The voltage, V, in volts, in an electrical outlet is given as a function of time, t, in seconds, by the function [tex]V = 156\cos(120 \pi t)[/tex].(adsbygoogle = window.adsbygoogle || []).push({});

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

(b) Is the rate of change ever zero? Explain.

(c) What is the maximum value of the rate of change?

Here's my attempts:

(a) This one is easy.

[tex]\frac{dV}{dt} = 156(-\sin(120 \pi *t))(120\pi)[/tex]

[tex]\frac{dV}{dt} = -18720\pi\sin(120 \pi *t) [/tex]

(b) This one is more tough.

The rate of change is zero at some times. (I don't know how to explain.).. I could probably just say that since the original function is a cosine function, at every peak and trough, the rate of change will be zero because the derivative at that point is zero.

(c) I'm not sure about this one. Wouldn't the maximum rate of change be the point of inflection?

So, we can just think of the graph of the original function, and since it has a period of 1/60, and a cosine function starts at (0,1) and goes downwards, we should be able to divide the period by 4 to get the point of inflection with negative slope. So, the point of inflection with positive slope would be the period times 3/4. So, one of the points of inflection is at x = 1/80.

Now we just need to find the rate of change at that point because it will be the maximum rate of change.

So, I suppose we could plug x = 1/80 into the derivative equation.

[tex]\frac{dV}{dt} = -18720\pi\sin{120\pi(\frac{1}{80})} = 18720\pi[/tex] or [tex]58810.6145[/tex] volts/second

Are these correct? I'm not sure about my approach to them.

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# Homework Help: Applications of Derivatives of Trig. Functions

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