APM3713 Assignment 6 2025

APM3713
Assignment 6
Unique No:
Due 2025

,APM3713

Assignment 6 (2025

Question 1(a)

We are given the parametrized surface:

𝑥(𝑢, 𝑣) = √ 𝑢2 + 1cos𝑣, 𝑦(𝑢, 𝑣) = √ 𝑢2 + 1sin𝑣, 𝑧(𝑢, 𝑣) = 𝑢

Let 𝑥1 = 𝑢, 𝑥 2 = 𝑣. We need to find the line element for the surface.

Step 1: Compute partial derivatives

∂𝐫 𝑢 𝑢
=( cos𝑣, sin𝑣, 1)
∂𝑢 √𝑢2 + 1 √𝑢2 + 1

∂𝐫
= (− √ 𝑢2 + 1sin𝑣, √ 𝑢2 + 1cos𝑣, 0)
∂𝑣



Step 2: Compute metric components

The metric is

∂𝐫 ∂𝐫
𝑔𝑖𝑗 = ⋅
∂𝑥 𝑖 ∂𝑥 𝑗

𝑢2 𝑢2 2𝑢 2 +1
 𝑔11 = 𝑢 2 +1 (cos 2 𝑣 + sin 2 𝑣) + 1 = 𝑢 2 +1
+1= 𝑢 2 +1


 𝑔22 = (𝑢 2 + 1)(sin2 𝑣 + cos 2 𝑣) = 𝑢 2 + 1

 𝑔12 = 𝑔 21 = 0 (since the dot product vanishes).




Step 3: Write line element

𝑑𝑠 2 = 𝑔11 𝑑𝑢 2 + 𝑔22 𝑑𝑣 2

, So,


2
2𝑢2 + 1
𝑑𝑠 = 2 𝑑𝑢 2 + (𝑢 2 + 1) 𝑑𝑣 2
𝑢 +1




Question 1(b)

We already found the line element:

2𝑢2 + 1
𝑑𝑠 2 = 𝑑𝑢 2 + (𝑢 2 + 1) 𝑑𝑣 2
𝑢2 + 1



Question 1(b): Metric tensor and dual metric tensor

Step 1: Identify the metric tensor 𝑔𝑖𝑗

From the line element, the metric is diagonal:

2𝑢 2 + 1
𝑔11 = 2 , 𝑔 22 = 𝑢 2 + 1, 𝑔 12 = 𝑔 21 = 0
𝑢 +1

So, the metric tensor is:

2𝑢2 + 1
𝑔𝑖𝑗 = [ 𝑢2 + 1 0 ]
2
0 𝑢 +1




Step 2: Compute the dual metric tensor 𝑔𝑖𝑗

The dual metric tensor is the inverse matrix of 𝑔𝑖𝑗 . Since it is diagonal, just invert the
diagonal entries:

𝑢2 + 1 1
𝑔11 = 2
, 𝑔 22 = 2 , 𝑔 12 = 𝑔 21 = 0
2𝑢 + 1 𝑢 +1

So,