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#1 2006-09-20 14:55:31

fusilli_jerry89
Member
Registered: 2006-06-23
Posts: 86

Graphing

Fine a parametrization for the curve:

a) the line segment with endpoints (-1,-3) and (4,1)
b) the lower half of the parabola x - 1 = y²

Last edited by fusilli_jerry89 (2006-09-20 16:38:15)

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#2 2006-09-20 15:50:07

Ricky
Moderator
Registered: 2005-12-04
Posts: 3,791

Re: Graphing

For any parametric equation, just let t be in place of the independant variable, and the equation (with t instead of x or y) be in place of the dependant variable.

a.

First, let's make a graph.  y + 3 = 4/5(x +1)

So y = 4/5x + 4/5 - 15/5 = 4/5x - 11/5

y is dependant on x, x is independant:

x1 = t
y1 = 4/5t - 11/5

b. x - 1 = y^2 so x = y^2 + 1

x is dependant on y, y is independant

y1 = t
x1 = t^2 + 1

And you're done.


"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."

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#3 2006-09-20 16:37:55

fusilli_jerry89
Member
Registered: 2006-06-23
Posts: 86

Re: Graphing

I already know how to get those answers, but the answers in the back of the book are as follows:

a) x = -1 + 5t, y = -3 + 4t, 0≤t≤1
b) x = t² + 1, y = t, t≤0

Last edited by fusilli_jerry89 (2006-09-21 03:16:24)

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