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Sketch the curve of the vector function r(t)=3ti+(3t+8)j,−1≤t≤2\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 the scalar tangential and normal components of acceleration of a particle with position vector r(t) =et{cos⁡6t,sin⁡6t,0}\mathbf { r } ( t ) = e ^ { t } \{ \cos 6 t , \sin 6 t , 0 \}


A) aT=37et,aN=637eta _ { \mathrm { T } } = \sqrt { 37 } e ^ { t } , a _ { \mathrm { N } } = 6 \sqrt { 37 } e ^ { t }
B) aT=6et,aN=37eta _ { \mathbf { T } } = 6 e ^ { t } , a _ { \mathrm { N } } = 37 e ^ { t }
C) aT=637et,aN=37eta _ { \mathrm { T } } = 6 \sqrt { 37 } e ^ { t } , a _ { \mathrm { N } } = \sqrt { 37 } e ^ { t }
D) aT=37et,aN=6eta _ { \mathbf { T } } = 37 e ^ { t } , a _ { \mathrm { N } } = 6 e ^ { t }

E) B) and C)
F) A) and C)

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A

Reparametrize the curve with respect to arc length measured from the point where t=0t = 0 in the direction of increasing tt . r(t) =(7+3t) i+(10+9t) j−(6t) k\mathbf { r } ( t ) = ( 7 + 3 t ) \mathbf { i } + ( 10 + 9 t ) \mathbf { j } - ( 6 t ) \mathbf { k }


A) r(t(s) ) =(7−3126s) i+(10+9126s) j−(6s126) k\mathbf { r } ( t ( s ) ) = \left( 7 - \frac { 3 } { \sqrt { 126 } } s \right) \mathbf { i } + \left( 10 + \frac { 9 } { \sqrt { 126 } } s \right) \mathbf { j } - \left( \frac { 6 s } { \sqrt { 126 } } \right) \mathbf { k }
B) r(t(s) ) =(7+3126s) i+(10−9126s) j−(6s126) k\mathbf { r } ( t ( s ) ) = \left( 7 + \frac { 3 } { \sqrt { 126 } } s \right) \mathbf { i } + \left( 10 - \frac { 9 } { \sqrt { 126 } } s \right) \mathbf { j } - \left( \frac { 6 s } { \sqrt { 126 } } \right) \mathbf { k }
C) r(t(s) ) =(7+3126s) i+(10+9126s) j−(6s126) k\mathbf { r } ( t ( s ) ) = \left( 7 + \frac { 3 } { \sqrt { 126 } } s \right) \mathbf { i } + \left( 10 + \frac { 9 } { \sqrt { 126 } } s \right) \mathbf { j } - \left( \frac { 6 s } { \sqrt { 126 } } \right) \mathbf { k }
D)
r(t(s) ) =(7+3126s) i+(10+9126s) j+(6s) k\mathbf { r } ( t ( s ) ) = \left( 7 + \frac { 3 } { \sqrt { 126 } } s \right) \mathbf { i } + \left( 10 + \frac { 9 } { \sqrt { 126 } } s \right) \mathbf { j } + ( 6 s ) \mathbf { k }
E)
r(t(s) ) =(7+3126s) i+(10+9126s) j−(6s) k\mathbf { r } ( t ( s ) ) = \left( 7 + \frac { 3 } { \sqrt { 126 } } s \right) \mathbf { i } + \left( 10 + \frac { 9 } { \sqrt { 126 } } s \right) \mathbf { j } - ( 6 s ) \mathbf { k }

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

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C

Let CC be a smooth curve defined by r(t) =2i+3tj+2t2k\mathbf { r } ( t ) = 2 \mathbf { i } + 3 t \mathbf { j } + 2 t ^ { 2 } \mathbf { k } , and let T(t) \mathrm { T } ( t ) and N(t) \mathrm { N } ( t ) be the unit tangent vector and unit normal vector to CC corresponding to tt . The plane determined by T\mathbf { T } and N\mathbf { N } is called the osculating plane. Find an equation of the osculating plane of the curve described by r(t) \mathbf { r } ( t ) at t=1t = 1 .


A) z=2z = 2
B) y=3y = 3
C) 2x+3y+2z=12 x + 3 y + 2 z = 1
D) x=2x = 2

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

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A projectile is fired with an initial speed of 700 m/s700 \mathrm {~m} / \mathrm { s } and angle of elevation 60∘60 ^ { \circ } . Find the range of the projectile.


A) d≈43.3 kmd \approx 43.3 \mathrm {~km}
B) d≈433 kmd \approx 433 \mathrm {~km}
C) d≈350 kmd \approx 350 \mathrm {~km}
D) d≈63.3 kmd \approx 63.3 \mathrm {~km}
E) d≈53.3 kmd \approx 53.3 \mathrm {~km}

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

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Find the velocity, acceleration, and speed of an object with position function r(t)=4sin⁡ti+cos⁡tj\mathbf { r } ( t ) = 4 \sin t \mathbf { i } + \cos t \mathbf { j } for t=π4t = \frac { \pi } { 4 } . Sketch the path of the object and its velocity and acceleration vectors.

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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) ) =(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 }
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 }
B)
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 }
C)
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 }
D)
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 }

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

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Find the speed of a particle with the given position function. r(t) =42ti+e4tj−e−4tk\mathbf { r } ( t ) = 4 \sqrt { 2 } t \mathbf { i } + e ^ { 4 t } \mathbf { j } - e ^ { - 4 t } \mathbf { k }


A) ∣v(t) ∣=4+4et+4e−t| v ( t ) | = \sqrt { 4 + 4 e ^ { t } + 4 e ^ { - t } }
B) ∣v(t) ∣=(e4t+e−4t) | v ( t ) | = \left( e ^ { 4 t } + e ^ { - 4 t } \right)
C) ∣v(t) ∣=4+e4t+e−40t| v ( t ) | = \sqrt { 4 + e ^ { 4 t } + e ^ { - 40 t } }
D) ∣v(t) ∣=4(e4t+e−4t) | v ( t ) | = 4 \left( e ^ { 4 t } + e ^ { - 4 t } \right)
E) ∣v(t) ∣=4(et+e−t) | v ( t ) | = 4 \left( e ^ { t } + e ^ { - t } \right)

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

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Sketch the curve of the vector function r(t)=5sin⁡ti+6cos⁡tjr ( t ) = 5 \sin t \mathbf { i } + 6 \cos t \mathbf { j } , and indicate the orientation of the curve.

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The torsion of a curve defined by r(t) \mathbf { r } ( t ) is given by τ=(rt×rtt) ⋅rt∣rt×rt∣2\tau = \frac { \left( \mathbf { r } ^ { t } \times \mathbf { r } ^ { tt } \right) \cdot \mathbf { r } ^ {t} } { \left| \mathbf { r } ^ { t } \times \mathbf { r } ^ { t } \right| ^ { 2 } } Find the torsion of the curve defined by r(t) =cos⁡2ti+sin⁡2tj+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) None of the above
F) C) and D)

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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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Find the limit. lim⁡t→0+⟨8cos⁡t,24sin⁡t,5tln⁡t⟩\lim _ { t \rightarrow 0 ^ { + } } \langle 8 \cos t , 24 \sin t , 5 t \ln t \rangle


A) r(t) =8i+24j+5k\mathbf { r } ( t ) = 8 \mathbf { i } + 24 \mathbf { j } + 5 \mathbf { k }
B) r(t) =8k\mathbf { r } ( t ) = 8 \mathbf { k }
C) r(t) =8i−5k\mathbf { r } ( t ) = 8 \mathbf { i } - 5 \mathbf { k }
D) r(t) =8j\mathbf { r } ( t ) = 8 \mathbf { j }
E) r(t) =8i\mathbf { r } ( t ) = 8 \mathbf { i }

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

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E

Find a vector function that represents the curve of intersection of the two surfaces: the top half of the ellipsoid x2+5y2+5z2=25x ^ { 2 } + 5 y ^ { 2 } + 5 z ^ { 2 } = 25 and the parabolic cylinder y=x2y = x ^ { 2 } .


A) r(t) =ti+t2j+25−t2−5t5k\mathrm { r } ( t ) = t \mathbf { i } + t ^ { 2 } \mathbf { j } + \sqrt { \frac { 25 - t ^ { 2 } - 5 t } { 5 } } \mathbf { k }
B) r(t) =ti−t2j−25−t2−5t5k\mathbf { r } ( t ) = t \mathbf { i } - t ^ { 2 } \mathbf { j } - \sqrt { \frac { 25 - t ^ { 2 } - 5 t } { 5 } } \mathbf { k }
C) r(t) =ti+t2j+5−t2−5t45k\mathbf { r } ( t ) = t \mathbf { i } + t ^ { 2 } \mathbf { j } + \sqrt { \frac { 5 - t ^ { 2 } - 5 t ^ { 4 } } { 5 } } \mathbf { k }
D) r(t) =ti+t4j+25−t2+5t5k\mathbf { r } ( t ) = t \mathbf { i } + t ^ { 4 } \mathbf { j } + \sqrt { \frac { 25 - t ^ { 2 } + 5 t } { 5 } } \mathbf { k }
E) r(t) =ti+t2j+25+t2−5t45k\mathbf { r } ( t ) = t \mathbf { i } + t ^ { 2 } \mathbf { j } + \sqrt { \frac { 25 + t ^ { 2 } - 5 t ^ { 4 } } { 5 } } \mathbf { k }

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

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 A particle moves with position function r(t)=5cos⁡ti+5sin⁡tj+5tk. \text { A particle moves with position function } \mathbf { r } ( t ) = 5 \cos t \mathbf { i } + 5 \sin t \mathbf { j } + 5 t \mathbf { k } \text {. } Find the normal component of the acceleration vector.

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Find the scalar tangential and normal components of acceleration of a particle with position vector r(t)=5ti+(t2+5)j\mathbf { r } ( t ) = 5 t \mathbf { i } + \left( t ^ { 2 } + 5 \right) \mathbf { j } .

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A particle moves with position function r(t) =(21t−7t3−5) 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) B) and C)
G) A) and E)

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Find the length of the curve r(t) =−2ti−tj+tk,−2≤t≤1\mathbf { r } ( t ) = - 2 t \mathbf { i } - t \mathbf { j } + t \mathbf { k } , - 2 \leq t \leq 1 .


A) 363 \sqrt { 6 }
B) 6\sqrt { 6 }
C) 262 \sqrt { 6 }
D) 666 \sqrt { 6 }

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

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The helix r1(t) =4cos⁡ti+sin⁡tj+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.


A) 0
B) π4\frac { \pi } { 4 }
C) π2\frac { \pi } { 2 }
D) π3\frac { \pi } { 3 }
E) None of these

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

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Find the scalar tangential and normal components of acceleration of a particle with position vector r(t) =3sin⁡ti+3cos⁡tj+5tk\mathbf { r } ( t ) = 3 \sin t \mathbf { i } + 3 \cos t \mathbf { j } + 5 t \mathbf { k } .


A) aT=0,aN=3a _ { \mathbf { T } } = 0 , a _ { \mathbf { N } } = 3
B) aT=5,aN=3a _ { \mathbf { T } } = 5 , a _ { \mathrm { N } } = 3
C) aT=0,aN=326a _ { \mathbf { T } } = 0 , a _ { \mathrm { N } } = 3 \sqrt { 26 }
D) aT=5,aN=326a _ { \mathbf { T } } = 5 , a _ { \mathrm { N } } = 3 \sqrt { 26 }

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

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A mortar shell is fired with a muzzle speed of 250 ft/sec.Find the angle of elevation of the mortar if the shell strikes a target located 1100 ft away.Round your answer to 2 decimal places.


A) 17.14∘17.14 ^ { \circ }
B) 14.69∘14.69 ^ { \circ }
C) 0.3∘0.3 ^ { \circ }
D) 0.9∘0.9 ^ { \circ }

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

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