Math Test (Calculus III)

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Mathematics

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Need help on Math exam, Calculus III.

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Explanation & Answer

Attached.

Problem 1:
We have

d
( 2sin t ) = 2 cos t
dt
d
y = ( 2sin 2t ) = 4 cos 2t
dt
d
z  = ( 2sin 3t ) = 6 cos 3t
dt
x =

(

)

At ( x, y, z ) = 1, 3, 2 , we can solve for t

 2 sint = 1

2sin 2t = 3
 2sin 3t = 2

1

 sint = 2

 sin 2t = 3 / 2
 sin 3t = 1


t =


3

Hence,



( x, y, z ) =  2 cos




 
  
, 4 cos  2  , 6 cos  3  
3
 3
 3 

 3
1 
 ( x, y, z  ) =  2
, 4  , 0 
2 
 2
 ( x, y, z  ) =

(

3,1, 0

)

Hence, the equation of the tangent line is

( x, y, z ) = (
Or

) (

3,1, 0 t + 1, 3, 2

)

x = 3t + 1
y=t+ 3
z=2

Attached.

Problem 4:
We have

P Q R
+
+
x y z



 divF = ( y 2 z 3 ) + ( 2 yz ) + ( 4 z 2 )
x
y
z
 divF = 0 + 2 z + 8 z
 divF = 10 z
divF =

By Divergence theorem, we have

 F .ndS =  divFdV =  10 zdV
2 3 9

=

   10 z  rdzdrd
0 0 r2

2 3

=

  5z

2 9
r2

 rdrd

0 0

2

=

3

 d  r ( 405 − 5r ) dr
4

0

0

= ( 2 )(1215 )
= 2430

Attached.

Problem 5:
We can calculated the value of b as follow

lim

x , y →0

3 xy
4 x2 + 4 y 2

= lim

x , y →0

= lim

=

3 x ( kx )
4 x 2 + 4 ( kx )

2

3kx 2

2 k 2 + 1x
3k
=
lim x = 0
2 k 2 + 1 x , y →0
x , y →0

Hence,

b=0
We have

 F .d r =
C

(

)

=  1, x + yz, z − xy ( 3,3,1) dt


Problem 3:
We want to minimize

D = f ( x, y , z ) = ( x − 4 ) + ( y − 2 ) + ( z − 0 )
2

2

2

 f ( x, y , z ) = ( x...


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