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Calculus problem

  1. Oct 16, 2007 #1
    The cost of producing x ounces of silver from a silver mine is c = f(x) dollars.

    a) What does the derivative of f(x) mean?

    Is the derivative the instantaneous cost?

    Is f(x) the average cost to get silver out of the ground or the change in cost over change in ounces?


    b) What does the statement f ' (600) = 25 mean?

    Does this mean that plugging in a x value 600 yields a y value (amount of instantaneous dollars) of 25?

    c) Do you think f ' (x) will increase/decrease in short run? Long run? Explain.

    I just would say if we go back to silver standard, it would decrease (economies of scale) in both long run and short run.
     
  2. jcsd
  3. Oct 16, 2007 #2

    EnumaElish

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    This isn't generic calculus; it is economics. What have you learned so far in Econ?
     
  4. Oct 16, 2007 #3
    lol. No it's not. I just have to come up with a function (wiggly stuff, exponential, etc.) and convince my teacher that it's ok. I have to come up with something to do part c.
     
  5. Oct 16, 2007 #4

    EnumaElish

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    Last edited: Oct 16, 2007
  6. Oct 16, 2007 #5
    It really said short term but I just put short run. It's basically the same thing. I can't do c without a graph.

    What about a and b?
     
  7. Oct 16, 2007 #6

    EnumaElish

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  8. Oct 16, 2007 #7
  9. Oct 16, 2007 #8

    EnumaElish

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    So, you looked at the Wiki link, and decided it isn't useful?
     
  10. Oct 16, 2007 #9
    Well, c is irrelevant to what a and b are asking. It's a math class (not business calc). She just wants us to come up with a graph to convince her on part c. Most likely she won't be picky.
     
  11. Oct 16, 2007 #10

    EnumaElish

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    I suggested the Wiki article for parts a and b.
     
  12. Oct 16, 2007 #11
    Oh, so the derivative of f(x) is the change in total cost / change in quantity.
     
  13. Oct 16, 2007 #12

    EnumaElish

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    Correct.
     
  14. Oct 16, 2007 #13
    Let's say there's a function y = f(x). and a tangent line at x = 3. Is that the instantaneous unit (whatever you're measuring)?
     
  15. Oct 16, 2007 #14

    EnumaElish

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    The slope of the tangent line is the change in y resulting from a very small change in x (from 3 to 3 + epsilon), divided by the change in x.
     
  16. Oct 16, 2007 #15
    What about the point of the tangent line that's touching the function?

    It passes through points (-2, 3) and (4, -1)

    I got y = -(2/3)x + 13/3 for the tangent line using basic algebra.

    It says plug in f ' (3) and f(3)

    How would I know the original equation y = f(x)?
     
  17. Oct 17, 2007 #16

    HallsofIvy

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    Do you not understand that the derivative of f is the "instantaneous rate of change" of f? If so then all of these should be easy. Certainly "Does this mean that plugging in a x value 600 (in f '(x)) yields a y value (amount of instantaneous dollars) of 25?" is not true: that's what f(x) gives you. f '(x) is not just another name for f(x)!
     
  18. Oct 17, 2007 #17

    EnumaElish

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    Ordinarily, a tangent line "barely touches" the function at the point where f' is evaluated (in your example, x = 3). That's different from intersecting the function, e.g. the way the normal line (perpendicular line) does.
    I am going to assume that means the tangent line intersects f(x) at points p = (-2, 3) and q = (4, -1). If that's what you mean, it doesn't mean anything special. A tangent line might intersect the function at an arbitrary number of points. What is special about the tangent line is that has the same slope as the function at the point where f' is evaluated. For any other point, the tangent line is an arbitrary graphical object.
    How did you derive the tangent line if you didn't know the function f?
     
  19. Oct 17, 2007 #18
    Well, actually I'm getting y = -2/3x + (5/3) for the tangent line now. I got it from those two points. Maybe f(x) is just a straight line and those 2 points can be used to find the whole line (slope).
     
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