# UNIT 4: CONTEXTUAL APPLICATIONS OF DIFFERENTIATION

# 1. INTRODUCTION

Forget the abstract formulas for a second. Unit 4 is where calculus actually "touches grass." We’re moving from the "how" of math to the "so what?"

Whether you're tracking how fast a viral TikTok trend is spreading, calculating the trajectory of a SpaceX booster, or figuring out if a company’s profit is about to peak, you're using the derivative. It's the ultimate tool for modeling change in real-time. By the end of this unit, you'll be able to solve "Related Rates" puzzles, use "Local Linearity" to predict the future without a calculator, and use "L’Hôpital’s Rule" to fix math problems that look broken. Let’s get into it.

# 2. ALL KEY CONCEPTS, TERMS, AND PRINCIPLES

# Foundational Terms

• Rate of Change: The speed at which a variable changes over a specific interval.
• Instantaneous Rate of Change: The rate of change at a specific moment in time, represented by the derivative $f'(c)$.
• Average Rate of Change: The slope of the secant line over an interval $[a, b]$, calculated as $\frac{f(b)-f(a)}{b-a}$.
• Contextual Units: If $y = f(x)$, where $y$ is measured in units $U_y$ and $x$ in units $U_x$, the units of $f'(x)$ are $\frac{U_y}{U_x}$.

# Kinematics (Motion) Concepts

• Position Function $s(t)$: The location of an object at time $t$.
• Velocity $v(t)$: The first derivative of position, $v(t) = s'(t)$. It indicates direction and speed.
• Acceleration $a(t)$: The derivative of velocity, $a(t) = v'(t) = s''(t)$.
• Speed: The absolute value of velocity, $|v(t)|$.
• Displacement: The net change in position, $s(t_2) - s(t_1)$.
• Total Distance Traveled: The sum of all distances moved, regardless of direction.

# Related Rates Principles

• Implicit Differentiation (Time-based): Differentiating both sides of an equation with respect to time ($t$), treating all geometric variables as functions of $t$.
• Chain Rule in Context: Using $\frac{dy}{dt} = \frac{dy}{dx} \cdot \frac{dx}{dt}$ to relate different rates.

# Approximation and Limits

• Linearization (Tangent Line Approximation): Using the equation of a tangent line at a known point $(a, f(a))$ to approximate the value of $f(x)$ for values of $x$ near $a$.
• Underestimate vs. Overestimate: Determining the accuracy of an approximation based on the concavity ($f''(x)$) of the function.
• L’Hôpital’s Rule: A method for evaluating limits of the indeterminate forms $\frac{0}{0}$ or $\frac{\infty}{\infty}$ by taking the derivative of the numerator and denominator separately.

# 3. IN-DEPTH EXPLANATION

# 3.1 Interpreting the Derivative in Context

In AP Calculus, you are often asked to "interpret the meaning of $f'(c)$ in the context of the problem." This requires a three-part answer: Value, Units, and Meaning.

If $W(t)$ represents the amount of water in a tank (gallons) at time $t$ (minutes), then $W'(5) = -3$ means: "At exactly $t=5$ minutes, the amount of water in the tank is decreasing at a rate of 3 gallons per minute."

# 3.2 Straight-Line Motion (Kinematics)

Motion along a horizontal or vertical line is a staple of Unit 4.

1. Direction:
   • If $v(t) > 0$, the particle moves right/up.
   • If $v(t) < 0$, the particle moves left/down.
   • If $v(t) = 0$, the particle is momentarily at rest.
2. Speeding Up vs. Slowing Down:
   • Speeding Up: $v(t)$ and $a(t)$ have the same sign (both positive or both negative).
   • Slowing Down: $v(t)$ and $a(t)$ have opposite signs.

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# 3.3 Related Rates

Related Rates problems involve finding the rate at which one quantity changes by relating it to other quantities whose rates of change are known.
The Procedure:

1. Identify Variables: List "Given" rates (e.g., $\frac{dr}{dt}$) and the "To Find" rate (e.g., $\frac{dV}{dt}$).
2. Write a Static Equation: Find a formula relating the variables (e.g., $V = \frac{4}{3}\pi r^3$).
3. Differentiate with respect to $t$: Remember to use the Chain Rule. Every variable $u$ becomes $\frac{du}{dt}$ when differentiated.
   • Example: $\frac{d}{dt}(V) = \frac{d}{dt}(\frac{4}{3}\pi r^3) \implies \frac{dV}{dt} = 4\pi r^2 \frac{dr}{dt}$.
4. Substitute and Solve: Plug in the specific values "at that moment." Never substitute constant values for variables before differentiating unless the variable is constant throughout the entire process.

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# 3.4 Linearization (Local Linearity)

Since a differentiable function looks like a straight line when zoomed in closely, we can use the tangent line $L(x)$ to approximate function values near the point of tangency $x=a$.
The formula is derived from the point-slope form $y - y_1 = m(x - x_1)$:
$$L(x) = f(a) + f'(a)(x - a)$$

Concavity and Error:

• If $f''(x) > 0$ (concave up), the tangent line lies below the curve; therefore, the linear approximation is an underestimate.
• If $f''(x) < 0$ (concave down), the tangent line lies above the curve; therefore, the linear approximation is an overestimate.

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# 3.5 L’Hôpital’s Rule

When evaluating $\lim_{x \to c} \frac{f(x)}{g(x)}$, if direct substitution results in $\frac{0}{0}$ or $\frac{\pm \infty}{\pm \infty}$, and $f$ and $g$ are differentiable near $c$:
$$\lim_{x \to c} \frac{f(x)}{g(x)} = \lim_{x \to c} \frac{f'(x)}{g'(x)}$$
Note: You must state that the individual limits of the numerator and denominator go to $0$ or $\infty$ before applying the rule. Do not use the quotient rule here; you are differentiating the top and bottom independently.

# 4. EXAMPLES

# Example 1: Interpreting Units

Let $C(p)$ be the cost in dollars to produce $p$ pounds of coffee. Interpret $C'(50) = 2.5$.
Solution: At the moment 50 pounds of coffee are produced, the cost is increasing at a rate of 2.5 dollars per pound.

# Example 2: Particle Motion (Direction)

A particle moves along the x-axis with position $x(t) = t^3 - 6t^2 + 9t$. When is the particle moving to the left?
Solution:
$v(t) = x'(t) = 3t^2 - 12t + 9 = 3(t^2 - 4t + 3) = 3(t-1)(t-3)$.
The particle moves left when $v(t) < 0$. By testing intervals, $v(t) < 0$ for $1 < t < 3$.

# Example 3: Speeding Up or Slowing Down

Using $x(t)$ from Example 2, is the particle speeding up or slowing down at $t = 1.5$?
Solution:
$v(1.5) = 3(1.5-1)(1.5-3) = 3(0.5)(-1.5) = -2.25$.
$a(t) = v'(t) = 6t - 12$.
$a(1.5) = 6(1.5) - 12 = 9 - 12 = -3$.
Since $v(1.5)$ and $a(1.5)$ are both negative, the particle is speeding up.

# Example 4: Related Rates (Expanding Circle)

The radius of a circular oil spill is increasing at $2 \text{ m/s}$. How fast is the area increasing when $r = 10 \text{ m}$?
Solution:

1. Given: $\frac{dr}{dt} = 2$. Find: $\frac{dA}{dt}$ when $r=10$.
2. Equation: $A = \pi r^2$.
3. Differentiate: $\frac{dA}{dt} = 2\pi r \frac{dr}{dt}$.
4. Solve: $\frac{dA}{dt} = 2\pi(10)(2) = 40\pi \text{ m}^2\text{/s}$.

# Example 5: Related Rates (Falling Ladder)

A $13\text{-ft}$ ladder leans against a wall. The bottom slides away at $3\text{ ft/s}$. How fast is the top sliding down when the bottom is $5\text{ ft}$ from the wall?
Solution:

1. Equation: $x^2 + y^2 = 13^2$.
2. Differentiate: $2x\frac{dx}{dt} + 2y\frac{dy}{dt} = 0$.
3. At $x=5$, $5^2 + y^2 = 169 \implies y = 12$.
4. Substitute: $2(5)(3) + 2(12)\frac{dy}{dt} = 0 \implies 30 + 24\frac{dy}{dt} = 0 \implies \frac{dy}{dt} = -\frac{30}{24} = -1.25\text{ ft/s}$.
   The top is sliding down at $1.25\text{ ft/s}$.

# Example 6: Linearization

Approximate $\sqrt{26}$ using a tangent line.
Solution:

1. Let $f(x) = \sqrt{x}$ and $a = 25$.
2. $f(25) = 5$.
3. $f'(x) = \frac{1}{2\sqrt{x}} \implies f'(25) = \frac{1}{10} = 0.1$.
4. $L(x) = 5 + 0.1(x - 25)$.
5. $L(26) = 5 + 0.1(26 - 25) = 5.1$.

# Example 7: L’Hôpital’s Rule (Basic)

Evaluate $\lim_{x \to 0} \frac{\sin x}{x}$.
Solution:
Since $\lim_{x \to 0} \sin x = 0$ and $\lim_{x \to 0} x = 0$, we use L’Hôpital’s:
$\lim_{x \to 0} \frac{\cos x}{1} = \cos(0) = 1$.

# Example 8: L’Hôpital’s Rule (Successive)

Evaluate $\lim_{x \to 0} \frac{1 - \cos x}{x^2}$.
Solution:
Form $\frac{0}{0}$. First derivative: $\lim_{x \to 0} \frac{\sin x}{2x}$.
Still form $\frac{0}{0}$. Second derivative: $\lim_{x \to 0} \frac{\cos x}{2} = \frac{1}{2}$.

# Example 9: Related Rates (Conical Tank)

Water is poured into a conical tank (height $10\text{ m}$, radius $5\text{ m}$) at $2\text{ m}^3\text{/min}$. How fast is the water level rising when $h = 4\text{ m}$?
Solution:

1. $V = \frac{1}{3}\pi r^2 h$.
2. By similar triangles: $\frac{r}{h} = \frac{5}{10} \implies r = \frac{h}{2}$.
3. Substitute: $V = \frac{1}{3}\pi (\frac{h}{2})^2 h = \frac{\pi}{12} h^3$.
4. Differentiate: $\frac{dV}{dt} = \frac{\pi}{4} h^2 \frac{dh}{dt}$.
5. $2 = \frac{\pi}{4} (4)^2 \frac{dh}{dt} \implies 2 = 4\pi \frac{dh}{dt} \implies \frac{dh}{dt} = \frac{1}{2\pi}\text{ m/min}$.

# Example 10: Linearization Over/Underestimate

If $f(x) = e^x$, is the linear approximation at $a=0$ an underestimate or overestimate?
Solution:
$f'(x) = e^x$, $f''(x) = e^x$. Since $e^x > 0$ for all $x$, the function is concave up. Therefore, the tangent line is below the curve, making it an underestimate.

# Example 11: Particle Motion (Total Distance)

Find total distance for $v(t) = t^2 - 4$ on $[0, 3]$.
Solution:
$v(t) = 0$ at $t = 2$.
$| \int_0^2 (t^2-4) dt | + | \int_2^3 (t^2-4) dt |$.
$= | [\frac{1}{3}t^3 - 4t]_0^2 | + | [\frac{1}{3}t^3 - 4t]_2^3 |$.
$= | \frac{8}{3} - 8 | + | (9-12) - (\frac{8}{3}-8) | = | -\frac{16}{3} | + | -3 + \frac{16}{3} | = \frac{16}{3} + \frac{7}{3} = \frac{23}{3}$.

# Example 12: Rate of Change of Volume

A balloon is inflating such that $V(t) = 100\sqrt{t}$. What is the rate of change of volume at $t=4$?
Solution:
$V'(t) = 100 \cdot \frac{1}{2\sqrt{t}} = \frac{50}{\sqrt{t}}$.
$V'(4) = \frac{50}{\sqrt{4}} = 25$.

# Example 13: L'Hôpital with Infinity

Evaluate $\lim_{x \infty} \frac{\ln x}{x^2}$.
Solution:
Form $\frac{\infty}{\infty}$.
$\lim_{x \infty} \frac{1/x}{2x} = \lim_{x \infty} \frac{1}{2x^2} = 0$.

# Example 14: Shadow Problem (Related Rates)

A $6\text{-ft}$ tall man walks away from a $15\text{-ft}$ lamp post at $5\text{ ft/s}$. How fast is his shadow length $s$ changing?
Solution:
By similar triangles: $\frac{s}{6} = \frac{x+s}{15}$ where $x$ is distance from pole.
$15s = 6x + 6s \implies 9s = 6x \implies s = \frac{2}{3}x$.
$\frac{ds}{dt} = \frac{2}{3} \frac{dx}{dt} = \frac{2}{3}(5) = \frac{10}{3}\text{ ft/s}$.

# Example 15: Tangent Line of a Parametric Curve (BC Only)

Given $x(t) = t^2$, $y(t) = t^3 - 3t$, find the slope of the tangent at $t=2$.
Solution:
$\frac{dy}{dx} = \frac{y'(t)}{x'(t)} = \frac{3t^2 - 3}{2t}$.
At $t=2$, $\frac{dy}{dx} = \frac{3(4)-3}{2(2)} = \frac{9}{4}$.

# Example 16: Differentiation of Inverse Functions (Application)

If $g$ is the inverse of $f$, find $g'(5)$ where $f(2)=5$ and $f'(2)=3$.
Solution:
$g'(5) = \frac{1}{f'(g(5))} = \frac{1}{f'(2)} = \frac{1}{3}$.

# Example 17: Implicit Related Rates (Sphere)

If $\frac{dV}{dt}$ is proportional to the surface area $S$, show $\frac{dr}{dt}$ is constant.
Solution:
$V = \frac{4}{3}\pi r^3 \implies \frac{dV}{dt} = 4\pi r^2 \frac{dr}{dt}$.
$S = 4\pi r^2$.
Given $\frac{dV}{dt} = kS \implies 4\pi r^2 \frac{dr}{dt} = k(4\pi r^2)$.
Therefore, $\frac{dr}{dt} = k$ (constant).

# Example 18: Non-existence of L’Hôpital’s

Evaluate $\lim_{x \to 0} \frac{e^x}{x}$.
Solution:
Substitution gives $\frac{1}{0}$, which is $\infty$. Do not use L’Hôpital’s; it is not an indeterminate form.

# Example 19: Population Growth

$P(t) = \frac{1000}{1 + 9e^{-0.5t}}$. Find the rate of change at $t=0$.
Solution:
$P'(t) = -1000(1+9e^{-0.5t})^{-2} \cdot (9e^{-0.5t} \cdot -0.5)$.
$P'(0) = -1000(10)^{-2} \cdot (-4.5) = -10 \cdot -4.5 = 45\text{ individuals/unit time}$.

# Example 20: Related Rates (Theta)

An observer 100m from a rocket launch site watches a rocket rise vertically. When the rocket is 100m high, its velocity is 20m/s. How fast is the angle of elevation changing?
Solution:
$\tan\theta = \frac{y}{100} \implies \sec^2\theta \frac{d\theta}{dt} = \frac{1}{100} \frac{dy}{dt}$.
When $y=100$, $\theta = 45^\circ$, $\sec^2(45^\circ) = 2$.
$2 \cdot \frac{d\theta}{dt} = \frac{1}{100}(20) \implies \frac{d\theta}{dt} = 0.1 \text{ rad/s}$.

# Example 21: Linearization for Trig

Approximate $\sin(0.1)$.
Solution:
$f(x) = \sin x$, $a=0$.
$f(0) = 0, f'(0) = \cos(0) = 1$.
$L(x) = 0 + 1(x-0) = x$.
$L(0.1) = 0.1$.

# Example 22: Identifying "Slowing Down" on a Graph

If the velocity graph is above the x-axis and decreasing, is the particle slowing down?
Solution:
$v > 0$ (above axis) and $v' = a < 0$ (decreasing). Opposite signs mean the particle is slowing down.

# Example 23: L'Hôpital with Transcendental Functions

$\lim_{x \to 1} \frac{\ln x}{x^2 - 1}$.
Solution:
$\frac{0}{0} \implies \lim_{x \to 1} \frac{1/x}{2x} = \frac{1}{2}$.

# Example 24: Related Rates (Area of Triangle)

A right triangle has legs $a$ and $b$. $a$ increases at $2\text{ cm/s}$ and $b$ decreases at $3\text{ cm/s}$. Rate of change of Area when $a=5, b=10$.
Solution:
$A = 0.5ab$.
$\frac{dA}{dt} = 0.5(a\frac{db}{dt} + b\frac{da}{dt})$.
$\frac{dA}{dt} = 0.5(5(-3) + 10(2)) = 0.5(-15 + 20) = 2.5\text{ cm}^2\text{/s}$.

# Example 25: Second Derivative and Motion

If $a(t) > 0$ for all $t$, can the particle's speed be decreasing?
Solution:
Yes, if $v(t) < 0$. If velocity is negative and acceleration is positive, the object is slowing down.

# Example 26: L'Hôpital form $\infty - \infty$

Evaluate $\lim_{x \to 1} (\frac{1}{\ln x} - \frac{1}{x-1})$.
Solution:
Combine into one fraction: $\lim_{x \to 1} \frac{x-1-\ln x}{(x-1)\ln x}$ (Form $\frac{0}{0}$).
Apply L'Hôpital: $\lim_{x \to 1} \frac{1 - 1/x}{\ln x + (x-1)/x} = \lim_{x \to 1} \frac{x-1}{x\ln x + x-1}$ (Form $\frac{0}{0}$).
Apply again: $\lim_{x \to 1} \frac{1}{\ln x + 1 + 1} = \frac{1}{2}$.

# Example 27: Related Rates (Cylinder)

A cylinder's radius stays $4\text{ cm}$ while its height increases at $2\text{ cm/s}$. Find $\frac{dV}{dt}$.
Solution:
$V = \pi r^2 h = 16\pi h$.
$\frac{dV}{dt} = 16\pi \frac{dh}{dt} = 16\pi(2) = 32\pi \text{ cm}^3\text{/s}$.

# Example 28: Linearization of $1/x$

Approximate $1/2.1$ using $a=2$.
Solution:
$f(x) = x^{-1}, f'(x) = -x^{-2}$.
$f(2) = 0.5, f'(2) = -0.25$.
$L(2.1) = 0.5 - 0.25(2.1 - 2) = 0.5 - 0.025 = 0.475$.

# Example 29: Vertical Motion

A ball is thrown up with $s(t) = -16t^2 + 48t + 160$. Max height occurs when?
Solution:
$v(t) = -32t + 48 = 0 \implies t = 1.5\text{ s}$.

# Example 30: Using Table Data for Rates

# Example 31: L'Hôpital for $0 \cdot \infty$

Evaluate $\lim_{x \to 0^+} x \ln x$.
Solution:
Rewrite: $\lim_{x \to 0^+} \frac{\ln x}{1/x}$ (Form $\frac{-\infty}{\infty}$).
$\lim_{x \to 0^+} \frac{1/x}{-1/x^2} = \lim_{x \to 0^+} (-x) = 0$.

# Example 32: Acceleration sign

If $v(t) = \cos(t)$, find $a(\pi)$.
Solution:
$a(t) = -\sin(t)$.
$a(\pi) = -\sin(\pi) = 0$.

# Example 33: Error Analysis

If $f(x) = \sqrt[3]{x}$, is $L(x)$ at $a=8$ an over or underestimate?
Solution:
$f'(x) = \frac{1}{3}x^{-2/3}$, $f''(x) = -\frac{2}{9}x^{-5/3}$.
At $x=8$, $f''(8) < 0$. Concave down means overestimate.

# Example 34: L'Hôpital BC Form ($1^\infty$)

Evaluate $\lim_{x \to \infty} (1 + \frac{1}{x})^x$.
Solution:
Let $y = (1 + 1/x)^x \implies \ln y = x \ln(1 + 1/x)$.
$\lim_{x \to \infty} \frac{\ln(1+1/x)}{1/x} = \frac{0}{0}$.
By L'Hôpital: $\lim_{x \to \infty} \frac{\frac{1}{1+1/x} \cdot (-1/x^2)}{-1/x^2} = \lim_{x \to \infty} \frac{1}{1+1/x} = 1$.
Since $\ln y \to 1$, $y \to e^1 = e$.

# Example 35: Related Rates (Cube)

Side of a cube increases at $1\text{ cm/s}$. Rate of surface area increase when $s=3$.
Solution:
$S = 6s^2 \implies \frac{dS}{dt} = 12s \frac{ds}{dt}$.
$\frac{dS}{dt} = 12(3)(1) = 36 \text{ cm}^2\text{/s}$.