parametrization

**terrytriangle** · October 26th 2009, 08:46 AM

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..

**Danny** · October 26th 2009, 08:55 AM

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 $t$ gives $y = 1 + \frac{1}{x}, 0 \le x \le 1$ . Why not choose $x = t \;\text{so}\; y = 1 + \frac{1}{t}$ .

**terrytriangle** · October 26th 2009, 08:58 AM

sorry but how did you eliminate t to get that?

**Danny** · October 26th 2009, 10:12 AM

Originally Posted by terrytriangle

sorry but how did you eliminate t to get that?

We have $x = e^{-t}, y = 1 + e^t$ . From the first we have $x = \frac{1}{e^t}$ or $e^t = \frac{1}{x}$ . Then substitute into the second giving $y = 1 + \frac{1}{x}.$

**terrytriangle** · October 26th 2009, 10:18 AM

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?

**HallsofIvy** · October 26th 2009, 12:26 PM

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 $x= e^{-t}$ , $e^t= \frac{1}{x}$ so $y= 1+ \frac{1}{x}$

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

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