Unizor - Derivatives - Exercises 3

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

Derivatives - Exercise 3

Exercise 3.1
Using the rules of taking a derivative, find the derivative of
f(x) = sin²(x) + cos²(x)
Why the answer is, what it is?

Exercise 3.2
Using the rules of taking a derivative, find the derivative of
f(x) = ln(e^x)
Why the answer is, what it is?

Exercise 3.3
Using the rules of taking a derivative, find the derivative of
f(x) = √x+√x

Answer
[1+1/(2√x)] /2√x+√x

Exercise 3.4
Using the rules of taking a derivative, find the derivative of
f(x) = arctan(1/x)

Answer
−1/(1+x²)

Exercise 3.5
Using the rules of taking a derivative, find the derivative of
f(x) = x^1/x

Answer
x^1/x·(1−ln(x))/x²

Exercise 3.6
Using the rules of taking a derivative, find the derivative of this implicitly defined function
y^x = x^y
(of course, the derivative will also be implicitly defined)

Answer
(y²−x·y·ln(y))/(x²−x·y·ln(x))

Unizor - Derivatives -Problems 3

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

Derivatives - Problems 3

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

Hint
Use L'Hopitale's rule.

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

Hint
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).

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 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 {V_x(t), V_y(t)} and absolute speed 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
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 {V_x(t), V_y(t)} and absolute speed V(t) during the time it's rising on its trajectory.

Solution
Horizontal component of the velocity V_x=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 V_y(t) would decrease from initial value v·sin(φ) by g=9.8m/sec² every second, so its value at time t isV_y(t)=v·sin(φ)−g·t.
Absolute speed is
V = √[V_x(t)]²+[V_y(t)]² =
= √[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 V_y(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) = V_x,
as expected.

Unizor - Derivatives - Differential Along a Curve

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

Differential Along the Curve

We will discuss curves on a plane defined parametrically as a pair of functions {x(t), y(t)}, where t is a parameter taking real values in some finite interval [a,b], while x(t) and y(t) are coordinates of points on a curve.
We assume that both coordinate functions are smooth functions of parameter t.

Our main task is to discuss the length that a point covers when it moves along this curve as parameter t changes its value within its domain.

Traditional approach to this task is to approximate the curve with a series of connected segments as presented on this picture.

The idea is to increase the number of these segments, while the length of each would become smaller, and consider the sum of the lengths of these segments to approximate the length of a curve. This approximation would be better if all segments are getting smaller, so we can assume that, if there is a limit of the sum of lengths of these segments, if they are infinitesimally small, this limit is, by definition, the length of a curve.

Let's assume that we deal with a motion of a particle along this curve. It's position {x(t), y(t)} is a function of time t and the length covered by this particle during its motion along the curve would be a function of time s(t).

During the time from t_n to t_n+1a particle moves along the curve from point {x(t_n), y(t_n)} to {x(t_n+1), y(t₊₁₁)}.
If all Δt_n+1=t_n+1−t_n are sufficiently small, the length along the curve between each pair of beginning and ending points can be approximated by the length of straight segment between them. The smaller the time intervals - the better is our approximation.

So, it's reasonable to assume that if the time increment is an infinitesimal variable, the difference between exact length along the curve and the length of a straight segment is an infinitesimal of a higher order, which in subsequent summation of all infinitesimals to a total length of a curve would not affect the limit.

Let Δx(t_n+1)=x(t_n+1)−x(t_n) and Δy(t_n+1)=y(t_n+1)−y(t_n).
Then the length of the straight segment between two points on a curve is
Δs(t_n)=√Δ²x(t_n)+Δ²y(t_n)

As customary, with all Δt_n converging to zero, we will use an infinitesimal variable dt to designate this increment of a parameter t.
With it, both Δx(t_n) and Δy(t_n)converge to zero and can be represented by infinitesimal variable increments dx and dy.
Analogously, increment of the length Δs(t_n) is an infinitesimal variable ds.

So, the formula for infinitesimal increment of the length of a curve, called a differential along the curve, looks like this:
ds(t) = √[dx(t)]²+[dy(t)]²

As we know, differentials dx(t)and dy(t) can be represented as
dx(t) = x'(t)·dt and
dy(t) = y'(t)·dt
That results in the following representation of a differential along the curve:
ds(t) = √[x'(t)]²+[y'(t)]² ·dt

Connection to Physics

In Physics of motion we know a concept of velocity - a vector with absolute value of a ratio between distance and time (speed), directed towards the direction of motion.

If our motion is not straight forward, but goes along some curve, the direction of motion at any time is along the tangential line to this curve.
If our motion is not constant in speed, we can talk about a very small interval of time during which we can assume the speed changes just a little and calculate an average speed during this time.

Approaching this from a more mathematical viewpoint, we consider a moment of time t, a distance covered by that time as a function s(t) and infinitesimal increment of time dt.
During this infinitesimal time increment of the distance covered would be
s(t+dt) − s(t) = ds(t) =
= √[x'(t)]²+[y'(t)]² ·dt

Then the average speed during this infinitesimal period of time dt equals to
|V(t)| = ds(t)/dt =
= √[x'(t)]²+[y'(t)]²

Derivative x'(t) represents the rate of change of X-coordinate of a particle moving along the curve, derivative y'(t) is the same for Y-coordinate.
What the formula for speed along a curve shows is that velocity of an object moving along a curve, as a vector, can be represented as a sum of two vectors - X-component, a vector directed along the X-axis having value V_x(t)=x(t)=x'(t)and Y-component, a vector directed along the Y-axis having value V_y(t)=y'(t).

In vector form it looks like
V(t) = V_x(t) + V_y(t)
where overline indicates that we deal with vectors and sign + means vector addition. Here V(t) is a velocity vector, V_x(t) is its X-component - projection of velocity onto X-axis and V_y(t)is its Y-component - projection of velocity onto Y-axis.

Unizor - Derivatives - Properties of Differential of a Function

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

Properties of Differential

Properties of differential of a function resemble properties of derivative and immediately follow from them.
Below are the proofs of all these properties, using Euler's notation for derivative D_x f(x).

Linear Combination
f(x) = a·g(x) + b·h(x)
df(x) = d(a·g(x) + b·h(x)) =
= D_x(a·g(x) + b·h(x))·dx =
= (a·D_xg(x) + b·D_xh(x))·dx =
= a·D_xg(x)·dx + b·D_xh(x)·dx =
= a·dg(x) + b·dh(x)

Product
f(x) = g(x)·h(x)
df(x) = d(g(x)·h(x)) =
= D_x(g(x)·h(x))·dx =
= (D_xg(x)·h(x) +
+ g(x)·D_xh(x))·dx =
= D_xg(x)·dx·h(x) +
+ g(x)·D_xh(x)·dx =
= dg(x)·h(x)+g(x)·dh(x)

Reciprocal
f(x) = 1/g(x)
df(x) = d(1/g(x)) =
= D_x(1/g(x))·dx =
= −[1/g²(x)]·D_xg(x)·dx =
= −[1/g²(x)]·dg(x)

Compound
f(x) = g(h(x))
df(x) = d(g(h(x))) =
= D_xg(h(x))·dx =
= D_yg(y)·D_xh(x)·dx =
= D_yg(y)·dh(x)
substituting y with h(x).
Let's illustrate this rule for compound functions.
g(x) = sin(x)
h(x) = ln(x)
f(x) = g(h(x)) = sin(ln(x))
First, let's calculate the differential directly by taking a derivative from f(x) and multiplying it by dx.
df(x) = D_x f(x)·dx =
= D_xsin(ln(x))·dx =
using the chain rule of differentiation of a compound function
= cos(ln(x))·(1/x)·dx
On the other hand, let's see what our expression for differential of a compound function gives.
df(x) = d(g(h(x))) =
= D_yg(y)·dh(x) =
= D_ysin(y)·dln(x) =
substituting y with ln(x).
= cos(y)·(1/x)·dx =
= cos(ln(x))·(1/x)·dx
As you see, the result is the same as with direct computing the differential.

Implicit
Implicitly defined differential can be calculated using the above rule for differential of a compound function.
Assume, f(x) = g(h(x)) and we need to find dh(x).
Since df(x) = D_yg(y)·dh(x),
we conclude that
dh(x) = df(x)/D_yg(y)
substituting y with h(x).
For example, we need to find dln(x) without the knowledge that D_xln(x) = 1/x, which would immediately give us
dln(x) = (1/x)·dx
Consider an equality
x = e^ln(x)
If functions are equal, their differentials at any point are also equal.
dx = de^ln(x)
Now we can use the expression for a differential of a compound function e^ln(x) where g(x)=e^xand h(x)=ln(x).
dx = de^ln(x) = D_ye^y·dln(x)
substituting y with ln(x).
Hence,
dx = e^y·dln(x) =
= e^ln(x)·dln(x) = x·dln(x)
Finally,
dln(x) = (1/x)·dx
which corresponds to the result of direct calculations if we use the known expression D_xln(x)=1/x.

Unizor - Derivatives - Concept of Differential

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

Concept of Differential

A concept of differential of a smooth function f(x) at point x=x₀ was briefly introduced when we defined a derivative of a function as a linear function of an infinitesimal increment of its argument with a coefficient of proportionality equal to a derivative of this function at point x=x₀.

The notation df(x₀)/dx=f ^I(x₀), which was introduced when we defined a concept of derivative, reflects this definition.
Here x₀ is any point in the domain of function f(x), dx is an infinitesimal increment of argument x from this point and df(x₀) is differential of function f(x) introduced above.

The notation that uses d instead of Δ implies that we are not talking about just any particular increment, but about a process of decreasing this increment to zero, thus making it an infinitesimal variable.

Using this type of notation, we can write the definition of a derivative as follows:
f ^I(x₀) =
= lim_dx→0[f(x₀+dx)−f(x₀)]/dx
This implies that
[f(x₀+dx)−f(x₀)]/dx = f ^I(x₀)+ε
where ε is another infinitesimal variable.
Next transformation:
f(x₀+dx)−f(x₀) =
= f ^I(x₀)·dx+ε·dx
or, using the definition of "littleo" as infinitesimal of higher order,
Δf(x₀) = f(x₀+Δx)−f(x₀) =
= f ^I(x₀)·dx+o(dx) =
= df(x₀) + o(dx)

Let me emphasize again that in the above statement dx is not just any increment of argument x, but an infinitesimal variable representing an increment converging to zero.
Similarly, df(x₀) is an infinitesimal variable representing an infinitesimal function increment during the process of an increment of an argument converging to zero.

From the definition of a differential
df(x₀)=f ^I(x₀)·dx
we see that differential is a function of two arguments: a fixed point x₀ within a domain of function f(x) and an infinitesimal increment of an argument dx.
Since x₀ is any fixed point, we can talk about a function differential at any value of argument x and use the notation df(x):
df(x)=f ^I(x)·dx

Here is an illustration of a concept of a differential.

The blue line represents a function, red line - a tangential to it at point A.
Segment BD=AC represents an increment of the argument.
Segment DE is an increment of a function.
Segment DF is function's differential - an increment of a value along the tangential line proportional to the increment of the argument.
When point C moves closer to point A, decreasing the increment of the argument, both segments DE and DF decrease as well, while points E and Fare getting closer to each other, illustrating that increment of a function and its differential are infinitesimals if the increment of the argument is infinitesimal.
What's extremely important is that these two infinitesimals are of the same order as an increment of the argument, while difference between them, segment EF, is an infinitesimal of a higher order.

Recall the Taylor series for function f(x) with expansion center x₀:
f(x)=Σ_n≥0[f ⁽ⁿ⁾(x₀)·(x−x₀)ⁿ/(n!)]

Let's set
dx=x−x₀,
Δf(x₀)=f(x)−f(x₀)
and present this series as follows:
Δf(x₀) =
= f ^I(x₀)·dx+f ^II(x₀)·(dx)²/2+...
According to our definition of the differential, this can be expressed as
Δf(x₀) = df(x₀) + o(dx)
which illustrates the same concept: increment of a function is of the same order as its differential and they differ by an infinitesimal of a higher order than increment of the argument.

It's appropriate to note here that the concept of a differential of a function justifies the Leibniz's notation for a derivative:
f ^I(x) = df(x)/dx

Conclusion

Differential df(x₀) of a function f(x) at some fixed point x₀ of its argument is an infinitesimal variable proportional to infinitesimal increment of the argument dx with a coefficient of proportionality equaled to a derivative of this function at chosen point x₀:
df(x₀) = f ^I(x₀)·dx

Differential differs from increment in a sense that increment is a fixed difference between two values, while differential is an infinitesimal variable.
Thus, Δx is a fixed number that is equal to a difference between the incremented value of argument x=x₁ and its base value x=x₀:
Δx = x₁ − x₀
But differential is an infinitesimal variable {x₁ − x₀} in the process of x₁→x₀.
As soon as we switch from a fixed Δx to a process by considering Δx→0, increment of an argument Δx becomes its differential dx.

Similarly, Δf(x) is a fixed number that is equal to a difference between the value of a function at incremented value of the argument and the value of a function at the base value of the argument:
Δf(x) = f(x₁) − f(x₀)
As x₁→x₀, function increment Δf(x) is getting smaller and relative difference between its values and corresponding values of differential df(x₀) are getting smaller as well in a sense that
lim_Δx→0[Δf(x)]/df(x) = 1
because
lim_Δx→0[Δf(x)]/Δx = f ^I(x)
while
df(x)/dx = f ^I(x)
and dx means the same as Δx→0, that is a process of infinitely decreasing increment of an argument.

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).
Answer
The maximum distance is reached when a cannonball is launched at an angle of π/4=45^o

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.
Answer
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.
Answer
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
F_v = F·sin(φ)
and horizontal component is
F_h = F·cos(φ)
Vertical component F_v reduces the weight and, therefore, reduces the friction. Therefore, the friction equals to
T = (P−F_v)·k
To pull an object with constant speed this friction force must be equal to horizontal component of the force applied to a rope F_h:
T = F_h
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.
F_h = (P−F_v)·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.

Unizor - Derivatives - Problem 1

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

Derivatives - Problems 1

Problem 1.1
Consider a sufficiently smooth function f(x).
Is condition f ^I(x₀) = 0
(a) necessary,
(b) sufficient or
(c) necessary and sufficient
for f(x₀) to have a local extremum (local maximum or local minimum) at point x=x₀?
Prove your answer.
Answer
(a) necessary
It is not sufficient because it might be an inflection point, like for f(x)=x³ at x=0.

Problem 1.2
Consider a sufficiently smooth function f(x).
Is condition f ^II(x₀) = 0
(a) necessary,
(b) sufficient or
(c) necessary and sufficient
for f(x₀) to have an inflection point at x=x₀?
Prove your answer.
Answer
(a) necessary
It is not sufficient because it might be a point of local extremum, like for f(x)=x⁴ at x=0.

Problem 1.3 - Derivative of the inverse function theorem
Consider a sufficiently smooth function y=f(x) with derivative f ^I(x) not equal to zero.
Prove that its inverse y=g(x) (that is, f(g(x))=x) has a derivative g ^I(x)such that
g ^I(x) = 1 / f ^I(g(x))
Example 1
y = f(x) = xⁿ
y = g(x) = x^1/n (inverse)
f ^I(x) = n·xⁿ⁻¹
g ^I(x) = (1/n)·x^(1/n)−1 =
= (1/n)·x^(1−n)/n
1 / f ^I(g(x)) = 1 / {n·[x^1/n]ⁿ⁻¹} =
= (1/n)·1/x^(n−1)/n =
= (1/n)·x^(1−n)/n = g ^I(x)
Example 2
y = f(x) = e^x
y = g(x) = ln(x) (inverse)
f ^I(x) = e^x
g ^I(x) = 1/x
1 / f ^I(g(x)) =
= 1 / e ^ln(x) = 1/x = g ^I(x)

Problem 1.4
Consider all possible regular square prisms with a given surface area.
Under what condition between the length of a base' side a and altitude h the volume of this prism will be minimum or maximum?

Solution
Volume V=a²h
Surface area S=2a²+4ah
Since S is given,
h=(S−2a²)/4a
Substitute it into an expression for volume:
V(a) = a²·(S−2a²)/4a =
= (1/4)·a·(S−2a²)
So, we have to find extremum(s) of function
V(a) = (1/4)·a·(S−2a²) =
= (1/4)·(−2a³+S·a)
This is a polynomial function of a, defined on an interval from 0 to a maximum value when the volume is still greater or equal to zero, that is satisfying the condition S−2a² ≥ 0.
Here is how this function looks on a graph (we have chosen S=6 in this case, so we have to consider this function only on an interval [0,√3]):

As seen from the graph, the extremum of our function within the specified domain is a local maximum.
To find its extremum(s), find the stationary points where derivative equals to zero:
dV/dx = (1/4)·(−6a²+S)
To make sure, we are dealing with a local maximum, we can take a second derivative, it's equal to −3a, and it is negative within a domain of our function, which confirms that a stationary point is a local maximum.
Set the first derivative to 0, getting an equation for variable a:
−6a²+S = 0
Its only root within the established domain is a = √S/6.
Now we can find the corresponding value of altitude h in terms of surface area S:
h = (S−2a²)/4a =
= [S−(S/3)]/(4√S/6) = √S/6.
As we see, h=a, which means that the maximum volume is reached when our prism is a cube.