*Notes to a video lecture on http://www.unizor.com*

__Derivatives - Problems 3__

**Problem 3.1**Prove that function

*is converging to zero as*

**y=x**^{n}·e^{−x}*.*

**x→∞***Hint*

Use L'Hopitale's rule.

**Problem 3.2**Prove that function

*is converging to zero as*

**y=ln(x)·x**^{−n}*.*

**x→∞***Hint*

Use L'Hopitale's rule.

**Problem 3.3**Taking into consideration that

*and using differential to approximate increment, calculate approximate value of*

**ln(10)=2.302585***.*

**ln(10.1)***Answer*

*2.312585*(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)***Answer*

*2.312535*

**Problem 3.5**A particle is moving along a circle of radius

*with a center at the origin of coordinates.*

**R**The angle from the positive direction of X-axis to a radius to a particle's position is a function

*of time*

**φ(t)***.*

**t**Determine the particle's components of velocity vector {

*} and absolute speed*

**V**_{x}(t), V_{y}(t)*.*

**V(t)***Solution*

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

**V**_{x}= X'(t) = −R·sin(φ(t))·φ'(t)

**V**_{y}= 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= 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

*, that corresponds to angle*

**s***, equals to*

**φ***, from which follows that absolute speed as a function of time is*

**R·φ***.*

**R·|φ'(t)|**F unction

*is called*

**φ'(t)***angular speed*of a particle.

If angle

*is changing proportionally to time, that is if*

**φ(t)***, where*

**φ(t)=C·t***is constant (positive for counterclockwise movement, negative for clockwise), angular speed*

**C***is constant too and, therefore, absolute speed of a particle moving along the circle is constant*

**φ'(t)=C***as well.*

**R·|C|**

**Problem 3.6**A projectile is launched at angle

*to horizon with initial absolute speed*

**φ***.*

**v**Determine the components of its velocity vector {

*} and absolute speed*

**V**_{x}(t), V_{y}(t)*during the time it's rising on its trajectory.*

**V(t)***Solution*

Horizontal component of the velocity

*remains the same since there are no forces acting in that direction. During the rising part of trajectory vertical component of the speed*

**V**_{x}=v·cos(φ)*would decrease from initial value*

**V**_{y}(t)*by*

**v·sin(φ)***every second, so its value at time*

**g=9.8m/sec²***is*

**t***.*

**V**_{y}(t)=v·sin(φ)−g·tAbsolute speed is

**V = √[**

= √[

= √v²−2v·sin(φ)·g·t+g²·t²**V**]²+[_{x}(t)**V**]² =_{y}(t)= √[

**v·cos(φ)**]²+[**v·sin(φ)−g·t**]² == √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

*from its initial value*

**V**_{y}(t)*to zero if it's diminishing by*

**v·sin(φ)***every unit of time.*

**g**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

*into a formula for absolute speed, we get:*

**t=v·sin(φ)/g***√*,

**v²−2v²·sin²(φ)+v²·sin²(φ)****=**

= v·cos(φ) = V= v·cos(φ) = V

_{x}as expected.

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