Thursday, January 26, 2017

Unizor - Derivatives -Problems 3

Notes to a video lecture on

Derivatives - Problems 3

Problem 3.1
Prove that function y=xn·e−x is converging to zero as x→∞.

Use L'Hopitale's rule.

Problem 3.2
Prove that function y=ln(x)·x−nis converging to zero as x→∞.

Use L'Hopitale's rule.

Problem 3.3
Taking into consideration that ln(10)=2.302585 and using differential to approximate increment, calculate approximate value of ln(10.1).

(more precise calculation gives value 2.312535)

Problem 3.4
As a continuation of the previous problem, using Taylor series up to second derivative, calculate approximate value of ln(10.1).


Problem 3.5
A particle is moving along a circle of radius R with a center at the origin of coordinates.
The angle from the positive direction of X-axis to a radius to a particle's position is a function φ(t) of time t.
Determine the particle's components of velocity vector {Vx(t), Vy(t)} and absolute speed V(t).

Position of a particle in Cartesian coordinates is as follows:
X(t) = R·cos(φ(t))
Y(t) = R·sin(φ(t))
The components of the velocity vector are
Vx = X'(t) = −R·sin(φ(t))·φ'(t)
Vy = Y'(t) = R·cos(φ(t))·φ'(t)
From this we can calculate the differential along the curve
ds = √[X'(t)]²+[Y'(t)]² ·dt =
= R|φ'(t)|·dt

Therefore, absolute speed is
V = ds/dt = R|φ'(t)|
This corresponds to another way to calculate the absolute speed, using our knowledge from geometry that the length of an arc s, that corresponds to angle φ, equals to R·φ, from which follows that absolute speed as a function of time is R·|φ'(t)|.
F unction φ'(t) is called angular speed of a particle.
If angle φ(t) is changing proportionally to time, that is ifφ(t)=C·t, where C is constant (positive for counterclockwise movement, negative for clockwise), angular speedφ'(t)=C is constant too and, therefore, absolute speed of a particle moving along the circle is constant R·|C| as well.

Problem 3.6
A projectile is launched at angleφ to horizon with initial absolute speed v.
Determine the components of its velocity vector {Vx(t), Vy(t)} and absolute speed V(t) during the time it's rising on its trajectory.

Horizontal component of the velocity Vx=v·cos(φ) remains the same since there are no forces acting in that direction. During the rising part of trajectory vertical component of the speed Vy(t) would decrease from initial value v·sin(φ) by g=9.8m/sec² every second, so its value at time t isVy(t)=v·sin(φ)−g·t.
Absolute speed is
V = √[Vx(t)]²+[Vy(t) =
= √[v·cos(φ)]²+[v·sin(φ)−g·t =
= √v²−2v·sin(φ)·g·t+g²·t²

Let's analyze this formula.
First, let's determine the time our projectile will rise. This is the time of diminishing its vertical component of the velocity Vy(t) from its initial value v·sin(φ) to zero if it's diminishing by g every unit of time.
This time, obviously, is v·sin(φ)/g.
At the end of this period vertical component of the velocity is zero and absolute speed should be equal to horizontal component, which retains its initial value v·cos(φ).
Substituting t=v·sin(φ)/g into a formula for absolute speed, we get:
v²−2v²·sin²(φ)+v²·sin²(φ) =
= v·cos(φ) = Vx
as expected.

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