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Math Help - slope

  1. #1
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    slope

    Okay I am getting lost.

    I have the IS curve

    Y= (1/ (1-c-d))* ( c0-cT+I-fi+G)

    Given a change in interest rate i the output is going to change in this way:

    delta Y =[ (- f)/( 1-c-d) ]* delta i

    given " i" on the y axis and Y on the x axis, if "f" is big/small the curve is steeper/flatter? Why? What is the right "couple" and why? I tried with numbers and calculator but what is the angle I shall consider?

    Thank you so so much.
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  2. #2
    MHF Contributor
    Joined
    Apr 2005
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    Quote Originally Posted by 0123 View Post
    Okay I am getting lost.

    I have the IS curve

    Y= (1/ (1-c-d))* ( c0-cT+I-fi+G)

    Given a change in interest rate i the output is going to change in this way:

    delta Y =[ (- f)/( 1-c-d) ]* delta i

    given " i" on the y axis and Y on the x axis, if "f" is big/small the curve is steeper/flatter? Why? What is the right "couple" and why? I tried with numbers and calculator but what is the angle I shall consider?

    Thank you so so much.
    I did not study any Business Math. I use only "common sense" or reasoning in trying to solve "business" problems.

    Okay. So my understanding is that c, d, O, T, I and G are all constants, the way you got your delta Y.

    Since i is vertical and Y is horizontal, then the slope of the graph is
    slope = (delta i) / (delta Y)
    slope = 1 / [(-f) /(1-c-d)]
    slope = -(1-c-d) / f

    Therefore,
    ---if f is big, then the slope is small, and the graph is flatter.
    ---if f is small, then the slope is big, and the graph is steeper.
    ---the negative sign only says that the graph is decreasing.

    "couple"?
    Beats me.

    angle?
    Could it be the arctan(slope)?
    Since slope is tan(theta), then theta = arctan(slope) = arctan[-(1-c-d)/f].
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