MHB Calculating the Slope of the IS-curve - Hint Included!

[mathmari]

Hey!

We have that the IS- curve is described by $$Y=C(Y)+I(r)+\overline{G}$$
I want to calculate the slope $\frac{dr}{dY}$ of the IS-curve.

How could we calculate that? Could you give me a hint? (Wondering)

 

[I like Serena]

Hey!

We have that the IS- curve is described by $$Y=C(Y)+I(r)+\overline{G}$$
I want to calculate the slope $\frac{dr}{dY}$ of the IS-curve.

How could we calculate that? Could you give me a hint? (Wondering)

Hi mathmari! (Smile)

Perhaps we can take the total differential?

That is:
$$dY=C'(Y)dY + I'(r)dr \quad\Rightarrow\quad
dY(1-C'(Y))=I'(r)dr \quad\Rightarrow\quad
\d r Y = \frac{1-C'(Y)}{I'(r)}
$$

 

[mathmari]

Perhaps we can take the total differential?

That is:
$$dY=C'(Y)dY + I'(r)dr \quad\Rightarrow\quad
dY(1-C'(Y))=I'(r)dr \quad\Rightarrow\quad
\d r Y = \frac{1-C'(Y)}{I'(r)}
$$

The total differential of $f$ is $$df=\sum_{i=1}^n\frac{\partial}{\partial{x_i}}dx_i$$

In ths case which is the function $f$ ? Do we define a function $f(Y,r)=Y-C(Y)-I(r)-\overline{G}$ ? Or is the function is this case $Y$ ? (Wondering)

How do we know that we have to use in this case the total differential? (Wondering)

 

[I like Serena]

The total differential of $f$ is $$df=\sum_{i=1}^n\frac{\partial}{\partial{x_i}}dx_i$$

In ths case which is the function $f$ ? Do we define a function $f(Y,r)=Y-C(Y)-I(r)-\overline{G}$ ? Or is the function is this case $Y$ ? (Wondering)

How do we know that we have to use in this case the total differential? (Wondering)

Let me put it differently.

Since we want the derivative of $r$ with respect to $Y$, we're talking about the function $r(Y)$.
That means we can write the equation as:
$$Y=C(Y)+I(r(Y))+\overline{G}$$
When we take the derivative of both sides with respect to $Y$ we get:
$$1=C'(Y)+I'(r(Y))\d r Y \quad\Rightarrow\quad \d r Y = \frac{1-C'(Y)}{I'(r(Y))} = \frac{1-C'(Y)}{I'(r)}$$

As for the total differential, we treat $Y$ as a function of all other variables, that is $Y=Y(r)$, and $r$ as a function of all other variables as well, that is $r=r(Y)$.
Note that $Y(r)$ is simply the inverse of $r(Y)$. (Thinking)

 

[mathmari]

Let me put it differently.

Since we want the derivative of $r$ with respect to $Y$, we're talking about the function $r(Y)$.
That means we can write the equation as:
$$Y=C(Y)+I(r(Y))+\overline{G}$$
When we take the derivative of both sides with respect to $Y$ we get:
$$1=C'(Y)+I'(r(Y))\d r Y \quad\Rightarrow\quad \d r Y = \frac{1-C'(Y)}{I'(r(Y))} = \frac{1-C'(Y)}{I'(r)}$$

As for the total differential, we treat $Y$ as a function of all other variables, that is $Y=Y(r)$, and $r$ as a function of all other variables as well, that is $r=r(Y)$.
Note that $Y(r)$ is simply the inverse of $r(Y)$. (Thinking)

I got it! Thank you so much! (Mmm)