## Friday, January 13, 2017

### Unizor - Derivatives - Problem 2

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

Derivatives - Problems 2

Problem 2.1
Artillery officer needs to choose an angle a cannonball should be launched to reach the maximum distance, provided its linear speed at the moment of launching is fixed.
Assume ideal physical conditions (no air resistance, gravity constant is not changed with the height and whatever else can be assumed to simplify the problem).
The maximum distance is reached when a cannonball is launched at an angle of π/4=45o

Problem 2.2
How many zero points does function f(x)=x³−3x+4 have?
Hint
Analyze intervals of monotonic behavior of this function and compare the signs of the function on each interval's ends.
This function has only one zero point.

Problem 2.3
Consider pulling an object by a rope on a surface with friction. If the rope is horizontal, the force needed to move it with constant speed should be equal to the force of friction. If a rope is at certain angle to horizon, the vertical component of the force applied to it partially neutralizes the friction.
What angle of a rope to horizon is needed to minimize the force applied to it and still to move forward with constant speed?
Consider ideal conditions and a coefficient of friction equal to k.
The angle or a rope to horizon should be equal to arctan(k).
Solution
Assume, the angle of a rope with horizon is φ (obviously, its range is from 0 to π/2), the weight of an object is P and the force applied to a rope is F.
Then vertical component of the pulling force is
Fv = F·sin(φ)
and horizontal component is
Fh = F·cos(φ)
Vertical component Fv reduces the weight and, therefore, reduces the friction. Therefore, the friction equals to
T = (P−Fv)·k
To pull an object with constant speed this friction force must be equal to horizontal component of the force applied to a rope Fh:
T = Fh
The above equation gives a dependency between the force applied to a rope F and angle of a rope to horizon φ. Having this as a function, we can minimize it and find an optimal angle of minimum force.
Fh = (P−Fv)·k
F·cos(φ) = [P−F·sin(φ)]·k
F·cos(φ)+F·sin(φ)·k = P·k
F = P·k/[cos(φ)+sin(φ)·k]
To minimize this function, let's take its derivative by φ and find where it's equal to zero.
dF/dφ = {−P·k/[cos(φ)+sin(φ)·k]²}·
·[-sin(φ)+cos(φ)·k]

Equation dF/dφ = 0
results in
-sin(φ)+cos(φ)·k = 0
which can be easily solved:
sin(φ) = cos(φ)·k
sin(φ)/cos(φ) = k
tan(φ) = k
φ = arctan(k)
This solution is independent of the weight of an object and means that the greater the friction - the more vertical should be an angle we pull the object to neutralize friction and minimize the force applied to a rope.
If we are talking about practical application of this, when a person pulls something by a rope, for lower friction coefficient we should use longer rope and for higher friction - shorter.