Related Rates

Related Rates: When Things Change Together 🎈

The Big Idea

Imagine you’re blowing up a balloon. As you blow air in, the balloon gets bigger. But here’s the cool part: the balloon’s width, height, and even how much air is inside are all changing at the same time!

Related Rates is like being a detective who figures out: “If I know how fast one thing is changing, how fast is the other thing changing?”

🎯 What Are Related Rates?

Think of a seesaw. When one side goes up, the other side goes down. They’re connected—they relate to each other.

In math, related rates means:

• Two or more things are changing over time
• They’re connected by an equation
• If we know how fast ONE changes, we can find how fast the OTHER changes

Real Life Examples

🔑 The Secret Formula

Here’s the magic trick:

Take the derivative of BOTH SIDES with respect to time (t)

When we see related rates, we always ask:

1. What equation connects the things?
2. What’s changing?
3. Take the derivative with respect to time!

🪜 The Sliding Ladder Problem

This is THE classic related rates problem!

The Story

Picture this: A 10-foot ladder is leaning against a wall. Suddenly, the bottom starts sliding away from the wall at 2 feet per second.

Question: How fast is the top of the ladder sliding DOWN the wall?

Setting It Up

graph TD
    A["Wall - vertical"] --> B["Ladder touches wall at height y"]
    B --> C["Ladder = 10 feet long"]
    C --> D["Bottom is x feet from wall"]
    D --> E["Pythagorean theorem connects them!"]

The ladder, wall, and ground form a right triangle!

The equation: $x^2 + y^2 = 10^2 = 100$

Finding the Answer

Step 1: Take derivative of both sides with respect to time

$\frac{d}{dt}(x^2) + \frac{d}{dt}(y^2) = \frac{d}{dt}(100)$

$2x \cdot \frac{dx}{dt} + 2y \cdot \frac{dy}{dt} = 0$

Step 2: Plug in what we know

• Bottom slides at 2 ft/sec → $\frac{dx}{dt} = 2$
• Say the bottom is 6 feet away → $x = 6$
• Using $x^2 + y^2 = 100$: $36 + y^2 = 100$ → $y = 8$

Step 3: Solve! $2(6)(2) + 2(8) \cdot \frac{dy}{dt} = 0$ $24 + 16 \cdot \frac{dy}{dt} = 0$ $\frac{dy}{dt} = -\frac{24}{16} = -1.5 \text{ ft/sec}$

The Answer: The top slides DOWN at 1.5 feet per second!

💡 The negative sign means y is decreasing (sliding down)

🎈 The Expanding Balloon

The Story

You’re inflating a spherical balloon at a rate of 100 cubic centimeters per second. How fast is the radius growing when the radius is 5 cm?

The Setup

What we know:

• Volume of sphere: $V = \frac{4}{3}\pi r^3$
• Air going in: $\frac{dV}{dt} = 100$ cm³/sec
• Current radius: $r = 5$ cm
• Find: $\frac{dr}{dt}$ = ?

The Solution

Step 1: Differentiate with respect to time

$\frac{dV}{dt} = \frac{4}{3}\pi \cdot 3r^2 \cdot \frac{dr}{dt}$

$\frac{dV}{dt} = 4\pi r^2 \cdot \frac{dr}{dt}$

Step 2: Plug in values

$100 = 4\pi (5)^2 \cdot \frac{dr}{dt}$

$100 = 100\pi \cdot \frac{dr}{dt}$

Step 3: Solve

$\frac{dr}{dt} = \frac{100}{100\pi} = \frac{1}{\pi} \approx 0.318 \text{ cm/sec}$

Answer: The radius grows at about 0.32 cm per second!

🌊 The Filling Cone (Water Tank)

The Story

Water pours into a cone-shaped tank (point down) at 2 cubic meters per minute. The cone is 10 meters tall and 4 meters wide at the top. How fast is the water level rising when the water is 5 meters deep?

The Setup

graph TD
    A["Cone Tank"] --> B["Height H = 10m"]
    A --> C["Top radius R = 2m"]
    D["Water at height h"] --> E["Water radius r"]
    B --> F["Similar triangles: r/h = 2/10 = 1/5"]

Key insight: Similar triangles give us $r = \frac{h}{5}$

Volume of cone: $V = \frac{1}{3}\pi r^2 h$

Substitute: $V = \frac{1}{3}\pi \left(\frac{h}{5}\right)^2 h = \frac{\pi h^3}{75}$

The Solution

Step 1: Differentiate

$\frac{dV}{dt} = \frac{\pi}{75} \cdot 3h^2 \cdot \frac{dh}{dt} = \frac{\pi h^2}{25} \cdot \frac{dh}{dt}$

Step 2: Plug in ($\frac{dV}{dt} = 2$, $h = 5$)

$2 = \frac{\pi (5)^2}{25} \cdot \frac{dh}{dt}$

$2 = \pi \cdot \frac{dh}{dt}$

Step 3: Solve

$\frac{dh}{dt} = \frac{2}{\pi} \approx 0.637 \text{ m/min}$

Answer: Water rises at about 0.64 meters per minute!

🚶 The Shadow Problem

The Story

A 6-foot person walks away from a 15-foot streetlight at 4 feet per second. How fast is their shadow growing?

The Setup

Using similar triangles: $\frac{15}{x + s} = \frac{6}{s}$

Where:

• $x$ = distance from light to person
• $s$ = shadow length

Cross multiply: $15s = 6(x + s) = 6x + 6s$

Simplify: $9s = 6x$ → $s = \frac{2x}{3}$

The Solution

Differentiate: $\frac{ds}{dt} = \frac{2}{3} \cdot \frac{dx}{dt} = \frac{2}{3} \cdot 4 = \frac{8}{3} \text{ ft/sec}$

Answer: The shadow grows at 2.67 feet per second!

📋 The 5-Step Recipe for ANY Related Rates Problem

⚠️ Important: Plug in numbers AFTER differentiating, not before!

🧠 Quick Tips to Remember

Chain Rule is Your Friend

When you differentiate $x^2$ with respect to time: $\frac{d}{dt}(x^2) = 2x \cdot \frac{dx}{dt}$

The $\frac{dx}{dt}$ appears because x changes with time!

Signs Matter

• Positive rate → thing is increasing
• Negative rate → thing is decreasing

Common Shapes & Formulas

🎉 You Did It!

Related rates might seem tricky at first, but remember:

1. Everything is connected - find the equation
2. Take derivatives with respect to time - use chain rule
3. Plug in the numbers - solve for what you need

You’re now ready to track how things change together in the real world! Whether it’s balloons inflating, ladders sliding, or shadows growing—you’ve got the tools to figure it all out.

🚀 Pro Tip: Practice with different scenarios. The more you do, the more patterns you’ll recognize!