You are not logged in.
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)
Offline
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..."
Offline
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)
Offline