Question 1

Find the limit  lim⁡t→0[(t2+3) i+cos⁡5tj−6k]\lim _ { t \rightarrow 0 } \left[ \left( t ^ { 2 } + 3 \right)  \mathbf { i } + \cos 5 t \mathbf { j } - 6 \mathbf { k } \right]limt→0​[(t2+3) i+cos5tj−6k]

Accepted Answer

A)    3i−6k3 i - 6 \mathbf { k }3i−6k 
B)    4i+j4 \mathbf { i } + \mathbf { j }4i+j 
C)    3i+j−6k3 i + j - 6 \mathbf { k }3i+j−6k 
D)    4i4 \mathbf { i }4i

Question 2

$$\text { Find equations of the normal plane to } x = t , y = t ^ { 2 } , z = t ^ { 3 } \text { at the point } ( 2,4,8 ) \text {. }$$

Accepted Answer

The answer of \[\text { Find equations of the normal...

Question 3

The curves $\mathbf { r } _ { 1 } ( t ) = \left\langle t , t ^ { 3 } , t ^ { 9 } \right\rangle$ and $\mathbf { r } _ { 2 } ( t ) = \langle \sin t , \sin 5 t , t \rangle$ intersects at the origin. Find their angle of intersection correct to the nearest degree.

Accepted Answer

The answer of The curves \(\mathbf { r } _...

Question 4

The curvature of the curve given by the vector function  rrr  is
 k(t) =∣rt(t) ×rtt(t) ∣∣rt(t) ∣3\mathrm { k } ( t )  = \frac { \left| \mathbf { r } ^ { t } ( t )  	imes \mathbf { r } ^ { tt } ( t )  ight| } { \left| \mathbf { r } ^ { t } ( t )  ight| ^ { 3 } }k(t) =∣rt(t) ∣3∣rt(t) ×rtt(t) ∣​ 
Use the formula to find the curvature of  r(t) =(19t,et,e−t}\mathbf { r } ( t )  = \left( \sqrt { 19 } t , e ^ { t } , e ^ { - t } ight\}r(t) =(19​t,et,e−t}  at the point  (0,1,1) ( 0,1,1 ) (0,1,1)  .

Accepted Answer

A)    21\sqrt { 21 }21c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5,-10,-9.5,-14c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54c44.2,-33.3,65.8,-50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0,35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429c69,-144,104.5,-217.7,106.5,-221l0 -0c5.3,-9.3,12,-14,20,-14H400000v40H845.2724s-225.272,467,-225.272,467s-235,486,-235,486c-2.7,4.7,-9,7,-19,7c-6,0,-10,-1,-12,-3s-194,-422,-194,-422s-65,47,-65,47zM834 80h400000v40h-400000z'/>​ 
B)    221\frac { \sqrt { 2 } } { 21 }212c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5,-10,-9.5,-14c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54c44.2,-33.3,65.8,-50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0,35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429c69,-144,104.5,-217.7,106.5,-221l0 -0c5.3,-9.3,12,-14,20,-14H400000v40H845.2724s-225.272,467,-225.272,467s-235,486,-235,486c-2.7,4.7,-9,7,-19,7c-6,0,-10,-1,-12,-3s-194,-422,-194,-422s-65,47,-65,47zM834 80h400000v40h-400000z'/>​​ 
C)    21221 \sqrt { 2 }212c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5,-10,-9.5,-14c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54c44.2,-33.3,65.8,-50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0,35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429c69,-144,104.5,-217.7,106.5,-221l0 -0c5.3,-9.3,12,-14,20,-14H400000v40H845.2724s-225.272,467,-225.272,467s-235,486,-235,486c-2.7,4.7,-9,7,-19,7c-6,0,-10,-1,-12,-3s-194,-422,-194,-422s-65,47,-65,47zM834 80h400000v40h-400000z'/>​ 
D)    212\frac { 21 } { \sqrt { 2 } }2c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5,-10,-9.5,-14c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54c44.2,-33.3,65.8,-50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0,35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429c69,-144,104.5,-217.7,106.5,-221l0 -0c5.3,-9.3,12,-14,20,-14H400000v40H845.2724s-225.272,467,-225.272,467s-235,486,-235,486c-2.7,4.7,-9,7,-19,7c-6,0,-10,-1,-12,-3s-194,-422,-194,-422s-65,47,-65,47zM834 80h400000v40h-400000z'/>​21​ 
E)    2121\frac { \sqrt { 21 } } { 21 }2121c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5,-10,-9.5,-14c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54c44.2,-33.3,65.8,-50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0,35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429c69,-144,104.5,-217.7,106.5,-221l0 -0c5.3,-9.3,12,-14,20,-14H400000v40H845.2724s-225.272,467,-225.272,467s-235,486,-235,486c-2.7,4.7,-9,7,-19,7c-6,0,-10,-1,-12,-3s-194,-422,-194,-422s-65,47,-65,47zM834 80h400000v40h-400000z'/>​​

Question 5

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

Accepted Answer

The answer of Find the derivative of the vector function....

Question 6

Find the unit tangent and unit normal vectors $\mathrm { T } ( t )$ and $\mathrm { N } ( t )$ for the curve $C$ defined by $\mathbf { r } ( t ) = t \mathbf { i } + 2 t ^ { 2 } \mathbf { j }$. Sketch the graph of $C$, and show $\mathrm { T } ( t )$ and $\mathrm { N } ( t )$ for $t = 1$.

Accepted Answer

The answer of Find the unit tangent and unit normal...

Question 7

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

Accepted Answer

A)    rtt(t) =−8sin⁡tj−cos⁡tk\mathrm { r } ^ { tt } ( t )  = - 8 \sin t \mathrm { j } - \cos t \mathrm { k }rtt(t) =−8sintj−costk 
B)    rtt(t) =−sin⁡tj−cos⁡tk\mathrm { r } ^ { tt} ( t )  = - \sin t \mathrm { j } - \cos t \mathrm { k }rtt(t) =−sintj−costk 
C)    rtt(t) =cos⁡tj−sin⁡tk\mathrm { r } ^ { tt } ( t )  = \cos t \mathrm { j } - \sin t \mathrm { k }rtt(t) =costj−sintk 
D)    rtt(t) =−sin⁡tj+8cos⁡tk\mathrm { r } ^ { tt } ( t )  = - \sin t \mathrm { j } + 8 \cos t \mathrm { k }rtt(t) =−sintj+8costk 
E)    rtt(t) =−8cos⁡tj−sin⁡tk\mathrm { r } ^ { tt } ( t )  = - 8 \cos t \mathrm { j } - \sin t \mathrm { k }rtt(t) =−8costj−sintk

Question 8

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

Accepted Answer

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

Question 9

Find the scalar tangential and normal components of acceleration of a particle with position vector  r(t) =4sin⁡ti+4cos⁡tj+3tk\mathbf { r } ( t )  = 4 \sin t \mathbf { i } + 4 \cos t \mathbf { j } + 3 t \mathbf { k }r(t) =4sinti+4costj+3tk

Accepted Answer

A)    aT=0,aN=410a _ { \mathrm { T } } = 0 , a _ { \mathrm { N } } = 4 \sqrt { 10 }aT​=0,aN​=410c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5,-10,-9.5,-14c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54c44.2,-33.3,65.8,-50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0,35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429c69,-144,104.5,-217.7,106.5,-221l0 -0c5.3,-9.3,12,-14,20,-14H400000v40H845.2724s-225.272,467,-225.272,467s-235,486,-235,486c-2.7,4.7,-9,7,-19,7c-6,0,-10,-1,-12,-3s-194,-422,-194,-422s-65,47,-65,47zM834 80h400000v40h-400000z'/>​ 
B)    aT=3,aN=410a _ { \mathrm { T } } = 3 , a _ { \mathrm { N } } = 4 \sqrt { 10 }aT​=3,aN​=410c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5,-10,-9.5,-14c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54c44.2,-33.3,65.8,-50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0,35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429c69,-144,104.5,-217.7,106.5,-221l0 -0c5.3,-9.3,12,-14,20,-14H400000v40H845.2724s-225.272,467,-225.272,467s-235,486,-235,486c-2.7,4.7,-9,7,-19,7c-6,0,-10,-1,-12,-3s-194,-422,-194,-422s-65,47,-65,47zM834 80h400000v40h-400000z'/>​ 
C)    aT=3,aN=4a _ { \mathrm { T } } = 3 , a _ { \mathrm { N } } = 4aT​=3,aN​=4 
D)    aT=0,aN=4a _ { T } = 0 , a _ { N } = 4aT​=0,aN​=4

Question 10

Find the position vector of a particle that has the given acceleration and the given initial velocity and position. $$\mathbf { a } ( t ) = - 4 \mathbf { k } , \mathbf { v } ( 0 ) = \mathbf { i } + \mathbf { j } - 20 \mathbf { k } , \mathbf { r } ( 0 ) = 4 \mathbf { i } + 9 \mathbf { j }$$

Accepted Answer

The answer of Find the position vector of a particle...

Question 11

Find the unit tangent vector for the curve given by  r(t) =⟨15t5,13t3,t⟩\mathbf { r } ( t )  = \left\langle \frac { 1 } { 5 } t ^ { 5 } , \frac { 1 } { 3 } t ^ { 3 } , t \right\rangler(t) =⟨51​t5,31​t3,t⟩ .

Accepted Answer

A)    ⟨t4,t2,1}t10+t4\frac { \left\langle t ^ { 4 } , t ^ { 2 } , 1 \right\} } { \sqrt { t ^ { 10 } + t ^ { 4 } } }t10+t4c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5,-10,-9.5,-14c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54c44.2,-33.3,65.8,-50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0,35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429c69,-144,104.5,-217.7,106.5,-221l0 -0c5.3,-9.3,12,-14,20,-14H400000v40H845.2724s-225.272,467,-225.272,467s-235,486,-235,486c-2.7,4.7,-9,7,-19,7c-6,0,-10,-1,-12,-3s-194,-422,-194,-422s-65,47,-65,47zM834 80h400000v40h-400000z'/>​⟨t4,t2,1}​ 
B)    ⟨t4,t2,1) 4t8+4t4+1\frac { \left\langle t ^ { 4 } , t ^ { 2 } , 1 \right)  } { \sqrt { 4 t ^ { 8 } + 4 t ^ { 4 } + 1 } }4t8+4t4+1c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5,-10,-9.5,-14c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54c44.2,-33.3,65.8,-50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0,35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429c69,-144,104.5,-217.7,106.5,-221l0 -0c5.3,-9.3,12,-14,20,-14H400000v40H845.2724s-225.272,467,-225.272,467s-235,486,-235,486c-2.7,4.7,-9,7,-19,7c-6,0,-10,-1,-12,-3s-194,-422,-194,-422s-65,47,-65,47zM834 80h400000v40h-400000z'/>​⟨t4,t2,1) ​ 
C)   ⟨t4,t2,1) t8+t4+1\frac { \left\langle t ^ { 4 } , t ^ { 2 } , 1 \right)  } { \sqrt { t ^ { 8 } + t ^ { 4 } + 1 } }t8+t4+1c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5,-10,-9.5,-14c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54c44.2,-33.3,65.8,-50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0,35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429c69,-144,104.5,-217.7,106.5,-221l0 -0c5.3,-9.3,12,-14,20,-14H400000v40H845.2724s-225.272,467,-225.272,467s-235,486,-235,486c-2.7,4.7,-9,7,-19,7c-6,0,-10,-1,-12,-3s-194,-422,-194,-422s-65,47,-65,47zM834 80h400000v40h-400000z'/>​⟨t4,t2,1) ​ 
D)   ⟨t4,t2,1⟩t8+t4\frac { \left\langle t ^ { 4 } , t ^ { 2 } , 1 \right\rangle } { \sqrt { t ^ { 8 } + t ^ { 4 } } }t8+t4c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5,-10,-9.5,-14c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54c44.2,-33.3,65.8,-50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0,35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429c69,-144,104.5,-217.7,106.5,-221l0 -0c5.3,-9.3,12,-14,20,-14H400000v40H845.2724s-225.272,467,-225.272,467s-235,486,-235,486c-2.7,4.7,-9,7,-19,7c-6,0,-10,-1,-12,-3s-194,-422,-194,-422s-65,47,-65,47zM834 80h400000v40h-400000z'/>​⟨t4,t2,1⟩​ 
E)   None of these

Question 12

At what point on the curve  x=t3,y=9t,z=t4x = t ^ { 3 } , y = 9 t , z = t ^ { 4 }x=t3,y=9t,z=t4  is the normal plane parallel to the plane  3x+9y−4z=4?3 x + 9 y - 4 z = 4 ?3x+9y−4z=4?

Accepted Answer

A)    (−1,−3,9) ( - 1 , - 3,9 ) (−1,−3,9)  
B)    (9,18,−2) ( 9,18 , - 2 ) (9,18,−2)  
C)    (−1,−9,1) ( - 1 , - 9,1 ) (−1,−9,1)  
D)    (−18,9,1) ( - 18,9,1 ) (−18,9,1)  
E)    (−9,1,1) ( - 9,1,1 ) (−9,1,1)

Question 13

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

Accepted Answer

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

Question 14

Find parametric equations for the tangent line to the curve with parametric equations  x=3tx = 3 tx=3t ,  y=5t2,z=4t3y = 5 t ^ { 2 } , z = 4 t ^ { 3 }y=5t2,z=4t3  at the point with  t=1t = 1t=1 .

Accepted Answer

A)    x=1+3t,y=1+10t,z=1+12tx = 1 + 3 t , y = 1 + 10 t , z = 1 + 12 tx=1+3t,y=1+10t,z=1+12t 
B)    x=3+3t,y=5+10t,z=4+12tx = 3 + 3 t , y = 5 + 10 t , z = 4 + 12 tx=3+3t,y=5+10t,z=4+12t 
C)    x=1+3t,y=1+5t,z=1+4tx = 1 + 3 t , y = 1 + 5 t , z = 1 + 4 tx=1+3t,y=1+5t,z=1+4t 
D)    x=3+3t,y=5+5t,z=4+4tx = 3 + 3 t , y = 5 + 5 t , z = 4 + 4 tx=3+3t,y=5+5t,z=4+4t

Question 15

Find the domain of the vector function  r(t) =⟨8t,1t−9,ln⁡t⟩\mathbf { r } ( t )  = \left\langle 8 \sqrt { t } , \frac { 1 } { t - 9 } , \ln t \right\rangler(t) =⟨8t​,t−91​,lnt⟩ .

Accepted Answer

A)    (0,9) ∪(9,∞) ( 0,9 )  \cup ( 9 , \infty ) (0,9) ∪(9,∞)  
B)    [0,8) ∪(8,9) ∪(9,∞) [ 0,8 )  \cup ( 8,9 )  \cup ( 9 , \infty ) [0,8) ∪(8,9) ∪(9,∞)  
C)    (0,8) ∪(8,∞) ( 0,8 )  \cup ( 8 , \infty ) (0,8) ∪(8,∞)  
D)    (0,8) ∪(8,9) ∪(9,∞) ( 0,8 )  \cup ( 8,9 )  \cup ( 9 , \infty ) (0,8) ∪(8,9) ∪(9,∞)

Question 16

Find a vector function that represents the curve of intersection of the two surfaces:
the top half of the ellipsoid  x2+7y2+7z2=49x ^ { 2 } + 7 y ^ { 2 } + 7 z ^ { 2 } = 49x2+7y2+7z2=49  and the parabolic cylinder  y=x2y = x ^ { 2 }y=x2 .

Accepted Answer

A)   r(t) =ti+t4j+49−t2+7t7k\mathbf { r } ( t )  = t \mathbf { i } + t ^ { 4 } \mathbf { j } + \sqrt { \frac { 49 - t ^ { 2 } + 7 t } { 7 } } \mathbf { k }r(t) =ti+t4j+749−t2+7t​l0 -0c4,-6.7,10,-10,18,-10 H400000v40H1013.1s-83.4,268,-264.1,840c-180.7,572,-277,876.3,-289,913c-4.7,4.7,-12.7,7,-24,7s-12,0,-12,0c-1.3,-3.3,-3.7,-11.7,-7,-25c-35.3,-125.3,-106.7,-373.3,-214,-744c-10,12,-21,25,-33,39s-32,39,-32,39c-6,-5.3,-15,-14,-27,-26s25,-30,25,-30c26.7,-32.7,52,-63,76,-91s52,-60,52,-60s208,722,208,722c56,-175.3,126.3,-397.3,211,-666c84.7,-268.7,153.8,-488.2,207.5,-658.5c53.7,-170.3,84.5,-266.8,92.5,-289.5zM1001 80h400000v40h-400000z'/>​k 
B)   r(t) =ti+t2j+49+t2−7t47k\mathbf { r } ( t )  = t \mathbf { i } + t ^ { 2 } \mathbf { j } + \sqrt { \frac { 49 + t ^ { 2 } - 7 t ^ { 4 } } { 7 } } \mathbf { k }r(t) =ti+t2j+749+t2−7t4​l0 -0c4,-6.7,10,-10,18,-10 H400000v40H1013.1s-83.4,268,-264.1,840c-180.7,572,-277,876.3,-289,913c-4.7,4.7,-12.7,7,-24,7s-12,0,-12,0c-1.3,-3.3,-3.7,-11.7,-7,-25c-35.3,-125.3,-106.7,-373.3,-214,-744c-10,12,-21,25,-33,39s-32,39,-32,39c-6,-5.3,-15,-14,-27,-26s25,-30,25,-30c26.7,-32.7,52,-63,76,-91s52,-60,52,-60s208,722,208,722c56,-175.3,126.3,-397.3,211,-666c84.7,-268.7,153.8,-488.2,207.5,-658.5c53.7,-170.3,84.5,-266.8,92.5,-289.5zM1001 80h400000v40h-400000z'/>​k 
C)   r(t) =ti+t2j+49−t2−7t7k\mathrm { r } ( t )  = t \mathbf { i } + t ^ { 2 } \mathbf { j } + \sqrt { \frac { 49 - t ^ { 2 } - 7 t } { 7 } } \mathbf { k }r(t) =ti+t2j+749−t2−7t​l0 -0c4,-6.7,10,-10,18,-10 H400000v40H1013.1s-83.4,268,-264.1,840c-180.7,572,-277,876.3,-289,913c-4.7,4.7,-12.7,7,-24,7s-12,0,-12,0c-1.3,-3.3,-3.7,-11.7,-7,-25c-35.3,-125.3,-106.7,-373.3,-214,-744c-10,12,-21,25,-33,39s-32,39,-32,39c-6,-5.3,-15,-14,-27,-26s25,-30,25,-30c26.7,-32.7,52,-63,76,-91s52,-60,52,-60s208,722,208,722c56,-175.3,126.3,-397.3,211,-666c84.7,-268.7,153.8,-488.2,207.5,-658.5c53.7,-170.3,84.5,-266.8,92.5,-289.5zM1001 80h400000v40h-400000z'/>​k 
D)   r(t) =ti+t2j+7−t2−7t47k\mathbf { r } ( t )  = t \mathbf { i } + t ^ { 2 } \mathbf { j } + \sqrt { \frac { 7 - t ^ { 2 } - 7 t ^ { 4 } } { 7 } } \mathbf { k }r(t) =ti+t2j+77−t2−7t4​l0 -0c4,-6.7,10,-10,18,-10 H400000v40H1013.1s-83.4,268,-264.1,840c-180.7,572,-277,876.3,-289,913c-4.7,4.7,-12.7,7,-24,7s-12,0,-12,0c-1.3,-3.3,-3.7,-11.7,-7,-25c-35.3,-125.3,-106.7,-373.3,-214,-744c-10,12,-21,25,-33,39s-32,39,-32,39c-6,-5.3,-15,-14,-27,-26s25,-30,25,-30c26.7,-32.7,52,-63,76,-91s52,-60,52,-60s208,722,208,722c56,-175.3,126.3,-397.3,211,-666c84.7,-268.7,153.8,-488.2,207.5,-658.5c53.7,-170.3,84.5,-266.8,92.5,-289.5zM1001 80h400000v40h-400000z'/>​k 
E)   r(t) =ti−t2j−49−t2−7t7k\mathbf { r } ( t )  = t \mathbf { i } - t ^ { 2 } \mathbf { j } - \sqrt { \frac { 49 - t ^ { 2 } - 7 t } { 7 } } \mathbf { k }r(t) =ti−t2j−749−t2−7t​l0 -0c4,-6.7,10,-10,18,-10 H400000v40H1013.1s-83.4,268,-264.1,840c-180.7,572,-277,876.3,-289,913c-4.7,4.7,-12.7,7,-24,7s-12,0,-12,0c-1.3,-3.3,-3.7,-11.7,-7,-25c-35.3,-125.3,-106.7,-373.3,-214,-744c-10,12,-21,25,-33,39s-32,39,-32,39c-6,-5.3,-15,-14,-27,-26s25,-30,25,-30c26.7,-32.7,52,-63,76,-91s52,-60,52,-60s208,722,208,722c56,-175.3,126.3,-397.3,211,-666c84.7,-268.7,153.8,-488.2,207.5,-658.5c53.7,-170.3,84.5,-266.8,92.5,-289.5zM1001 80h400000v40h-400000z'/>​k

Question 17

If  r(t) =(t,t5,tı⟩\mathbf { r } ( t )  = \left( t , t ^ { 5 } , t ^ { \imath } \right\rangler(t) =(t,t5,t⟩ , find  rtt(t) \mathbf { r } ^ { tt } ( t ) rtt(t)

Accepted Answer

A)    (0,42t4,20t2⟩\left( 0,42 t ^ { 4 } , 20 t ^ { 2 } \right\rangle(0,42t4,20t2⟩ 
B)    (0,42t5,20t3) \left( 0,42 t ^ { 5 } , 20 t ^ { 3 } \right) (0,42t5,20t3)  
C)    (0,20t3,42t5) \left( 0,20 t ^ { 3 } , 42 t ^ { 5 } \right) (0,20t3,42t5)  
D)    ⟨1,5t6,7t8}\left\langle 1,5 t ^ { 6 } , 7 t ^ { 8 } \right\}⟨1,5t6,7t8} 
E)    (0,20t2,42t4) \left( 0,20 t ^ { 2 } , 42 t ^ { 4 } \right) (0,20t2,42t4)

Question 18

$$\text { Find the integral } \int \left( \sin 7 t \mathbf { i } + \cos 7 t \mathbf { j } + e ^ { - t / 5 } \mathbf { k } \right) d t \text {. }$$

Accepted Answer

The answer of \[\text { Find the integral } \int...

Question 19

Use Simpson's Rule with  n=4\mathrm { n } = 4n=4  to estimate the length of the arc of the curve with equations  x=t,y=4t,z=t2+1x = \sqrt { t } , y = \frac { 4 } { t } , z = t ^ { 2 } + 1x=t​,y=t4​,z=t2+1 , from  (1,4,2) ( 1,4,2 ) (1,4,2)   to  (2,1,17) ( 2,1,17 ) (2,1,17)  . Round your answer to four decimal places.

Accepted Answer

A)    14.82414.82414.824 
B)    7.20417.20417.2041 
C)    15.830615.830615.8306 
D)    6.57066.57066.5706 
E)   None of these

Question 20

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}t0.511.522.5​x5.812.625.639.242.4​y9.114.921.239.542.4​z4.316.829.437.943​​

Find the average velocity over the time interval  [1,2][ 1,2 ][1,2] .

Accepted Answer

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

Exam 13: Vector Functions

If $\mathbf { r } ( t ) = \left( t , t ^ { 5 } , t ^ { \imath } \right\rangle$ , find $\mathbf { r } ^ { tt } ( t )$

Correct Answer
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The curves $\mathbf { r } _ { 1 } ( t ) = \left\langle t , t ^ { 3 } , t ^ { 9 } \right\rangle$ and $\mathbf { r } _ { 2 } ( t ) = \langle \sin t , \sin 5 t , t \rangle$ intersects at the origin. Find their angle of intersection correct to the nearest degree.

Correct Answer
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Find the unit tangent vector for the curve given by $\mathbf { r } ( t ) = \left\langle \frac { 1 } { 5 } t ^ { 5 } , \frac { 1 } { 3 } t ^ { 3 } , t \right\rangle$ .

Correct Answer
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Find the domain of the vector function $\mathbf { r } ( t ) = \left\langle 8 \sqrt { t } , \frac { 1 } { t - 9 } , \ln t \right\rangle$ .

Correct Answer
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Find the limit $\lim _ { t \rightarrow 0 } \left[ \left( t ^ { 2 } + 3 \right) \mathbf { i } + \cos 5 t \mathbf { j } - 6 \mathbf { k } \right]$

Correct Answer
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Find the integral $\int \left( 4 t \mathbf { i } + 6 t ^ { 2 } \mathbf { j } + 4 \mathbf { k } \right) d t$ .

Correct Answer
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Find parametric equations for the tangent line to the curve with parametric equations $x = 3 t$ , $y = 5 t ^ { 2 } , z = 4 t ^ { 3 }$ at the point with $t = 1$ .

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

Correct Answer
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Correct Answer
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Find a vector function that represents the curve of intersection of the two surfaces: the top half of the ellipsoid $x ^ { 2 } + 7 y ^ { 2 } + 7 z ^ { 2 } = 49$ and the parabolic cylinder $y = x ^ { 2 }$ .

Correct Answer
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At what point on the curve $x = t ^ { 3 } , y = 9 t , z = t ^ { 4 }$ is the normal plane parallel to the plane $3 x + 9 y - 4 z = 4 ?$

Correct Answer
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Find the unit tangent and unit normal vectors $\mathrm { T } ( t )$ and $\mathrm { N } ( t )$ for the curve $C$ defined by $\mathbf { r } ( t ) = t \mathbf { i } + 2 t ^ { 2 } \mathbf { j }$ . Sketch the graph of $C$ , and show $\mathrm { T } ( t )$ and $\mathrm { N } ( t )$ for $t = 1$ .

Correct Answer
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Correct Answer
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Find $\mathbf { r } ^ { \tt } ( t )$ for the function given. $\mathbf { r } ( t ) = 8 \mathbf { i } + \sin t \mathbf { j } + \cos t \mathbf { k }$

Correct Answer
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Find the curvature of the curve $\mathbf { r } ( t ) = 3 \sin 4 t \mathbf { i } + 3 \cos 4 t \mathbf { j } + 3 t \mathbf { k }$ .

Correct Answer
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Use Simpson's Rule with $\mathrm { n } = 4$ to estimate the length of the arc of the curve with equations $x = \sqrt { t } , y = \frac { 4 } { t } , z = t ^ { 2 } + 1$ , from $( 1,4,2 )$ to $( 2,1,17 )$ . Round your answer to four decimal places.

Correct Answer
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$\text { Find equations of the normal plane to } x = t , y = t ^ { 2 } , z = t ^ { 3 } \text { at the point } ( 2,4,8 ) \text {. }$

Correct Answer
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$\text { Find the integral } \int \left( \sin 7 t \mathbf { i } + \cos 7 t \mathbf { j } + e ^ { - t / 5 } \mathbf { k } \right) d t \text {. }$

Correct Answer
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Find the position vector of a particle that has the given acceleration and the given initial velocity and position. $\mathbf { a } ( t ) = - 4 \mathbf { k } , \mathbf { v } ( 0 ) = \mathbf { i } + \mathbf { j } - 20 \mathbf { k } , \mathbf { r } ( 0 ) = 4 \mathbf { i } + 9 \mathbf { j }$

Correct Answer
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Find the scalar tangential and normal components of acceleration of a particle with position vector $\mathbf { r } ( t ) = 4 \sin t \mathbf { i } + 4 \cos t \mathbf { j } + 3 t \mathbf { k }$

Correct Answer
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Exam 13: Vector Functions

If r(t) =(t,t5,tı⟩\mathbf { r } ( t ) = \left( t , t ^ { 5 } , t ^ { \imath } \right\rangler(t) =(t,t5,t⟩ , find rtt(t) \mathbf { r } ^ { tt } ( t ) rtt(t)

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Correct AnswerverifiedShow Answer

Find the unit tangent vector for the curve given by r(t) =⟨15t5,13t3,t⟩\mathbf { r } ( t ) = \left\langle \frac { 1 } { 5 } t ^ { 5 } , \frac { 1 } { 3 } t ^ { 3 } , t \right\rangler(t) =⟨51​t5,31​t3,t⟩ .

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Find the domain of the vector function r(t) =⟨8t,1t−9,ln⁡t⟩\mathbf { r } ( t ) = \left\langle 8 \sqrt { t } , \frac { 1 } { t - 9 } , \ln t \right\rangler(t) =⟨8t​,t−91​,lnt⟩ .

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Find the limit lim⁡t→0[(t2+3) i+cos⁡5tj−6k]\lim _ { t \rightarrow 0 } \left[ \left( t ^ { 2 } + 3 \right) \mathbf { i } + \cos 5 t \mathbf { j } - 6 \mathbf { k } \right]limt→0​[(t2+3) i+cos5tj−6k]

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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∫(4ti+6t2j+4k) dt .

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Find parametric equations for the tangent line to the curve with parametric equations x=3tx = 3 tx=3t , y=5t2,z=4t3y = 5 t ^ { 2 } , z = 4 t ^ { 3 }y=5t2,z=4t3 at the point with t=1t = 1t=1 .

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

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Correct AnswerverifiedShow Answer

Find a vector function that represents the curve of intersection of the two surfaces: the top half of the ellipsoid x2+7y2+7z2=49x ^ { 2 } + 7 y ^ { 2 } + 7 z ^ { 2 } = 49x2+7y2+7z2=49 and the parabolic cylinder y=x2y = x ^ { 2 }y=x2 .

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At what point on the curve x=t3,y=9t,z=t4x = t ^ { 3 } , y = 9 t , z = t ^ { 4 }x=t3,y=9t,z=t4 is the normal plane parallel to the plane 3x+9y−4z=4?3 x + 9 y - 4 z = 4 ?3x+9y−4z=4?

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Correct AnswerverifiedShow Answer

Correct AnswerverifiedShow Answer

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

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

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Correct AnswerverifiedShow Answer

Find equations of the normal plane to x=t,y=t2,z=t3 at the point (2,4,8). \text { Find equations of the normal plane to } x = t , y = t ^ { 2 } , z = t ^ { 3 } \text { at the point } ( 2,4,8 ) \text {. } Find equations of the normal plane to x=t,y=t2,z=t3 at the point (2,4,8).

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Find the integral ∫(sin⁡7ti+cos⁡7tj+e−t/5k)dt. \text { Find the integral } \int \left( \sin 7 t \mathbf { i } + \cos 7 t \mathbf { j } + e ^ { - t / 5 } \mathbf { k } \right) d t \text {. } Find the integral ∫(sin7ti+cos7tj+e−t/5k)dt.

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Correct AnswerverifiedShow Answer

Find the scalar tangential and normal components of acceleration of a particle with position vector r(t) =4sin⁡ti+4cos⁡tj+3tk\mathbf { r } ( t ) = 4 \sin t \mathbf { i } + 4 \cos t \mathbf { j } + 3 t \mathbf { k }r(t) =4sinti+4costj+3tk

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If $\mathbf { r } ( t ) = \left( t , t ^ { 5 } , t ^ { \imath } \right\rangle$ , find $\mathbf { r } ^ { tt } ( t )$

Correct Answer
verified

Correct Answer
verified

Find the unit tangent vector for the curve given by $\mathbf { r } ( t ) = \left\langle \frac { 1 } { 5 } t ^ { 5 } , \frac { 1 } { 3 } t ^ { 3 } , t \right\rangle$ .

Correct Answer
verified

Find the domain of the vector function $\mathbf { r } ( t ) = \left\langle 8 \sqrt { t } , \frac { 1 } { t - 9 } , \ln t \right\rangle$ .

Correct Answer
verified

Find the limit $\lim _ { t \rightarrow 0 } \left[ \left( t ^ { 2 } + 3 \right) \mathbf { i } + \cos 5 t \mathbf { j } - 6 \mathbf { k } \right]$

Correct Answer
verified

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

Correct Answer
verified

Find parametric equations for the tangent line to the curve with parametric equations $x = 3 t$ , $y = 5 t ^ { 2 } , z = 4 t ^ { 3 }$ at the point with $t = 1$ .

Correct Answer
verified

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

Correct Answer
verified

Correct Answer
verified

Find a vector function that represents the curve of intersection of the two surfaces: the top half of the ellipsoid $x ^ { 2 } + 7 y ^ { 2 } + 7 z ^ { 2 } = 49$ and the parabolic cylinder $y = x ^ { 2 }$ .

Correct Answer
verified

At what point on the curve $x = t ^ { 3 } , y = 9 t , z = t ^ { 4 }$ is the normal plane parallel to the plane $3 x + 9 y - 4 z = 4 ?$

Correct Answer
verified

Correct Answer
verified

Correct Answer
verified

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

Correct Answer
verified

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

Correct Answer
verified

Correct Answer
verified

$\text { Find equations of the normal plane to } x = t , y = t ^ { 2 } , z = t ^ { 3 } \text { at the point } ( 2,4,8 ) \text {. }$

Correct Answer
verified

$\text { Find the integral } \int \left( \sin 7 t \mathbf { i } + \cos 7 t \mathbf { j } + e ^ { - t / 5 } \mathbf { k } \right) d t \text {. }$

Correct Answer
verified

Correct Answer
verified

Find the scalar tangential and normal components of acceleration of a particle with position vector $\mathbf { r } ( t ) = 4 \sin t \mathbf { i } + 4 \cos t \mathbf { j } + 3 t \mathbf { k }$

Correct Answer
verified