APM3713 Assignment 6 |COMPLETE CALCULATIONS| 2025

APM3713
Assignment 6
2025

, ASSIGNMENT 6

Question 1

This question considers a surface parameterized by the equations:

• 𝑥(𝑢, 𝑣) = √𝑢2 + 1𝑐𝑜𝑠 𝑣

• 𝑦(𝑢, 𝑣) = √𝑢2 + 1𝑠𝑖𝑛 𝑣

• 𝑧(𝑢, 𝑣) = 𝑢

We are given that 𝑥1 = 𝑢 and 𝑥 2 = 𝑣 are the coordinates for the surface.



(a) Find the line element for the surface.

To find the line element, we first calculate the partial derivatives of the position vector
𝑟⃗ = (𝑥, 𝑦, 𝑧)with respect to the parameters 𝑢 and 𝑣:

⃗
∂𝑟⃗ 𝑢 𝑢
• =( cos𝑣, sin𝑣, 1)
∂𝑢 √𝑢 2 +1 √𝑢 2 +1

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


⃗ ∂𝑟⃗
∂𝑟⃗ ⃗
The metric tensor components 𝑔𝑖𝑗 are defined by 𝑔𝑖𝑗 = ⋅ .
∂𝑥 𝑖 ∂𝑥 𝑗


⃗ ∂𝑟⃗
∂𝑟⃗ ⃗ 𝑢 2 𝑢 2 𝑢2
• 𝑔11 = ⋅ =( ) cos 2 𝑣 + ( ) sin2 𝑣 + 12 𝑔11 = (cos 2 𝑣 + sin 2 𝑣) +
∂𝑢 ∂𝑢 √𝑢 2 +1 √𝑢 2 +1 𝑢 2 +1

𝑢2 𝑢 2 +𝑢 2 +1 2𝑢 2 +1
1= +1= =
𝑢 2 +1 𝑢 2 +1 𝑢 2 +1


⃗ ∂𝑟⃗
∂𝑟⃗ ⃗
• 𝑔22 = ⋅ = (−√𝑢2 + 1sin𝑣) 2 + (√𝑢2 + 1cos𝑣) 2 + 02 𝑔22 = (𝑢 2 + 1)sin2 𝑣 +
∂𝑣 ∂𝑣

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

⃗ ∂𝑟⃗
∂𝑟⃗ ⃗ 𝑢 𝑢
• 𝑔12 = 𝑔 21 = ⋅ =( cos𝑣) (−√𝑢2 + 1sin𝑣) + ( sin𝑣) (√𝑢2 + 1cos𝑣) +
∂𝑢 ∂𝑣 √𝑢 2 +1 √𝑢 2 +1

(1)(0) 𝑔12 = −𝑢cos𝑣sin𝑣 + 𝑢sin𝑣cos𝑣 = 0

The line element, 𝑑𝑠 2 , is given by 𝑑𝑠 2 = 𝑔𝑖𝑗 𝑑𝑥 𝑖 𝑑𝑥 𝑗 .