Question
Show that the curve with parametric equations $x=t^2$, $y=1-3 t, z=1+t^3$ passes through the points $(1,4,0)$ and $(9,-8,28)$ but not through the point $(4,7,-6)$.
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Show that the curve with parametric equations $ x = t^2 $, $ y = 1 - 3t $, $ z = 1 + t^3 $ passes through the points $ (1, 4, 0) $ and $ (9, -8, 28) $ but not through the point $ (4, 7, -6) $.
Vector Functions
Vector Functions and Space Curves
Consider the curve with parametric equations x = t2 , y = 1-3t, z = 1+t3 . Does the curve pass through the point (1,4,0)? does the curve pass through the point (4,7,-6)?
Show that the curve with parametric equations x = t^2, y = 3 - 2t, z = 1 + t^3 passes through the points (1, 5, 0) and (16, -5, 65) but not through the point (4, 7, -6). If x = 1, y = 5, z = 0, then the curve passes through the point (1, 5, 0). This occurs at t = If x = 16, y = -5, z = 65, then the curve passes through the point (16, -5, 65). This occurs at t = For the point (4, 7, -6) to be on the curve, we require y = 3 - 2t = 7 → t = . But then z = 1 + t^3 = 1 + ( )^3 = ≠ -6, so (4, 7, -6) is not on the curve.
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