B.4. EXERCISES 345

e2 (x) = x1 cos(x2)cos

(x3)i1

+x1 cos(x2)sin

(x3)i2− x1 sin

(x2)i3,

e3 (x) =−x1 sin(x2)sin

(x3)i1 + x1 sin

(x2)cos

(x3)i2 +0i3.

It follows the metric tensor is

G =

 1 0 00(x1)2 0

0 0(x1)2 sin2 (x2

)= (gi j) = (ei ·e j) . (2.18)

Therefore, by Theorem B.1.6G−1 =

(gi j)

=(ei,e j)=

 1 0 00(x1)−2 0

0 0(x1)−2 sin−2 (x2

) .

To obtain the dual basis, use Theorem B.1.6 to write

e1 (x) = g1 je j (x) = e1 (x)

e2 (x) = g2 je j (x) =(x1)−2

e2 (x)

e3 (x) = g3 je j (x) =(x1)−2

sin−2 (x2)e3 (x) .

Note that ∂y

∂yk ≡ ek (y) = ik = ik where, as described,(

y1 · · · yn)

are the rectan-gular coordinates of the point in Rn.

B.4 Exercises1. Let  y1

y2

y3

=

 x1 +2x2

x2 + x3

x1−2x2

where the yi are the rectangular coordinates of the point. Find ei,ei, i = 1,2,3, andfind (gi j)(x) and

(gi j (x)

).

2. Let y = y (x,t) where t signifies time and x ∈ U ⊆ Rm for U an open set, whiley ∈ Rn and suppose x is a function of t. Physically, this corresponds to an objectmoving over a surface in Rn which may be changing as a function of t. The pointy = y (x(t) , t) is the point in Rn corresponding to t. For example, consider the pen-dulum

m

lθ