Consider the curve $\vec { r } ( t ) = ( t + 1 ) \vec { i } + ( 2 - t ) \vec { j } + \left( 2 t ^ { 2 } - t + 3 \right) \vec { k } , 0 \leq t \leq 1$ .
(a)Find a unit vector tangent to the curve at the point (1,2,3).
(b)Show that the curve lies on the surface $z = x ^ { 2 } + y ^ { 2 } - y$ .

Question

Quizplus · Accepted Answer

The Answer of Consider the curve $\vec { r } ( t ) = ( t + 1 ) \vec { i } + ( 2 - t ) \vec { j } + \left( 2 t ^ { 2 } - t + 3 \right) \vec { k } , 0 \leq t \leq 1$ .
(a)Find a unit vector tangent to the curve at the point (1,2,3).
(b)Show that the curve lies on the surface $z = x ^ { 2 } + y ^ { 2 } - y$ .

Consider the Curve r⃗(t)=(t+1)i⃗+(2−t)j⃗+(2t2−t+3)k⃗,0≤t≤1\vec { r } ( t ) = ( t + 1 ) \vec { i } + ( 2 - t ) \vec { j } + \left( 2 t ^ { 2 } - t + 3 \right) \vec { k } , 0 \leq t \leq 1r(t)=(t+1)i+(2−t)j​+(2t2−t+3)k,0≤t≤1

Consider the Curve $\vec { r } ( t ) = ( t + 1 ) \vec { i } + ( 2 - t ) \vec { j } + \left( 2 t ^ { 2 } - t + 3 \right) \vec { k } , 0 \leq t \leq 1$