Begin2.DVI

I8-6. (i)r=

√

x^2 +y^2 +z^2 and

∇rn=

∂rn

∂x

eˆ 1 +∂r

n

∂y

ˆe 2 +∂r

n

∂z

ˆe 3

=nrn−^1

[

∂r

∂x

ˆe 1 +∂r

∂y

ˆe 2 +∂r

∂z

eˆ 3

]

=nrn−^1

[x

r

ˆe 1 +y

r

eˆ 2 +z

r

ˆe 3

]

=nrn−^2 ~r

(iii)

∇f(r) =∂f

∂r

∂r

∂x

ˆe 1 +∂f

∂r

∂r

∂y

ˆe 2 +∂f

∂r

∂r

∂z

ˆe 3

=f′(t)

[x

r

ˆe 1 +y

r

ˆe 2 +z

r

ˆe 3

]

=f′(r)~r

|~r|

=f′(r)~r

r

I8-7. Let~r 1 (τ) = (τ−1)ˆe 1 + (16−τ)ˆe 2 + (2τ−2)ˆe 3 and~r 2 (t) =−tˆe 1 + 2tˆe 2 + 3tˆe 3 and

define vector from line 1 to line 2 as~r 2 −~r 1. Minimize the distance squared

f(t,τ) =|~r 2 −~r 1 |^2 = (−t−τ+ 1)^2 + (2t+τ−16)^2 + (3t− 2 τ+ 2)^2

by examining the critical points where∂f

∂τ

= 0and∂f

∂t

= 0and show minimum occurs

wheret= 3andτ= 5giving minimum distance

√

75

I8-8. Method I: Normal to plane isN~=ˆe 1 +ˆe 2 +ˆe 3 and unit normal to plane is

ˆeN=eˆ^1 + ˆ√e^23 + ˆe^3. Consider vector~r 1 =eˆ 1 projected ontoˆeN to obtain distance√^13

Method II:f=d^2 =x^2 +y^2 +z^2 is distance from origin to(x,y,z)squared. Here

z= 1−x−yso thatf=d^2 =x^2 +y^2 + (1−x−y)^2 Showf has critical points atx= 1/ 3

,y= 1/ 3 andz= 1/ 3 givingd^2 = 1/ 3 ord=√^13

I8-9. (iii) dφ
dn

= [(3x^2 y+y^2 )ˆe 1 + (x^3 + 2xy)ˆe 2 ]·eˆnwhere

on bottom of square ˆen=−ˆe 2 , y= 0 =⇒ dφ

dn

=−x^3

on right side of square ˆen=eˆ 1 , x= 1 =⇒ dφ

dn

= 3y+y^2

on top of square ˆen=ˆe 2 , y= 1 =⇒ dφ

dn

=x^3 + 2x

on left side of square ˆen=−ˆe 2 , x= 0 =⇒ dφ

dn

= 0

I8-10. (ii)z= (x−2)^2 −(y−3)^2 has critical points atx= 2andy= 3. HereA=∂

(^2) z
∂x^2 = 2,
C=∂∂y^2 z 2 =− 2 andB=∂x ∂y∂^2 z = 0givesAC−B^2 =− 4 < 0 so there is a saddle point at
the critical value.
Solutions Chapter 8