Applications: Rate of Change, Monotonicity, Maxima and Minima

In this section, we will explore the practical applications of derivatives in understanding the behavior of functions. Derivatives, which represent the instantaneous rate of change, provide powerful tools to analyze how a function changes, whether it is increasing or decreasing, and where it reaches its highest or lowest points. This knowledge is crucial in various fields, from physics and engineering to economics and biology.

Rate of Change

The derivative of a function $f(x)$ with respect to $x$, denoted as $f'(x)$ or $\frac{dy}{dx}$, fundamentally represents the rate of change of the dependent variable $y$ with respect to the independent variable $x$. If we have a quantity $y$ that depends on another quantity $x$, then $\frac{dy}{dx}$ tells us how fast $y$ is changing as $x$ changes.

Example: Velocity and Acceleration

Consider the motion of a particle. If the position of the particle at time $t$ is given by a function $s(t)$, then the velocity $v(t)$ of the particle is the rate of change of its position with respect to time. Therefore, velocity is the first derivative of the position function: $v(t) = \frac{ds}{dt} = s'(t)$

Similarly, acceleration $a(t)$ is the rate of change of velocity with respect to time. Thus, acceleration is the first derivative of the velocity function, or the second derivative of the position function: $a(t) = \frac{dv}{dt} = v'(t) = \frac{d^2s}{dt^2} = s''(t)$

Example: Area of a Circle

Let $A$ be the area of a circle and $r$ be its radius. The formula for the area is $A = \pi r^2$. If the radius is changing with time, say $r(t)$, then the area is also changing with time, $A(t) = \pi [r(t)]^2$. To find the rate at which the area is changing with respect to time, we differentiate $A$ with respect to $t$: $\frac{dA}{dt} = \frac{d}{dt}(\pi r^2) = \pi \cdot 2r \cdot \frac{dr}{dt}$ This equation tells us that the rate of change of the area depends on the current radius and the rate at which the radius is changing.

Example: Population Growth

Suppose the population of a city at time $t$ is given by $P(t)$. The rate of population growth is then given by the derivative $\frac{dP}{dt}$. If a model suggests that the population grows at a rate proportional to its current size, this can be expressed as a differential equation: $\frac{dP}{dt} = kP$ where $k$ is the constant of proportionality. The solution to this equation reveals exponential growth.

Monotonicity of a Function (Increasing and Decreasing Functions)

The derivative of a function provides a direct way to determine whether the function is increasing or decreasing over an interval.

• A function $f(x)$ is said to be increasing on an interval if for any two points $x_1$ and $x_2$ in the interval such that $x_1 < x_2$, we have $f(x_1) < f(x_2)$.
• A function $f(x)$ is said to be decreasing on an interval if for any two points $x_1$ and $x_2$ in the interval such that $x_1 < x_2$, we have $f(x_1) > f(x_2)$.

First Derivative Test for Monotonicity

Let $f(x)$ be a differentiable function on an open interval $(a, b)$.

• If $f'(x) > 0$ for all $x$ in $(a, b)$, then $f(x)$ is strictly increasing on $(a, b)$.
• If $f'(x) < 0$ for all $x$ in $(a, b)$, then $f(x)$ is strictly decreasing on $(a, b)$.
• If $f'(x) = 0$ for all $x$ in $(a, b)$, then $f(x)$ is constant on $(a, b)$.

If $f'(x) \ge 0$ on $(a, b)$, $f(x)$ is non-decreasing. If $f'(x) \le 0$ on $(a, b)$, $f(x)$ is non-increasing.

To find the intervals where a function is increasing or decreasing, we first find the critical points, which are the points where $f'(x) = 0$ or $f'(x)$ is undefined. These critical points divide the domain of the function into intervals. We then test the sign of $f'(x)$ in each of these intervals.

Example: Intervals of Monotonicity for $f(x) = x^3 - 3x^2 + 5$

Step 1: Find the derivative of the function. $f'(x) = \frac{d}{dx}(x^3 - 3x^2 + 5) = 3x^2 - 6x$

Step 2: Find the critical points by setting $f'(x) = 0$. $3x^2 - 6x = 0$ $3x(x - 2) = 0$ The critical points are $x = 0$ and $x = 2$.

Step 3: These critical points divide the number line into three intervals: $(-\infty, 0)$, $(0, 2)$, and $(2, \infty)$.

Step 4: Test the sign of $f'(x)$ in each interval.

• Interval $(-\infty, 0)$: Choose a test value, say $x = -1$. $f'(-1) = 3(-1)^2 - 6(-1) = 3(1) + 6 = 9$. Since $f'(-1) > 0$, the function is increasing on $(-\infty, 0)$.
• Interval $(0, 2)$: Choose a test value, say $x = 1$. $f'(1) = 3(1)^2 - 6(1) = 3 - 6 = -3$. Since $f'(1) < 0$, the function is decreasing on $(0, 2)$.
• Interval $(2, \infty)$: Choose a test value, say $x = 3$. $f'(3) = 3(3)^2 - 6(3) = 3(9) - 18 = 27 - 18 = 9$. Since $f'(3) > 0$, the function is increasing on $(2, \infty)$.

Conclusion: $f(x)$ is increasing on $(-\infty, 0]$ and $[2, \infty)$, and decreasing on $[0, 2]$.

Maxima and Minima (Extrema)

Maxima and minima refer to the highest and lowest values of a function, either over its entire domain (absolute extrema) or over a specific interval (local or relative extrema). These points are often critical for optimization problems.

Definitions

• A function $f(x)$ has a local maximum at $x=c$ if $f(c) \ge f(x)$ for all $x$ in some open interval containing $c$.
• A function $f(x)$ has a local minimum at $x=c$ if $f(c) \le f(x)$ for all $x$ in some open interval containing $c$.
• A function $f(x)$ has an absolute maximum at $x=c$ if $f(c) \ge f(x)$ for all $x$ in the domain of $f$.
• A function $f(x)$ has an absolute minimum at $x=c$ if $f(c) \le f(x)$ for all $x$ in the domain of $f$.

Critical Points and Extrema

Fermat's Theorem states that if $f(x)$ has a local extremum at $x=c$, and if $f'(c)$ exists, then $f'(c) = 0$. This means that local extrema can only occur at critical points (where $f'(x) = 0$ or $f'(x)$ is undefined). However, not all critical points correspond to local extrema (e.g., points of inflection where $f'(x)=0$ but the function does not change direction).

First Derivative Test for Local Extrema

Let $c$ be a critical point of a continuous function $f(x)$.

• If $f'(x)$ changes from positive to negative at $c$ (i.e., $f$ is increasing to the left of $c$ and decreasing to the right), then $f$ has a local maximum at $c$.
• If $f'(x)$ changes from negative to positive at $c$ (i.e., $f$ is decreasing to the left of $c$ and increasing to the right), then $f$ has a local minimum at $c$.
• If $f'(x)$ does not change sign at $c$ (i.e., it is positive on both sides or negative on both sides), then $f$ has neither a local maximum nor a local minimum at $c$.

Example: Local Extrema for $f(x) = x^3 - 3x^2 + 5$

From the previous example, we found the critical points are $x=0$ and $x=2$.

• At $x=0$: $f'(x)$ changes from positive (increasing) to negative (decreasing). Therefore, $f(x)$ has a local maximum at $x=0$. The value is $f(0) = 0^3 - 3(0)^2 + 5 = 5$.
• At $x=2$: $f'(x)$ changes from negative (decreasing) to positive (increasing). Therefore, $f(x)$ has a local minimum at $x=2$. The value is $f(2) = 2^3 - 3(2)^2 + 5 = 8 - 12 + 5 = 1$.

Second Derivative Test for Local Extrema

This test can be used when $f'(c) = 0$. Let $f(x)$ be a twice-differentiable function and $c$ be a critical point such that $f'(c) = 0$.

• If $f''(c) < 0$, then $f$ has a local maximum at $x=c$. (The curve is concave down).
• If $f''(c) > 0$, then $f$ has a local minimum at $x=c$. (The curve is concave up).
• If $f''(c) = 0$, the test is inconclusive. We must use the First Derivative Test.

Example: Local Extrema using Second Derivative Test for $f(x) = x^3 - 3x^2 + 5$

We have $f'(x) = 3x^2 - 6x$. Step 1: Find the second derivative. $f''(x) = \frac{d}{dx}(3x^2 - 6x) = 6x - 6$

Step 2: Evaluate $f''(x)$ at the critical points $x=0$ and $x=2$.

• At $x=0$: $f''(0) = 6(0) - 6 = -6$. Since $f''(0) < 0$, $f$ has a local maximum at $x=0$.
• At $x=2$: $f''(2) = 6(2) - 6 = 12 - 6 = 6$. Since $f''(2) > 0$, $f$ has a local minimum at $x=2$.

This matches the result from the First Derivative Test.

Finding Absolute Extrema on a Closed Interval $[a, b]$

For a continuous function $f(x)$ on a closed interval $[a, b]$, the absolute maximum and absolute minimum values occur either at the critical points within the interval $(a, b)$ or at the endpoints $a$ and $b$.

Procedure:

1. Find all critical points of $f$ in the open interval $(a, b)$.
2. Evaluate $f$ at these critical points.
3. Evaluate $f$ at the endpoints $a$ and $b$.
4. The largest value among those found in steps 2 and 3 is the absolute maximum, and the smallest value is the absolute minimum.

Example: Absolute Extrema for $f(x) = x^3 - 3x^2 + 5$ on $[0, 3]$

The interval is $[0, 3]$. Critical points are $x=0$ and $x=2$. Both are within or at the boundary of the interval $[0, 3]$.

Step 1 & 2: Evaluate $f$ at critical points within $(0, 3)$. The only critical point in $(0, 3)$ is $x=2$. $f(2) = 1$. (Note: $x=0$ is an endpoint, so we will evaluate it in the next step).

Step 3: Evaluate $f$ at the endpoints $a=0$ and $b=3$. $f(0) = 0^3 - 3(0)^2 + 5 = 5$ $f(3) = 3^3 - 3(3)^2 + 5 = 27 - 27 + 5 = 5$

Step 4: Compare the values: $f(2)=1$, $f(0)=5$, $f(3)=5$.

Conclusion: The absolute maximum value is 5, which occurs at $x=0$ and $x=3$. The absolute minimum value is 1, which occurs at $x=2$.

Applications in Optimization

Derivatives are fundamental to solving optimization problems. These problems aim to find the best possible outcome (maximum profit, minimum cost, maximum area, etc.) under given constraints.

Example: Maximizing Area of a Rectangular Field

A farmer wants to fence a rectangular field bordering a straight river. He has 2400 meters of fencing. What are the dimensions of the field that has the largest area?

Let the dimensions of the rectangle be length $l$ and width $w$. The side along the river does not need fencing. The total fencing available is 2400 m. So, $l + 2w = 2400$. The area of the field is $A = l \times w$.

We need to maximize $A$. First, express $A$ in terms of a single variable. From the perimeter equation, $l = 2400 - 2w$. Substitute this into the area formula: $A(w) = (2400 - 2w)w = 2400w - 2w^2$.

Now, we find the critical points of $A(w)$ by taking the derivative with respect to $w$ and setting it to zero. $A'(w) = \frac{d}{dw}(2400w - 2w^2) = 2400 - 4w$. Set $A'(w) = 0$: $2400 - 4w = 0$ $4w = 2400$ $w = 600$ meters.

Now we need to confirm this gives a maximum. We can use the second derivative test. $A''(w) = \frac{d}{dw}(2400 - 4w) = -4$. Since $A''(600) = -4 < 0$, the area is maximized when $w=600$.

Find the corresponding length $l$: $l = 2400 - 2w = 2400 - 2(600) = 2400 - 1200 = 1200$ meters.

The dimensions that maximize the area are 1200 meters by 600 meters. The maximum area is $A = 1200 \times 600 = 720,000$ square meters.

Example: Minimizing Surface Area of a Cylindrical Can

A company needs to design a cylindrical can with a volume of 1000 cubic centimeters. What dimensions (radius $r$ and height $h$) minimize the surface area of the can?

The volume of a cylinder is $V = \pi r^2 h$. We are given $V = 1000$, so $\pi r^2 h = 1000$. The surface area of a closed cylinder is $A = 2\pi r^2 + 2\pi rh$.

Express $A$ in terms of a single variable. From the volume equation, $h = \frac{1000}{\pi r^2}$. Substitute this into the surface area formula: $A(r) = 2\pi r^2 + 2\pi r \left(\frac{1000}{\pi r^2}\right)$ $A(r) = 2\pi r^2 + \frac{2000}{r}$.

Find the critical points of $A(r)$. $A'(r) = \frac{d}{dr}\left(2\pi r^2 + \frac{2000}{r}\right) = 4\pi r - \frac{2000}{r^2}$. Set $A'(r) = 0$: $4\pi r - \frac{2000}{r^2} = 0$ $4\pi r = \frac{2000}{r^2}$ $4\pi r^3 = 2000$ $r^3 = \frac{2000}{4\pi} = \frac{500}{\pi}$ $r = \sqrt[3]{\frac{500}{\pi}}$ cm.

Use the second derivative test to confirm it's a minimum. $A''(r) = \frac{d}{dr}\left(4\pi r - 2000r^{-2}\right) = 4\pi - 2000(-2)r^{-3} = 4\pi + \frac{4000}{r^3}$. Since $r > 0$, $r^3 > 0$, so $A''(r) = 4\pi + \frac{4000}{r^3}$ is always positive. Thus, this value of $r$ minimizes the surface area.

Calculate the corresponding height $h$: $h = \frac{1000}{\pi r^2}$. We know $r^3 = \frac{500}{\pi}$, so $\pi r^3 = 500$. $h = \frac{1000}{\pi r^2} = \frac{1000 r}{\pi r^3} = \frac{1000 r}{500} = 2r$. So, the height should be twice the radius for minimum surface area.

The dimensions are $r = \sqrt[3]{\frac{500}{\pi}}$ cm and $h = 2\sqrt[3]{\frac{500}{\pi}}$ cm.