Mechanics and Calculus of Variations APM3712 – University of South Africa (UNISA) 2025 | Complete Assignment Memos and Exam Preparation Pack

APM3712/202/0/2025




Tutorial letter 202/0/2025

MECHANICS AND CALCULUS OF
VARIATIONS
APM3712

Year module


Department of Mathematical Sciences


This tutorial letter contains Solutions for assignment 01.




BARCODE




university
Define tomorrow. of south africa

, STUDY GUIDE: CHAPTERS 1 and 2
We marked Questions 2, 4 and 7. Q 2: 10 marks Q 4: 20 marks and Q7 : 10 marks.

Question 1



Question 2

A mass M is constrained to slide without friction on the track AB as shown in the following Fig.
A mass m is connected to M by a massless inextensible string. Determine the Lagrangian function




of the two masses.
Solution
Use coordinates as shown in Figure, M and m have coordinates

(x, 0), (X = x + b sin θ, Y = −b cos θ)

respectively.

The kinetic energy of the two masses are
1 1 1
T1 = M ẋ2 ; T2 = m(Ẋ 2 + Ẏ 2 ) = m(ẋ2 + b2 θ̇2 + 2bẋθ̇ cos θ).
2 2 2
Also we can then write the potential energy of the two masses as

V1 = 0; V2 = −mg(b cos θ).

The Lagrangian of the system then takes the form
1 1
L = T1 + T2 − V1 − V2 = M ẋ2 + m(ẋ2 + b2 θ̇2 + 2bẋθ̇ cos θ) + mg(b cos θ)
2 2




2

, APM3712/202/0/2025


Question 3



Question 4

Consider the variational problem with Lagrangian function

L(t, x, y, ẋ, ẏ) = ẋ2 + ẏ 2 − 32xy.

Obtain the Euler-Lagrange equations and solve them.
Solution: The Euler-Lagrange equations are
   
d ∂L ∂L d ∂L ∂L
− = 0, and − = 0.
dt ∂ ẋ ∂x dt ∂ ẏ ∂y

So  
∂L ∂L d ∂L
= −16yand = 2ẋ =⇒ = 2ẍ.
∂x ∂ ẋ dt ∂ ẋ
Therefore ẍ + 16y = 0. Similarly,
 
∂L ∂L d ∂L
= −16x, and = 2ẏ =⇒ = 2ÿ.
∂y ∂ ẏ dt ∂ ẏ

Therefore ÿ + 16x = 0. And finally as required, we have

ẍ + 16y = 0, ÿ + 16x = 0.

Differentiate the first equation twice with respect to t and substitute in the second equation to get
1 (4)
x(4) + 16ÿ = 0 ⇔ ÿ = − x .
16
then
1 (4)
− x + 16x = 0 ⇔ x(4) − 162 x = 0.
16
We can solve this equation using various methods. The characteristic equation is λ4 − 162 = 0, which
gives
λ = ±4i or λ = ±4.
Hence
x(t) = Ae4t + Be−4t + C cos 4t + D sin 4t.
Now
1
y(t) = − ẍ(t) = −Ae4t − Be−4t + C cos 4t + D sin 4t.
16


Question 5




3

, Question 6



Question 7

Exercise 2.5.19.14: √
1 + ẋ2
L (x, ẋ) = .
x

Solution
Using Lemma 2.4.1 on page 37 in the study guide, i.e.,
∂L
ẋ − L = c.
∂ ẋ
where c is a constant, and then simplifying we obtain
√
ẋ 1 + ẋ2 (1 + ẋ2 − ẋ2
ẋ √ − =c→ √ = c.
x 1 + ẋ2 x x 1 + ẋ2
√ 1 1 1
cx 1 + ẋ2 = 1 → x = √ → x 2 c2 = 2
→ 1 + ẋ2 = 2 2
c 1 + ẋ 2 1 + ẋ xc
r √
1 1 − c2 x 2 dx 1 − c2 x 2
ẋ2 = 2 2 − 1 → ẋ = → =
xc c2 x 2 dt cx
cx 1√ 1 1
√ dx = dt → − 1 − c2 x2 = t + d → (t + d)2 = 2 (1 − c2 x2 ) → x2 + (t + d)2 = 2 .
1 − c2 x 2 c c c




4