Page 64 - Intro to Tensor Calculus

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2
1
3
i
i
i
I 12. For ~ = y b e i where y = y (x ,x ,x ), i =1, 2, 3 we have by definition
r
∂~ r ∂y i ∂x m
~
~ m
E j = = b e i . From this relation show that E = b e j
∂x j ∂x j ∂y j
and consequently
i
∂y m ∂y m ij ∂x ∂x j
~ i ~ j
~
~
g ij = E i · E j = , and g = E · E = , i,j,m =1,... , 3
i
∂x ∂x j ∂y m ∂y m
I 13. Consider the set of all coordinate transformations of the form
i
j
i
y = a x + b i
j
i
i
i
where a and b are constants and the determinant of a is different from zero. Show this set of transforma-
j j
tions forms a group.
i
I 14. For α i , β i constants and t a parameter, x = α i + tβ i ,i =1, 2, 3 is the parametric representation of
a straight line. Find the parametric equation of the line which passes through the two points (1, 2, 3) and
(14, 7, −3). What does the vector d~ r represent?
dt
I 15. A surface can be represented using two parameters u, v by introducing the parametric equations
i
i
x = x (u, v), i =1, 2, 3, a < u < b and c<v < d.
The parameters u, v are called the curvilinear coordinates of a point on the surface. A point on the surface
3
1
2
r
can be represented by the position vector ~ = ~(u, v)= x (u, v) b e 1 + x (u, v) b e 2 + x (u, v) b e 3 . The vectors ∂~ r
r
∂u
r
and ∂~ r are tangent vectors to the coordinate surface curves ~(u, c 2 )and ~(c 1 ,v) respectively. An element of
r
∂v
surface area dS on the surface is defined as the area of the elemental parallelogram having the vector sides
∂~ r ∂~ r
∂u du and ∂v dv. Show that
r
∂~ ∂~ r p 2
dS = | × | dudv = g 11 g 22 − (g 12 ) dudv
∂u ∂v
where
r
∂~ ∂~ ∂~ r ∂~ r ∂~ ∂~
r
r
r
g 11 = · g 12 = · g 22 = · .
∂u ∂u ∂u ∂v ∂v ∂v
~
~
~
~
~
~ 2
Hint: (A × B) · (A × B)= |A × B| See Exercise 1.1, problem 9(c).
I 16.
(a) Use the results from problem 15 and find the element of surface area of the circular cone
x = u sin α cos v y = u sin α sin v z = u cosα
α a constant 0 ≤ u ≤ b 0 ≤ v ≤ 2π
(b) Find the surface area of the above cone.