200πt + π/4)
need to know the time for voltage to first reach a maximum

Answer 1

can you use derivatives?

Answer 2

what is the maximum value of sin(x)?
what is x when sin(x) is at that maximum?

Answer 3

tried this but think it is wrong
dv/dt = (2 x 104 π) cos (200 π t + π/4)

maximum occurs when dv/dt = 0 and d2v/dt2 is negative

d2v/dt2 = -(4 x 106 π2) sin (200 π t + π/4)

cos (200 π t + π/4) = 0 when t = 0 and when t = 0 d2v/dt2 is negative

when t = 0, v = 100 as sin (200 π + π/t) = 1
dv/dt is a maximum when cos (200 π t + π/4) is either 1 or -1
which occurs when 200 π t + π/4 = 0 or π radians

t is either = -1/800 or 1/800 seconds

t = -0.00125 or 0.00125 seconds

Answer 4

too much work
max of sine is one when input is \(\frac{\pi}{2}\) set the inside part equal to \(\frac{\pi}{2}\) and solve for \(t\)

Answer 5

\[ \large \frac{dV}{dt}=100\cos(200\pi t+\pi/4)200\pi \]

Answer 6

@tomtom - do you understand the reasoning used by @phi and @satellite73 ?

Answer 7

not really

Answer 8

A sine curve looks like this: http://www.wolframalpha.com/input/?i=y%3Dsin%28x%29

notice its value always ranges between 1 and -1

Answer 9

so its maximum value is 1 - agreed?

Answer 10

yes

Answer 11

now what would x need to be for sin(x) to be equal to 1?

Answer 12

are you aware of the fact that \(\sin(90^0)=1\)?

Answer 13

and that \(90^0=\frac{\pi}{2}\) radians

Answer 14

yes

Answer 15

ok, so now all you need to do is to find what value of t will make this equal to 1:\[\sin(200\pi t+\frac{\pi}{4})=1\]which means we need to solve:\[200\pi t+\frac{\pi}{4}=\frac{\pi}{2}\]understand?

Answer 16

yes -wow thank you

Answer 17

yw :)