The radius of curvature of $\mathbf { r } ( t ) = \cos t \mathbf { i } + \sin t \mathbf { j } + t \mathbf { k }$ at the point corresponding to $t = \frac { \pi } { 2 }$ is

A) $\frac { 2 } { 17 \sqrt { 17 } }$
B) $\frac { 1 } { 2 \sqrt { 2 } }$
C)2
D) $\frac { 101 \sqrt { 101 } } { 12 }$
E) $\frac { 7 } { 125 }$

Question

The radius of curvature of $\mathbf { r } ( t ) = \cos t \mathbf { i } + \sin t \mathbf { j } + t \mathbf { k }$ at the point corresponding to $t = \frac { \pi } { 2 }$ is

A) $\frac { 2 } { 17 \sqrt { 17 } }$
B) $\frac { 1 } { 2 \sqrt { 2 } }$
C)2
D) $\frac { 101 \sqrt { 101 } } { 12 }$
E) $\frac { 7 } { 125 }$

Quizplus · Accepted Answer

The Answer of The radius of curvature of $\mathbf { r } ( t ) = \cos t \mathbf { i } + \sin t \mathbf { j } + t \mathbf { k }$ at the point corresponding to $t = \frac { \pi } { 2 }$ is

A) $\frac { 2 } { 17 \sqrt { 17 } }$
B) $\frac { 1 } { 2 \sqrt { 2 } }$
C)2
D) $\frac { 101 \sqrt { 101 } } { 12 }$
E) $\frac { 7 } { 125 }$

The Radius of Curvature Of r(t)=cos⁡ti+sin⁡tj+tk\mathbf { r } ( t ) = \cos t \mathbf { i } + \sin t \mathbf { j } + t \mathbf { k }r(t)=costi+sintj+tk

The Radius of Curvature Of $\mathbf { r } ( t ) = \cos t \mathbf { i } + \sin t \mathbf { j } + t \mathbf { k }$