Trigonometric Derivative Help!?

Question

The voltage, V, in volts, in an electrical outlet is given as a function of time, t, in seconds, by the function V = 156cos(120phit).

a. give an expression for the rate of change of voltage with respect
to time.
b. is the rate of change ever zero? why?
c. What is the maximum value of the rate of change?

Answer 1

The derivative of cosine is -sine. Since the argument of cosine is not t, but t times a coefficient, you have to use the chain rule. So the time derivative of V is 120phi*(-156)sin(120phi*t). I'll leave that little bit of multiplication to you. That's your answer to a. For b, you just need to see if this expression is ever equal to zero. Since the sine function does indeed have zeroes at 0 and pi, you should be able to solve easily for values of t where the argument of sine is equal to those zeroes. For c, you instead need to maximize the function. You may recall that sine has a maximum value of 1 at pi/2, so again you need to solve for the argument of sine having that value.

Answer 2

The derivative is -156*w*sin(w*t) where w = 120*pi. The derivative will be zero for t = 0 and t = n*T/2 (whenever sin(w*t) =0) where T is the period (for this function the period is 1/60 s) and n is an integer. The maximum value for the rate of change is 156*w, or simply the amplitude. This will occur at t = n*T/4 where n is odd (whenever sin(w*t) =1). Also note that cos(w*t - pi/2) = sin(w*t).

Answer 3

a)V=156 cos(120 phi t)
dV/dt =156{ - sin(120 phi t)(120 phi)
=- 18720(phi )sin(120 phi t)....................i
b)The right hand side is zero when t=0,1,2,3.........ii
because 120 phi =60(2 phi) and sin(2pi)=0
This means that dV/dt is always/ever zero
c) Find second derivative of V
d^2V/dt^2 = -18720(phi){120 phi) cos(120 phi t)
for rate of chage maximum, put d^2V/dt^2=0
therefore t=0,1,2,.............After every second it has maximum value

Answer 4

is phit supposed to be pi*t ?