# How to graph the parametric equation: $y=e^{-t}$ and $x=e^{t}$

question

How to graph the parametric equation: $y=e^{-t}$ and $x=e^{t}$

I preferable would like to cancel out the t's on both sides so then I instinctively ln both sides but that gives me ln(x)=t and ln(y)=-t to devolve into me adding the two equations to finally devolve into the following:

x=y which doesn't seem like the real solution/equation to the problem

• Notice that $x y=1$ – Claude Leibovici Feb 22 '17 at 7:45
• Adding the equations gives $\ln(x)= -\ln(y),$ not $\ln(x)=\ln(y).$ Then this can be solved to give what Claude says – spaceisdarkgreen Feb 22 '17 at 7:55
• Also note that $x$ and $y$ have to be positive. So the correct graph is only the 1st-quadrant part of $y = 1/x.$ – B. Goddard Feb 22 '17 at 12:00

$y = e^{-t} \implies y = \frac{1}{e^t} = \frac{1}{x}$ as $x = e^t$. So the equation is is $y = \frac{1}{x}$.
The way you have solved it (by taking ln) gives $ln(x) +ln(y) =0 \implies ln(xy) = 0$ or $xy =1$.
In your substitution that gave you $x=y$ you seem to have made a mistake. Here is a correct substitution, $$x = e^t\ \ \implies\ \ \ln(x) = t$$ $$\therefore\ \ y=e^{-t}=e^{-\ln(x)}=e^{\ln{\frac{1}{x}}}=\frac{1}{x}$$
Or as stated in the comments, $$xy = 1$$ which was kinda obvious if you just multiplied $x$ and $y$ from the start.
Your intuition to use $\ln$ was good, but remember that $$-\ln{z}=\ln{\frac{1}{z}}$$ and more importantly that, $$e^{-\ln(z)} \neq -z$$