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I'm confused by what my solution manual is telling me.

Part a of a problem I'm doing is: Sketch the plane curve with the given vector equation.

$r(t) = <t-2, t^2+1>, t = -1$

The solution manual basically says:

Since $(x+2)^2 = t^2 = y-1 => y = (x+2)^2 - 1$, the curve is a parabola.

How is $(x+2)^2 = t^2 = y-1?$

Another problem $r(t) = sinti + 2costj, t = pi/4$

I thought I was supposed to sketch it by plugging in values for t, but the solution manual says it's a parabola by doing this:

$x = sint, y = 2cost$ so $x^2 + (y/2)^2 = 1$ and the curve is an ellipse.

How come you are allowed to just square both x and y to get the ellipse?

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up vote 0 down vote accepted

You probably mean $r(t) = <t-2, t^2 +1, -1>$. $r(t) = <x, y, z>$ where $x = t-2$, $y=t^2+1$ and $z = -1$ hence $(x+2)^2 = t^2$ etc.

See parametric equations in Wikipedia.

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Makes sense thanks responder. just put in terms of x,y,z and then compare to come up with an equation that meets all parameters. – user67075 Mar 17 '13 at 3:53

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