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Applications of Derivatives of Trig. Functions

  1. Oct 19, 2006 #1
    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].

    (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.
    Last edited: Oct 19, 2006
  2. jcsd
  3. Oct 19, 2006 #2
    I am assuming that this is [tex] V = 156\cos{120 \pi t} [/tex]

    (b) Solve [tex] \frac{dV}{dt} = 0 [/tex]

    (c) Set [tex] \frac{d^{2} V}{dt^{2}} = 0 [/tex] to find critical points.
    Last edited: Oct 19, 2006
  4. Oct 19, 2006 #3
    All right.. so assuming part a is correct:

    (b) [tex]-18720 \pi \sin(120 \pi t) = 0[/tex]
    [tex] \sin(120 \pi t) = 0 [/tex]

    I forgot how to solve this (using the inverse sine but i forget exactly).

    (c) [tex] \frac{d^2 V}{dt^2} = -18720 \pi \cos(120 \pi t) * 120 \pi [/tex]
    [tex] \frac{d^2 V}{dt^2} = -18720(120) \pi ^2 \cos(120 \pi t) [/tex]
    [tex] -18720(120) \pi ^2 \cos(120 \pi t) = 0 [/tex]
    [tex] \cos(120 \pi t) = 0 [/tex]

    Again, I forgot how to solve this.

    Thanks for the help in advance.
  5. Oct 19, 2006 #4
    [tex] \sin(120 \pi t) = 0 [/tex]

    [tex] \arcsin(\sin(120\pi t) = \arcsin(0) [/tex]

    [tex] 120\pi t = 0 + n\pi [/tex]

    [tex] \arccos(\cos(120\pi t) = \arccos(0) [/tex]

    [tex] 120\pi t = \frac{\pi}{2} + n\pi [/tex]
  6. Oct 19, 2006 #5
    All right, I'll continue..

    (b) [tex] \sin(120 \pi t) = 0 [/tex]
    [tex] \arcsin (\sin (120 \pi t) = \arcsin (0) [/tex]
    [tex] 120 \pi t = 0 = n \pi [/tex]
    [tex] t = \frac{n \pi }{120 \pi } = \frac{n}{120} [/tex]
    Yes, it is zero at all times [tex] t = \frac{n}{120} [/tex]. The rate of change is zero at these points because the derivative at the points is zero.

    (c) [tex] \arccos(\cos(120\pi t) = \arccos(0) [/tex]
    [tex] 120\pi t = \frac{\pi}{2} + n\pi [/tex]
    [tex] t = \frac{1}{240} + \frac{n}{120} [/tex]
    So, the rate of change is max at all these points. But, they are asking for the maximum rate of change, so i'll try 2 values, since one will be the minimum rate of change, and the next will be the maximum rate of change.

    [tex]V'(0) = -18720\pi \sin(120 \pi * (\frac{1}{240} + \frac{0}{120})) = -18720 \pi \approx -58810.6145[/tex] <---- this is the minimum rate of change

    So, I'll try the next number to get the maximum rate of change.
    [tex]V'(1) = -18720\pi \sin(120 \pi * (\frac{1}{240} + \frac{1}{120})) = 18720 \pi \approx 58810.6145[/tex] <---- this is the maximum rate of change

    Looks good now? Same answers as before lol, just a lot more work :biggrin:
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