Question 1

The helix  r1(t) =8cos⁡ti+sin⁡tj+tk\mathbf { r } _ { 1 } ( t )  = 8 \cos t \mathbf { i } + \sin t \mathbf { j } + t \mathbf { k }r1​(t) =8costi+sintj+tk  intersects the curve  r2(t) =(8+t) i+10t2j+9t3k\mathbf { r } _ { 2 } ( t )  = ( 8 + t )  \mathbf { i } + 10 t ^ { 2 } \mathbf { j } + 9 t ^ { 3 } \mathbf { k }r2​(t) =(8+t) i+10t2j+9t3k  at the point  (8,0,0) ( 8,0,0 ) (8,0,0)   . Find the angle of intersection.

Accepted Answer

A)    π3\frac { \pi } { 3 }3π​ 
B)    π4\frac { \pi } { 4 }4π​ 
C)    π2\frac { \pi } { 2 }2π​ 
D)    000 
E)   None of these

Question 2

Find the point(s) on the graph of $y = e ^ { - 11 x ^ { 2 } }$ at which the curvature is zero.

Accepted Answer

The answer of Find the point(s) on the graph of...

Question 3

Find the unit tangent vector  T(t) T ( t ) T(t)   .  r(t) =⟨2sin⁡t,4t,2cos⁡t⟩\mathbf { r } ( t )  = \langle 2 \sin t , 4 t , 2 \cos t \rangler(t) =⟨2sint,4t,2cost⟩

Accepted Answer

A)    ⟨cos⁡t25,25,−sin⁡t25⟩\left\langle\frac { \cos t } { 2 \sqrt { 5 } } , \frac { 2 } { \sqrt { 5 } } , - \frac { \sin t } { 2 \sqrt { 5 } } \right\rangle⟨25c-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'/>​cost​,5c-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'/>​2​,−25c-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'/>​sint​⟩ 
B)    ⟨3cos⁡t,6,−3sin⁡t⟩\langle3 \cos t , 6 , - 3 \sin t \rangle⟨3cost,6,−3sint⟩ 
C)    ⟨cos⁡t5,25,−sin⁡t5⟩\left\langle\frac { \cos t } { \sqrt { 5 } } , \frac { 2 } { \sqrt { 5 } } , - \frac { \sin t } { \sqrt { 5 } } \right\rangle⟨5c-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'/>​cost​,5c-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'/>​2​,−5c-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'/>​sint​⟩ 
D)    ⟨35cos⁡t,6,−35sin⁡t⟩\langle3 \sqrt { 5 } \cos t , 6 , - 3 \sqrt { 5 } \sin t \rangle⟨35c-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'/>​cost,6,−35c-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'/>​sint⟩ 
E)    ⟨−cos⁡t25,425,3sin⁡t25⟩\left\langle- \frac { \cos t } { 2 \sqrt { 5 } } , \frac { 4 } { 2 \sqrt { 5 } } , \frac { 3 \sin t } { 2 \sqrt { 5 } } \right\rangle⟨−25c-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'/>​cost​,25c-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'/>​4​,25c-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'/>​3sint​⟩

Question 4

Find the acceleration of a particle with the given position function.  r(t) =9sin⁡ti+10tj−8cos⁡tk\mathbf { r } ( t )  = 9 \sin t \mathbf { i } + 10 t \mathbf { j } - 8 \cos t \mathbf { k }r(t) =9sinti+10tj−8costk

Accepted Answer

A)    a(t) =−9sin⁡ti−9cos⁡tj\mathbf { a } ( t )  = - 9 \sin t \mathbf { i } - 9 \cos t \mathbf { j }a(t) =−9sinti−9costj 
B)    a(t) =−9sin⁡ti+9cos⁡tj\mathbf { a } ( t )  = - 9 \sin t \mathbf { i } + 9 \cos t \mathbf { j }a(t) =−9sinti+9costj 
C)    a(t) =−9sin⁡ti+10cos⁡tk\mathbf { a } ( t )  = - 9 \sin t \mathbf { i } + 10 \cos t \mathbf { k }a(t) =−9sinti+10costk 
D)    a(t) =−9sin⁡ti+8cos⁡tk\mathbf { a } ( t )  = - 9 \sin t \mathbf { i } + 8 \cos t \mathbf { k }a(t) =−9sinti+8costk 
E)    a(t) =9sin⁡ti−10tk\mathbf { a } ( t )  = 9 \sin t \mathbf { i } - 10 t \mathbf { k }a(t) =9sinti−10tk

Question 5

For the curve given by  r(t) =(4sin⁡t,5t,4cost) \mathbf { r } ( t )  = (4 \sin t , 5 t , 4 \mathrm { cos } t ) r(t) =(4sint,5t,4cost)   , find the unit normal vector.  a(t) =2k,v(0) =10i−9j\mathbf { a } ( t )  = 2 \mathbf { k } , \mathbf { v } ( 0 )  = 10 \mathbf { i } - 9 \mathbf { j }a(t) =2k,v(0) =10i−9j

Accepted Answer

A)    (2sin⁡t,5,−2cos⁡t) ( 2 \sin t , 5 , - 2 \cos t) (2sint,5,−2cost)  
B)    (−2sin⁡t,0,−2cos⁡t) (- 2 \sin t , 0 , - 2 \cos t ) (−2sint,0,−2cost)  
C)    (−sin⁡t29,0,−cos⁡t29) \left( - \frac { \sin t } { \sqrt { 29 } } , 0 , - \frac { \cos t } { \sqrt { 29 } } \right) (−29c-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'/>​sint​,0,−29c-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'/>​cost​)  
D)    (29sin⁡t,0,−29cost) ( \sqrt { 29 } \sin t , 0 , - \sqrt { 29 } \mathrm { cost } ) (29c-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'/>​sint,0,−29c-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'/>​cost)  
E)   None of these

Question 6

Find a vector function that represents the curve of intersection of the two surfaces: The circular cylinder  x2+y2=9x ^ { 2 } + y ^ { 2 } = 9x2+y2=9  and the parabolic cylinder  z=xyz = x yz=xy  .

Accepted Answer

A)    r(t) =cos⁡ti+sin⁡tj−9cos⁡2tk\mathbf { r } ( t )  = \cos t \mathbf { i } + \sin t \mathbf { j } - 9 \cos 2 t \mathbf { k }r(t) =costi+sintj−9cos2tk 
B)    r(t) =3cos⁡(t) i+3sin⁡(t) j+9sin⁡(t) cos⁡(t) k\mathbf { r } ( t )  = 3 \cos ( t )  \mathbf { i } + 3 \sin ( t )  \mathbf { j } + 9 \sin ( t )  \cos ( t )  \mathbf { k }r(t) =3cos(t) i+3sin(t) j+9sin(t) cos(t) k 
C)    r(t) =3cos⁡(t) i+3sin⁡(t) j−sin⁡(t) cos⁡(t) k\mathbf { r } ( t )  = 3 \cos ( t )  \mathbf { i } + 3 \sin ( t )  \mathbf { j } - \sin ( t )  \cos ( t )  \mathbf { k }r(t) =3cos(t) i+3sin(t) j−sin(t) cos(t) k 
D)    r(t) =cos⁡(t) i+sin⁡(t) j+9sin⁡(t) cos⁡(t) k\mathbf { r } ( t )  = \cos ( t )  \mathbf { i } + \sin ( t )  \mathbf { j } + 9 \sin ( t )  \cos ( t )  \mathbf { k }r(t) =cos(t) i+sin(t) j+9sin(t) cos(t) k 
E)    r(t) =9cos⁡ti+9tj+9cos⁡2tk\mathbf { r } ( t )  = 9 \cos t \mathbf { i } + 9 t \mathbf { j } + 9 \cos ^ { 2 } t \mathbf { k }r(t) =9costi+9tj+9cos2tk

Question 7

Find the point of intersection of the tangent lines to the curve $\mathbf { r } ( t ) = ( \sin \pi t , 5 \sin \pi t , \cos \pi t )$ , at the points where $t = 0$ and $t = 0.5$ .

Accepted Answer

The answer of Find the point of intersection of the...

Question 8

Find the limit $\lim _ { t \rightarrow \infty } \left\langle e ^ { - 4 t } , \frac { 1 } { t } , \frac { 7 t ^ { 2 } } { t ^ { 2 } + 1 } \right\rangle$ .

Accepted Answer

The answer of Find the limit \(\lim _ { t...

Question 9

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

Accepted Answer

A)    363 \sqrt { 6 }36c-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)    6\sqrt { 6 }6c-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)    262 \sqrt { 6 }26c-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)    666 \sqrt { 6 }66c-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 10

The curvature of the curve given by the vector function  rrr  is  k(t) =∣r′(t) ×r′′(t) ∣∣r′(t) ∣3k ( t )  = \frac { \left| \mathbf { r } ^ { \prime } ( t )  	imes \mathbf { r } ^ { \prime \prime } ( t )  ight| } { \left| \mathbf { r } ^ { \prime } ( t )  ight| ^ { 3 } }k(t) =∣r′(t) ∣3∣r′(t) ×r′′(t) ∣​  Use the formula to find the curvature of  r(t) =⟨13t,et,e−t⟩\mathbf { r } ( t )  = \left\langle \sqrt { 13 } t , e ^ { t } , e ^ { - t } ightangler(t) =⟨13​t,et,e−t⟩ 
 at the point  (0,1,1) ( 0,1,1 ) (0,1,1)   .

Accepted Answer

A)    15\sqrt { 15 }15c-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)    215\frac { \sqrt { 2 } } { 15 }152c-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)    15215 \sqrt { 2 }152c-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)    152\frac { 15 } { \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'/>​15​ 
E)    1515\frac { \sqrt { 15 } } { 15 }1515c-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 11

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 }r(t) =3sinti+3costj+5tk

Accepted Answer

A)    aT=0a _ { \mathbf { T } } = 0aT​=0   aN=3a _ { \mathrm { N } } = 3aN​=3 
B)    aT=5a _ { \mathbf { T } } = 5aT​=5   aN=3a _ { \mathrm { N } } = 3aN​=3 
C)    aT=0a _ { \mathbf { T } } = 0aT​=0   aN=326a _ { \mathrm { N } } = 3 \sqrt { 26 }aN​=326c-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)    aT=5a _ { \mathbf { T } } = 5aT​=5   aN=326a _ { \mathrm { N } } = 3 \sqrt { 26 }aN​=326c-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 12

A particle moves with position function $\mathbf { r } ( t ) = 4 \sqrt { 2 } t \mathbf { i } + e ^ { 4 t } \mathbf { j } + e ^ { - 4 t } \mathbf { k }$ . Find the acceleration of the particle.

Accepted Answer

The answer of A particle moves with position function \(\mathbf...

Question 13

Sketch the curve of the vector function $\mathbf { r } ( t ) = \left\langle t ^ { 2 } , t ^ { 3 } , t \right\rangle$ $t \geq 0$ and indicate the orientation of the curve.

Accepted Answer

The answer of Sketch the curve of the vector function...

Question 14

Find $\mathbf { r } ^ { \prime } ( t )$ and $\mathbf { r } ^ { \prime \prime } ( t )$ for $\mathbf { r } ( t ) = 5 t \mathbf { i } + 4 t ^ { 2 } \mathbf { j } + 5 t ^ { 3 } \mathbf { k }$

Accepted Answer

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

Question 15

Find the curvature of the curve  r(t) =2ti+6tj+9k\mathbf { r } ( t )  = 2 t \mathbf { i } + 6 t \mathbf { j } + 9 \mathbf { k }r(t) =2ti+6tj+9k  .

Accepted Answer

A)   1
B)    2102 \sqrt { 10 }210c-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)   11
D)   0

Question 16

Let  r(t) =⟨6−t,(et−2) t,ln⁡(t+1) ⟩\mathbf { r } ( t )  = \left\langle \sqrt { 6 - t } , \frac { \left( e ^ { t } - 2 \right)  } { t } , \ln ( t + 1 )  \right\rangler(t) =⟨6−t​,t(et−2) ​,ln(t+1) ⟩  . Find the domain of  rrr  .

Accepted Answer

A)    (−2,6]( - 2,6 ](−2,6] 
B)    (−2,0) ∪(0,6]( - 2,0 )  \cup ( 0,6 ](−2,0) ∪(0,6] 
C)    (6,∞]( 6 , \infty ](6,∞] 
D)    (−∞,−2) ( - \infty , - 2 ) (−∞,−2)  
E)    [6,0) ∪(0,2) [ 6,0 )  \cup ( 0,2 ) [6,0) ∪(0,2)

Question 17

Find  r(t) \mathbf { r } ( t ) r(t)   satisfying the conditions for  rt(t) =9e9ti+9e−tj+etk\mathbf { r } ^ { t } ( t )  = 9 e ^ { 9 t } \mathbf { i } + 9 e ^ { - t } \mathbf { j } + e ^ { t } \mathbf { k }rt(t) =9e9ti+9e−tj+etk   r(0) =i−j+9k\mathbf { r } ( 0 )  = \mathbf { i } - \mathbf { j } + 9 \mathbf { k }r(0) =i−j+9k

Accepted Answer

A)    (e9t+1) i−(9e−t+1) j+(et+9) k\left( e ^ { 9 t } + 1 \right)  \mathbf { i } - \left( 9 e ^ { - t } + 1 \right)  \mathbf { j } + \left( e ^ { t } + 9 \right)  \mathbf { k }(e9t+1) i−(9e−t+1) j+(et+9) k 
B)    9e9ti−(9e−t+8) j+(et+8) k9 e ^ { 9 t } \mathbf { i } - \left( 9 e ^ { - t } + 8 \right)  \mathbf { j } + \left( e ^ { t } + 8 \right)  \mathbf { k }9e9ti−(9e−t+8) j+(et+8) k 
C)    e9ti−(9e−t−8) j+(et+8) ke ^ { 9 t } \mathbf { i } - \left( 9 e ^ { - t } - 8 \right)  \mathbf { j } + \left( e ^ { t } + 8 \right)  \mathbf { k }e9ti−(9e−t−8) j+(et+8) k 
D)    (9e9t+1) i−(9e−t−1) j+(et+9) k\left( 9 e ^ { 9 t } + 1 \right)  \mathbf { i } - \left( 9 e ^ { - t } - 1 \right)  \mathbf { j } + \left( e ^ { t } + 9 \right)  \mathbf { k }(9e9t+1) i−(9e−t−1) j+(et+9) k

Question 18

Find the domain of the vector function  r(t) =8ti+1t−4j\mathbf { r } ( t )  = 8 t \mathbf { i } + \frac { 1 } { t - 4 } \mathbf { j }r(t) =8ti+t−41​j  .

Accepted Answer

A)    (−∞,−4) ∪(−4,∞) ( - \infty , - 4 )  \cup ( - 4 , \infty ) (−∞,−4) ∪(−4,∞)  
B)    (−∞,8) ∪(8,∞) ( - \infty , 8 )  \cup ( 8 , \infty ) (−∞,8) ∪(8,∞)  
C)    (−∞,−8) ∪(−8,∞) ( - \infty , - 8 )  \cup ( - 8 , \infty ) (−∞,−8) ∪(−8,∞)  
D)    (−∞,4) ∪(4,∞) ( - \infty , 4 )  \cup ( 4 , \infty ) (−∞,4) ∪(4,∞)

Question 19

Find $\mathbf { r } ^ { \prime } ( t )$ and $\mathbf { r } ^ { \prime \prime } ( t )$ for $\mathbf { r } ( t ) = \langle t \cos 3 t - \sin 3 t , t \sin 4 t + \cos 4 t \rangle$

Accepted Answer

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

Question 20

Two points A and B are located 100 ft apart on a straight line. A particle moves from A toward B with an initial velocity of 9 ft/sec and an acceleration of $\frac { 1 } { 2 }$ $\mathrm { ft } / \mathrm { sec } ^ { 2 }$ . Simultaneously, a particle moves from B toward A with an initial velocity of 2 ft/sec and an acceleration of $\frac { 3 } { 4 }$ $\mathrm { ft } / \mathrm { sec } ^ { 2 }$ . When will the two particles collide? At what distance from A will the collision take place?

Accepted Answer

The answer of Two points A and B are located...

Exam 13: Vector Functions

Find the curvature of the curve $\mathbf { r } ( t ) = 2 t \mathbf { i } + 6 t \mathbf { j } + 9 \mathbf { k }$ .

Correct Answer
verified

Find the length of the curve $\mathbf { r } ( t ) = - 2 t \mathbf { i } - t \mathbf { j } + t \mathbf { k }$ $- 2 \leq t \leq 1 .$

Correct Answer
verified

Find the unit tangent vector $T ( t )$ . $\mathbf { r } ( t ) = \langle 2 \sin t , 4 t , 2 \cos t \rangle$

Correct Answer
verified

Find the acceleration of a particle with the given position function. $\mathbf { r } ( t ) = 9 \sin t \mathbf { i } + 10 t \mathbf { j } - 8 \cos t \mathbf { k }$

Correct Answer
verified

Correct Answer
verified

Find the point(s) on the graph of $y = e ^ { - 11 x ^ { 2 } }$ at which the curvature is zero.

Correct Answer
verified

Find $\mathbf { r } ^ { \prime } ( t )$ and $\mathbf { r } ^ { \prime \prime } ( t )$ for $\mathbf { r } ( t ) = \langle t \cos 3 t - \sin 3 t , t \sin 4 t + \cos 4 t \rangle$

Correct Answer
verified
\[\begin{array} { l }
\mathbf { r } ^ {...
View Answer

Find the domain of the vector function $\mathbf { r } ( t ) = 8 t \mathbf { i } + \frac { 1 } { t - 4 } \mathbf { j }$ .

Correct Answer
verified

Find the limit $\lim _ { t \rightarrow \infty } \left\langle e ^ { - 4 t } , \frac { 1 } { t } , \frac { 7 t ^ { 2 } } { t ^ { 2 } + 1 } \right\rangle$ .

Correct Answer
verified

For the curve given by $\mathbf { r } ( t ) = (4 \sin t , 5 t , 4 \mathrm { cos } t )$ , find the unit normal vector. $\mathbf { a } ( t ) = 2 \mathbf { k } , \mathbf { v } ( 0 ) = 10 \mathbf { i } - 9 \mathbf { j }$

Correct Answer
verified

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

Correct Answer
verified

Correct Answer
verified

Let $\mathbf { r } ( t ) = \left\langle \sqrt { 6 - t } , \frac { \left( e ^ { t } - 2 \right) } { t } , \ln ( t + 1 ) \right\rangle$ . Find the domain of $r$ .

Correct Answer
verified

Find $\mathbf { r } ( t )$ satisfying the conditions for $\mathbf { r } ^ { t } ( t ) = 9 e ^ { 9 t } \mathbf { i } + 9 e ^ { - t } \mathbf { j } + e ^ { t } \mathbf { k }$ $\mathbf { r } ( 0 ) = \mathbf { i } - \mathbf { j } + 9 \mathbf { k }$

Correct Answer
verified

Find $\mathbf { r } ^ { \prime } ( t )$ and $\mathbf { r } ^ { \prime \prime } ( t )$ for $\mathbf { r } ( t ) = 5 t \mathbf { i } + 4 t ^ { 2 } \mathbf { j } + 5 t ^ { 3 } \mathbf { k }$

Correct Answer
verified
\[\begin{array} { l }
\mathbf { r } ^ {...
View Answer

The helix $\mathbf { r } _ { 1 } ( t ) = 8 \cos t \mathbf { i } + \sin t \mathbf { j } + t \mathbf { k }$ intersects the curve $\mathbf { r } _ { 2 } ( t ) = ( 8 + t ) \mathbf { i } + 10 t ^ { 2 } \mathbf { j } + 9 t ^ { 3 } \mathbf { k }$ at the point $( 8,0,0 )$ . Find the angle of intersection.

Correct Answer
verified

A particle moves with position function $\mathbf { r } ( t ) = 4 \sqrt { 2 } t \mathbf { i } + e ^ { 4 t } \mathbf { j } + e ^ { - 4 t } \mathbf { k }$ . Find the acceleration of the particle.

Correct Answer
verified

Find a vector function that represents the curve of intersection of the two surfaces: The circular cylinder $x ^ { 2 } + y ^ { 2 } = 9$ and the parabolic cylinder $z = x y$ .

Correct Answer
verified

Find the point of intersection of the tangent lines to the curve $\mathbf { r } ( t ) = ( \sin \pi t , 5 \sin \pi t , \cos \pi t )$ , at the points where $t = 0$ and $t = 0.5$ .

Correct Answer
verified

Sketch the curve of the vector function $\mathbf { r } ( t ) = \left\langle t ^ { 2 } , t ^ { 3 } , t \right\rangle$ $t \geq 0$ and indicate the orientation of the curve.

Correct Answer
verified

Exam 13: Vector Functions

Find the curvature of the curve r(t) =2ti+6tj+9k\mathbf { r } ( t ) = 2 t \mathbf { i } + 6 t \mathbf { j } + 9 \mathbf { k }r(t) =2ti+6tj+9k .

Correct AnswerverifiedShow Answer

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

Correct AnswerverifiedShow Answer

Find the unit tangent vector T(t) T ( t ) T(t) . r(t) =⟨2sin⁡t,4t,2cos⁡t⟩\mathbf { r } ( t ) = \langle 2 \sin t , 4 t , 2 \cos t \rangler(t) =⟨2sint,4t,2cost⟩

Correct AnswerverifiedShow Answer

Find the acceleration of a particle with the given position function. r(t) =9sin⁡ti+10tj−8cos⁡tk\mathbf { r } ( t ) = 9 \sin t \mathbf { i } + 10 t \mathbf { j } - 8 \cos t \mathbf { k }r(t) =9sinti+10tj−8costk

Correct AnswerverifiedShow Answer

Correct AnswerverifiedShow Answer

Find the point(s) on the graph of y=e−11x2y = e ^ { - 11 x ^ { 2 } }y=e−11x2 at which the curvature is zero.

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Find r′(t)\mathbf { r } ^ { \prime } ( t )r′(t) and r′′(t)\mathbf { r } ^ { \prime \prime } ( t )r′′(t) for r(t)=⟨tcos⁡3t−sin⁡3t,tsin⁡4t+cos⁡4t⟩\mathbf { r } ( t ) = \langle t \cos 3 t - \sin 3 t , t \sin 4 t + \cos 4 t \rangler(t)=⟨tcos3t−sin3t,tsin4t+cos4t⟩

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Find the domain of the vector function r(t) =8ti+1t−4j\mathbf { r } ( t ) = 8 t \mathbf { i } + \frac { 1 } { t - 4 } \mathbf { j }r(t) =8ti+t−41​j .

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Find the limit lim⁡t→∞⟨e−4t,1t,7t2t2+1⟩\lim _ { t \rightarrow \infty } \left\langle e ^ { - 4 t } , \frac { 1 } { t } , \frac { 7 t ^ { 2 } } { t ^ { 2 } + 1 } \right\ranglelimt→∞​⟨e−4t,t1​,t2+17t2​⟩ .

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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 }r(t) =3sinti+3costj+5tk

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Let r(t) =⟨6−t,(et−2) t,ln⁡(t+1) ⟩\mathbf { r } ( t ) = \left\langle \sqrt { 6 - t } , \frac { \left( e ^ { t } - 2 \right) } { t } , \ln ( t + 1 ) \right\rangler(t) =⟨6−t​,t(et−2) ​,ln(t+1) ⟩ . Find the domain of rrr .

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Find r′(t)\mathbf { r } ^ { \prime } ( t )r′(t) and r′′(t)\mathbf { r } ^ { \prime \prime } ( t )r′′(t) for r(t)=5ti+4t2j+5t3k\mathbf { r } ( t ) = 5 t \mathbf { i } + 4 t ^ { 2 } \mathbf { j } + 5 t ^ { 3 } \mathbf { k }r(t)=5ti+4t2j+5t3k

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A particle moves with 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 }r(t)=42​ti+e4tj+e−4tk . Find the acceleration of the particle.

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Find a vector function that represents the curve of intersection of the two surfaces: The circular cylinder x2+y2=9x ^ { 2 } + y ^ { 2 } = 9x2+y2=9 and the parabolic cylinder z=xyz = x yz=xy .

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Find the point of intersection of the tangent lines to the curve r(t)=(sin⁡πt,5sin⁡πt,cos⁡πt)\mathbf { r } ( t ) = ( \sin \pi t , 5 \sin \pi t , \cos \pi t )r(t)=(sinπt,5sinπt,cosπt) , at the points where t=0t = 0t=0 and t=0.5t = 0.5t=0.5 .

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Sketch the curve of the vector function r(t)=⟨t2,t3,t⟩\mathbf { r } ( t ) = \left\langle t ^ { 2 } , t ^ { 3 } , t \right\rangler(t)=⟨t2,t3,t⟩ t≥0t \geq 0t≥0 and indicate the orientation of the curve.

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Find the curvature of the curve $\mathbf { r } ( t ) = 2 t \mathbf { i } + 6 t \mathbf { j } + 9 \mathbf { k }$ .

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Find the length of the curve $\mathbf { r } ( t ) = - 2 t \mathbf { i } - t \mathbf { j } + t \mathbf { k }$ $- 2 \leq t \leq 1 .$

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Find the unit tangent vector $T ( t )$ . $\mathbf { r } ( t ) = \langle 2 \sin t , 4 t , 2 \cos t \rangle$

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Find the acceleration of a particle with the given position function. $\mathbf { r } ( t ) = 9 \sin t \mathbf { i } + 10 t \mathbf { j } - 8 \cos t \mathbf { k }$

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Find the point(s) on the graph of $y = e ^ { - 11 x ^ { 2 } }$ at which the curvature is zero.

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Find $\mathbf { r } ^ { \prime } ( t )$ and $\mathbf { r } ^ { \prime \prime } ( t )$ for $\mathbf { r } ( t ) = \langle t \cos 3 t - \sin 3 t , t \sin 4 t + \cos 4 t \rangle$

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\[\begin{array} { l }
\mathbf { r } ^ {...
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Find the domain of the vector function $\mathbf { r } ( t ) = 8 t \mathbf { i } + \frac { 1 } { t - 4 } \mathbf { j }$ .

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Find the limit $\lim _ { t \rightarrow \infty } \left\langle e ^ { - 4 t } , \frac { 1 } { t } , \frac { 7 t ^ { 2 } } { t ^ { 2 } + 1 } \right\rangle$ .

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Correct Answer
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Find the scalar tangential and normal components of acceleration of a particle with position vector $\mathbf { r } ( t ) = 3 \sin t \mathbf { i } + 3 \cos t \mathbf { j } + 5 t \mathbf { k }$

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Let $\mathbf { r } ( t ) = \left\langle \sqrt { 6 - t } , \frac { \left( e ^ { t } - 2 \right) } { t } , \ln ( t + 1 ) \right\rangle$ . Find the domain of $r$ .

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Correct Answer
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Find $\mathbf { r } ^ { \prime } ( t )$ and $\mathbf { r } ^ { \prime \prime } ( t )$ for $\mathbf { r } ( t ) = 5 t \mathbf { i } + 4 t ^ { 2 } \mathbf { j } + 5 t ^ { 3 } \mathbf { k }$

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\[\begin{array} { l }
\mathbf { r } ^ {...
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A particle moves with position function $\mathbf { r } ( t ) = 4 \sqrt { 2 } t \mathbf { i } + e ^ { 4 t } \mathbf { j } + e ^ { - 4 t } \mathbf { k }$ . Find the acceleration of the particle.

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Find a vector function that represents the curve of intersection of the two surfaces: The circular cylinder $x ^ { 2 } + y ^ { 2 } = 9$ and the parabolic cylinder $z = x y$ .

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Find the point of intersection of the tangent lines to the curve $\mathbf { r } ( t ) = ( \sin \pi t , 5 \sin \pi t , \cos \pi t )$ , at the points where $t = 0$ and $t = 0.5$ .

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Sketch the curve of the vector function $\mathbf { r } ( t ) = \left\langle t ^ { 2 } , t ^ { 3 } , t \right\rangle$ $t \geq 0$ and indicate the orientation of the curve.

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