Consider the curve $\vec { r } ( t ) = \left( 2 t ^ { 2 } + 1 \right) \vec { i } + \left( t ^ { 2 } + 1 \right) \vec { j } + t \vec { k }$ Does the curve lie on the parametric surface $x = s ^ { 2 } + t ^ { 2 } , y = s ^ { 2 } , z = t ?$

Question

Quizplus · Accepted Answer

The curve lies on the surface because the coordinates of the curve satisfy the equations of the surface.

Consider the Curve r⃗(t)=(2t2+1)i⃗+(t2+1)j⃗+tk⃗\vec { r } ( t ) = \left( 2 t ^ { 2 } + 1 \right) \vec { i } + \left( t ^ { 2 } + 1 \right) \vec { j } + t \vec { k }r(t)=(2t2+1)i+(t2+1)j​+tk

Consider the Curve $\vec { r } ( t ) = \left( 2 t ^ { 2 } + 1 \right) \vec { i } + \left( t ^ { 2 } + 1 \right) \vec { j } + t \vec { k }$