The radius of curvature of $\mathbf { r } ( t ) = e ^ { t } \mathbf { i } + e ^ { - t } \mathbf { j } + \mathbf { k }$ at the point corresponding to t = 0 is

A) $4 \sqrt { 2 }$
B) $\frac { 1 } { 2 \sqrt { 2 } }$
C)2
D) $\sqrt { 2 }$
E) $\frac { 7 } { 125 }$

Question

The radius of curvature of $\mathbf { r } ( t ) = e ^ { t } \mathbf { i } + e ^ { - t } \mathbf { j } + \mathbf { k }$ at the point corresponding to t = 0 is

A) $4 \sqrt { 2 }$
B) $\frac { 1 } { 2 \sqrt { 2 } }$
C)2
D) $\sqrt { 2 }$
E) $\frac { 7 } { 125 }$

Quizplus · Accepted Answer

The radius of curvature $ R $ is given by $ R = \frac{|\mathbf{v}|^3}{|\mathbf{v} \times \mathbf{a}|} $, where $\mathbf{v}$ is the velocity and $\mathbf{a}$ is the acceleration. At $ t = 0 $, $\mathbf{v} = \mathbf{i} - \mathbf{j}$ and $\mathbf{a} = \mathbf{i} + \mathbf{j}$. Thus, $|\mathbf{v}| = \sqrt{2}$ and $|\mathbf{v} \times \mathbf{a}| = 2$. Therefore, $ R = \frac{(\sqrt{2})^3}{2} = \sqrt{2} $.

The Radius of Curvature Of r(t)=eti+e−tj+k\mathbf { r } ( t ) = e ^ { t } \mathbf { i } + e ^ { - t } \mathbf { j } + \mathbf { k }r(t)=eti+e−tj+k

The Radius of Curvature Of $\mathbf { r } ( t ) = e ^ { t } \mathbf { i } + e ^ { - t } \mathbf { j } + \mathbf { k }$