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Math Help - parametrization

  1. #1
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    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..
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  2. #2
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    Quote Originally Posted by terrytriangle View Post
    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}.
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  3. #3
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    sorry but how did you eliminate t to get that?
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    Quote Originally Posted by terrytriangle View Post
    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}.
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  5. #5
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    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?
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  6. #6
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    Quote Originally Posted by terrytriangle View Post
    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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