MAC 2313 Florida International University Calculus III Worksheet

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MAC 2313

Florida International University

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MAC 2313 - Calculus III - Worksheet 2 Name_________________________ Complete the worksheet on a separate sheet of paper. Solve each question below. The worksheet worth 100 points. Show all work. Please print these pages and attach then, staple, to the front of your work. 1. Given ! A = (1, 4,5 ) and ! B = ( 4,−2, 7 ) find: (a) the distance between the points ! A and ! B . !!!" !!!" (Aside: The same as ! AB .) (b) the vector ! AB . (c) the midpoint of the line segment ! AB . 2. Find the center and radius of the sphere ! x 2 + y 2 + z 2 − 6y + 8z = 0 . ! ! ! ! ! ! 3. Given ! u = 2i + 2 j + k and ! v = 2i + 10 j − 11k find: (a) ! u ⋅ v , ! u , and ! v . (b) the angle between ! ! ! ! ! u and ! v . (c) the vector ! projv! u . (d) the vector ! proju! v . ! ! ! ! ! ! 4. Given ! u = 2,−2, 4 and ! v = 1,1,−2 , find: (a) the vector ! u × v . (b) the unit vector for ! v × u . 5. Find the line ! L1 , in Vector Form and Parametric Form, through the point ! ( 3,−2,1) and parallel to the line ! L2 : ! x = 1+ 2t , ! y = 2 − t , ! z = 3t . 6. Find the line ! L , in Vector Form and Symmetric Form, through the point ! ( 2, 3,0 ) and ! ! perpendicular to the vectors ! u = i + 2 j + 3k and ! v = 3i + 4 j + 5k . 7. Find the point of intersection of the lines ! L1 : ! x, y, z = 0,2, ! x, y, z = 3 + t 2,−15,6 and ! L2 : 5 4 3 13 , , + s 2, 4,2 . 3 2 5 8. Determine if the lines ! L1 : ! x, y, z = 2,0,−4 + t −3,1,0 and ! L2 : ! x, y, z = −2,0, 4 + s −1,0, 3 are parallel, intersecting, or skewed. 9. Find the plane ! P1 , in Linear Form, containing the point ! (1,−1, 3) and parallel to the plane ! P2 ! 3x + y + z = 7 . 10. Find the plane ! P , in Linear Form, containing the points ! (1,1,−1) , ! ( 2,0,2 ) , and ! ( 0,−2,1) . 11. Find the plane ! P , in Linear Form, containing the points ! ( 0,0,0 ) , ! ( 0,−1,−1) , ! ( 2,0,−2 ) , and ⎛ 3 ⎞ ! ⎜ 1, ,0 ⎟ . (Hint: Finding a plane requires at least three points.) ⎝ 4 ⎠ 12. Find the line ! L of intersection and the angle ! θ of intersection of the planes ! P1 : ! −x + z = 2 3 1 5 and ! P2 : ! x − 2y + z = 4 . 2 2 13. Find the point of intersection of the lines ! L1 : ! x = 2t + 1 , ! y = 3t + 2 , ! z = 4t + 3 and ! L2 : ! x = s + 2 , ! y = 2s + 4 , ! z = −4s − 1 , and then find the plane ! P , in Linear Form, determined by these lines. (Aside: The plane ! P containing the point of intersection and vectors parallel to each line.) ! 14. Find the natural domain of the space curve ! r ( t ) = ! 15. Find the limit of the space curve ! r ( t ) = lim t→∞ ( ) 5 − t i + ( t + 1) 1/4 ( ) j + log 3 ( t 2 ) k . 3e2t − 1 t +1 1 1 , , + t 2t 2 ( e − 2e ) t + 9t 2 − 1 t 2 − 1 t ! !" ! ! ! 16. Given the space curve ! r ( t ) = e2t ,e−2t ,te2t , find ! T ( 0 ) , ! r '' ( 0 ) , and ! r ' ( t ) ⋅ r '' ( t ) . ! 17. Find the tangent line to the space curve ! r ( t ) = ! ! 18. Find the space curve ! r ( t ) , given that ! r ' ( t ) = This is an explicit differential equation.) 3 2 t 2 + 3,ln ( t 2 + 3) ,t at the point ! ( 2,ln ( 4 ) ,1) . (t + 1) 1 2 i + e −t j + ! 1 k and ! r ( 0 ) = k . (Aside: t +1
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Explanation & Answer

Attached.

(a) By the Pythagorean theorem we get
AB = (4 − 1) 2 + (−2 − 4) 2 + (7 − 5) 2 = 9 + 36 + 4 = 49 = 7 .

(b) The vector is
AB = (4 − 1)i + ( −2 − 4) j + (7 − 5)k = 3i − 6 j + 2k .

(c) The midpoint C  AB has coordinates
 1+ 4 4 − 2 5 + 7   5

C =
,
,
 =  ,1, 6  .
2
2  2
 2


Write the equation in an appropriate form
x2 + y 2 + z 2 − 6 y + 8z = 0
x 2 + ( y 2 − 6 y + 9 ) + ( z 2 + 8 z + 16 ) − 9 − 16 = 0
x 2 + ( y − 3) 2 + ( z + 4) 2 = 25

.

( x − 0) 2 + ( y − 3) 2 + ( z − (−4)) 2 = 52
So the center is O = (0,3, −4) , the radius equals R = 5 .

(a) u  v = 2  2 + 2 10 + 1 (−11) = 4 + 20 − 11 = 13 , u = u  u = 22 + 22 + 12 = 9 = 3 , and
v = v  v = 22 + 102 + 112 = 225 = 15

(b) Let  =  ( u, v ) be the angle. We have
cos  =

u v
13
13
=
= .
u  v 3 15 45

13
 73.21 .
45
u v
13
26
26 143
i+
j−
k,
(c) projv u = 2 v =
( 2i + 10 j − 11k ) =
225
225 45
225
v

So  = cos −1

proju v =

u v
u

2

u=

13
26 26 13
( 2i + 2 j + 1k ) = i + j + k
3
3
3
3

1

(a)
i j k
−2 4
2 4
2 −2
u  v = 2 −2 4 =
i−
j+
k = 8 j + 4k .
1 −2
1 −2
1 1
1 1 −2

(b) The unit vector for v  u = −8 j − 4k
v u
−8j − 4k
8
4
2
1
=
=−
j−
k=−
j−
k.
2
2
v u
80
80
5
5
8 +4

The line has a directive vector v = 2i − j + 3k . A Vector form for the line L...


Anonymous
Excellent! Definitely coming back for more study materials.

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