How do you write #x = e^t-1# and #y=e^2t# as a cartesian equation and then sketch the curve?

1 Answer
Feb 25, 2015

First of all I think that, maybe, there is a little mistake in your writing.

Doesn't the function is:

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

instead of yours?

If I am right, than:

#x+1=e^trArrt=ln(x-1)# and

#y=e^(2ln(x+1))rArry=e^ln((x+1)^2)rArry=(x+1)^2#

(this is because: #e^lnf(x)=f(x)#)

And this is the equation of a parabola centered in #V(-1,0)# and that passes from the points #(0,+1)#, as you can see:

graph{(x+1)^2 [-10, 10, -5, 5]}

I hope that my hypotesis (of your little error of writing) is correct.