Duke Vuong
MEAE 7010
Homework # 6
44. The temperature at any point in space is given by T
 xy
 yz
 zx a) Find the direction cosines of the direction in which the temperature changes most rapidly with distance from the point (1, 1, 1), and determine the maximum rate of change. b) Find the derivative of T in the direction of the vector 3 i – 4 k at the point (1, 1, 1).
Solution: a) The direction cosines l

1
3 m

1
3 n

1
3
The maximum rate of change grad T
 y i
 z j
 x k
 li
 mj
 nk

 grad T

3
3
60. For each of the following vector functions, determine whether the equation
  
F possesses a solution, and determine that solution exists: a) b)
F
F


2 xyz
3 i
2 xy i


 x
2
 x

2 z
2
3 yz

2 j y

 j y

2
3
 x
2
1
 yz k
2 k
Solution: a) If
  
F then
 
F

0
From Equation (68b)
 
F





F y z


F y
 z

 i




F x
 z
 
F


3 x
2 z
2 
3 x
2 z
2


F z
 x
6 xyz
2

 j




6 xyz
2

F y
 x

No Solution


F x
 y

 k
2 xz
3 
2 xz
3
 k

0

b) From Equation (68b)
 
F


2 y

2 y
 
0

0
 
2 x

2 x
 k

0
From Equation (95) d
 
2 xy dx

 x
2
We have
 
 x
 
 y

2 xy
 x
2 
2 yz

2 yz
 dy

 y
2 
1
 dz
(1)
(2)
 
 z
 y
2 
1 (3)
Integrate (1), holding y constant
  x
2 y
 f ( y ) (4)
Substitute (4) into (2) to determine f ( y )
 
 y
 x
2  f (
 y y )
 x
2 
2 yz (5)
Integration gives, f ( y )
 y
2 z
 g ( z ) (6)
Substitute (4) and (6) into (3), gives
 
 z
 y 2  g ( z )
 z
 y 2 
1 g ( z )
 z

C
The solution is
  x
2 y
 y
2 z
 z

C