Parametric graph

1. May 28, 2010

nameVoid

x=e^(2t)

y=t+1

t= ( lnx ) / 2

y= ( lnx ) / 2 + 1

or

blah

Last edited: May 29, 2010
2. May 28, 2010

Staff: Mentor

Why did you add +- ?

3. May 28, 2010

nameVoid

what i meant was for t<0 how to represent the graph in terms of y and x

4. May 29, 2010

nameVoid

x=e^(-2t)
y=t+1

-2t=lnx
y=1-lnx/2
i suppose

5. May 29, 2010

HallsofIvy

Staff Emeritus
Is this the same question? How did "$x= e^{2t}$" become $x= e^{-2t}$?

Assuming you really did mean $x= e^{-2t}$, yes, that is correct.