# Interpreting how graphs of Cobb-Douglas utility functions. How does MRS vary as we vary $\alpha$?

Suppose I have the following Cobb-Douglas function $$U(x,y) = x^\alpha y^{1-\alpha} = 1$$ where $\alpha \in [0,1]$.

$$MRS = -\frac{U_x}{U_y} = - \frac{\alpha}{1-\alpha} \frac{y}{x}$$ $$\frac{\partial MRS}{\partial \alpha} = -\frac{1}{(1-\alpha)^2}\frac{y}{x}$$

So suppose I have the following set of graphs:

Here I just picked some values. Red is $\alpha = \frac{1}{4}$, blue is $\alpha = \frac{1}{2}$, black is $\alpha = \frac{3}{4}$

I understand how the steepness and flatness of the different curves change as I vary $\alpha$. But I am confused as to what $\frac{\partial MRS}{\partial \alpha}$ tells me about the graph. In particular, $\frac{\partial MRS}{\partial \alpha}$ is dependent on $\alpha$. So even though I know $\frac{\partial MRS}{\partial \alpha}$ tells me how much $MRS$ changes as I change alpha....changing $\alpha$ changes $\frac{\partial MRS}{\partial \alpha}$...so I am very confused!

My Question

What is $\frac{\partial MRS}{\partial \alpha}$ telling me about the graph? Since $\frac{\partial MRS}{\partial \alpha}$ is dependent on $\alpha$, how does this affect things?

I think it is important to note that $MRS(x,y)$ is a function. There is exactly one indifference curve passing through $(x,y)$. $MRS(x,y)$ shows the steepness of this curve at point $(x,y)$. Then $\frac{\partial MRS(x,y)}{\partial \alpha}$ would show how much steeper the indifference curve passing through $(x,y)$ gets at this point if you change the parameter $\alpha$.
As YM'fr already stated in his answer, the interpretation of this is how the marginal substition rate (the rate at which she would be willing to trade) of a consumer in possesion of the goods package $(x,y)$ would change.
If $\alpha$ rises, the utility puts more weights to the $x$. Then you must give up more $x$ for one $y$ (for the same utility). Your MRS, for a given $y$, increases in absolute value. Graphically:
If you set $y$ and $x$, the slope is lower for higher $\alpha$ (be careful to only change one thing at a time).
Concerning your "loop" problem, look at the function $1/x$, the first derivative is also a function of $x$, like the 2nd and so on. What matter is the value at one particular x.