# Average value of a function and such.

1. Mar 2, 2009

### hellothere123

1. The problem statement, all variables and given/known data
The voltage V in an electrical outlet is given as a function of time, t, by the function V = V0 cos(120πt), where V is in volts and t is in seconds, and V0 is a positive constant representing the maximum voltage.
(a) What is the average value of the voltage over 3 seconds?
(b) Engineers do not use the average voltage. They use the root mean square voltage defined by V = √avg of V^2. Find V in terms of V0. (Take the average over 3 seconds.)
(c) The standard voltage in an American house is 110 volts, meaning that V = 110. What is V0?

the V0 is really V_o. and that is pi up there.

any help is appreciated. this is my set up:

(V_o)/3 [sin(120 pi t)/(120pi)][0 to 3]

2. Mar 2, 2009

### csprof2000

(a) The average over 3 seconds can be found by taking the integral of V0 cos(120 PI t) from t=0 to t=3, and dividing your answer by 3. NOTE: you divide by the change in time, not by the change in argument to the cosine function. NOTE: this should come out to be very close to zero.

(b) Same as (a), but now the function you're integrating from 0 to 3 seconds is:
(V0)^2 cos^2(120 PI t)
You may want to use the fact that cos(2x) = cos^2(x) - sin^2(x) = 2cos^2(x) - 1.
Once you get the integral, then just divide by 3, and that's <V^2>.
Then take the sqrt and you're done.

(c) Well,

110 = <V^2> = f(V0)

is the equation you need to solve. You found f(V0) in part (b).