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Provide an appropriate response. -The following equations each describe the motion of a particle. For which path is the particle's speed constant? (1) r(t) =t3i+t2jr ( t ) = t ^ { 3 } i + t ^ { 2 } j (2) r(t) =cos(8t) i+sin(6t) jr ( t ) = \cos ( 8 t ) i + \sin ( 6 t ) j (3) r(t) =ti+tjr ( t ) = t i + t j (4) r(t) =cos(5t2) i+sin(5t2) jr ( t ) = \cos \left( - 5 t ^ { 2 } \right) \mathbf { i } + \sin \left( - 5 t ^ { 2 } \right) \mathbf { j }


A) Path (4) and Path (2)
B) Path (1)
C) Path (2) and Path (3)
D) Path (3)

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

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For the curve r(t) , find an equation for the indicated plane at the given value of t. - r(t) =409(1+t) 3/2i+409(1t) 3/2j+103tkr ( t ) = \frac { 40 } { 9 } ( 1 + t ) ^ { 3 / 2 } \mathbf { i } + \frac { 40 } { 9 } ( 1 - t ) ^ { 3 / 2 } \mathbf { j } + \frac { 10 } { 3 } t \mathbf { k } ; osculating plane att=0a t \mathrm { t } = 0


A) xy=0x - y = 0
B) xy=409x - y = - \frac { 40 } { 9 }
C) xy=409x - y = \frac { 40 } { 9 }
D) xy=809x - y = \frac { 80 } { 9 }

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

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Find the curvature of the space curve. - r(t) =5ti+(4+8cos32t) j+(4+8sin32) k\mathbf { r } ( \mathrm { t } ) = 5 \mathrm { ti } + \left( 4 + 8 \cos \frac { 3 } { 2 } \mathrm { t } \right) \mathbf { j } + \left( 4 + 8 \sin \frac { 3 } { 2 } \right) \mathbf { k }


A) κ=1813\kappa = \frac { 18 } { 13 }
B) κ=12169\kappa = \frac { 12 } { 169 }
C) κ=18169\kappa = \frac { 18 } { 169 }
D) κ=1213\kappa = \frac { 12 } { 13 }

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

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Find the curvature of the curve r(t) . -r(t) = (8 + cos 9t - sin 9t) i + (5 + sin 9t + cos 9t) j + 7k


A) κ=22\kappa = \frac { \sqrt { 2 } } { 2 }
B) κ=2\kappa = \sqrt { 2 }
C) κ=92\kappa = \frac { 9 } { 2 }
D) κ=922\kappa = \frac { 9 } { 2 } \sqrt { 2 }

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

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Find the unit tangent vector of the given curve. - r(t) =(5sin39t) i+(5cos39t) jr ( t ) = \left( 5 \sin ^ { 3 } 9 t \right) i + \left( 5 \cos ^ { 3 } 9 t \right) \mathbf { j }


A) T = (5 sin 9t) i - (5 cos 9t) j
B) T = (sin 9t) i - (cos 9t) j
C) T = (135 sin 9t) i -(135 cos 9t) j
D) T = (5 cos 9t) i - (5 sin 9t) j

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

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The position vector of a particle is r(t) . Find the requested vector. -The velocity at t=0t = 0 for r(t) =ln(t33t2+4) i2t+36j5cos(t) k\mathbf { r } ( \mathrm { t } ) = \ln \left( \mathrm { t } ^ { 3 } - 3 \mathrm { t } ^ { 2 } + 4 \right) \mathbf { i } - \sqrt { 2 \mathrm { t } + 36 } \mathbf { j } - 5 \cos ( \mathrm { t } ) \mathbf { k }


A) v(0) =14i16j\mathbf { v } ( 0 ) = \frac { 1 } { 4 } \mathbf { i } - \frac { 1 } { 6 } \mathrm { j }
B) v(0) =16j\mathbf { v } ( 0 ) = \frac { 1 } { 6 } \mathbf { j }
C) v(0) =16j\mathbf { v } ( 0 ) = - \frac { 1 } { 6 } \mathrm { j }
D) v(0) =14i112j+5k\mathbf { v } ( 0 ) = \frac { 1 } { 4 } \mathbf { i } - \frac { 1 } { 12 } \mathbf { j } + 5 \mathbf { k }

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

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Provide an appropriate response. -A human cannonball is to be fired with an initial speed of 80ft/sec80 \mathrm { ft } / \mathrm { sec } . The circus performer hopes to land on a cushion located 190 feet downrange at the same height as the muzzle of the cannon. The circus is being held in a large room with a flat ceiling 33 feet higher than the muzzle. Can the performer be fired to the cushion without striking the ceiling? If so, what is the proper firing angle? ( g=32ft/sec2\mathrm { g } = 32 \mathrm { ft } / \sec ^ { 2 } )

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Solve the initial value problem. -Differential Equation: d2rdt2=(8t3) i\frac { \mathrm { d } ^ { 2 } \mathbf { r } } { \mathrm { dt } ^ { 2 } } = ( 8 \mathrm { t } - 3 ) \mathbf { i } Initial Conditions: drdtr=0=k,r(0) =8i+7j+6k\left. \frac { \mathrm { d } \mathbf { r } } { \mathrm { dt } } \right| _ { \mathrm { r } = 0 } = - \mathbf { k } , \mathrm { r } ( 0 ) = 8 \mathbf { i } + 7 \mathbf { j } + 6 \mathbf { k }


A) r(t) =(83t33t2+8) i+7j+(6t) kr ( t ) = \left( \frac { 8 } { 3 } t ^ { 3 } - 3 t ^ { 2 } + 8 \right) i + 7 j + ( 6 - t ) \mathbf { k }
B) r(t) =(83t3+32t2+8) i7j+(t6) kr ( t ) = \left( \frac { 8 } { 3 } t ^ { 3 } + \frac { 3 } { 2 } t ^ { 2 } + 8 \right) i - 7 j + ( t - 6 ) \mathbf { k }
C) r(t) =(43t332t2+8) i+7j+(6t) k r ( t ) = \left( \frac { 4 } { 3 } t ^ { 3 } - \frac { 3 } { 2 } t ^ { 2 } + 8 \right) i + 7 j + ( 6 - t ) k
D) r(t) =(43t3+32t2+8) i7j+(6t) kr ( t ) = \left( \frac { 4 } { 3 } t ^ { 3 } + \frac { 3 } { 2 } t ^ { 2 } + 8 \right) i - 7 j + ( 6 - t ) k

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

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Find the principal unit normal vector N for the curve r(t) . -r(t) = (10 sin t) i + (10 cos t) j + 9k


A) N = (sin t) i + (cos t) j
B) N = (-sin t) i + (cos t) j
C) N = (-sin t) i - (cos t) j
D) N = (-cos t) i - (sin t) j

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

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Solve the problem. -Show that a planet in a circular orbit moves with constant speed.

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Kepler's first law says that the eccentr...

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Calculate the arc length of the indicated portion of the curve r(t) . -Following the curve r(t) =5ti+(12cost) j+(12sint) kr ( t ) = 5 t i + ( 12 \cos t ) j + ( 12 \sin t ) k in the direction of increasing of arc length, find the point that lies 26π26 \pi units away from the point where t=0t = 0 .


A) (10π,0,0) ( 10 \pi , 0,0 )
B) (10,12,0) ( 10,12,0 )
C) (10π,12,0) ( 10 \pi , 12,0 )
D) (10π,0,12) ( 10 \pi , 0,12 )

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

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Solve the problem. Unless stated otherwise, assume that the projectile flight is ideal, that the launch angle is measured from the horizontal, and that the projectile is launched from the origin over a horizontal surface -A projectile is fired at a speed of 720 m/sec720 \mathrm {~m} / \mathrm { sec } at an angle of 4545 ^ { \circ } . How long will it take to get 18 km18 \mathrm {~km} downrange? Round your answer to the nearest whole number.


A) 33 sec
B) 35 sec
C) 37 sec
D) It will never get that far downrange.

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

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Provide an appropriate response. -Derive the equations x=x0+v0k(1ekt)cosαy=y0+v0k(1ekt)sinα+gk2(1ktekt)\begin{array} { l } x = x _ { 0 } + \frac { v _ { 0 } } { k } \left( 1 - e ^ { - k t } \right) \cos \alpha \\y = y _ { 0 } + \frac { v _ { 0 } } { k } \left( 1 - e ^ { - k t } \right) \sin \alpha + \frac { g } { k ^ { 2 } } \left( 1 - k t - e ^ { - k t } \right)\end{array} by solving the following initial value problem for a vector r\mathbf { r } in the plane.  Differential equation d2rdt2=gjkv=gjkdrdt Initial conditions: r(0)=x0i+y0jdrdt(0)=v0=(v0cosα)i+(v0sinα)j\begin{aligned}\text { Differential equation } \frac { \mathrm { d } ^ { 2 } \mathbf { r } } { \mathrm { dt } ^ { 2 } } & = - g \mathbf { j } - \mathrm { k } \mathbf { v } = - \mathrm { g } \mathbf { j } - \mathrm { k } \frac { \mathrm { d } \mathbf { r } } { \mathrm { dt } } \\\text { Initial conditions: } \quad \mathrm { r } ( 0 ) & = \mathrm { x } _ { 0 } \mathbf { i } + \mathrm { y } _ { 0 } \mathbf { j } \\& \frac { \mathrm { d } \mathbf { r } } { \mathrm { dt } } ( 0 ) = \mathbf { v } _ { 0 } = \left( \mathrm { v } _ { 0 } \cos \alpha \right) \mathbf { i } + \left( \mathrm { v } _ { 0 } \sin \alpha \right) \mathbf { j }\end{aligned} The drag coefficient k\mathrm { k } is a positive constant representing resistance due to air density, vo and α\alpha are the projectile's initial speed and launch angle, and gg is the acceleration of gravity.

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Solve the problem. -The orbit of a satellite had a semimajor axis of a=6955 kma = 6955 \mathrm {~km} . Calculate the period of the satellite. (Earth's mass =5.975×1024 kg= 5.975 \times 10 ^ { 24 } \mathrm {~kg} and G=6.6720×1011Nm2 kg2\mathrm { G } = 6.6720 \times 10 ^ { - 11 } \mathrm { Nm } ^ { 2 } \mathrm {~kg} - 2 ) .


A) 0.18 sec
B) 96.2 min
C) 124 min
D) 9250 hr

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

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For the curve r(t) , find an equation for the indicated plane at the given value of t. - r(t) =(3sin53t+7) i+(3cos20t) 6) j+12kk\left. \mathbf { r } ( \mathrm { t } ) = \left( 3 \sin \frac { 5 } { 3 } \mathrm { t } + 7 \right) \mathbf { i } + ( 3 \cos 20 \mathrm { t } ) - 6 \right) \mathbf { j } + 12 \mathrm { k } \mathbf { k } ; osculating plane at t=2.5πt = 2.5 \pi .


A) 12169(y6) +60169(z30π) =0\frac { 12 } { 169 } ( y - 6 ) + \frac { 60 } { 169 } ( z - 30 \pi ) = 0
B) 1213(y+6) +6013(z+30) =0\frac { 12 } { 13 } ( y + 6 ) + \frac { 60 } { 13 } ( z + 30 ) = 0
C) 1213(y+6) +6013(z30π) =0\frac { 12 } { 13 } ( y + 6 ) + \frac { 60 } { 13 } ( z - 30 \pi ) = 0
D) 1213(x7) +6013(z30π) =0\frac { 12 } { 13 } ( x - 7 ) + \frac { 60 } { 13 } ( z - 30 \pi ) = 0

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

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 For the curve r(t) , write the acceleration in the form aTT + anN. \text { For the curve } r ( t ) \text {, write the acceleration in the form aTT } + \text { anN. } -r(t) = (6t sin t + 6 cos t) i + (6t cos t - 6 sin t) j + 3k


A) a=6T+6tNa = 6 T + 6 t N
B) a=16tNa = \frac { 1 } { 6 t } N
C) a=6T+16tNa = 6 T + \frac { 1 } { 6 t } N
D) a=6ta = 6 \mathrm { t }

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

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For the curve r(t) , find an equation for the indicated plane at the given value of t. - r(t) =(t+6) i+(ln(cost) 8) j+7k,π/2<t<π2;\mathbf { r } ( \mathrm { t } ) = ( \mathrm { t } + 6 ) \mathbf { i } + ( \ln ( \cos \mathrm { t } ) - 8 ) \mathbf { j } + 7 \mathbf { k } , - \pi / 2 < \mathrm { t } < \pi \cdot 2 ; rectifying plane att=π4\mathrm { at } \mathrm {} \mathrm { t } = \frac { \pi } { 4 } .


A) 22(x(π4+6) ) 22(y(ln(22) 8) ) =0\frac { \sqrt { 2 } } { 2 } \left( x - \left( \frac { \pi } { 4 } + 6 \right) \right) - \frac { \sqrt { 2 } } { 2 } \left( y - \left( \ln \left( \frac { \sqrt { 2 } } { 2 } \right) - 8 \right) \right) = 0
B) 22(x(π4+6) ) +22(y(ln(22) 8) ) =0- \frac { \sqrt { 2 } } { 2 } \left( x - \left( \frac { \pi } { 4 } + 6 \right) \right) + \frac { \sqrt { 2 } } { 2 } \left( y - \left( \ln \left( \frac { \sqrt { 2 } } { 2 } \right) - 8 \right) \right) = 0
C) 22(x(1+6) ) =0\frac { \sqrt { 2 } } { 2 } ( x - ( 1 + 6 ) ) = 0
D) 22(x(π4+6) ) ) +22(y(ln(22) 8) ) =0\left. \frac { \sqrt { 2 } } { 2 } \left( x - \left( \frac { \pi } { 4 } + 6 \right) \right) \right) ^ { \prime } + \frac { \sqrt { 2 } } { 2 } \left( y - \left( \ln \left( \frac { \sqrt { 2 } } { 2 } \right) - 8 \right) \right) = 0

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

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 For the curve r(t) , write the acceleration in the form aTT + anN. \text { For the curve } r ( t ) \text {, write the acceleration in the form aTT } + \text { anN. } -r(t) = (t + 10) i + (ln(cos t) - 6) j + 2k


A) a = (sec t tan t) T + (sec t) N
B) a = (sec t tan t) T - (sec t) N
C) a = (-sec t tan t) T + (sec t) N
D) a = (-sec t tan t) T - (sec t) N

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

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Calculate the arc length of the indicated portion of the curve r(t) . - r(t) =(2cos35t) j+(2sin35t) k;35πt710πr ( t ) = \left( 2 \cos ^ { 3 } 5 t \right) \mathbf { j } + \left( 2 \sin ^ { 3 } 5 t \right) \mathbf { k } ; \frac { 3 } { 5 } \pi \leq t \leq \frac { 7 } { 10 } \pi


A) 1
B) 3
C) 0
D) 6

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

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Solve the problem. Unless stated otherwise, assume that the projectile flight is ideal, that the launch angle is measured from the horizontal, and that the projectile is launched from the origin over a horizontal surface -An athlete puts a 16-lb shot at an angle of 3939 ^ { \circ } to the horizontal from 6.1ft6.1 \mathrm { ft } above the ground at an initial speed of 47ft/sec47 \mathrm { ft } / \mathrm { sec } . How far forward does the shot travel before it hits the ground? Round your answer to the nearest tenth.


A) 2 ft
B) 74.4 ft
C) 227.8 ft
D) 6.8 ft

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

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