Rewrite the following parametric equations by solving for y

x(t) = e^-t

y(t) = 3e^2t

a) y=1/3x^2,x>0

b)3/x^2, x>0

c) y= 3e^t, x>0

d) y=4e^t, x>0

2 answers

x(t) = e^-t

x^-2 = (e^-t)^-2 = e^2t

3 x^-2 = 3 e^2t = 3/x^2 = y
x(t) = e^-t

ln x = - t

t = - ln x

y = 3 e ^ ( 2 t )

y = 3 e ^ [ 2 ∙ ( - ln x ) ]

y = 3 e ^ [ - ( 2 ∙ ln x ) ]

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Remark:
e ^ [ ( 2 ∙ ln x ) ] = x ^ 2

e ^ [ ( - 2 ∙ ln x ) ] = 1 / x ^ 2
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So:

y = 3 e ^ [ - ( 2 ∙ ln x ) ]

y = 3 / x ^ 2

x > 0