APM3713 Assignment 6 2025

APM3713
Assignment 6
Unique No:
Due 2025

,APM3713 — Assignment 6 (2025)

Question 1(a) — Line element for the surface

We are given the parametrised surface

𝐫(𝑢, 𝑣) = ቀඥ𝑢2 + 1 cos𝑣, ඥ𝑢2 + 1 sin𝑣, 𝑢ቁ,

with coordinates 𝑥 1 = 𝑢, 𝑥 2
= 𝑣. Find the line element 𝑑𝑠 2 .

Step 1 — Partial derivatives

Compute ∂𝐫/ ∂𝑢 and ∂𝐫/ ∂𝑣.

First note ξ 𝑢2 + 1 = (𝑢 2 + 1)1/2 . Its derivative w.r.t. 𝑢 is

𝑑 𝑢
ඥ𝑢2 + 1 = .
𝑑𝑢 ξ 𝑢2 + 1

Now

∂𝐫 𝑢 𝑢
𝐫𝑢 = =൬ cos𝑣, sin𝑣, 1൰,
∂𝑢 ξ 𝑢2 + 1 ξ 𝑢2 + 1
∂𝐫
𝐫𝑣 = = ቀ−ඥ𝑢2 + 1 sin𝑣, ඥ𝑢2 + 1 cos𝑣, 0ቁ.
∂𝑣

Step 2 — Metric components 𝑔𝑖𝑗 = 𝐫𝑖 ⋅ 𝐫𝑗

Compute the dot products.

1. 𝑔11 = 𝐫𝑢 ⋅ 𝐫𝑢 :

𝑢 𝑢
𝑔11 = ( cos𝑣) 2 + ( sin𝑣) 2 + 12
2
ξ𝑢 +1 2
ξ𝑢 +1
2
𝑢
= 2 (cos 2 𝑣 + sin 2 𝑣) + 1
𝑢 +1
𝑢2 𝑢2 + (𝑢 2 + 1) 2𝑢 2 + 1
= 2 +1= = 2 .
𝑢 +1 𝑢2 + 1 𝑢 +1

2. 𝑔22 = 𝐫𝑣 ⋅ 𝐫𝑣 :

, 𝑔22 = (− ඥ𝑢2 + 1sin𝑣) 2 + (ඥ𝑢2 + 1cos𝑣) 2 + 02
= (𝑢 2 + 1)(sin2 𝑣 + cos 2 𝑣) = 𝑢 2 + 1.

3. 𝑔12 = 𝑔 21 = 𝐫𝑢 ⋅ 𝐫𝑣 :

𝑢 𝑢
𝑔12 = cos𝑣 ⋅ ቀ−ඥ𝑢2 + 1sin𝑣ቁ+ sin𝑣 ⋅ ቀඥ𝑢2 + 1cos𝑣ቁ+ 1 ⋅ 0
ξ 𝑢2 + 1 ξ 𝑢2 + 1
= −𝑢cos𝑣sin𝑣 + 𝑢sin𝑣cos𝑣 + 0 = 0.

So the metric is diagonal with entries found above.

Step 3 — Line element

The line element is

2𝑢 2 + 1
𝑑𝑠 2 = 𝑔11 𝑑𝑢 2 + 𝑔22 𝑑𝑣 2 = 𝑑𝑢 2 + (𝑢 2 + 1) 𝑑𝑣 2 .
𝑢2 + 1




Question 1(b) — Metric tensor and inverse (dual) metric

From the line element we read off the metric tensor 𝑔𝑖𝑗 in the (𝑢, 𝑣) coordinates:

2𝑢2 + 1
𝑔𝑖𝑗 = ቌ 𝑢2 + 1 0 ቍ.
2
0 𝑢 +1

Because the matrix is diagonal, the inverse metric 𝑔𝑖𝑗 is obtained by inverting the
diagonal entries:

𝑢2 + 1
2
0
𝑔𝑖𝑗 = ൮2𝑢 + 1 ൲.
1
0 2
𝑢 +1