- **Calculus & Beyond Homework**
(*http://www.physicsforums.com/forumdisplay.php?f=156*)

- - **Drawing the lattice curves of a transformation**
(*http://www.physicsforums.com/showthread.php?t=577353*)

Drawing the lattice curves of a transformation1. The problem statement, all variables and given/known dataDraw the lattice lines for the following transformation. [itex]x=se^t[/itex] [itex]y=se^-t[/itex] 3. The attempt at a solutionTrying to draw this transformation. So first thing I did was let t be constant. I then used substitution to show that [itex]x=ye^{2c}[/itex]. This is the set of all lines (as C is changed) with a positive slope that go through the origin, if I am not mistaken, which is the transform of the 'S' lattice of the grid. For the next step I let S be constant, and used substitution to get what I think should be... [itex]x=Ce^{-\frac{lny}{lnC}}[/itex], and this would be the transform of the 't' lattice, if I did that right. What am I looking at here? I used wolframalpha to graph several instances of this, with C = 1,2 and 3, and seemed to give me wildly different graphs each time. Did I do that right? It seems like C=1,2 are hyperbolas with their asymptotes along the x and y axes, but C = 3 gives me something I've never seen before. Sort of like a hyperbola, but one of the sides has been flipped into a different quadrant. Any push in the right direction would be helpful, thanks. EDIT: I just realized that C can't be 1 right? because ln C would be 0, and that would make the denominator 0 in the exponent part of the equation...? Also I tried to graph it with my TI-84, but it only gave me the portion of the graph in the first quadrant. Why is that? EDIT: Also, my graphing calculator gets funky when I graph this equation for C=1/4 |

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