Hi, in session 2 clip 2 ('harder problem'), I still don't get why you can just swap xo for yo and x for y so that y=2yo just like x=2xo. I know it has something to do with 1/x being symmetric around the diagonal (since y=1/x and x=1/y are equivalent), but I don't understand why this implies that every x (xo) and y (yo) can be exchanged in the equation for the tangent line. Can someone clarify this for me? Thanks!

Question

phi · Answer

The best I can say is the the curve  xy=1 is symmetric with respect to the x and y axes. If we swap the axes, we get the same picture.

However, I would take his observation more as a guideline (be on the look-out for symmetry), and solve the problem using algebra.  i.e. unless I am completely sure, I verify the answer using algebra.

anonymous · Answer

Thanks for your answer!

anonymous · Answer

To put it another way, we look at an equation, such as the one we saw in that clip,

$$Y = \frac{ 1 }{ X }$$

and we test it for symmetry by simply trading all the values of X and Y. In this example, we simply swap the two letters, X and Y

$$X = \frac{ 1 }{ Y }$$

Here is where I get slightly confused in how to explain it. I can see it but can't think of the words. The two equations look identical to each other, even after we traded them. 
When X changed in the first equation, Y was its inverse, and when X changed, Y was its inverse. So since they are the same, we have symmetry.

REREAD the "Accompanying" PDF files and I bet you'll get it. (I print mine for review).

Hope this helped expand on the prior person's great answer.

John (only up to Session 2 so far, but I still remember my college calculus back when I had a slide rule, before calculators came out)

anonymous · Answer

I might add that I don't remember his saying anything about swapping 
$$X _{0}$$ and $$Y _{0}$$ I think we only do this trading of X and Y in our original equation, the one that only has X and Y, or X and f(x).

Sorry I forgot to mention that above.