# Help with parametric curves

1. Mar 22, 2016

### Frankenstein19

1. The problem statement, all variables and given/known data
Consider the parametric curve
x=ln(t) and
y= 1+t^2

i need to eliminate the parameter to find the cartesian equation

2. Relevant equations

3. The attempt at a solution

if i solve for t using x I get that t=e^x
but if i solve for t using y i get t=(y-1)^(1/2)

when i plug in x into y i get that y=1+e^2x

but plugging y into x i get that x= ln(y-1)^(1/2)

both of which have different domains of parameterization and i need to find the domain of parameterization

how can i know which to use? which is the correct one?

then part b says that: "since dx/dt=______ it is _____ along the domain of parameterization, the curve goes through ________________" I HAVE NO IDEA WHAT TO PUT THERE

because i thought i had to just put like dx/dt= 1/t and then i thought i had to put concave up or down but if i use y=1+e^x^2's domain i cant do that

Last edited: Mar 22, 2016
2. Mar 23, 2016

### andrewkirk

I don't think they do.

Check your work. If, having done that, you still think that the two approaches give different domains, post your reasons why.

3. Mar 23, 2016

### HallsofIvy

Staff Emeritus
Don't solve the second equation for t- just replace the "t" in the second equation by e^x.

You mean y= 1+ e^(2x) but, yes, that is correct.

You get the "domains of parameterization from the original parametric equations, not from the Cartesian equations.
x= ln(t) so t must be positive.

[/quote]then part b says that: "since dx/dt=______ it is _____ along the domain of parameterization, the curve goes through ________________" I HAVE NO IDEA WHAT TO PUT THERE

because i thought i had to just put like dx/dt= 1/t and then i thought i had to put concave up or down but if i use y=1+e^x^2's domain i cant do that[/QUOTE]
Does "it" refer to the graph of the Cartesian equation? You had previously determined that that function was y= 1+ e^(2x), NOT y= 1+ e^(x^2).

Last edited by a moderator: Mar 23, 2016