Find the length of the curve $\mathbf { r } ( t ) = \left\langle 2 t ^ { \frac { 3 } { 2 } } , 2 t + 1 , \sqrt { 5 } t \right\rangle 0 \leq t \leq 3$
A) $3 \sqrt { 21 }$
B)6
C)8
D)10
E)189
F)14
G)16
H)18

Question

Find the length of the curve $\mathbf { r } ( t ) = \left\langle 2 t ^ { \frac { 3 } { 2 } } , 2 t + 1 , \sqrt { 5 } t \right\rangle 0 \leq t \leq 3$ 
A) $3 \sqrt { 21 }$
B)6
C)8
D)10
E)189
F)14
G)16
H)18

Quizplus · Accepted Answer

The length of the curve is found by integrating the magnitude of the derivative of $\mathbf{r}(t)$. The derivative is $\mathbf{r}'(t) = \left\langle 3t^{\frac{1}{2}}, 2, \sqrt{5} \right\rangle$. The magnitude is $\sqrt{(3t^{\frac{1}{2}})^2 + 2^2 + (\sqrt{5})^2} = \sqrt{9t + 4 + 5} = \sqrt{9t + 9} = 3\sqrt{t + 1}$. Integrating from 0 to 3 gives $\int_0^3 3\sqrt{t+1} \, dt = 14$.

Find the Length of the Curve r(t)=⟨2t32,2t+1,5t⟩0≤t≤3\mathbf { r } ( t ) = \left\langle 2 t ^ { \frac { 3 } { 2 } } , 2 t + 1 , \sqrt { 5 } t \right\rangle 0 \leq t \leq 3r(t)=⟨2t23​,2t+1,5​t⟩0≤t≤3

Find the Length of the Curve $\mathbf { r } ( t ) = \left\langle 2 t ^ { \frac { 3 } { 2 } } , 2 t + 1 , \sqrt { 5 } t \right\rangle 0 \leq t \leq 3$