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Find the Unit Tangent Vector of the Given Curve r(t)=(2cos6t)i+(2sin6t)j5tkr ( t ) = ( 2 \cos 6 t ) i + ( 2 \sin 6 t ) j - 5 t k

Question 49

Multiple Choice

Find the unit tangent vector of the given curve.
- r(t) =(2cos6t) i+(2sin6t) j5tkr ( t ) = ( 2 \cos 6 t ) i + ( 2 \sin 6 t ) j - 5 t k


A) T=(1213sin6t) i+(1213cos6t) j513k\mathbf { T } = \left( - \frac { 12 } { 13 } \sin 6 \mathrm { t } \right) \mathbf { i } + \left( \frac { 12 } { 13 } \cos 6 \mathrm { t } \right) \mathbf { j } - \frac { 5 } { 13 } \mathbf { k }
B) T=(1213cos6t) i+(1213sin6t) j513kT = \left( - \frac { 12 } { 13 } \cos 6 t \right) i + \left( \frac { 12 } { 13 } \sin 6 t \right) j - \frac { 5 } { 13 } \mathbf { k }
C) T=(12169cos6t) i+(12169sin6t) j5169k\mathrm { T } = \left( - \frac { 12 } { 169 } \cos 6 \mathrm { t } \right) \mathrm { i } + \left( \frac { 12 } { 169 } \sin 6 \mathrm { t } \right) \mathrm { j } - \frac { 5 } { 169 } \mathbf { k }
D) T=(213sin6t) i+(213cos6t) j513kT = \left( - \frac { 2 } { 13 } \sin 6 t \right) i + \left( \frac { 2 } { 13 } \cos 6 t \right) j - \frac { 5 } { 13 } \mathbf { k }

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