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This was the first problem on my test

  1. Jun 23, 2009 #1
    1. The problem statement, all variables and given/known data

    Find the average rate of change of function g(t)=3+tan(t) under interval [-pi/4;pi/4]

    2. Relevant equations



    3. The attempt at a solution

    I am stuck at (-tan(-pi/4))/pi
     
  2. jcsd
  3. Jun 23, 2009 #2

    CompuChip

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    What are sin(pi/4) and cos(pi/4)?
     
  4. Jun 23, 2009 #3
    Same. So tan(pi/4) is 1?
     
  5. Jun 23, 2009 #4

    Mark44

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    Yes.
     
  6. Jun 23, 2009 #5

    HallsofIvy

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    Of course, you say your difficulty is with
    [tex]-tan(-\pi/4)/\pi= -\frac{sin(-\pi/4)}{cos(-\pi/4)}[/tex]
    What is that?
     
    Last edited: Jun 24, 2009
  7. Jun 23, 2009 #6

    CompuChip

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    So my next question is: what happens to the sine and the cosine when you put a minus in the argument. In other words: how does sin(-x) relate to sin(x) and how does cos(-x) relate to cos(x)?
     
  8. Jun 23, 2009 #7
    It is the inverse
     
  9. Jun 23, 2009 #8

    rock.freak667

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    -x => you measure clockwise from the horizontal axis, draw for quadrants and draw an acute angle clockwise from the positive horizontal axis. What quadrant does it lie in? Is sin(x) positive or negative in this quadrant?
     
  10. Jun 23, 2009 #9
    it lies on IV quadrant. sin is only positive in quadrant II, so it is negative.
     
  11. Jun 23, 2009 #10

    rock.freak667

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    thus sin(-x)=-sinx, do the same for cos(x)
     
  12. Jun 24, 2009 #11

    HallsofIvy

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    ??? Sine is positive in the I and III quadrants. It is negative in the II and IV quadrants.
     
  13. Jun 24, 2009 #12

    HallsofIvy

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    ??? Sine is positive in the I and III quadrants. It is negative in the II and IV quadrants.

    But, as Rockfreak667 pointed out, it is much simpler to use the fact that sine is an odd function, cosine an even function, that to worry about "quadrants".
     
  14. Jun 24, 2009 #13

    Mark44

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    That's half right. Sine is positive in the quadrants I and II, and negative in quadrants III and IV. Cosine is positive in quadrants I and IV, and negative in quadrants II and III. As a result, tangent is positive in quadrants I and III, and negative in quadrants II and IV.
     
  15. Jun 24, 2009 #14
    So the answer is -1/pi?

    Got 88 on my first calculus test. If I had this problem correct would have gotten an A.
     
  16. Jun 24, 2009 #15

    Mark44

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    I don't think so. Your answer should be positive. The average rate of change of a function f on an interval [a, b] is
    [tex]\frac{f(b) - f(a)}{b - a}[/tex]
     
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