Conservation of energy A projectile with mass m is launched into the air on a parabolic trajectory. For t ≥ 0, its horizontal and vertical coordinates are x ( t ) = u 0 t and y ( t ) = − 1 2 g t 2 + v 0 t , respectively. where u 0 is the initial horizontal velocity, v 0 is the initial vertical velocity, and g is the acceleration due to gravity. Recalling that u ( t ) = x ′ ( t ) and v ( t ) = y ′ ( t ) are the components of the velocity, the energy of the projectile (kinetic plus potential) is E ( t ) = 1 2 m ( u 2 + v 2 ) + m g y . Use the Chain Rule to compute E ′ ( t ) and show that E ′ ( t ) = 0 , for all t ≥ 0. Interpret the result.
Conservation of energy A projectile with mass m is launched into the air on a parabolic trajectory. For t ≥ 0, its horizontal and vertical coordinates are x ( t ) = u 0 t and y ( t ) = − 1 2 g t 2 + v 0 t , respectively. where u 0 is the initial horizontal velocity, v 0 is the initial vertical velocity, and g is the acceleration due to gravity. Recalling that u ( t ) = x ′ ( t ) and v ( t ) = y ′ ( t ) are the components of the velocity, the energy of the projectile (kinetic plus potential) is E ( t ) = 1 2 m ( u 2 + v 2 ) + m g y . Use the Chain Rule to compute E ′ ( t ) and show that E ′ ( t ) = 0 , for all t ≥ 0. Interpret the result.
Solution Summary: The author calculates the value of Eprime (t) based on the energy of the projectile.
Conservation of energy A projectile with mass m is launched into the air on a parabolic trajectory. For t ≥ 0, its horizontal and vertical coordinates are x(t) = u0t and
y
(
t
)
=
−
1
2
g
t
2
+
v
0
t
, respectively. where u0 is the initial horizontal velocity, v0 is the initial vertical velocity, and g is the acceleration due to gravity. Recalling that
u
(
t
)
=
x
′
(
t
)
and
v
(
t
)
=
y
′
(
t
)
are the components of the velocity, the energy of the projectile (kinetic plus potential) is
E
(
t
)
=
1
2
m
(
u
2
+
v
2
)
+
m
g
y
.
Use the Chain Rule to compute
E
′
(
t
)
and show that
E
′
(
t
)
=
0
, for all t ≥ 0. Interpret the result.
The OU process studied in the previous problem is a common model for interest rates.
Another common model is the CIR model, which solves the SDE:
dX₁ = (a = X₁) dt + σ √X+dWt,
-
under the condition Xoxo. We cannot solve this SDE explicitly.
=
(a) Use the Brownian trajectory simulated in part (a) of Problem 1, and the Euler
scheme to simulate a trajectory of the CIR process. On a graph, represent both the
trajectory of the OU process and the trajectory of the CIR process for the same
Brownian path.
(b) Repeat the simulation of the CIR process above M times (M large), for a large
value of T, and use the result to estimate the long-term expectation and variance
of the CIR process. How do they compare to the ones of the OU process?
Numerical application: T = 10, N = 500, a = 0.04, x0 = 0.05, σ = 0.01, M = 1000.
1
(c) If you use larger values than above for the parameters, such as the ones in Problem
1, you may encounter errors when implementing the Euler scheme for CIR. Explain
why.
#8 (a) Find the equation of the tangent line to y = √x+3 at x=6
(b) Find the differential dy at y = √x +3 and evaluate it for x=6 and dx = 0.3
Q.2 Q.4 Determine ffx dA where R is upper half of the circle shown below.
x²+y2=1
(1,0)
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, calculus and related others by exploring similar questions and additional content below.