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anonymous
 2 years ago
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!
anonymous
 2 years ago
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!

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phi
 2 years ago
Best ResponseYou've already chosen the best response.0The 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 lookout for symmetry), and solve the problem using algebra. i.e. unless I am completely sure, I verify the answer using algebra.

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0Thanks for your answer!

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0To 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
 one year ago
Best ResponseYou've already chosen the best response.0I 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.
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