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

Question

Ezra Adams · Accepted Answer

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.

Isabella Ackerman · Answer

undefined To write the equations in Cartesian form, express $x$ and $y$ in terms of each other. Since $x = e^t - 1$ and $y = e^{2t}$, solve the first equation for $t$ and substitute it into the second equation for $y$. This yields:

$$y = e^{2(\ln(x+1))}$$

To sketch the curve, note that the function $y = e^{2\ln(x+1)}$ can be simplified using properties of logarithms:

$$y = e^{\ln((x+1)^2)} = (x+1)^2$$

This is the Cartesian equation of the curve. To sketch it, you can plot points on the curve by choosing values of $x$, calculating the corresponding values of $y$, and then connecting the points to form the graph.