Can you show me why the parametric equation x=t^2,y=t^4,z=t^6 lies entirely in the first quadrant?

Question

Why doesn't this equation extend into the negative xy and zx plane?

but if I let z=x^3, because x=t^(1/2) then I've got something odd.

Answer 1

For it to extend outside the first octant (not quadrant, this is a parametric equation in three dimensions, not two), there would have to be some value of t for which at least one of x, y, or z was negative. Now, x, y, and z are all t to some even power, and a real number to an even power can never be negative, so that is impossible. Thus, the curve is confined entirely to the first octant.

Edit: Yes, z=x³, but the only way x³ can be negative is if x is negative, and is x negative for any value of t? No. You cannot simply write that t=√x, because for negative values of x, there IS no √x (in the real numbers).

Answer 2

Any value of " t " will result in a positive solution.
However, if t<0, the point will not be in the first quadrant, it will be in the 2nd quadrant of the x plane, y plane or z plane.

Answer 3

Because if you let t be any value, then t^2 is positive. Since t^4=(t^2)^2 and t^6=(t^2)^3, these are also positive. Hence x,y,z are all positive for any choice of t.