Financial Mathematics (2024 – Brusov, Filatova & Orekhova) | Exercises & Solutions eBook PDF

ALL CHAPTERS COVERED

, 1
Solutions to Chapter 1 Student Exercises




1.1
Think back to how you derived the equations of motion for 1D motion with
constant velocity and 1D motion with constant acceleration. Now derive the
equations of motion for rotational motion with constant angular velocity and
rotational motion with constant angular acceleration, shown in Table 1.1. Also,
construct graphs for the position, velocity, and acceleration (both angular and
linear) as functions of time for all combinations of positive and negative veloc-
ity, and positive and negative acceleration. Compare the graphs for the linear
and angular quantities as functions of time. Why should you not be surprised
that the graphs look the same?

Let’s start with the equations for rotational motion with constant angular
⃗ . Because the angular velocity is constant, then the average angular
velocity ω
velocity ω ⃗ . That is:
⃗ avg is the same as the instantaneous angular velocity ω

∆⃗θ
⃗ =ω
ω ⃗ avg = . (1.1)
∆t
Realizing that when a quantity is constant its average and instantaneous values
are equal is the foundation of such derivations where we do not use calculus.
Now, we can easily see that
∆⃗θ = ω
⃗ ∆t, (1.2)
which is the equation of motion with constant angular velocity.

Of course, we could derive the same equation using integration. By defini-
tion, we know that
d⃗θ
⃗ =
ω ⇒ d⃗θ = ω
⃗ dt. (1.3)
dt

1

,2 Solutions to Chapter 1 Student Exercises

We can integrate both sides of this equation having ⃗θ as the variable of integra-
tion on the left-hand side while t is the variable of integration on the right-hand
side. Also, we are assuming that at the initial time t0 the initial angular position
is ⃗θ0 while at some arbitrary time t the angular position is ⃗θ. We have:
Z ⃗θ Z t
d⃗θ = ⃗ dt
ω
⃗θ0 t0
Z t (1.4)
⇒ ⃗θ − ⃗θ0 = ω⃗ dt.
t0

⃗ is constant, we have taken it out of the integral on the right-
where, since ω
hand side. Therefore, we can now also evaluate the integral on the right-hand
side:
⃗θ − ⃗θ0 = ω
⃗ (t − t0 )
(1.5)
⇒ ∆⃗θ = ω ⃗ ∆t,
which is, of course, the same as Equation 1.2.

We can also derive the equations for rotational motion with constant accel-
⃗ using calculus. Again, by definition, we know that
eration α
ω
d⃗
⃗=
α ⃗ dt.
ω=α
⇒ d⃗ (1.6)
dt
We again integrate both sides of this equation having ω ⃗ as the variable of inte-
gration on the left-hand side while t is the variable of integration on the right-
hand side. Also, we are assuming that at the initial time t0 the initial angular
⃗ 0 while at some arbitrary time t the angular velocity is ω
velocity is ω ⃗ . We have:
Z ω⃗ Z t
ω=
d⃗ ⃗ dt
α
⃗0
ω t0
Z t (1.7)
⇒ω ⃗ −ω ⃗0 = α⃗ dt.
t0

⃗ is constant, we have taken it out of the integral on the right-hand
where, since α
side. We can now evaluate the integral on the right-hand side:
⃗ −ω
ω ⃗0 = α
⃗ (t − t0 )
(1.8)
⃗ =ω
⇒ω ⃗0 + α
⃗ ∆t,
which is the equation for the angular velocity as a function of time when the
angular acceleration is constant.

Given the definition of Equation 1.3, we can now use our result of Equation

, Solutions to Chapter 1 Student Exercises 3

1.8 to get an equation for the angular position as a function of time when the
acceleration is constant. We again make the assumptions that at times t0 and t
the angular positions are ⃗θ0 and ⃗θ, respectively. We have:

d⃗θ
⃗ =
ω
dt
⇒ d⃗θ = ω ⃗ dt
Z ⃗θ Z t
(1.9)
⇒ d⃗θ = ⃗ dt
ω
⃗θ0 t0
Z ⃗θ Z t
⇒ d⃗θ = ⃗ ∆t)dt.
ω0 + α
(⃗
⃗θ0 t0

Both integrals can be evaluated easily since ω⃗ 0 and α
⃗ are constants and we
obtain:
1
∆⃗θ = ω
⃗ 0 ∆t + α ⃗ ∆t2 . (1.10)
2
Figure 1.1 shows the graphs for the displacement, velocity, and acceleration
as functions of time when we have rotational motion with constant angular
velocity and when we have linear motion with constant linear velocity. If we
assume that the initial position is zero (i.e., ⃗θ0 = 0 and ⃗x0 = 0, as we have
done here) then the displacement is the same as the position at time t. Figure
1.1a shows the graphs when the velocity is positive while Figure 1.1b shows
the graphs when the velocity is negative. As expected, the graphs are identical
since the positions, velocities, and acceleration, whether angular or linear, are
described by equations of the same form.
Similarly, Figure 1.2 shows the graphs for the position, velocity, and ac-
celeration as functions of time when we have rotational motion with constant
angular acceleration and when we have linear motion with constant linear ac-
celeration. Again, if we assume that the initial position is zero (i.e., ⃗θ0 = 0
and ⃗x0 = 0, as we have done here) then the displacement is the same as the
position at time t. We present four scenarios: a) ω ⃗ 0, α
⃗ > 0, which is the same
as when ⃗v0 , ⃗a > 0, b) ω
⃗ 0 > 0, α
⃗ < 0, which is the same as when ⃗v0 > 0, ⃗a < 0,
⃗ 0 < 0, α
c) ω ⃗ > 0, which is the same as when ⃗v0 < 0, ⃗a > 0, and d) ω ⃗ 0, α
⃗ < 0,
which is the same as when ⃗v0 , ⃗a < 0. As expected, in each case the rotational
motion graphs are identical to the corresponding linear motion graphs since the
positions, velocities, and acceleration, whether angular or linear, are described
by equations of the same form.
Of course, if θ0 , 0 and x0 , 0 then the graphs of position as a function
of time will be shifted up or down, depending on the sign of ⃗θ0 and ⃗x0 . In