Related Rates

∫ Related Rates

Part 1 of 7 — Introduction to Related Rates

Topics in This Part

🔑 Key Concept: When two or more quantities change with time and are connected by an equation, their rates of change are related through implicit differentiation with respect to $t$.

📖 What Are Related Rates?

In a related rates problem, quantities change over time ($t$), and we use known rates to find unknown rates.

The Framework

1. Identify all variables and their rates ($\frac{dx}{dt}$, $\frac{dy}{dt}$, etc.)
2. Find an equation relating the variables
3. Differentiate both sides with respect to $t$
4. Substitute known values and solve for the unknown rate

Classic Example: Expanding Circle

A stone dropped in a pond creates a circular ripple. The radius expands at $2$ ft/s. How fast is the area increasing when $r = 10$ ft?

Given: $\frac{dr}{dt} = 2$ ft/s, $r = 10$ ft

Find: $\frac{dA}{dt}$

Equation: $A = \pi r^2$

Differentiate: $\frac{dA}{dt} = 2\pi r \frac{dr}{dt}$

Substitute: $\frac{dA}{dt} = 2\pi(10)(2) = 40\pi$ $ft^{2}/s$

$\boxed{\frac{dA}{dt} = 40\pi \approx 125.7 \text{ ft}^2\text{/s}}$

Key Insight: Even though $\frac{dr}{dt}$ is constant, $\frac{dA}{dt}$ increases as $r$ grows — the circle adds more area per second as it gets bigger.

📌 Basic Examples

Example 1: Expanding Sphere

A balloon is inflated at $100$ $cm^{3}/s$. Find $\frac{dr}{dt}$ when $r = 5$ cm.

$V = \frac{4}{3}\pi r^3$

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

$100 = 4\pi(25)\frac{dr}{dt}$

$\frac{dr}{dt} = \frac{100}{100\pi} = \frac{1}{\pi}$ cm/s

Example 2: Sliding Ladder

A 13-ft ladder leans against a wall. The base slides away at 2 ft/s. How fast is the top sliding down when the base is 5 ft from the wall?

$x^2 + y^2 = 169$

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

When $x = 5$: $y = \sqrt{169 - 25} = 12$

$2(5)(2) + 2(12)\frac{dy}{dt} = 0$

$\frac{dy}{dt} = -\frac{20}{24} = -\frac{5}{6}$ ft/s

The negative sign indicates the top is sliding down.

Example 3: Filling a Cone

Water fills a conical tank (radius 3 m, height 6 m) at 2 $m^{3}/min$.

Similar triangles: $r = h/2$

$V = \frac{1}{3}\pi r^2 h = \frac{1}{3}\pi \frac{h^2}{4} h = \frac{\pi h^3}{12}$

$\frac{dV}{dt} = \frac{\pi h^2}{4}\frac{dh}{dt}$

When $h = 4$: $2 = \frac{16\pi}{4}\frac{dh}{dt} = 4\pi\frac{dh}{dt}$

$\frac{dh}{dt} = \frac{1}{2\pi}$ m/min

Related Rates Basics 🎯

Setting Up Related Rates Problems

The 5-Step Method

Common Geometric Formulas

AP Tip: AP Calculus provides no formula sheet, so memorize these geometry formulas.

Setup & Solve 🧮

1) Circle with $A = \pi r^2$. $dr/dt = 3$, $r = 4$. $dA/dt =$ (give exact answer with pi, like 24pi)

2) Sphere: $V = \frac{4}{3}\pi r^3$. $dV/dt = 36\pi$, $r = 3$. $dr/dt =$

3) Ladder: 25 ft, $dx/dt = 3$ ft/s, $x = 7$. $dy/dt =$ (give as fraction)

Problem Setup 🔽

Exit Quiz — Part 1 ✅

∫ Related Rates — Part 2

Part 2 of 7 — Pythagorean Theorem & Distance Problems

📖 Pythagorean Theorem in Related Rates

Many problems involve right triangles where sides change with time.

The Sliding Ladder (Extended)

A 10-ft ladder leans against a wall. The base moves away at $1$ ft/s.

$x^2 + y^2 = 100$

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

The top accelerates as it falls! As $y \to 0$, $dy/dt \to -\infty$.

Distance Problems

Two Ships

Ship A heads north at 15 km/h. Ship B heads east at 20 km/h. Both start from the same port.

After $t$ hours:

• Ship A: $y = 15t$ north
• Ship B: $x = 20t$ east
• Distance: $D = \sqrt{x^2 + y^2}$

$D^2 = x^2 + y^2$

$2D\frac{dD}{dt} = 2x\frac{dx}{dt} + 2y\frac{dy}{dt}$

At $t = 2$: $x = 40$, $y = 30$, $D = 50$

$2(50)\frac{dD}{dt} = 2(40)(20) + 2(30)(15) = 1600 + 900 = 2500$

$\frac{dD}{dt} = 25 \text{ km/h}$

Moving Along Curves

A particle moves along $y = x^2$. When $x = 3$, $dx/dt = 4$. Find $dD/dt$ where $D$ is distance from origin.

$D^2 = x^2 + y^2 = x^2 + x^4$

$2D\frac{dD}{dt} = (2x + 4x^3)\frac{dx}{dt}$

At $x = 3$: $D = \sqrt{9 + 81} = \sqrt{90} = 3\sqrt{10}$

$2(3\sqrt{10})\frac{dD}{dt} = (6 + 108)(4) = 456$

$\frac{dD}{dt} = \frac{456}{6\sqrt{10}} = \frac{76}{\sqrt{10}} = \frac{76\sqrt{10}}{10} = 7.6\sqrt{10}$

Pythagorean & Distance 🎯

📌 Advanced Scenarios

Approaching & Separating

Car A is 120 mi west of an intersection, driving east at 60 mph; Car B is at the intersection, driving north at 45 mph. Let $x = 120 - 60t$ be A's distance from the intersection (decreasing, so $\frac{dx}{dt} = -60$) and $y = 45t$ be B's ($\frac{dy}{dt} = 45$).

$D^2 = x^2 + y^2 \implies \frac{dD}{dt} = \frac{x\frac{dx}{dt} + y\frac{dy}{dt}}{D}$

At $t = 1$: $x = 60$, $y = 45$, $D = 75$, so

$\frac{dD}{dt} = \frac{60(-60) + 45(45)}{75} = \frac{-3600 + 2025}{75} = -21$ mph

The distance between the cars is shrinking at 21 mph. The sign of each rate, not the compass direction, decides whether $D$ grows.

Shadow Problems

A 6-ft person walks away from a 15-ft lamppost at 4 ft/s. How fast does the shadow tip move?

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

$15s = 6(x + s) = 6x + 6s$

$9s = 6x \implies s = \frac{2x}{3}$

$\frac{ds}{dt} = \frac{2}{3}\frac{dx}{dt} = \frac{2}{3}(4) = \frac{8}{3}$ ft/s

Shadow tip speed: $\frac{d(x+s)}{dt} = 4 + \frac{8}{3} = \frac{20}{3}$ ft/s

Distance Rate Practice 🧮

1) 20-ft ladder. $x = 12$, $dx/dt = 3$. $dy/dt =$ (fraction)

2) Ships: A north at 10 km/h, B east at 24 km/h. After 1 hr, $dD/dt =$ (integer)

Problem Classification 🔽

Exit Quiz — Part 2 ✅

∫ Related Rates — Part 3

Part 3 of 7 — Volume & Surface Area Problems

📖 Volume Related Rates

Shrinking Sphere

A snowball melts at $12$ $cm^{3}/min$. How fast does the radius shrink when $r = 5$?

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

$-12 = 4\pi(25)\frac{dr}{dt}$

$\frac{dr}{dt} = \frac{-12}{100\pi} = \frac{-3}{25\pi}$ cm/min

Conical Tank (Detailed)

A tank: top radius 4 m, height 8 m. Water drains at 3 $m^{3}/min$.

Similar triangles: $\frac{r}{h} = \frac{4}{8} = \frac{1}{2} \implies r = \frac{h}{2}$

$V = \frac{1}{3}\pi r^2 h = \frac{1}{3}\pi \frac{h^2}{4} h = \frac{\pi h^3}{12}$

$\frac{dV}{dt} = \frac{\pi h^2}{4}\frac{dh}{dt}$

When $h = 4$: $-3 = \frac{16\pi}{4}\frac{dh}{dt} = 4\pi\frac{dh}{dt}$

$\frac{dh}{dt} = \frac{-3}{4\pi}$ m/min

Cylinder with Changing Dimensions

A cylinder: $r$ grows at 2 cm/s, $h$ shrinks at 3 cm/s. Find $dV/dt$ when $r = 5$, $h = 10$.

$V = \pi r^2 h$

$\frac{dV}{dt} = \pi(2rh\frac{dr}{dt} + r^2\frac{dh}{dt})$

$= \pi(2(5)(10)(2) + 25(-3)) = \pi(200 - 75) = 125\pi$ $cm^{3}/s$

Surface Area Problems

Expanding Sphere Surface Area

A sphere expands at $\frac{dr}{dt} = 0.5$ cm/s. Find $\frac{dSA}{dt}$ when $r = 10$.

$SA = 4\pi r^2$

$\frac{dSA}{dt} = 8\pi r \frac{dr}{dt} = 8\pi(10)(0.5) = 40\pi$ $cm^{2}/s$

Volume & Surface Area Connected

A sphere inflates at $100$ $cm^{3}/s$. How fast does surface area increase when $r = 5$?

First find $dr/dt$: $\frac{dV}{dt} = 4\pi r^2 \frac{dr}{dt}$

$100 = 4\pi(25)\frac{dr}{dt} \implies \frac{dr}{dt} = \frac{1}{\pi}$

Then: $\frac{dSA}{dt} = 8\pi r \frac{dr}{dt} = 8\pi(5)(1/\pi) = 40$ $cm^{2}/s$

Alternative: Use $SA = 4\pi r^2$ and $V = \frac{4}{3}\pi r^3$, so $SA = \frac{3V}{r}$... but the two-step approach above is simpler.

Volume & Surface Area 🎯

📌 Trough Problems

Triangular Cross-Section Trough

A trough is 10 ft long. Its cross-section is an equilateral triangle with side 2 ft. Water fills at 1 $ft^{3}/min$.

For equilateral triangle with water depth $h$: width $= \frac{2h}{\sqrt{3}}$

Area of water cross-section: $A = \frac{1}{2} \cdot \frac{2h}{\sqrt{3}} \cdot h = \frac{h^2}{\sqrt{3}}$

$V = 10A = \frac{10h^2}{\sqrt{3}}$

$\frac{dV}{dt} = \frac{20h}{\sqrt{3}}\frac{dh}{dt}$

$1 = \frac{20h}{\sqrt{3}}\frac{dh}{dt}$

When $h = 0.5$: $\frac{dh}{dt} = \frac{\sqrt{3}}{10} \approx 0.173$ ft/min

Key Insight for Trough Problems

Troughs have uniform cross-sections along their length $L$: $V = L \times A(h)$ where $A(h)$ is the cross-sectional area as a function of water depth.

Volume Rate Problems 🧮

1) Sphere: $dV/dt = 36\pi$, $r = 3$. $dr/dt =$ (integer)

2) Cone ($r = h/3$, filling at $2\pi$ m³/min, $h = 3$). $dh/dt =$ (integer)

Volume Problem Setup 🔽

Exit Quiz — Part 3 ✅

∫ Related Rates — Part 4

Part 4 of 7 — Trigonometric Related Rates

📖 Trigonometric Models

Many related rates problems involve angles. The key relationships:

$\tan\theta = \frac{\text{opposite}}{\text{adjacent}}, \quad \sin\theta = \frac{\text{opposite}}{\text{hypotenuse}}$

Lighthouse Problem

A lighthouse rotates at $2$ rev/min = $4\pi$ rad/min. The beam hits a straight shoreline 1 km away.

If $\theta$ is the angle from perpendicular, and $x$ is the distance along shore:

$\tan\theta = x/1 \implies x = \tan\theta$

$\frac{dx}{dt} = \sec^2\theta \cdot \frac{d\theta}{dt}$

At $\theta = \pi/4$: $\frac{dx}{dt} = (\sec^2(\pi/4))(4\pi) = 2 \cdot 4\pi = 8\pi$ km/min

At $\theta = \pi/3$: $\frac{dx}{dt} = (\sec^2(\pi/3))(4\pi) = 4 \cdot 4\pi = 16\pi$ km/min

The beam sweeps faster along the shore as the angle increases!

Angle of Elevation Problems

Rocket Launch

A rocket rises vertically at 600 ft/s. An observer stands 1000 ft from the launch pad.

$\tan\theta = \frac{h}{1000}$

$\sec^2\theta \cdot \frac{d\theta}{dt} = \frac{1}{1000}\frac{dh}{dt}$

When $h = 1000$: $\theta = \pi/4$, so $\sec^2\theta = 2$

$2\frac{d\theta}{dt} = \frac{600}{1000} = 0.6$

$\frac{d\theta}{dt} = 0.3$ rad/s

Alternative: Use $\theta = \arctan(h/1000)$

$\frac{d\theta}{dt} = \frac{1}{1 + (h/1000)^2} \cdot \frac{1}{1000} \cdot \frac{dh}{dt}$

At $h = 1000$:

$\frac{d\theta}{dt} = \frac{1}{1+1} \cdot \frac{600}{1000} = \frac{1}{2} \cdot 0.6 = 0.3$ rad/s ✓

Kite Problem

A kite flies at a constant height of 100 ft and drifts horizontally away from you at 8 ft/s. Let $x$ be the horizontal distance and $\theta$ the angle between the string and the ground. How fast is $\theta$ changing when $x = 100$ ft?

$\tan\theta = \frac{100}{x}$, so $\sec^2\theta\,\frac{d\theta}{dt} = -\frac{100}{x^2}\frac{dx}{dt}$.

At $x = 100$: $\theta = \frac{\pi}{4}$ and $\sec^2\theta = 2$, so $2\frac{d\theta}{dt} = -\frac{100}{10000}(8) = -0.08$ and $\frac{d\theta}{dt} = -0.04$ rad/s (the angle is decreasing).

Trigonometric Related Rates 🎯

📌 Baseball Diamond Problem

A runner goes from first to second base at 24 ft/s. How fast does the distance to home plate change when the runner is 30 ft from first base?

Baseball diamond: 90 ft sides, home plate to second base is the diagonal.

Let $x$ = distance from first base toward second base.

Distance from home plate: use the law of cosines (angle at first base = 90°) or coordinates.

Using coordinates: Home = $(0,0)$, First = $(90, 0)$, Second = $(90, 90)$.

Runner position: $(90, x)$ where $x$ goes from $0$ to $90$.

$D^2 = 90^2 + x^2 = 8100 + x^2$

$2D\frac{dD}{dt} = 2x\frac{dx}{dt}$

$\frac{dD}{dt} = \frac{x \cdot dx/dt}{D}$

At $x = 30$: $D = \sqrt{8100 + 900} = \sqrt{9000} = 30\sqrt{10}$

$\frac{dD}{dt} = \frac{30 \cdot 24}{30\sqrt{10}} = \frac{24}{\sqrt{10}} = \frac{24\sqrt{10}}{10} \approx 7.59$ ft/s

Trig Rate Practice 🧮

1) $\tan\theta = h/500$, $dh/dt = 200$, $h = 500$. $d\theta/dt =$ (enter as decimal)

2) Lighthouse 2 km from shore, rotates at $\pi$ rad/min. At $\theta = 0$, $dx/dt =$ (give answer with pi like 2pi)

Trig Setup 🔽

Exit Quiz — Part 4 ✅

∫ Related Rates — Part 5

Part 5 of 7 — Area & Perimeter Rate Problems

📖 Area Rate Problems

Expanding Rectangle

Length grows at 3 cm/s, width at 2 cm/s. Find $dA/dt$ when $l = 10$, $w = 6$.

$A = lw$

$\frac{dA}{dt} = l\frac{dw}{dt} + w\frac{dl}{dt} = 10(2) + 6(3) = 38$ $cm^{2}/s$

Triangle with Changing Dimensions

A triangle: base $b$ increases at 1 cm/s, height $h$ decreases at 2 cm/s. Is area increasing or decreasing when $b = 20$, $h = 15$?

$A = \frac{1}{2}bh$

$\frac{dA}{dt} = \frac{1}{2}(b\frac{dh}{dt} + h\frac{db}{dt}) = \frac{1}{2}(20(-2) + 15(1)) = \frac{1}{2}(-40 + 15) = -12.5$ $cm^{2}/s$

Area is decreasing at 12.5 $cm^{2}/s$.

Circular Sector

Radar sweeps: $r = 100$ m (constant), $\theta$ increases at $\pi/3$ rad/s.

$A = \frac{1}{2}r^2\theta = \frac{1}{2}(10000)\theta = 5000\theta$

$\frac{dA}{dt} = 5000\frac{d\theta}{dt} = 5000 \cdot \frac{\pi}{3} = \frac{5000\pi}{3}$ $m^{2}/s$

Perimeter & Circumference Rates

Growing Circle

If $A = \pi r^2$ and $dA/dt = 6\pi$, find $dC/dt$ when $r = 3$.

Step 1: Find $dr/dt$: $6\pi = 2\pi r \cdot dr/dt = 6\pi \cdot dr/dt \implies dr/dt = 1$

Step 2: $C = 2\pi r$ so $dC/dt = 2\pi \cdot dr/dt = 2\pi$

Rectangle: Perimeter Constraint

A rectangle has fixed perimeter $P = 40$ cm. Length increases at 2 cm/s.

$P = 2l + 2w = 40 \implies l + w = 20$

$\frac{dl}{dt} + \frac{dw}{dt} = 0 \implies \frac{dw}{dt} = -2$ cm/s

So width decreases at 2 cm/s (the perimeter constraint links the rates).

Area: $A = l(20 - l) = 20l - l^2$

$\frac{dA}{dt} = (20 - 2l)\frac{dl}{dt}$

When $l = 8$: $dA/dt = (20-16)(2) = 8$ $cm^{2}/s$ (increasing) When $l = 10$: $dA/dt = (20-20)(2) = 0$ (max area!) When $l = 12$: $dA/dt = (20-24)(2) = -8$ $cm^{2}/s$ (decreasing)

Insight: Maximum area of a rectangle with fixed perimeter is a square ($l = w = 10$). At that moment, $dA/dt = 0$.

Area & Perimeter Rates 🎯

📌 Applied Scenarios

Oil Spill

Oil spreads as a circular slick. Volume of oil is constant $V_0$, thickness $h$ decreases.

$V_0 = \pi r^2 h \implies h = \frac{V_0}{\pi r^2}$

$\frac{dh}{dt} = -\frac{2V_0}{\pi r^3}\frac{dr}{dt}$

If $dr/dt = 0.5$ m/min: $dh/dt < 0$ (thickness decreases as it spreads).

Two Related Quantities from a Graph

If $A = f(x)$ and we know $dA/dt$ and $dx/dt$:

$\frac{dA}{dt} = f'(x) \cdot \frac{dx}{dt}$

This is just the chain rule! If the particle moves along $y = x^3$ with $dx/dt = 2$:

$\frac{dy}{dt} = 3x^2 \cdot \frac{dx}{dt} = 3x^2 \cdot 2 = 6x^2$

At $x = 4$: $dy/dt = 6(16) = 96$ units/s

AP Connection: "A particle moves along the curve $y = f(x)$" is a standard AP setup for applying the chain rule as a related rate.

Area Rate Calculations 🧮

1) Rectangle: $l = 8$, $w = 5$, $dl/dt = 3$, $dw/dt = -1$. $dA/dt =$ (integer)

2) Circle: $dA/dt = 12\pi$, $r = 6$. $dr/dt =$ (integer)

Setup Classification 🔽

Exit Quiz — Part 5 ✅

∫ Related Rates — Part 6

Part 6 of 7 — Multi-Step & Combined Problems

📖 Multi-Step Related Rates

Some problems require finding an intermediate rate before the final answer.

Balloon & Observer

A balloon rises at 10 ft/s from a point 100 ft from an observer. Find how fast the angle of elevation changes when the balloon is 100 ft high.

$\tan\theta = \frac{h}{100}$

$\sec^2\theta \cdot \frac{d\theta}{dt} = \frac{1}{100} \cdot 10 = 0.1$

When $h = 100$: $\theta = \pi/4$, $\sec^2\theta = 2$

$\frac{d\theta}{dt} = \frac{0.1}{2} = 0.05$ rad/s

Now find how fast the distance is changing:

$D = \sqrt{100^2 + h^2}$.

At $h = 100$: $D = 100\sqrt{2}$

$\frac{dD}{dt} = \frac{h \cdot dh/dt}{D} = \frac{100(10)}{100\sqrt{2}} = \frac{10}{\sqrt{2}} = 5\sqrt{2}$ ft/s

Combined Equations

Street Intersection Problem

Car A: 300 ft south, heading north at 20 ft/s Car B: 400 ft east, heading west at 15 ft/s

Distance: $D^2 = a^2 + b^2$ where $a = 300 - 20t$ and $b = 400 - 15t$

$2D\frac{dD}{dt} = 2a\frac{da}{dt} + 2b\frac{db}{dt}$

At $t = 0$: $a = 300$, $b = 400$, $D = 500$

$\frac{da}{dt} = -20$ (decreasing), $\frac{db}{dt} = -15$ (decreasing)

$2(500)\frac{dD}{dt} = 2(300)(-20) + 2(400)(-15) = -12000 - 12000 = -24000$

$\frac{dD}{dt} = -24$ ft/s

They're approaching each other at 24 ft/s.

When Do They Stop Approaching?

They stop approaching when $dD/dt = 0$:

$a \cdot da/dt + b \cdot db/dt = 0$

$(300 - 20t)(-20) + (400 - 15t)(-15) = 0$

$-6000 + 400t - 6000 + 225t = 0$

$625t = 12000 \implies t = 19.2$ seconds

At $t = 19.2$: They're at minimum distance, then separate.

Multi-Step Problems 🎯

📌 Table-Based Related Rates (AP Style)

The AP exam often gives a table of values rather than formulas.

Example

Find $dD/dt$ where $D = \sqrt{x^2 + y^2}$ at $t = 1$.

$\frac{dD}{dt} = \frac{x \cdot x'(t) + y \cdot y'(t)}{\sqrt{x^2 + y^2}}$

At $t = 1$: $\frac{dD}{dt} = \frac{5(1) + 3(-2)}{\sqrt{25 + 9}} = \frac{5 - 6}{\sqrt{34}} = \frac{-1}{\sqrt{34}}$

AP Tip: When given a table, you DON'T need a formula for $x(t)$ or $y(t)$. Just read off the values and plug in.

Multi-Step Practice 🧮

1) From the table above: $dD/dt$ at $t = 0$ = (enter as fraction like a/b)

2) Cars: $a = 200$, $b = 150$, $da/dt = -30$, $db/dt = -40$. $dD/dt =$ (enter integer)

Problem Analysis 🔽

Exit Quiz — Part 6 ✅

∫ Related Rates — Part 7

Part 7 of 7 — AP Exam Free Response Strategies

📖 AP Free Response Strategy

Related rates appears regularly in both multiple choice and free response.

What a Complete Answer Shows

• The equation relating the variables
• That equation differentiated with respect to $t$ (the chain-rule step)
• The substitution of values after differentiating
• The answer, with units when the question asks for them

A diagram and a list of givens help you set up the problem, but the calculus steps above are what must appear in your work.

Structure Your Solution

1. "Let..." Define variables clearly
2. "Given:" List rates and values
3. "Find:" State what you need
4. "Equation:" Write the relationship
5. "Differentiate:" Show the chain rule step
6. "Substitute:" Plug in values
7. "Therefore:" Box your answer (with units when asked)

Full AP Free Response Walkthrough

AP-Style Problem

"A conical pile of sand has height equal to twice its radius. Sand is added at 5 cubic feet per minute. How fast is the height increasing when $h = 6$ ft?"

Solution:

Let $r$ = radius, $h$ = height. Given: $h = 2r$, so $r = h/2$.

$V = \frac{1}{3}\pi r^2 h = \frac{1}{3}\pi\left(\frac{h}{2}\right)^2 h = \frac{\pi h^3}{12}$

Differentiate with respect to $t$:

$\frac{dV}{dt} = \frac{\pi h^2}{4} \cdot \frac{dh}{dt}$

Substitute $dV/dt = 5$ and $h = 6$:

$5 = \frac{\pi(36)}{4} \cdot \frac{dh}{dt} = 9\pi \cdot \frac{dh}{dt}$

$\frac{dh}{dt} = \frac{5}{9\pi} \approx 0.177 \text{ ft/min}$

$\boxed{\frac{dh}{dt} = \frac{5}{9\pi} \text{ ft/min}}$

Note: Show the differentiated equation and the substitution; include units when the question asks for them.

AP Strategy 🎯

📌 Most Common AP Related Rates Setups

The Big Five

1. Ladder sliding — Pythagorean theorem: $x^2 + y^2 = L^2$
2. Expanding/contracting shapes — Area/volume formulas
3. Conical tanks — Similar triangles + $V = \frac{1}{3}\pi r^2 h$
4. Angle of elevation — $\tan\theta = h/d$
5. Particles on curves — $y = f(x)$, use chain rule: $dy/dt = f'(x) \cdot dx/dt$

Quick Decision Chart

Mistakes That Cost Points

1. Substituting values BEFORE differentiating
2. Forgetting negative signs for decreasing quantities
3. Not finding intermediate values ($y$ from $x$ in ladder problems)
4. Missing units on the final answer
5. Not eliminating a variable (e.g., in cone problems)

Full AP Problems 🧮

1) Spherical balloon: $dV/dt = 100\pi$, $r = 5$. $dr/dt =$ (integer)

2) Cone: $h = 3r$, $dV/dt = 27\pi$, $h = 9$. $dh/dt =$ (integer)

3) Ladder 25 ft, $dx/dt = 2$, $x = 7$. $dy/dt =$ (fraction)

Problem Classification 🔽

Final Exit Quiz — Related Rates ✅