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Find r(t) \mathbf { r } ( t ) satisfying the conditions for rt(t) =6e6ti+7etj+etk,r(0) =ij+5k\mathbf { r } ^ { t } ( t ) = 6 e ^ { 6 t } \mathbf { i } + 7 e ^ { - t } \mathbf { j } + e ^ { t } \mathbf { k } , \quad \mathbf { r } ( 0 ) = \mathbf { i } - \mathbf { j } + 5 \mathbf { k }


A) 6e6ti(7et+6) j+(et+4) k6 e ^ { 6 t } \mathbf { i } - \left( 7 e ^ { - t } + 6 \right) \mathbf { j } + \left( e ^ { t } + 4 \right) \mathbf { k }
B) e6ti(7et6) j+(et+4) ke ^ { 6 t } \mathbf { i } - \left( 7 e ^ { - t } - 6 \right) \mathbf { j } + \left( e ^ { t } + 4 \right) \mathbf { k }
C) (6e6t+1) i(7et1) j+(et+5) k\left( 6 e ^ { 6 t } + 1 \right) \mathbf { i } - \left( 7 e ^ { - t } - 1 \right) \mathbf { j } + \left( e ^ { t } + 5 \right) \mathbf { k }
D) (e6t+1) i(7et+1) j+(et+5) k\left( e ^ { 6 t } + 1 \right) \mathbf { i } - \left( 7 e ^ { - t } + 1 \right) \mathbf { j } + \left( e ^ { t } + 5 \right) \mathbf { k }

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

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If r(t) =7i+tcosπj+4sinπtk\mathbf { r } ( t ) = 7 \mathbf { i } + t \cos \pi \mathbf { j } + 4 \sin \pi t \mathbf { k } , evaluate 01r(t) dt\int _ { 0 } ^ { 1 } r ( t ) d t


A) 7i2π2j1πk7 \mathbf { i } - \frac { 2 } { \pi ^ { 2 } } \mathbf { j } - \frac { 1 } { \pi } \mathbf { k }
B) 7i2π2j+8πk7 \mathbf { i } - \frac { 2 } { \pi ^ { 2 } } \mathbf { j } + \frac { 8 } { \pi } \mathbf { k }
C) i+2π2j8πk\mathbf { i } + \frac { 2 } { \pi ^ { 2 } } \mathbf { j } - \frac { 8 } { \pi } \mathbf { k }
D) 2π2j+1πk\frac { 2 } { \pi ^ { 2 } } \mathbf { j } + \frac { 1 } { \pi } \mathbf { k }
E) 7i+2π2j+8πk7 \mathbf { i } + \frac { 2 } { \pi ^ { 2 } } \mathbf { j } + \frac { 8 } { \pi } \mathbf { k }

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

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The torsion of a curve defined by r(t) \mathbf { r } ( t ) is given by τ=(rt×rtt) rtttrt×rtt2\tau = \frac { \left( \mathbf { r } ^ { t } \times \mathbf { r } ^ {tt } \right) \cdot \mathbf { r } ^ { ttt } } { \left| \mathbf { r } ^ {t } \times \mathbf { r } ^ {t t } \right| ^ { 2 } } Find the torsion of the curve defined by r(t) =cos2ti+sin2tj+5tk\mathbf { r } ( t ) = \cos 2 t \mathbf { i } + \sin 2 t \mathbf { j } + 5 t \mathbf { k } .


A) 2729\frac { 27 } { 29 }
B) 1029\frac { 10 } { 29 }
C) 5029\frac { 50 } { 29 }
D) 729\frac { 7 } { 29 }

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

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 Find an expression for ddt[x(t).(y(t)×z(t))]\text { Find an expression for } \frac { d } { d t } [ x ( t ) . ( y ( t ) \times z ( t ) ) ] \text {. }

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A ball is thrown at an angle of 4545 ^ { \circ } to the ground. If the ball lands 90 m90 \mathrm {~m} away, what was the initial speed of the ball? Let g=9.82 m/sg = 9.82 \mathrm {~m} / \mathrm { s } .

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Find the curvature of the curve r(t) =3sin4ti+3cos4tj+3tkr ( t ) = 3 \sin 4 t \mathbf { i } + 3 \cos 4 t \mathbf { j } + 3 t \mathbf { k } .


A) 43\frac { 4 } { 3 }
B) 34\frac { 3 } { 4 }
C) 5116\frac { 51 } { 16 }
D) 1651\frac { 16 } { 51 }

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

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The following table gives coordinates of a particle moving through space along a smooth curve. txyz0.55.89.14.3112.614.916.81.525.621.229.4239.239.537.92.542.442.443\begin{array}{|c|c|c|c|}\hline \mathbf{t} & x & y & z \\\hline 0.5 & 5.8 & 9.1 & 4.3 \\\hline 1 & 12.6 & 14.9 & 16.8 \\\hline 1.5 & 25.6 & 21.2 & 29.4 \\\hline 2 & 39.2 & 39.5 & 37.9 \\\hline 2.5 & 42.4 & 42.4 & 43 \\\hline\end{array} Find the average velocity over the time interval [1, 2].


A) v=21.1i+24.6j+26.6k\mathbf { v } = 21.1 \mathbf { i } + 24.6 \mathbf { j } + 26.6 \mathbf { k }
B) v=21.1i+26.6j+21.1k\quad \mathbf { v } = 21.1 \mathbf { i } + 26.6 \mathbf { j } + 21.1 \mathbf { k }
C) v=24.6i+24.6j+21.1k\quad \mathbf { v } = 24.6 \mathbf { i } + 24.6 \mathbf { j } + 21.1 \mathbf { k }
D) v=26.6i+24.6j+21.1k\mathbf { v } = 26.6 \mathbf { i } + 24.6 \mathbf { j } + 21.1 \mathbf { k }
E) v=21.1i+26.6j+26.6k\mathbf { v } = 21.1 \mathbf { i } + 26.6 \mathbf { j } + 26.6 \mathbf { k }

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

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Find the derivative of the vector function. r(t)=a+tb+t2c\mathbf { r } ( t ) = a + t b + t ^ { 2 } c

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Sketch the curve of the vector function r(t)=3ti+(3t+8)j,1t2\mathbf { r } ( t ) = 3 t \mathbf { i } + ( 3 t + 8 ) \mathbf { j } , - 1 \leq t \leq 2 , and indicate the orientation of the curve.

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 Find rt(t) and rtt(t) for r(t)=tcos9tsin9t,tsin10t+cos10t\text { Find } \mathbf { r } ^ { t } ( t ) \text { and } \mathbf { r } ^ {tt } ( t ) \text { for } \mathbf { r } ( t ) = \langle t \cos 9 t - \sin 9 t , t \sin 10 t + \cos 10 t \rangle

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\[\begin{array} { l }
\mathbf { r } ^ {...

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Reparametrize the curve with respect to arc length measured from the point where t=0t = 0 in the direction of increasing tt . r(t) =(5+3t) i+(8+9t) j(6t) k\mathbf { r } ( t ) = ( 5 + 3 t ) \mathbf { i } + ( 8 + 9 t ) \mathbf { j } - ( 6 t ) \mathbf { k }


A) r(t(s) ) =(53126s) i+(8+9126s) j(6s126) k\mathbf { r } ( t ( s ) ) = \left( 5 - \frac { 3 } { \sqrt { 126 } } s \right) \mathbf { i } + \left( 8 + \frac { 9 } { \sqrt { 126 } } s \right) \mathbf { j } - \left( \frac { 6 s } { \sqrt { 126 } } \right) \mathbf { k }
B)
r(t(s) ) =(5+3126s) i+(89126s) j(6s126) k\mathbf { r } ( t ( s ) ) = \left( 5 + \frac { 3 } { \sqrt { 126 } } s \right) \mathbf { i } + \left( 8 - \frac { 9 } { \sqrt { 126 } } s \right) \mathbf { j } - \left( \frac { 6 s } { \sqrt { 126 } } \right) \mathbf { k }
C)
r(t(s) ) =(5+3126s) i+(8+9126s) j+(6s) k\mathbf { r } ( t ( s ) ) = \left( 5 + \frac { 3 } { \sqrt { 126 } } s \right) \mathbf { i } + \left( 8 + \frac { 9 } { \sqrt { 126 } } s \right) \mathbf { j } + ( 6 s ) \mathbf { k }
D)
r(t(s) ) =(5+3126s) i+(8+9126s) j(6s126) k\mathbf { r } ( t ( s ) ) = \left( 5 + \frac { 3 } { \sqrt { 126 } } s \right) \mathbf { i } + \left( 8 + \frac { 9 } { \sqrt { 126 } } s \right) \mathbf { j } - \left( \frac { 6 s } { \sqrt { 126 } } \right) \mathbf { k }
E)
r(t(s) ) =(5+3126s) i+(8+9126s) j(6s) k\mathbf { r } ( t ( s ) ) = \left( 5 + \frac { 3 } { \sqrt { 126 } } s \right) \mathbf { i } + \left( 8 + \frac { 9 } { \sqrt { 126 } } s \right) \mathbf { j } - ( 6 s ) \mathbf { k }

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

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A particle moves with position function r(t) =(21t7t35) i+21t2j\mathbf { r } ( t ) = \left( 21 t - 7 t ^ { 3 } - 5 \right) \mathbf { i } + 21 t ^ { 2 } \mathbf { j } . Find the tangential component of the acceleration vector.


A) aT=5ta _ { T } = 5 t
B) aT=542ta _ { T } = 542 t
C) aT=42t+5a _ { T } = 42 t + 5
D) aT=55ta _ { T } = - 55 t
E) aT=42ta _ { T } = 42 t

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

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Find the velocity of a particle that has the given acceleration and the given initial velocity. a(t)=3k,v(0)=18i7j\mathbf { a } ( t ) = 3 \mathrm { k } , \mathbf { v } ( 0 ) = 18 \mathbf { i } - 7 \mathbf { j }

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Find r(t)\mathbf { r } ^ { \prime \prime } ( t ) for the function given. r(t)=8i+sintj+costk\mathbf { r } ( t ) = 8 \mathbf { i } + \sin t \mathbf { j } + \cos t \mathbf { k }

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Sketch the curve of the vector function r(t)=5sinti+4costj\mathbf { r } ( t ) = 5 \sin t \mathbf { i } + 4 \cos t \mathbf { j } , and indicate the orientation of the curve.

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The torsion of a curve defined by r(t)r ( t ) is given by τ=(rt×rtt)rtttrt×rtt2\tau = \frac { \left( \mathbf { r } ^ { t } \times \mathbf { r } ^ { tt } \right) \cdot \mathbf { r } ^ { ttt } } { \left| \mathbf { r } ^ { t } \times \mathbf { r } ^ { tt} \right| ^ { 2 } } Find the torsion of the curve defined by r(t)=cos7ti+sin7tj+4tk\mathbf { r } ( t ) = \cos 7 t \mathbf { i } + \sin 7 t \mathbf { j } + 4 t \mathbf { k } .

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The helix r1(t)=4costi+sintj+tk\mathbf { r } _ { 1 } ( t ) = 4 \cos t \mathbf { i } + \sin t \mathbf { j } + t \mathbf { k } intersects the curve r2(t)=(4+t)i+6t2j+5t3k\mathbf { r } _ { 2 } ( t ) = ( 4 + t ) \mathbf { i } + 6 t ^ { 2 } \mathbf { j } + 5 t ^ { 3 } \mathbf { k } at the point (4,0,0)( 4,0,0 ) . Find the angle of intersection.

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The position function of a particle is given by r(t)=5t2,5t,5t2100t}\mathbf { r } ( t ) = \left\langle 5 t ^ { 2 } , 5 t , 5 t ^ { 2 } - 100 t \right\} When is the speed a minimum?

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Find the velocity and position vectors of an object with acceleration a(t)=4i72j+(72t+4)k\mathbf { a } ( t ) = 4 \mathbf { i } - 72 \mathbf { j } + ( 72 t + 4 ) \mathbf { k } , initial velocity v(0)=i+k\mathbf { v } ( 0 ) = \mathbf { i } + \mathbf { k } , and initial position r(0)=j+3k\mathbf { r } ( 0 ) = \mathbf { j } + 3 \mathbf { k } .

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Find the integral (4ti+6t2j+4k) dt\int \left( 4 t \mathbf { i } + 6 t ^ { 2 } \mathbf { j } + 4 \mathbf { k } \right) d t


A) 4i+12tj+C4 \mathbf { i } + 12 t \mathbf { j } + \mathbf { C }
B) 4ti+6t2j+4k+C4 t \mathbf { i } + 6 t ^ { 2 } \mathbf { j } + 4 \mathbf { k } + \mathbf { C }
C) 2t2i+2t3j+4tk+C2 t ^ { 2 } \mathbf { i } + 2 t ^ { 3 } \mathbf { j } + 4 t \mathbf { k } + \mathbf { C }
D) 4t2i+6t3j+4tk+C4 t ^ { 2 } \mathbf { i } + 6 t ^ { 3 } \mathbf { j } + 4 t \mathbf { k } + \mathbf { C }

E) None of the above
F) All of the above

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