Consider the parametric surface $\vec { r } ( s , t ) = s \sin \left( \frac { \pi } { 2 } \right) \vec { i } + s \cos \left( \frac { \pi } { 2 } \right) \vec { j } + 4 t \vec { k }$ Does it contain the x-axis?

Question

Quizplus · Accepted Answer

The x-axis is defined by the equation $y = 0$ and $z = 0$. In the given parametric surface, setting $y = 0$ and $z = 0$ would require $s \cos \left( \frac { \pi } { 2 } \right) = 0$ and $4t = 0$, which are always true for any $s$ because $\cos \left( \frac { \pi } { 2 } \right) = 0$, and $t = 0$. However, the x-component is $s \sin \left( \frac { \pi } { 2 } \right) = s$, which means the surface can have any value along the x-axis depending on $s$, not specifically the x-axis itself, which requires $y = 0$ and $z = 0$ for all points.

Consider the Parametric Surface r⃗(s,t)=ssin⁡(π2)i⃗+scos⁡(π2)j⃗+4tk⃗\vec { r } ( s , t ) = s \sin \left( \frac { \pi } { 2 } \right) \vec { i } + s \cos \left( \frac { \pi } { 2 } \right) \vec { j } + 4 t \vec { k }r(s,t)=ssin(2π​)i+scos(2π​)j​+4tk

Consider the Parametric Surface $\vec { r } ( s , t ) = s \sin \left( \frac { \pi } { 2 } \right) \vec { i } + s \cos \left( \frac { \pi } { 2 } \right) \vec { j } + 4 t \vec { k }$