Simple fraction query - is this right?

Time Magazine

[latex]\displaystyle \alpha = \frac{0.5 - a}{0.5 - b}[/latex]

I want to see how alpha changes as a or b change. So here's what I did.

First, assume a=0.3, graph b over the range of [0, 0.5). I get a standard upward-sloping asymptote. Fine.

Now, assume b=0.3, graph a over the range of [0, 0.5). I get a downward-sloping linear relationship.


So when you change the numerator you get a linear relationship, but when you equivalently change the denominator you get a non-linear relationship. Is this correct, or have I made a mistake in Excel?

Can anyone provide an intuition for this?

Somebody correct me if I'm wrong, but I'm sure this is how it works:

I'll take the case for b=0.3 first. If b=0.3, the bottom half of the fraction stays constant at 0.2. So, you have a basic linear relationship that is [latex]\displaystyle\alpha = \frac{1}{0.2}\ \mbox{a}[/latex]. You can see that this is a linear relationship because it's just stating that one variable equals 5 times another.

In the first case, the top half will be kept constant at 0.2, but the varible part of it, b, increases (assuming that b increases and doesn't decrease), making the whole fraction tend towards negative infinity, producing an asymptote. If b decreases and a is kept constant the whole fraction will tend towards some constant.

RoundTower

what JammyDodger said, except I think he meant + infinity

Time Magazine

Thanks for the replies guys.

I understood the mechanics of it alright, my intuition was that the inverse of something would take the same general functional form, and that's what prompted the thread.

Yakuza

I'd approach it using calculus.

assuming b is constant, then alpha is proportional to -a

The first derivative of alpha w.r.t. a is -1, so you get an decreasing linear relationship.

In the case where a is contstant, then alpha is proportional to 1/b (or b^(-1))

The derivative of 1/b is -1/b², so your formula for the rate of change of alpha w.r.t. B is an inverse square.

(These aren't exact derivatives, just showing the type of function you're dealing with).

To find the exact derivatives, you'll need to use the chain and quotient rules respectively.