Applications of Differentiation

Maximum and Minimum Values

Some of the most important applications of differential calculus are optimization problems, in which we are required to find the optimal (best) way of doing something.

Absolute and Local Extreme Values

An absolute maximum or minimum is sometimes called a global maximum or minimum. The maximum and minimum values of $f$ are called extreme values of $f$.

The Extreme Value Theorem is illustrated in Figure 8.

Critical Numbers and the Closed Interval Method

PROOF: Suppose that $f$ has a local maximum at $c$. Then accoding to the definition of a local maximum $f(c) \geq f(x)$ if $x$ is sufficiently close to $c$. This implies that if $h$ is sufficiently close to $0$, with $h$ being positive or negative, then:

$$ f(c) \geq f(c + h) $$

and therefore

$$ f(c + h) - f(c) \leq 0 $$

We can divide both sides of an inequality by a positive number $h$, then

$$ \frac{f(c + h) - f(c)}{h} \leq 0 $$

Taking the right-hand limit of both sides we get

$$ \lim_{h \rightarrow 0^+} \frac{f(c + h) - f(c)}{h} \leq \lim_{h \rightarrow 0^+} = 0 $$

Since $f’(0)$ exists, we have

$$ \lim_{h \rightarrow 0} \frac{f(c + h) - f(c)}{h} = \lim_{h \rightarrow 0^+} \frac{f(c + h) - f(c)}{h} $$

and so we have shown that $f’(c) \leq 0$. If $h < 0$, then

$$ \frac{f(c + h) - f(c)}{h} \geq 0 $$

So taking the left-hand limit, we have

$$ \lim_{h \rightarrow 0} \frac{f(c + h) - f(c)}{h} = \lim_{h \rightarrow 0^-} \frac{f(c + h) - f(c)}{h} \geq 0 $$

We have shown that $f’(c) \geq 0$ and $f’(c) \leq 0$ so the only possibility is that $f’(c) = 0$.

We can’t expect to locate extreme values simply by setting $f’(x) = 0$ and solving for $x$. The fact that $f’(x) = 0$ simply means that the curve $f$ has a horizontal tangent at $(x, f(x))$. In other words, the converse of Fermat’s Theorem is false in general. However, Fermat’s Theorem does suggest that we should at least start looking for extreme values of $f$ at the number $c$ where $f’(c) = 0$ or where $f’(c)$ does not exist.

In terms of critical numbers, Fermat’s Theorem can be rephared as follows:

The Mean Value Theorem

Rolle’s Theorem

PROOF: There are three cases:

Case 1 $f(x) = k$, a constant. Then $f’(x) = 0$, so the number $c$ can be taken to be any numer in $(a, b)$.

Case 2 $f(x) > f(a)$ for some $x \in (a, b)$. Then, by the Extreme Value Theorem (which we can apply by hypothesis 1), $f$ has a maximum value somewhere in $[a, b]$. Since $f(a) = f(b)$ it must attain this maximum value at a number $c$ in the open interval $(a, b)$. Then $f$ has a local maximum at $c$, and by hypothesis 2, $f$ is differentiable at $c$. Therefore $f’(c) = 0$ by Fermat’s Theorem.

Case 3 $f(x) < f(a)$ for some $x \in (a, b)$. Then, by the Extreme Value Theorem (which we can apply by hypothesis 1), $f$ has a minimum value somewhere in $[a, b]$. Since $f(a) = f(b)$ it must attain this minimum value at a number $c$ in the open interval $(a, b)$. Then $f$ has a local minimum at $c$, and by hypothesis 2, $f$ is differentiable at $c$. Therefore $f’(c) = 0$ by Fermat’s Theorem.

Figure 1 shows the graphs of four such functions.

The Mean Value Theorem

The Mean Value Theorem says that there is at least one point $P(c, f(c))$ on the graph where the slope of the tangent line is the same as the slope of the secant line $A, B$ (see Figures 3 and 4).

PROOF We see the the equation of the line $AB$ can be written as:

$$ y - f(a) = m_{AB} (x - a) $$

where $m_{AB} = \frac{f(b) - f(a)}{b - a}$, thus

$$ y = f(a) + \frac{f(b) - f(a)}{b - a}(x - a) $$

So, as we can see on Figure 5:

$$ h(x) = f(x) - y $$$$ h(x) = f(x) - f(a) - \frac{f(b) - f(a)}{b - a}(x - a) $$

First, we must verify that $h$ satisfies the three hypotheses of Rolle’s Theorem:

1. The function $h$ is continuous on $[a, b]$ because it is the sum of $f$ and a first-degree polynomial both of which are continuous.
2. The function $h$ is differentiable because both $f$ and the first degree polynomial are differentiable. Thus

$$ h'(x) = f'(x) - \frac{f(b) - f(a)}{b - a} $$

3. We must show that $h(a) = h(b)$

$$ h(a) = f(a) - f(a) - \frac{f(b) - f(a)}{b - a}(a - a) = 0 $$$$ h(b) = f(b) - f(a) - \frac{f(b) - f(a)}{b - a}(b - a) $$$$ = f(b) - f(a) - f(b) + f(a) = 0 $$

Which means $h(a) = h(b) = 0$.

Since $h$ satisfies all the hypotheses of Rolle’s Theorem, there exists a number $c$ in $(a, b)$ such that $h’(c) = 0$. Therefore

$$ h'(c) = f'(c) - \frac{f(b) - f(a)}{b - a} = 0 $$

So

$$ f'(c) = \frac{f(b) - f(a)}{b - a} $$

In general, the Mean Value Theorem can be interpreted as saying that there is a number at which the instantaneous rate of change is equal to the average rate of change over an interval.

The Mean Value Theorem can be used to establish some of the basic facts of differential calculus.

PROOF: Let $x_1$ and $x_2$ be any two numbers in $(a, b)$ with $x_1 < x_2$. Since $f$ is differentiable on $(a, b)$ it must be differentiable on $(x_1, x_2)$ and continuous on $[x_1, x_2]$. By applying the Mean Value Theorem to $f$ on the interval $[x_1, x_2]$, we get a number $c$ such that $x_1 < c < x_2$ and

$$ f(x_2) - f(x_1) = f'(c) (x_2 - x_1) $$

Since $f’(x) = 0$ for all $x$, then $f’(c) = 0$ and therefore

$$ f(x_2) - f(x_1) = 0 \leftrightarrow f(x_2) = f(x_1) $$

Thus $f$ has the same value at any two numbers $x_1$ and $x_2$ in $(a, b)$. This means that $f$ is constant on $(a, b)$.

PROOF: Let $F(x) = f(x) - g(x)$. Then

$$ F'(x) = f'(x) - g'(x) = 0 $$

for all $x \in (a, b)$. Thus by the previous theorem $F$ is constant, that is $f - g$ is constant.

This corollary says that if two functions have the same derivatives on an interval, then their graphs must be vertical trasnlation of each other.

What Derivatives Tell Us about the Shape of a Graph

What Does $f’$ say about $f$?

See Figure 1 for a graphical representation.

PROOF: Without loss of generality we let $x_1$ and $x_2$ be any two numbers on the interval with $x_1 < x_2$. According to the definition of an increasing function we have to show that $f(x_1) < f(x_2)$.

Because we are given that $f’(x) > 0$, we know that $f$ is differentiable on $[x_1, x_2]$. So, by the Mean Value Theorem, there is a number $c$ between $x_1$ and $x_2$ such that

$$ f(x_2) - f(x_1) = f'(c)(x_2 - x_1) $$

We know that $f’(c) > 0$ by our initial assumption and $x_2 - x_1$ because $x_1 < x_2$, thus the right side of the previous equation must be positive:

$$ f(x_2) - f(x_1) > 0 $$

Which is equivalent to

$$ f(x_2) > f(x_1) $$

This shows that $f$ is increasing.

The First Derivative Test

The First Derivative Test is illustrated in Figure 4.

What does $f’’$ say about $f’$?

See Figure 7 for an illustration on the concavity of a function.

In Figure 7(a) we see that the slope of the tangent increases. This means that the derivative $f’$ is an increasing function and therefore its derivative $f’’$ is positive. Likewise, in Figure 7(b) the slope of the tangent decreases, so $f’$ decreases and therefore $f’’$ is negative.

The Second Derivative Test

Note that The Second Derivative Test is inconclusive when $f’’(c) = 0$ as there might be a maximum, a minimum or neither in that point. This test also fails when $f’’(c)$ does not exist.

Indeterminate Forms and l’Hospital’s Rule

Indeterminate Forms (Types $\frac{\infty}{\infty}$, $\frac{0}{0}$)

In general, if we have a limit of the form:

$$ \lim_{x \to a} \frac{f(x)}{g(x)} $$

where both $f(x) \to 0$ and $g(x) \to 0$ as $x \to a$, then this limit may or may not exist, and is called an indeterminate form of type $\frac{0}{0}$.

In general, if we have a limit of the form:

$$ \lim_{x \to a} \frac{f(a)}{g(a)} $$

where both $f(x) \to \infty$ (or $-\infty$) and $g(x) \to \infty$ (or $-\infty$) as $x \to a$, then this limit may or may not exist, and is called an indeterminate form of type $\frac{\infty}{\infty}$.

L’Hospital’s Rule

We introduce a systematic method for the evaluation of indeterminate forms of type $\frac{0}{0}$ or type $\frac{\infty}{\infty}$.

Note:

1. L’Hospital’s Rule says that the limit of a quotient of function is equal to the limit of the quotient of their derivatives, provided that the conditions are satisfied.
2. L’Hospital’s Rule is also valid for one-sided limit, and for limits at infinity or negative infinity.
3. For the special case in which $f(a) = g(a) = 0$, $f’$ and $g’$ are continuous and $g’(a) \neq 0$, it is easy to show why l’Hospital’s Rule is true:

Given $f’$ and $g’$ are continuous:

$$ \lim_{x \to a} \frac{f'(x)}{g'(x)} = \frac{f'(a)}{g'(a)} $$

By the definition of a derivative:

$$ \frac{f'(a)}{g'(a)} = \frac{\lim_{x \to a} \frac{f(x) - f(a)}{x - a}}{\lim_{x \to a} \frac{g(x) - g(a)}{x - a}} $$

By the properties of limits:

$$ \frac{\lim_{x \to a} \frac{f(x) - f(a)}{x - a}}{\lim_{x \to a} \frac{g(x) - g(a)}{x - a}} = \lim_{x \to a} \frac{\frac{f(x) - f(a)}{x - a}}{ \frac{g(x) - g(a)}{x - a}} $$$$ \lim_{x \to a} \frac{\frac{f(x) - f(a)}{x - a}}{ \frac{g(x) - g(a)}{x - a}} = \lim_{x \to a} \frac{f(x) - f(a)}{g(x) - g(a)} $$

Because $f(a) = g(a) = 0$

$$ \lim_{x \to a} \frac{f(x) - f(a)}{g(x) - g(a)} = \lim_{x \to a} \frac{f(x)}{g(x)} $$

It is more difficult to prove the general version of l’Hospital’s Rule.

Indeterminate Products (Type $0 \cdot \infty$)

If $\lim_{x \to a} f(x) = 0$ and $\lim_{x \to a} g(x) = \infty$ or ($- \infty$), then it isn’t clear what the value of $\lim_{x \to a} [f(x)g(x)]$ will be.

This kind of limit is called an indeterminate form of type $0 \cdot \infty$. We can deal with it by writing the product $fg$ as a quotient:

$$ fg = \frac{f}{\frac{1}{g}} \text{ or } fg = \frac{g}{\frac{1}{f}} $$

This converts the given limit into an indeterminate form of type $\frac{0}{0}$ or $\frac{\infty}{\infty}$ so that we can use l’Hospital’s Rule.

Indeterminate Differences (Type $\infty - \infty$)

If $\lim_{x \to a} f(x) = \infty$ and $\lim_{x \to a} g(x) = \infty$, then the limit:

$$ \lim_{x \to a} [f(x) - g(x)] $$

is called an indeterminate form of type $\infty - \infty$. There is a contest between $f$ and $g$. Will the answer be $\infty$ ($f$ wins), $-\infty$ ($g$ wins) or will they compromise on a finite number? To find out, we try to convert the difference into a quotient so that we can have an indeterminate form of type $\frac{0}{0}$ or $\frac{\infty}{\infty}$.

Indeterminate Powers (Types $0^0, \infty^0, 1^{\infty}$)

Several indeterminate forms arise from the limit:

$$ \lim_{x \to a} [f(x)]^{g(x)} $$

1. $\lim_{x \to a} f(x) = 0$ and $\lim_{x \to a} g(x) = 0$, type $0^0$
2. $\lim_{x \to a} f(x) = \infty$ and $\lim_{x \to a} g(x) = 0$, type $\infty^0$
3. $\lim_{x \to a} f(x) = 1$ and $\lim_{x \to a} g(x) = \pm \infty$, type $1^{\infty}$

Each of these three cases can be treated by taking the natural logarithm as follows:

$$ \text{let } y = [f(x)]^{g(x)} \text{, then } \ln y = g(x) \ln f(x) $$

Now we calculate:

$$ \lim_{x \to a} g(x) \ln f(x) $$

This limit is always an indeterminate product (indeterminate form of type $0 \cdot \infty$), and can be solved as one, which will require the application of L’Hospital’s Rule.

Once we get the value of the limit for $\ln y$, we can obtain the limit of $y$ by writing the logarithm of $y$ as an exponential:

$$ \lim_{x \to a} y = \lim_{x \to a} [f(x)]^{g(x)} = \lim_{x \to a} e^{\ln y} = \lim_{x \to a} e^{g(x) \ln f(x)} = e^{\lim_{x \to a} g(x) \ln f(x)} $$

Summary of Curve Sketching

Guidelines for Sketching a Curve

The following checklist is intended as a guide to sketching a curve $y = f(x)$.

1. Domain: Start by determining the domain $D$ of $f$.
2. Intercepts:
   • Find where the curve intersects the $y$-axis by evaluating $f(0)$.
   • Find where the curve intersects the $x$-axis by solving $f(x) = 0$ for $x$.
3. Symmetry:
   • If $f(x) = f(-x)$ for all $x \in D$, then $f$ is an even function and the curve is symmetric about the $y$-axis.
   • If $-f(x) = f(-x)$ for all $x \in D$, then $f$ is an odd function and the curve is symmetric about the origin.
   • If $f(x) = f(x + p)$ for all $x \in D$, where $p$ is a positive constant then $f$ is a periodic function and the smallest such number $p$ is called the period.
4. Asymptotes:
   • Horizontal asymptotes: If either $\lim_{x \to \infty} f(x) = L$ or $\lim_{x \to -\infty} f(x) = L$, then the line $y = L$ is a horizontal asymptote of the curve $y = f(x)$.
   • Vertical asymptotes: The line $x = a$ is a vertical asymptote if either $\lim_{x \to a^+} f(x) = \infty$ and $\lim_{x \to a^-} f(x) = \infty$ or $\lim_{x \to a^+} f(x) = -\infty$ and $\lim_{x \to a^-} f(x) = -\infty$.
   • Slant asymptotes: these will be discussed later on.
5. Intervals of Increase or Decrease: Use the Increasing/Decreasing Test to find the intervals in which $f$ increases or decreases.
6. Local Maximum or Minimum Values: Find the critical number of $f$, then use the First Derivative Test to determine whether the critical numbers are maximum or minimum values for $f$.
7. Concavity and Points of Inflection: Compute $f’’(x)$ and use the Concavity Test to determine where $f$ is concave upward and where $f$ is concave downward.
8. Sketch the Curve: Using the information from the previous points, draw the graph of $f$.

Slant Asymptotes

Some curve have asymptotes that are oblique. If:

$$ \lim_{x \to \infty} [f(x) - (mx + b)] = 0 $$

where $m \neq 0$, then the line $y = mx + b$ is called a slant asymptote because the vertical distance between the curve $y = f(x)$ and the line $y = mx + b$ approaches $0$ (see Figure 12). A similar situation exists is we let $x \to - \infty$.

In the case of rational functions, slant asymptotes occur when the degree of the numerator is one more than the degree of the denominator. In such a case the equation of the slant asymptote can be found by long division.

Example

Given a function $f(x) = xe^{-\frac{1}{x}}$, we obtain its slant asymptote, denoted by $y = mx + b$ by first finding the slope $m$ as:

$$ m = \lim_{x \to \pm \infty} \frac{f(x)}{x} = \frac{xe^{-\frac{1}{x}}}{x} = e^{-\frac{1}{x}} = e^{0} = 1 $$

Thus $m = 1$. Now we compute the value for the intercept $b$ as:

$$ b = \lim_{x \to \pm \infty} (f(x) - mx) = xe^{-\frac{1}{x}} - x = x \left(e^{-\frac{1}{x}} - 1\right) = \infty \cdot (1 - 1) = \infty \cdot 0 $$

We apply L’Hopital’s Rule:

$$ = \lim_{x \to \pm \infty} \frac{e^{- \frac{1}{x}} - 1}{\frac{1}{x}} = -\frac{e^{- \frac{1}{x}}x^{-2}}{x^{-2}} = -e^{- \frac{1}{x}} = e^{0} = -1 $$

Such that the slant asymptote is given by $y = mx + b = x - 1$.

Practical questions

How to prove that a function is not periodic

A function $f(x)$ is periodic if it exists $T > 0$ such that:

$$ f(x + T) = f(x) \forall x $$

There are a few ways in which we could prove $f$ is not periodic:

1. For simple functions you can directly check whether a period $T > 0$ can exist by solving:

$$ f(x + T) = f(x) $$

2. Find a contradiction with specific values.
   • Pick a generic $x_1$ and try to find a generic $T$ that satisfies $f(x + T) = f(x)$, if there is not such $T$, then $f$ is not periodic.
   • If there is such a $T$, repeat the previous step with another generic $x_2$, repeat until you can disprove the periodicity (try with a sensible number of points).
3. If the function increases or decreases without bound (i.e. it tends to $\pm \infty$), then we can assume there is no periodicity because the values of $f$ keep getting larger and larger and do not follow a pattern.
4. If it is a composite function, where one of the components is a non-periodic function, then the composite function is also non-periodic.

How to prove that a function does not have asymptotes

To show there exist no vertical asymptotes for a function $f(x)$ we need to check limits near critical points:

• For a rational function, find $x$ such that the denominator is $0$.
• For logarithms, the argument of the function must be postive. So boundary points like $x = 0, \text{for} \ln(x)$ are candidates.
• For square roots or even roots, its argument must be non-negative. So boundary points like $x < 0 \text{for} \sqrt{x}$ are candidates.
• Any other undefined points.

This is to say that if

1. If $f$ is defined everywhere, then there is no vertical asymptote.
2. If the limits on critical points where $f$ would be undefined are finite, than there is no vertical asymptote.

To show there exist no horizontal asymptotes for a function $f(x)$ we have to study how the function behaves when $x$ increases or decreases without bound. If $f$ is infinite or does not exist when $x$ increases or decreases without bound, then there are no horizontal asymptotes.

To show there exist no oblique asymptotes for a function $f(x)$ we must show that the limit:

$$ \lim_{x \to \infty} \frac{f(x)}{x} = m $$

or

$$ \lim_{x \to \infty} (f(x) - mx) = b $$

either do not exist, or one of them is infinite.

Graphing with Calculus and Techonology

Optimization Problems

Proof (Absolute Minimum)

Suppose $f$ is a continuous differentiable function on an interval $I$, and let $c \in I$ be a critical point of $f$. Let us assume:

$$ \forall x \in I, \text{ where } x < c, f'(x) < 0 \text{ and } \forall x \in I, \text{ where } x > c, f'(x) > 0 $$

(1) Given any $x \in I$ with $x < c$, by The Mean Value Theorem there exists some $\xi \in (x, c)$ such that

$$ f'(\xi) = \frac{f(c) - f(x)}{c - x} $$

Because $\xi < c$ we know that $f’(\xi) < 0$ by assumption. Also $c > x \leftrightarrow c - x > 0$. Hence

$$ f(c) - f(x) < 0 \leftrightarrow f(c) < f(x) $$

That is, for every point $x$ to the left of $c$, $f(x)$ is bigger than $f(c)$.

(2) Given any $x \in I$ with $x > c$, by The Mean Value Theorem there exists some $\nu \in (c, x)$ such that

$$ f'(\nu) = \frac{f(x) - f(c)}{x - c} $$

Because $\nu > c$ we know that $f’(\nu) > 0$ by assumption. Also $x > c \leftrightarrow x - c > 0$. Hence

$$ f(x) - f(c) > 0 \leftrightarrow f(x) > f(x) $$

That is, for every point $x$ to the right of $c$, $f(x)$ is bigger than $f(c)$.

(3) By combining (1) and (2) we show that for all points on the interval $I$ that are not $c$, its value under $f$ is bigger than $f(c)$ thus we conclude that $f(c)$ is the abolute minimum of $f$ on the interval $I$.

Newton’s Method

Suppose we want to solve an equation $f(x)$, then we need to find the $x$-intercepts of $f$ (labelled $r$ in the following figure). Newton’s Method allows us to find an approximation of $r$ by following these steps:

1. We start by defining a first approximation $x_1$ (by guesssing, from a sketch of the graph of $f$, etc.)
2. Then, we find the tanget line of $f$ on $(x_1, f(x_1))$, that we will denote as $L$. The idea behind Newton’s Method is that as $L$ is close to the curve, then its $x$-intercept, $x_2$, should be close to the $x$-intercept of $f$. And so we use $x_2$ as the next approximation to $r$.
3. We keep repeting this process to obtain a sequence of approximations $x_1, x_2, \cdots$

Because $L$ is a line, finding its $x$-intercept is easy. Let’s find an expression that defines $x_2$ in terms of $x_1$. To do so we use the equation of a line:

$$ y - f(x_1) = f'(x_1)(x - x_1) $$

Where $f’(x_1)$ is the slope of $L$. Because we know that $(x_2, 0)$ is on $L$, then:

$$ 0 - f(x_1) = f'(x_1)(x_2 - x_1) \Leftrightarrow -f(x_1) = f'(x_1)x_2 - f'(x_1) x_1 $$$$ f'(x_1)x_2 = f'(x_1) x_1 - f(x_1) $$$$ x_2 = x_1 - \frac{f(x_1)}{f'(x_1)} $$

If the approximations $x_n$ become closer and closer to $r$ as $n$ becomes larger and larger, then we say the sequence converges to $r$, that is:

$$ \lim_{n \to \infty} x_n = r $$

However, in certain circumstances, the sequence may not converge. See the following figure, where you can see that the approximation $x_2$ is worse than $x_1$. This is likely to be the case when $f’(x_n)$ is close to zero, as the slope becomes smaller and the $x$-intercept for the linear approximation of $f$ moves far away.

Then Newton’s Method fails and a better approximation $x_1$ should be chosen.

Suppose we want to achieve a given accuracy of $k$ decimals, how do we know when to stop? The rule of thumb is to keep iterating over Newton’s Method until $x_n$ and $x_{n + 1}$ agree up to $k$ decimal places.

Basins of Attraction: Newton’s Fractals

Reference: Newton’s fractal (which Newton knew nothing about)

In simpler words: A basin of attraction is like a “territory” in the plane. Wherever you start inside it, you’ll be “pulled” to the same root.

On the previous image you can see that the complexity is introduced at the boundaries of each basin. Let us define a boundary: Given any set (like a basin), a boundary point is one where, for any infinitely small circle, some points on the circle are inside the set while some other points on the circle are outside the set.

So, why do these fractal boundaries appear? One property that these boundaries must satisfy is: for any starting point that ends up at this boundary, it is not decisively attracted to one root. Instead, infinitely close neighbors can be pulled to wildly different roots. This means that even the tiniest perturbation flips its destiny. That’s why the boundaries look chaotic — they are made up of points that can be “attracted” to any of the possible roots.

We can visualize this phenomenon by:

1. Picking a cluster of nearby starting points. For the phenomenon to take place the cluster of points shows be on the fractal boundaries.
2. Applying Newton’s method step by step.
3. Watching them scatter: some head to one root, others to another, depending on tiny differences in the start.

Antiderivatives

But is $F$ the only antiderivative for $f$? By the corollary of The Mean Value Function, given two functions $f$ and $g$, such that $f’(x) = g’(x)$, then

$$ f'(x) = g'(x) \leftrightarrow H(x) = f'(x) - g'(x) = 0 \leftrightarrow \\[5pt] H \text{ is a constant function, such that } H(x) = C, \text{ where } C \text{ is a constant} $$

So, we would say that the most general antiderivative for a function $f$ is defined as:

$$ F(x) + C $$

where $F$ is an antiderivative of $f$ on an interval $I$. Note that when solving integrals or differential equations, we will always add $+C$ to account for all possible solutions.

Antidifferentiation formulas

On the following table we list some popular antiderivatives.

What does it mean to solve a differential equation? A solution to a differential equation is a function for which the differential equation is satisfied. The general solution is the form of any solution to the differential equation. For instance, the general solution to

$$ \frac{\delta y}{\delta x} = e^x $$

is

$$ f(x) = e^x + C $$

Graphing Antiderivatives

Given the graph of $f$ we should be able to sketch the graph for the general antiderivate $F$. To sketch $F$ we follow the shape of $f$ and what it tells us:

1. Is $f$ increasing or decreasing based on the sign of $f$
2. If $f$ has any root, then $F$ has critical points on those roots
3. If $f$ has local maximums or local minimums, then $F$ has inflections points

In the following figures we show an example on how to graph $F$ given the graph of $f$ and $F(0) = 2$.

Note that the general form of the antiderivative is $F(x) + C$, that is, if we do not know any $(x, y)$ on $F$ there are infinitely possible graphs of $F$ given $f$, as $f$ only tells us the shape of $F$.

Linear Motion

Given the position function $s(t)$, we know that the following relationships hold:

1. The velocity function is defined as $v(t) = s’(t)$
2. The acceleration function is defined as $a(t) = v’(t)$

Therefore:

1. The velocity function is the antiderivative for the acceleration function.
2. The position function is the antiderivative for the velocity function.

Numerical Approximations

Calculus provides us with the powerful definite integral, $\int_{a}^{b} f(x) dx$, which geometrically represents the exact area under a curve.

While the Fundamental Theorem of Calculus (FTC) gives us an exact solution, it relies on being able to find an antiderivative $F(x)$ for the function $f(x)$. For many functions, especially those encountered in real-world data or complex models, finding an explicit antiderivative is impossible or extremely difficult.

This is where Riemann Sums, the Trapezoid Rule, and Simpson’s Rule come in. These are all methods of numerical approximation (or numerical integration), which means they calculate the definite integral using arithmetic, not by finding an antiderivative. They break the complex shape (the area under the curve) into simpler, measurable geometric shapes (rectangles, trapezoids, or parabolic segments) to get a result that is close to the true value.

Similarly, in Differential Equations (DEs), we seek the unknown function $y(x)$ that satisfies the diferential equation. As many DEs are too complex to solve analytically (with exact formulas), we use numerical methods for ODEs. This is the role of Euler’s Method. It’s a numerical technique that approximates the solution function y(x)

Riemann Sum

Riemann sums are a method for approximating the area under a curve (a definite integral) by dividing the area into a large number of simple shapes, usually rectangles, and summing their areas.

To do so we would divide the interval $[a,b]$ into n subintervals of equal width, $\Delta x = \frac{b -a }{n}$. In each subinterval, construct a rectangle whose height is determined by the function’s value at a specific point within that subinterval.

The area would be computed as:

$$ \text{Area} \approx \sum_{i=1}^n f(x_i^*)\Delta x $$

where $x_i^*$ is the chosen sample point (e.g., left endpoint, right endpoint, midpoint) in the $i$-th subinterval.

Trapezoid Rule

The Trapezoid Rule is an alternative to Riemann sums that usually provides a more accurate approximation for the definite integral. Instead of using rectangles (constant height), the area under the curve is approximated by dividing it into trapezoids. This means connecting the function values $f(x_i)$ and $f(x_{i + 1})$ with a straight line across each subinterval $\Delta x$.

Given that the area of a trapezoid with height $h$ and parallel sides $b_1$ and $b_2$ is

$$ \frac{1}{2} h(b_1 + b_2). $$

Here $h = \Delta x$ and the bases are $f(x_i)$ and $f(x_{i + 1})$. Thus the area under the curve would be approximated as:

$$ \int_{a}^{b} f(x) dx \approx \frac{\Delta x}{2} [f(x_0) + 2f(x_1) + 2f(x_2) + \cdots 2f(x_{n - 1}) + f(x_n)] $$

where $\Delta x = \frac{b - a}{n}$.

Simpson’s Rule

Simpson’s Rule offers an even higher degree of accuracy by approximating the curve with parabolas (second-degree polynomials) instead of straight lines or constants. Instead of straight lines, the curve is approximated by quadratic functions over pairs of subintervals.

Then the area under the curve would be approximated as:

$$ \int_{a}^{b} f(x) dx \approx \frac{\Delta x}{3} [f(x_0) + 4f(x_1) + 2f(x_2) + 4f(x_3) + \cdots + 4f(x_{n - 1}) + f(x_n)] $$

where $\Delta x = \frac{b - a}{n}$

Euler’s Method

To solve a differential equation means to find the unknown function $y(x)$ (or whatever the dependent variable is) that satisfies the equation. For example, for the differential equation

$$ \frac{dy}{dx} = 2x $$

The general solution is found by integrating $y(x)$

$$ y(x) = x^2 + C $$

If the initial condition is $y(0) = 5$, then the particular solution is

$$ y(x) = x^2 + 5 $$

Euler’s method is a numerical technique used to approximate the solution to a first-order ordinary differential equation (ODE) with a given initial condition. It’s often called the tangent line method.

We have an ODE of the form $\frac{dy}{dx} = f(x, y)$ and an initial point $(x_0, y_0)$, and we want to estimate the value of $y$ at subsequent $x$ values.

The idea is to use the DE to find the slope ($\frac{dy}{dx}$) at our current point. We then assume this slope remains constant over a small distance (the step size, $h$) and use the tangent line to predict the next point.

Thus we start at a point $(x_i, y_i)$, and the next point $(x_{i + 1}, y_{i + 1})$ is calculated as:

$$ x_{i + 1} = x_i + h $$$$ y_{i + 1} = y_i + h \cdot f(x_i, y_i) $$