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Tutorial 9

Q1. The air pressure in a particular room varies from point to point and is givenby () = 32+22 exp(−2)where () describes a general point(with reference to a Cartesian coordinate system).

(a) At the point (1 0 2) calculate the rate of change of pressure per unitdistance in the direction of the vector i−2 j + 2k

(b) At (1 0 2) find all directions along which the rate of change of pressureper unit distance is zero. (Give the directions using vectors in the formi +  j + k)

(c) At (1 0 2) find the direction along which the rate of increase in thepressure per unit distance is the greatest. What is the greatest rate of 

increase in the pressure per unit distance at (1 0 2)?

Solution . We need ∇ |()=(102) in all the parts.

∇  =

i +

 j +

k

= [6 + 22 exp(−2)]i− 222 exp(−2) j + 22 exp(−2)k

= 14i− 8 j + 4k at () = (1 0 2)

(a) The vector i−2 j+2k has magnitudep 

12 + (−2)2 + 22 = 3 The requiredrate is given by

¡14i− 8 j + 4k

¢•µ

13i− 23

 j + 23k¶

= 383

(b) The rate of change of pressure in the direction of the unit magnitudevector n is given by n•∇  and is zero if n is perpendicular to ∇ 

At (1 0 2) ∇  is given by 14i− 8 j + 4kThus, for the directions, we need to find all vectors perpendicular to 14i−

8 j + 4kLet i +  j + k be a vector perpendicular to 14i− 8 j + 4k It follows that

14− 8 + 4 = 0The directions are given by vectors i +  j+ k such that 14− 8 + 4 = 0

(with 2 + 2 + 2 6= 0 so that we won’t have = = = 0)(c) For maximum rate, we need to maximise n•

∇  The rate n•

∇  is max-

imum if n and ∇  are in the same direction. Thus, at (1 0 2) the direction of 14i− 8 j + 4k gives the maximum rate of change of pressure. The maximum isgiven by

n •∇  = |n||∇ | cos(0) = |∇ | =¯̄14i− 8 j + 4k

¯̄=

p 142 + (−8)2 + 42 = 2

√ 69

1

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Q2. Find the straight line which is normal to the surface = 1 + 22 + 32

at the point (1 2 15)(Hint . Firstfi

nd a vector which is parallel to thestraight line.)

Solution .Any vector which is normal to the surface at (1 2 15) should be parallel to

the straight line.A normal vector to the surface at (1 2 15) is given by

∇( − 22 − 32)¯̄()=(1215)

=¡−4i− 6 j + k

¢¯̄()=(1215)

= −4i− 12 j + k

Thus, the required straight line is given by the parametric equations

= 1− 4 = 2− 12

= 15 +

where is a free parameter.

Q3. Past year examination question, semester 1 2004/2005. If F and G arevector functions of  and such that their components have first orderpartial derivatives, prove or disprove that

div(F ×G) = G • curl(F) −F • curl(G).

Solution . Let F= i +   j + k and G= i +  j + kIt follows that

F ×G = ( − )i + (− ) j + ( − )k

and

div(F ×G) = 

( − ) +

 

(− ) +

 

( − )

=  

+

 

+

+

+

+

−

−

− 

−

−  

−

 

= [

−  

] + [

−

] + [

 

−

]

− [

−

]−  [

−

]− [

−

]

2

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= [] • [

−  

−

 

−

]

−[  ] • [

−

−

−

]

Now from the picture for working out ∇× [  ] that is,

we have

curl(F) = [

−  

−

 

−

]

curl(G) = [

−

−

−

]

It follow then that

div(F ×G) = G • curl(F) −F • curl(G).

Q4. If the velocity field of a fluid flow is given by q= ∇, where is a scalarfunction of  and , show that the flow is irrotational, that is, showthat the fluid particles have zero rotational velocity. (Note . As pointedout during the lecture, ∇ × q gives the rotational velocity of the fluidparticles. You may assume that the function has second order partialderivatives.)

Solution . The rotational velocity is given by∇× q = ∇× (∇)

= [

−  

−

 

−

]

where =   = and = (if we follow the picture inQuestion 3).

3

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Thus,

∇× q = ∇× (∇)

= [ 

(

)−  

(

)

 

(

)−  

(

)

 

(

)−  

(

)]

= [0 0 0] (after using the results 2

=

 2

etc etc).

5. Find value(s) of the constant such that () = (2 + 2 + 2) 

satisfies ∇2 = 0 at all points () except possibly at (0 0 0)

Solution . ∇2 = 0 can be written out in full as

 2

2+

 2

2+

 2

2= 0

From () = (2 + 2 + 2)  we obtain

= 2 (2 + 2 + 2) −1

= 2 (2 + 2 + 2) −1

= 2 (2 + 2 + 2) −1

and

 2

2= 4 ( 

−1)2(2 + 2 + 2) −2 + 2 (2 + 2 + 2) −1

 2

2= 4 ( − 1)2(2 + 2 + 2) −2 + 2 (2 + 2 + 2) −1

 2

2= 4 ( − 1)2(2 + 2 + 2) −2 + 2 (2 + 2 + 2) −1

Thus, ∇2 = 0 gives

4 ( − 1)2(2 + 2 + 2) −2 + 2 (2 + 2 + 2) −1

+4 ( − 1)2(2 + 2 + 2) −2 + 2 (2 + 2 + 2) −1

+4 ( − 1)2(2 + 2 + 2) −2 + 2 (2 + 2 + 2) −1

= 0

which can simplified to

[4 ( − 1) + 6 ](2 + 2 + 2) −1 = 0

The above equation is true for and (except possibly at (0 0 0)) if the constant is selecetd such that

4 ( − 1) + 6  = 0 ⇒ (4  + 2) = 0 ⇒   = 0 = −12

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6. If F and G are vector functions of  and such that their components

havefi

rst order partial derivatives, prove or disprove that(F×∇) ×G

= ( 11

+  2

2

+  3

3

)i + ( 1

1

+  2

2

+  3

3

) j

+( 11

+  2

2

+  3

3

)k−F(∇ •G)

(if F = [ 1  2  3] and G = [1 2 3])

(Note . F×∇ is a vector diff erential operator.)

Solution . First work out the operator (F ×∇) (draw a picture). Youshould get

F×∇ = [ 2 

−  3

 

 3 

−  1

 

 1 

−  2

 

]

Next work out (F×∇) ×G (draw a picture). You should get

(F×∇) ×G

= [( 3 

−  1

 

)3 − ( 1

 

−  2

 

)2

( 1 

−  2

 

)1 − ( 2

 

−  3

 

)3

( 2 

−  3

 

)2 − ( 3

 

−  1

 

)1]

= [ 11

+  2

2

+  3

3

−  1

1

−  1

2

−  1

3

 11

+  2

2

+  3

3

−  2

1

−  2

2

−  2

3

 11

+  2

2

+  3

3

−  3

1

−  3

2

−  3

3

]

= [( 11

+  2

2

+  3

3

)( 1

1

+  2

2

+  3

3

)

( 11

+  22

+  33

)]

−( 1  2  3)(1

+

2

+

3

)

= ( 11

+  2

2

+  3

3

)i + ( 1

1

+  2

2

+  3

3

) j

+( 11

+  2

2

+  3

3

)k−F(∇ •G)

5

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7. Past year examination question, semester 2 2004/2005. The vector func-

tions F( ) and G( ) are related to the scalar function ( ) by

F( ) =

i−

k and G( ) = −  j

Assume that the first and second order partial derivatives of  exist. (Notethat all the functions involved are independent of  .)

(a) Evaluate div(F).

(b) If curl(F) = −G show that div(grad( )) =

Solution . (a)

div(F) =

 

(

) +

 

(0) +

 

(−

)

= 2 

−  2 

= 0

(b) From definition of curl, we have

curl(F) = ( 

(−

)−  

(0))i + (

 

(

)−  

(−

)) j

+(0−  

(

))k

= 0i + ( 2 

2+

 2 

2) j + 0k

since  

(

) = 0 and  

(

) = 0

as is independent of 

From curl(F) = −G we conclude that

 2 

2+

 2 

2= −(− )

Now

div(grad( )) = [ 

 

 

] · [

]

= [  

 

 

] · [

0

]

= 

(

) +

 

(0) +

 

(

)

= 2 

2+

 2 

2

=

6

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8. Past year examination question, semester 1 2005/2006. Let and be

scalar functions of the Cartesian coordinates and Assuming that and are partially diff erentiable (at least twice) with respect to theCartesian coordinates, prove or disprove:

curl( grad()) = grad() × grad ()

Solution . Now

grad() = [

]

From definition of curl (draw a picture, if you cannot see it), we find that

curl( grad())

=∇

× [

]

= [ 

(

)−  

(

)

 

(

)−  

(

)

 

(

)−  

(

)]

= [ 2

+

−

 2

−

 2

+

−

 2

−

 2

+

−

 2

−

]

= [

−

−

−

]

If we work out grad()×grad() (draw a picture), we find that

grad() × grad () = [

] × [

]

= [

−

−

−

]

= curl( grad())

7