# Rate of change problem

• Jan 13th 2009, 08:46 PM
VkL
Rate of change problem
The voltage, V , in volts, is an electrical outlet given as a function of time, t, in seconds, by the function V = 156cos(120[pie]t)

a.) Give an expression for the rate of change of the voltage with respect to time.

b.) What is the maximum value of the rate of change.
• Jan 13th 2009, 09:14 PM
ThePerfectHacker
Quote:

Originally Posted by VkL
The voltage, V , in volts, is an electrical outlet given as a function of time, t, in seconds, by the function V = 156cos(120[pie]t)

a.) Give an expression for the rate of change of the voltage with respect to time.

The rate of change is given by $V ' (t) = 156 (120\pi t)' [ - \sin (120 \pi t) ] = - 150\cdot 120 \pi \sin (120 \pi t)$

Quote:

b.) What is the maximum value of the rate of change.
The maximum value of $- 150\cdot 120 \pi \sin (120 \pi t)$ occurs when $\sin (120 \pi t) = -1$ (that is the smallest value that sine takes) and so the max values is $(150)(120 \pi)$. (The reason why we want $\sin (120 \pi t) = -1$ and not $\sin (120 t) = 1$ is because we have a negative sign infront and so if we used $\sin (120 \pi t) = -1$ that would give us $-(150)(120 \pi)$ i.e. the minimum value).