# slope

• Oct 24th 2007, 12:16 PM
0123
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.
• Oct 24th 2007, 02:56 PM
ticbol
Quote:

Originally Posted by 0123
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].