Solved

Find the Derivative drds\frac { d \vec { r } } { d s }

Question 2

Multiple Choice

Find the derivative drds\frac { d \vec { r } } { d s } r(t) =t,t3,t,t=(s2+1) 3\vec { r } ( t ) = \left\langle t , t ^ { 3 } , \sqrt { t } \right\rangle , t = \left( s ^ { 2 } + 1 \right) ^ { 3 }


A) r(s) =1,3(s2+1) 6,12(s2+1) 32\vec { r } ^ { \prime } ( s ) = \left\langle 1,3 \left( s ^ { 2 } + 1 \right) ^ { 6 } , \frac { 1 } { 2 } \left( s ^ { 2 } + 1 \right) ^ { - \frac { 3 } { 2 } } \right\rangle
B) r(s) =1,3t2,12t12\vec { r } ^ { \prime } ( s ) = \left\langle 1,3 t ^ { 2 } , \frac { 1 } { 2 } t ^ { - \frac { 1 } { 2 } } \right\rangle
C) r(s) =1,3(s2+1) 6,12(s2+1) 32>6s(s2+1) 2\vec { r } ^ { \prime } ( s ) = \left\langle 1,3 \left( s ^ { 2 } + 1 \right) ^ { 6 } , \frac { 1 } { 2 } \left( s ^ { 2 } + 1 \right) ^ { - \frac { 3 } { 2 } } > 6 s \left( s ^ { 2 } + 1 \right) ^ { 2 } \right.
D) r(s) =t,t3,t6s(s2+1) \vec { r } ^ { \prime } ( s ) = \left\langle t , t ^ { 3 } , \sqrt { t } \right\rangle 6 s \left( s ^ { 2 } + 1 \right)

Correct Answer:

verifed

Verified

Related Questions