Question

For the parametric curve of x = 1 + e^2t, y = e^t how do you find the interval on which the graph starts after you convert to a Cartesian equation?

Answer 1

Graph lies in the interval `[1,oo)`

Explanation:

We convert the parametric equation `x=1+e^(2t)`, `y=e^t` by eliminating `t`,

which can be done by putting the value `e^t=y` into first equation

i.e. `x=1+y^2`, which is the equation of a parabola.

Observe that as we have `y^2` on the RHS

the minimum value of `x=1`, when `y=0`

and as `y` moves on either side `x` only rises

Hence the graph lies in the interval `[1,oo)`

graph{x=1+y^2 [-5.375, 14.625, -4.84, 5.16]}