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Find the curvature K of the curve given below. r(t) =ti+2t2j+2tk\mathbf { r } ( t ) = t \mathbf { i } + 2 t ^ { 2 } \mathbf { j } + 2 t \mathbf { k }


A) 5(5+16t2) 3\frac { 5 } { \sqrt { \left( 5 + 16 t ^ { 2 } \right) ^ { 3 } } }
B) 45(5+4t2) 3/2\frac { 4 \sqrt { 5 } } { \left( 5 + 4 t ^ { 2 } \right) ^ { 3 / 2 } }
C) 2(5+16t2) 3\frac { 2 } { \sqrt { \left( 5 + 16 t ^ { 2 } \right) ^ { 3 } } }
D) 4(5+16t2) 3\frac { 4 } { \sqrt { \left( 5 + 16 t ^ { 2 } \right) ^ { 3 } } }
E) 5(5+16t2) 34\sqrt [ 4 ] { \frac { 5 } { \left( 5 + 16 t ^ { 2 } \right) ^ { 3 } } }

F) B) and E)
G) A) and D)

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Find the principle unit normal vector to the curve given below at the specified point. r(t) =ti+4tj,t=3\mathbf { r } ( t ) = t \mathbf { i } + \frac { 4 } { t } \mathbf { j } , \quad t = 3


A) N(3) =497i−997jN ( 3 ) = \frac { 4 } { \sqrt { 97 } } \mathbf { i } - \frac { 9 } { \sqrt { 97 } } \mathbf { j }
B) N(3) =497i+997jN ( 3 ) = \frac { 4 } { \sqrt { 97 } } \mathbf { i } + \frac { 9 } { \sqrt { 97 } } \mathbf { j }
C) N(3) =41297i−91297jN ( 3 ) = \frac { 4 } { \sqrt { 1297 } } \mathbf { i } - \frac { 9 } { \sqrt { 1297 } } \mathbf { j }
D) N(3) =41297i+91297jN ( 3 ) = \frac { 4 } { \sqrt { 1297 } } \mathbf { i } + \frac { 9 } { \sqrt { 1297 } } \mathbf { j }
E) N(3) =−497i−997jN ( 3 ) = \frac { - 4 } { \sqrt { 97 } } \mathbf { i } - \frac { 9 } { \sqrt { 97 } } \mathbf { j }

F) B) and C)
G) None of the above

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 Find r(t)  given the following. \text { Find } \mathbf { r } ( t ) \text { given the following. } rt(t) =18t5j+6tk,r(0) =2i+18j\mathbf { r } ^ { t } ( t ) = 18 t ^ { 5 } \mathbf { j } + 6 t \mathbf { k } , \mathbf { r } ( 0 ) = 2 \mathbf { i } + 18 \mathbf { j }


A) r(t) =2i+(18+3t6) j+3t2k\mathbf { r } ( t ) = 2 \mathbf { i } + \left( 18 + 3 t ^ { 6 } \right) \mathbf { j } + 3 t ^ { 2 } \mathbf { k }
B) r(t) =2i+3t2j+(18+3t6) k\mathbf { r } ( t ) = 2 \mathbf { i } + 3 t ^ { 2 } \mathbf { j } + \left( 18 + 3 t ^ { 6 } \right) \mathbf { k }
C) r(t) =(18+3t6) j+3t2k\mathbf { r } ( t ) = \left( 18 + 3 t ^ { 6 } \right) \mathbf { j } + 3 t ^ { 2 } \mathbf { k }
D) r(t) =2i+(18−3t6) j−3t2k\mathbf { r } ( t ) = 2 \mathbf { i } + \left( 18 - 3 t ^ { 6 } \right) \mathbf { j } - 3 t ^ { 2 } \mathbf { k }
E) r(t) =18i+(2+3t6) j+3t2k\mathbf { r } ( t ) = 18 \mathbf { i } + \left( 2 + 3 t ^ { 6 } \right) \mathbf { j } + 3 t ^ { 2 } \mathbf { k }

F) D) and E)
G) C) and E)

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 Find rt(t) ⋅rtt(t)  given the following vector function. \text { Find } \mathbf { r } ^ { t } ( t ) \cdot \mathbf { r } ^ { tt } ( t ) \text { given the following vector function. } r(t) =3cos⁡ti+5sin⁡tj\mathbf { r } ( t ) = 3 \cos t \mathbf { i } + 5 \sin t \mathbf { j }


A) cos⁡tsin⁡t\cos t \sin t
B) −34cos⁡tsin⁡t- 34 \cos t \sin t
C) 34cos⁡tsin⁡t34 \cos t \sin t
D) −16sin⁡tcos⁡t- 16 \sin \mathrm { t } \cos \mathrm { t }
E) 3cos⁡tsin⁡t3 \cos t \sin t

F) A) and E)
G) B) and E)

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Evaluate the definite integral below. ∫02(−6t2i+10t4j+15t4k) dt\int _ { 0 } ^ { 2 } \left( - 6 t ^ { 2 } \mathbf { i } + 10 t ^ { 4 } \mathbf { j } + 15 t ^ { 4 } \mathbf { k } \right) d t


A) −16i+64j+64k- 16 \mathbf { i } + 64 \mathbf { j } + 64 \mathbf { k }
B) 64i+64j+96k64 \mathbf { i } + 64 \mathbf { j } + 96 \mathbf { k }
C) 64i+128j+64k64 \mathbf { i } + 128 \mathbf { j } + 64 \mathbf { k }
D) −16i+128j+96k- 16 \mathbf { i } + 128 \mathbf { j } + 96 \mathbf { k }
E) −16i+64j+96k- 16 \mathbf { i } + 64 \mathbf { j } + 96 \mathbf { k }

F) None of the above
G) C) and E)

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Dt[2r(t) −5u(t) ] given the following D _ { t } [ 2 \mathbf { r } ( t ) - 5 \mathbf { u } ( t ) ] \text { given the following } Use the properties of the derivative to find vector-valued functions. r(t) =2ti+6t3j+2t2ku(t) =2i+4t2j+3t3k\begin{array} { l } \mathbf { r } ( t ) = 2 t \mathbf { i } + 6 t ^ { 3 } \mathbf { j } + 2 t ^ { 2 } \mathbf { k } \\\mathbf { u } ( t ) = 2 \mathbf { i } + 4 t ^ { 2 } \mathbf { j } + 3 t ^ { 3 } \mathbf { k }\end{array}


A) 4i+(36t3−40t) j+(8t−45t2) k4 \mathbf { i } + \left( 36 t ^ { 3 } - 40 t \right) \mathbf { j } + \left( 8 t - 45 t ^ { 2 } \right) \mathbf { k }
B) 4i+(36t2−40t) j+(8t2−45t) k4 \mathbf { i } + \left( 36 t ^ { 2 } - 40 t \right) \mathbf { j } + \left( 8 t ^ { 2 } - 45 t \right) \mathbf { k }
C) 4ti+(36t2+40t) j−(8t+45t2) k4 t \mathbf { i } + \left( 36 t ^ { 2 } + 40 t \right) \mathbf { j } - \left( 8 t + 45 t ^ { 2 } \right) \mathbf { k }
D) 4i+(36t2+40t) j+(8t+45t2) k4 \mathbf { i } + \left( 36 t ^ { 2 } + 40 t \right) \mathbf { j } + \left( 8 t + 45 t ^ { 2 } \right) \mathbf { k }
E) 4i+(36t2−40t) j+(8t−45t2) k4 \mathbf { i } + \left( 36 t ^ { 2 } - 40 t \right) \mathbf { j } + \left( 8 t - 45 t ^ { 2 } \right) \mathbf { k }

F) A) and D)
G) All of the above

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The quarterback of a football team releases a pass at a height of 8 feet above the playing field, and the football is caught by a receiver 38 yards directly downfield at a height of 4 feet. The pass is released at an angle of 45∘ with the horizontal. Find the speed of the football when it is 45 ^ { \circ } \text { with the horizontal. Find the speed of the football when it is } released. Round your answer to three decimal places.


A) 33.169 feet per second
B) 59.366 feet per second
C) 19.462 feet per second
D) 118.732 feet per second

E) C) and D)
F) B) and C)

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Find the curvature of the plane curve y=3x2+4 at x=−2y = 3 x ^ { 2 } + 4 \text { at } x = - 2 . Round your answer to three decimal places.


A) 0.498
B) 0.332
C) 0.015
D) 0.005
E) 0.003

F) All of the above
G) B) and E)

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Find the unit tangent vector to the curve given below at the specified point. r(2) =5cos⁡ti+4sin⁡tj,t=2π3\mathbf { r } ( 2 ) = 5 \cos t \mathbf { i } + 4 \sin t \mathbf { j } , t = \frac { 2 \pi } { 3 }


A) T(2π3) =5391i−491j\mathrm { T } \left( \frac { 2 \pi } { 3 } \right) = \frac { 5 \sqrt { 3 } } { \sqrt { 91 } } \mathbf { i } - \frac { 4 } { \sqrt { 91 } } \mathbf { j }
B) T(2π3) =5391i+491j\mathrm { T } \left( \frac { 2 \pi } { 3 } \right) = \frac { 5 \sqrt { 3 } } { \sqrt { 91 } } \mathbf { i } + \frac { 4 } { \sqrt { 91 } } \mathbf { j }
C) T(2π3) =−4391i−591j\mathbf { T } \left( \frac { 2 \pi } { 3 } \right) = - \frac { 4 \sqrt { 3 } } { \sqrt { 91 } } \mathbf { i } - \frac { 5 } { \sqrt { 91 } } \mathbf { j }
D) T(2π3) =−5391i+491j\mathrm { T } \left( \frac { 2 \pi } { 3 } \right) = - \frac { 5 \sqrt { 3 } } { \sqrt { 91 } } \mathbf { i } + \frac { 4 } { \sqrt { 91 } } \mathbf { j }
E) T(2π3) =−5391i−491j\mathrm { T } \left( \frac { 2 \pi } { 3 } \right) = - \frac { 5 \sqrt { 3 } } { \sqrt { 91 } } \mathbf { i } - \frac { 4 } { \sqrt { 91 } } \mathbf { j }

F) B) and D)
G) A) and E)

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Evaluate the limit given below. lim⁡t→2(e−3ti+2t2t2+6tj+2tk) \lim _ { t \rightarrow 2 } \left( e ^ { - 3 t } i + \frac { 2 t ^ { 2 } } { t ^ { 2 } + 6 t } j + \frac { 2 } { t } k \right)


A) r(t) =i+2j\mathbf { r } ( t ) = i + 2 j
B) r(t) =ir ( t ) = i
C) r(t) =2j\mathbf { r } ( t ) = 2 j
D) r(t) =0\mathbf { r } ( t ) = 0
E) The limit does not exist.

F) B) and D)
G) C) and E)

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C

 Find aN at time t=2 for the space curve r(t) =(5t−7) i+t2j−7tk. Round your \text { Find } a _ { \mathbb { N } } \text { at time } t = 2 \text { for the space curve } \mathbf { r } ( t ) = ( 5 t - 7 ) \mathbf { i } + t ^ { 2 } \mathbf { j } - 7 t \mathbf { k } \text {. Round your } answer to three decimal places.


A) 0.970
B) 0.465
C) 0.930
D) 1.814
E) 1.060

F) B) and D)
G) A) and B)

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Find the principle unit normal vector to the curve given below at the specified point. r(t) =5cos⁡ti+5sin⁡tj,t=5π3\mathbf { r } ( t ) = 5 \cos t \mathbf { i } + 5 \sin t \mathbf { j } , \quad t = \frac { 5 \pi } { 3 }


A) N=(32,12}\mathbf { N } = \left( \frac { \sqrt { 3 } } { 2 } , \frac { 1 } { 2 } \right\}
B) N=(32,−12) \mathbf { N } = \left( \frac { \sqrt { 3 } } { 2 } , - \frac { 1 } { 2 } \right)
C) N=(12,−32) \mathbf { N } = \left( \frac { 1 } { 2 } , - \frac { \sqrt { 3 } } { 2 } \right)
D) N=(−12,32) \mathbf { N } = \left( - \frac { 1 } { 2 } , \frac { \sqrt { 3 } } { 2 } \right)
E) N=(−12,−32) \mathrm { N } = \left( - \frac { 1 } { 2 } , - \frac { \sqrt { 3 } } { 2 } \right)

F) A) and C)
G) C) and E)

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Use the properties of the derivative to find Dt(r(t) ×u(t) )  given the following D _ { t } ( \mathbf { r } ( t ) \times \mathbf { u } ( t ) ) \text { given the following } vector-valued functions. r(t) =4cos⁡4tj+4sin⁡4tku(t) =5ti+4cos⁡4tj+4sin⁡4tk\begin{array} { l } \mathbf { r } ( t ) = 4 \cos 4 t \mathbf { j } + 4 \sin 4 t \mathbf { k } \\\mathbf { u } ( t ) = \frac { 5 } { t } \mathbf { i } + 4 \cos 4 t \mathbf { j } + 4 \sin 4 t \mathbf { k }\end{array}


A) (80cos⁡(4t) t−20sin⁡(4t) t2) j+(20cos⁡(4t) t2−80sin⁡(4t) t) k\left( \frac { 80 \cos ( 4 t ) } { t } - \frac { 20 \sin ( 4 t ) } { t ^ { 2 } } \right) \mathbf { j } + \left( \frac { 20 \cos ( 4 t ) } { t ^ { 2 } } - \frac { 80 \sin ( 4 t ) } { t } \right) \mathbf { k }
B) (80cos⁡(4t) t+20sin⁡(4t) t2) j+(20cos⁡(4t) t2+80sin⁡(4t) t) k\left( \frac { 80 \cos ( 4 t ) } { t } + \frac { 20 \sin ( 4 t ) } { t ^ { 2 } } \right) \mathbf { j } + \left( \frac { 20 \cos ( 4 t ) } { t ^ { 2 } } + \frac { 80 \sin ( 4 t ) } { t } \right) \mathbf { k }
C) (80cos⁡(4t) t2−20sin⁡(4t) t) j+(20cos⁡(4t) t2+80sin⁡(4t) t) k\left( \frac { 80 \cos ( 4 t ) } { t ^ { 2 } } - \frac { 20 \sin ( 4 t ) } { t } \right) \mathbf { j } + \left( \frac { 20 \cos ( 4 t ) } { t ^ { 2 } } + \frac { 80 \sin ( 4 t ) } { t } \right) \mathbf { k }
D) (80cos⁡(4t) t−20sin⁡(4t) t2) i+(20cos⁡(4t) t2−80sin⁡(4t) t) k\left( \frac { 80 \cos ( 4 t ) } { t } - \frac { 20 \sin ( 4 t ) } { t ^ { 2 } } \right) \mathbf { i } + \left( \frac { 20 \cos ( 4 t ) } { t ^ { 2 } } - \frac { 80 \sin ( 4 t ) } { t } \right) \mathbf { k }
E) (80cos⁡(4t) t−20sin⁡(4t) t2) j+(20cos⁡(4t) t2+80sin⁡(4t) t) k\left( \frac { 80 \cos ( 4 t ) } { t } - \frac { 20 \sin ( 4 t ) } { t ^ { 2 } } \right) \mathbf { j } + \left( \frac { 20 \cos ( 4 t ) } { t ^ { 2 } } + \frac { 80 \sin ( 4 t ) } { t } \right) \mathbf { k }

F) B) and E)
G) D) and E)

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 The position vector r(t) =4ti+2tj+16t2k describes the path of an object moving in \text { The position vector } \mathbf { r } ( t ) = 4 t \mathbf { i } + 2 t \mathbf { j } + \frac { 1 } { 6 } t ^ { 2 } \mathbf { k } \text { describes the path of an object moving in } space. Find the acceleration a(t) \mathbf { a } ( t ) of the object.


A) a(t) =16k\mathbf { a } ( t ) = \frac { 1 } { 6 } \mathbf { k }
B) a(t) =4i+2j\mathbf { a } ( t ) = 4 \mathbf { i } + 2 \mathbf { j }
C) a(t) =4i+2j+13k\mathbf { a } ( t ) = 4 \mathbf { i } + 2 \mathbf { j } + \frac { 1 } { 3 } \mathbf { k }
D) a(t) =4i+2j+2k\mathbf { a } ( t ) = 4 \mathbf { i } + 2 \mathbf { j } + 2 \mathbf { k }
E) a(t) =13k\mathbf { a } ( t ) = \frac { 1 } { 3 } \mathbf { k }

F) A) and E)
G) A) and B)

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 The position vector r(t) =8ti++4tj+5tk describes the path of an object moving in \text { The position vector } \mathbf { r } ( t ) = 8 t \mathbf { i } + + 4 t \mathbf { j } + 5 t \mathbf { k } \text { describes the path of an object moving in } space. Find the speed s(t) s ( t ) of the object.


A) s(t) =1/73s ( t ) = 1 / 73
B) s(t) =105s ( t ) = 105
C) s(t) =1/105s ( t ) = 1 / 105
D) s(t) =1/85s ( t ) = 1 / 85
E) s(t) =17s ( t ) = 17

F) A) and D)
G) A) and B)

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Given the vector-valued function below, evaluate r(11+Δt) −r(11) \mathbf { r } ( 11 + \Delta t ) - \mathbf { r } ( 11 ) r(t) =ln⁡ti+8tj+4tk\mathbf { r } ( t ) = \ln t \mathbf { i } + \frac { 8 } { t } \mathbf { j } + 4 t \mathbf { k }


A) ln⁡(1+Δt11) i−8Δt(11+Δt) 11j+4k\ln \left( 1 + \frac { \Delta t } { 11 } \right) \mathbf { i } - \frac { 8 \Delta t } { ( 11 + \Delta t ) 11 } \mathbf { j } + 4 \mathbf { k }
B) ln⁡(1+Δt11) i+8Δt(11+Δt) 11j+4Δtk\ln \left( 1 + \frac { \Delta t } { 11 } \right) \mathbf { i } + \frac { 8 \Delta t } { ( 11 + \Delta t ) 11 } \mathbf { j } + 4 \Delta t \mathbf { k }
C) ln⁡(1+Δt11) i−8Δt(11+Δt) 11j+4Δtk\ln \left( 1 + \frac { \Delta t } { 11 } \right) \mathbf { i } - \frac { 8 \Delta t } { ( 11 + \Delta t ) 11 } \mathbf { j } + 4 \Delta t \mathbf { k }
D) ln⁡(1+Δt11) i−8Δt(11+Δt) j+4Δtk\ln \left( 1 + \frac { \Delta t } { 11 } \right) \mathbf { i } - \frac { 8 \Delta t } { ( 11 + \Delta t ) } \mathbf { j } + 4 \Delta t \mathbf { k }
E) ln⁡(1−Δt11) i−8Δt(11+Δt) 11j+4Δtk\ln \left( 1 - \frac { \Delta t } { 11 } \right) \mathbf { i } - \frac { 8 \Delta t } { ( 11 + \Delta t ) 11 } \mathbf { j } + 4 \Delta t \mathbf { k }

F) D) and E)
G) A) and D)

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 Find aT at time t=2 for the space curve r(t) =(6t−7) i+t2j−10tk. Round your \text { Find } a _ { \mathbf { T } } \text { at time } t = 2 \text { for the space curve } \mathbf { r } ( t ) = ( 6 t - 7 ) \mathbf { i } + t ^ { 2 } \mathbf { j } - 10 t \mathbf { k } \text {. Round your } answer to three decimal places.


A) 0.781
B) 0.724
C) 0.343
D) 0.686
E) 0.649

F) B) and C)
G) None of the above

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A particle moves in the yz-plane along the curve represented by the vector-valued function r(t) =(4cos⁡t) j+(9sin⁡t) k\mathbf { r } ( t ) = ( 4 \cos t ) \mathbf { j } + ( 9 \sin t ) \mathbf { k } . Find the minimum value of ∥rt∥\left\| \mathbf { r } ^ {t } \right\|


A) 6
B) 4
C) 7
D) 9
E) 8

F) C) and D)
G) A) and E)

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B

Find the domain of the vector-valued function given below. r(t) =16−t2i+1t+6j+(t−4) k\mathbf { r } ( t ) = \sqrt { 16 - t ^ { 2 } } \mathbf { i } + \frac { 1 } { t + 6 } \mathbf { j } + ( t - 4 ) \mathbf { k }


A) [−∞,4) [ - \infty , 4 )
B) [0,4][ 0,4 ]
C) [−4,4][ - 4,4 ]
D) (−∞,∞) ( - \infty , \infty )
E) [−4,∞) [ - 4 , \infty )

F) D) and E)
G) A) and D)

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Because of a storm, ground controllers instruct the pilot of a plane flying at an altitude of 1.6 miles to make a 90∘ 90^{\circ} turn and climb to an altitude of 1.8 miles. The model for the path of the plane during this maneuver is r(t) =⟨4cos⁡4πt,4sin⁡4πt,1.6+1.6t⟩,0≤t≤18 where t is the \mathbf { r } ( t ) = \langle 4 \cos 4 \pi t , 4 \sin 4 \pi t , 1.6 + 1.6 t \rangle , 0 \leq t \leq \frac { 1 } { 8 } \text { where } t \text { is the } time in hours and is the distance in miles. Determine the speed of the plane. Round your answer to The nearest integer.


A) 50 mi/h
B) 16 mi/h
C) 54 mi/h
D) 34 mi/h
E) 79 mi/h

F) A) and B)
G) C) and D)

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A

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