Why does this graph appear as it does ?

Question

After graphing x = e^t , y = e^2t , why does it appear as it does ?

Answer 1

Since you don't show your graphs, its a bit difficult to understand what you mean. The difficulty with these are that they are functions of (t, x) and (t, y). You are used to (x, y) where x is the independent variable and y is the dependent variable. x = e^t has t as the independent variable and x as the dependent variable. So, if you were going to graph it, you would put "t" on the horizontal axis (the one you are used to having "x" on) and "x" on the vertical axis. So you would get:

Graphing y = e^2t would result in this:

To get a "regular" x-y plot, we would need to solve one of these parametric equations for "t" and plug it into the other. Since x = e^t, then ln(x) = ln(e^t) or t = ln(x). We can then plug "ln(x)" into the other equation (y = e^2t) where we see "t". So it becomes: y = e^2ln(x) or y = [e^ln(x)]² which simplifies to y = x² which would give us an x-y plot that would be a parabola (only the positive half though because x cannot be negative in this scenario)