The radius of curvature of $\mathbf { r } ( t ) = 2 t \mathbf { i } + \left( t ^ { 2 } - 1 \right) \mathbf { j }$ at the point corresponding to t = 1 is

A) $4 \sqrt { 2 }$
B) $\frac { 1 } { 2 \sqrt { 2 } }$
C)2
D) $\frac { 101 \sqrt { 101 } } { 12 }$
E) $\frac { 7 } { 125 }$

Question

The radius of curvature of $\mathbf { r } ( t ) = 2 t \mathbf { i } + \left( t ^ { 2 } - 1 \right) \mathbf { j }$ at the point corresponding to t = 1 is

A) $4 \sqrt { 2 }$
B) $\frac { 1 } { 2 \sqrt { 2 } }$
C)2
D) $\frac { 101 \sqrt { 101 } } { 12 }$
E) $\frac { 7 } { 125 }$

Quizplus · Accepted Answer

Explanation:The radius of curvature is given by the formula$$\kappa = \frac{| \mathbf { r } ' ( t ) \times \mathbf { r } '' ( t ) |}{| \mathbf { r } ' ( t ) |^3}$$So, we first find the first and second derivatives of $\mathbf { r } ( t )$:$$\mathbf { r } ' ( t ) = 2 \mathbf { i } + 2 t \mathbf { j }$$$$\mathbf { r } '' ( t ) = 2 \mathbf { j }$$Then, we evaluate these derivatives at t = 1:$$\mathbf { r } ' ( 1 ) = 2 \mathbf { i } + 2 \mathbf { j }$$$$\mathbf { r } '' ( 1 ) = 2 \mathbf { j }$$Next, we compute the cross product of these two vectors:$$\mathbf { r } ' ( 1 ) \times \mathbf { r } '' ( 1 ) = 4 \mathbf { k }$$Finally, we plug these values into the formula for the radius of curvature:$$\kappa = \frac{| 4 \mathbf { k } |}{| 2 \mathbf { i } + 2 \mathbf { j } |^3} = \frac{4}{\sqrt{8}^3} = \frac{4}{8\sqrt{2}} = \frac{1}{2\sqrt{2}}$$Therefore, the radius of curvature at the point corresponding to t = 1 is $\frac{1}{2\sqrt{2}}$.

The Radius of Curvature Of r(t)=2ti+(t2−1)j\mathbf { r } ( t ) = 2 t \mathbf { i } + \left( t ^ { 2 } - 1 \right) \mathbf { j }r(t)=2ti+(t2−1)j

The Radius of Curvature Of $\mathbf { r } ( t ) = 2 t \mathbf { i } + \left( t ^ { 2 } - 1 \right) \mathbf { j }$