1. ## parametrization

show that f(t)= (e^-t, 1+e^t), 0< or equal to t < infinity
g(t)= (t^3 -2, (t^3-1)/(t^3-2)), cube root of 2 < t < or equal to cube root of 3

parametrize the same curve.

how do i go about doing this? i tried plugging in points, but that didn't really work out..

2. Originally Posted by terrytriangle
show that f(t)= (e^-t, 1+e^t), 0< or equal to t < infinity
g(t)= (t^3 -2, (t^3-1)/(t^3-2)), cube root of 2 < t < or equal to cube root of 3

parametrize the same curve.

how do i go about doing this? i tried plugging in points, but that didn't really work out..
You'll notice eliminating $\displaystyle t$ gives $\displaystyle y = 1 + \frac{1}{x}, 0 \le x \le 1$. Why not choose $\displaystyle x = t \;\text{so}\; y = 1 + \frac{1}{t}$.

3. sorry but how did you eliminate t to get that?

4. Originally Posted by terrytriangle
sorry but how did you eliminate t to get that?
We have $\displaystyle x = e^{-t}, y = 1 + e^t$. From the first we have $\displaystyle x = \frac{1}{e^t}$ or $\displaystyle e^t = \frac{1}{x}$. Then substitute into the second giving $\displaystyle y = 1 + \frac{1}{x}.$

5. thanks that makes sense. do i do the same for g(t)? t^3= x+2 so y= 1 +1/x
and then what? did i prove it since the y's are equal?

6. Originally Posted by terrytriangle
thanks that makes sense. do i do the same for g(t)? t^3= x+2 so y= 1 +1/x
and then what? did i prove it since the y's are equal?
Yes, if the same x gives the same y, then the two curves are the same.

Here is more detail:
Since $\displaystyle x= e^{-t}$, $\displaystyle e^t= \frac{1}{x}$ so $\displaystyle y= 1+ \frac{1}{x}$

If $\displaystyle x= t^3- 2$, then $\displaystyle t^3= x+ 2$ and [tex]t^3- 1= x+ 1[tex]. That means that $\displaystyle y= \frac{x+1}{x+2}= 1+ \frac{1}{x}$