Introduction

This topic focuses on solving complete related rates problems by finding unknown rates of change using known rates. The essential knowledge tells us that the derivative can be used to solve related rates problems; that is, finding a rate at which one quantity is changing by relating it to other quantities whose rates of change are known. This builds on the foundation from Topic 4.4 to solve real-world applications.

The Complete Related Rates Strategy

Step-by-Step Problem-Solving Process

Step 1: Read and Understand

Step 2: Draw a Diagram

Step 3: Write an Equation

Step 4: Differentiate

Step 5: Substitute Known Information

Step 6: Interpret the Answer

Geometric Related Rates Problems

Expanding Circles

Example 1:

The radius of a circle is increasing at a rate of 3 centimeters per second. How fast is the area increasing when the radius is 10 centimeters?

Step 1: Given $\frac{dr}{dt}=3cm/s$, find $\frac{dA}{dt}$ when $r=10\mathrm{cm}$

Step 2: Circle with radius $r$ and area $A$

Step 3: $A=\pi r^2$

Step 4: Differentiate: $\frac{dA}{dt}=2\pi r\frac{dr}{dt}$

Step 5: Substitute: $\frac{dA}{dt}=2\pi (10)(3)=60\pi \mathrm{c}{\mathrm{m}}^2/s$

Step 6: The area is increasing at $60\pi \approx 188.5\mathrm{c}{\mathrm{m}}^2/s$ when the radius is 10 cm.

Expanding Spheres

Example 2:

A spherical balloon is being inflated so that its radius increases at a rate of 2 inches per minute. How fast is the volume increasing when the radius is 6 inches?

Step 1: Given $\frac{dr}{dt}=2in/min$, find $\frac{dV}{dt}$ when $r=6\mathrm{in}$

Step 2: Sphere with radius $r$ and volume $V$

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

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

Step 5: Substitute: $\frac{dV}{dt}=4\pi (6{)}^2(2)$ $\frac{dV}{dt}=4\pi (36)(2)$

$\frac{dV}{dt}=288\pi \mathrm{i}{\mathrm{n}}^3/min$

Step 6: The volume is increasing at $288\pi \approx 904.8\mathrm{i}{\mathrm{n}}^3/min$.

Changing Rectangles

Example 3:

The length of a rectangle is increasing at 5 feet per second while the width is decreasing at 2 feet per second. When the length is 20 feet and the width is 10 feet, how fast is the area changing?

Step 1: Given $\frac{dL}{dt}=5ft/s$, $\frac{dW}{dt}=-2ft/s$, find $\frac{dA}{dt}$ when $L=20\mathrm{ft}$ and $W=10\mathrm{ft}$

Step 2: Rectangle with length $L$, width $W$, and area $A$

Step 3: $A=LW$

Step 4: Differentiate using product rule: $\frac{dA}{dt}=L\frac{dW}{dt}+W\frac{dL}{dt}$

Step 5: Substitute: $\frac{dA}{dt}=20(-2)+10(5)$

$\frac{dA}{dt}=-40+50$

$\frac{dA}{dt}=10\mathrm{f}{\mathrm{t}}^2/s$

Step 6: The area is increasing at 10 square feet per second.

Ladder Problems

Classic Sliding Ladder

Example 4:

A 13-foot ladder is leaning against a wall. The bottom of the ladder is sliding away from the wall at 2 feet per second. How fast is the top of the ladder sliding down the wall when the bottom is 5 feet from the wall?

Step 1: Given $\frac{dx}{dt}=2ft/s$, find $\frac{dy}{dt}$ when $x=5\mathrm{ft}$

Step 2: Right triangle with ladder as hypotenuse (13 ft), horizontal distance $x$, vertical height $y$

Step 3: By Pythagorean theorem: $x^2+y^2=13^2=169$

Step 4: Differentiate: $2x\frac{dx}{dt}+2y\frac{dy}{dt}=0$

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

Step 5: When $x=5$: $25+y^2=169$, so $y^2=144$ and $y=12$

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

$10+12\frac{dy}{dt}=0$

$\frac{dy}{dt}=-\frac{10}{12}=-\frac{5}{6}ft/s$

Step 6: The top of the ladder is sliding down at $\frac{5}{6}$ feet per second (negative indicates downward motion).

Distance and Shadow Problems

Moving Objects and Shadows

Example 5:

A 6-foot-tall person walks away from a 15-foot-tall streetlight at 4 feet per second. How fast is the length of the person's shadow increasing?

Step 1: Given $\frac{dx}{dt}=4ft/s$, find $\frac{ds}{dt}$

Step 2: Draw streetlight (15 ft), person (6 ft) at distance $x$ from light, shadow length $s$

Step 3: By similar triangles: $\frac{15}{x+s}=\frac{6}{s}$

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

$15s-6s=6x$

$9s=6x$

$s=\frac{2x}{3}$

Step 4: Differentiate: $\frac{ds}{dt}=\frac{2}{3}\frac{dx}{dt}$

Step 5: Substitute: $\frac{ds}{dt}=\frac{2}{3}(4)=\frac{8}{3}ft/s$

Step 6: The shadow is lengthening at $\frac{8}{3}\approx 2.67$ feet per second.

Conical Tank Problems

Filling or Draining Cones

Example 6:

Water is draining from a conical tank at a rate of 5 cubic feet per minute. The tank has height 10 feet and top radius 5 feet. How fast is the water level dropping when the water is 6 feet deep?

Step 1: Given $\frac{dV}{dt}=-5\mathrm{f}{\mathrm{t}}^3/min$, find $\frac{dh}{dt}$ when $h=6\mathrm{ft}$

Step 2: Cone with water depth $h$, radius of water surface $r$, volume $V$

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

From similar triangles: $\frac{r}{h}=\frac{5}{10}=\frac{1}{2}$, so $r=\frac{h}{2}$

Substitute: $V=\frac{1}{3}\pi {\left(\frac{h}{2}\right)}^{2}h=\frac{1}{3}\pi \cdot \frac{h^2}{4}\cdot h=\frac{\pi h^3}{12}$

Step 4: Differentiate: $\frac{dV}{dt}=\frac{\pi }{12}\cdot 3h^2\frac{dh}{dt}=\frac{\pi h^2}{4}\frac{dh}{dt}$

Step 5: Substitute: $-5=\frac{\pi (6{)}^2}{4}\frac{dh}{dt}$ $-5=\frac{36\pi }{4}\frac{dh}{dt}$ $-5=9\pi \frac{dh}{dt}$ $\frac{dh}{dt}=-\frac{5}{9\pi }$ ft/min

Step 6: The water level is dropping at $\frac{5}{9\pi }\approx 0.177$ feet per minute.

Right Triangle Problems

Pythagorean Relationship Applications

Example 7:

Two cars start from the same point. One travels north at 50 mph and the other travels east at 60 mph. How fast is the distance between them increasing after 2 hours?

Step 1: Given $\frac{dy}{dt}=50$ mph (north), $\frac{dx}{dt}=60$ mph (east), find $\frac{dz}{dt}$ when $t=2$ hours

Step 2: Right triangle with north distance $y$, east distance $x$, separation distance $z$

Step 3: $x^2+y^2=z^2$

Step 4: Differentiate: $2x\frac{dx}{dt}+2y\frac{dy}{dt}=2z\frac{dz}{dt}$

$x\frac{dx}{dt}+y\frac{dy}{dt}=z\frac{dz}{dt}$

Step 5: After 2 hours: $x=60(2)=120$ miles, $y=50(2)=100$ miles

Find $z$: $z^2=120^2+100^2=14400+10000=24400$

$z=\sqrt{24400}=20\sqrt{61}$ miles

Substitute: $120(60)+100(50)=20\sqrt{61}\frac{dz}{dt}$

$7200+5000=20\sqrt{61}\frac{dz}{dt}$

$12200=20\sqrt{61}\frac{dz}{dt}$

$\frac{dz}{dt}=\frac{12200}{20\sqrt{61}}=\frac{610}{\sqrt{61}}=10\sqrt{61}$ mph

Step 6: The distance between the cars is increasing at $10\sqrt{61}\approx 78.1$ miles per hour.

Rate Problems with Trigonometry

Angles and Trigonometric Relationships

Example 8:

A camera is 100 feet from a rocket launch pad. A rocket is launched vertically and is rising at 300 feet per second when it is 400 feet above the ground. How fast is the angle of elevation of the camera changing at that moment?

Step 1: Given $\frac{dy}{dt}=300$ ft/s, find $\frac{d\theta }{dt}$ when $y=400$ ft

Step 2: Right triangle with horizontal distance 100 ft, vertical height $y$, angle of elevation $\theta$

Step 3: $\tan (\theta )=\frac{y}{100}$

Step 4: Differentiate: ${\sec }^2(\theta )\frac{d\theta }{dt}=\frac{1}{100}\frac{dy}{dt}$

Step 5: When $y=400$, find ${\sec }^2(\theta )$:

Distance to rocket: $z=\sqrt{100^2+400^2}=\sqrt{10000+160000}=\sqrt{170000}=100\sqrt{17}$

$\cos (\theta )=\frac{100}{100\sqrt{17}}=\frac{1}{\sqrt{17}}$

${\sec }^2(\theta )=\frac{1}{{\cos }^2(\theta )}=17$

Substitute: $17\frac{d\theta }{dt}=\frac{1}{100}(300)$ $17\frac{d\theta }{dt}=3$ $\frac{d\theta }{dt}=\frac{3}{17}$ radians/s

Step 6: The angle of elevation is increasing at $\frac{3}{17}\approx 0.176$ radians per second.

Volume and Flow Rate Problems

Cylindrical Containers

Example 9:

Water flows into a cylindrical tank at 8 cubic feet per minute. The tank has radius 4 feet. How fast is the water level rising?

Step 1: Given $\frac{dV}{dt}=8\mathrm{f}{\mathrm{t}}^3/min$, radius $r=4\mathrm{ft}$ (constant), find $\frac{dh}{dt}$

Step 2: Cylinder with radius 4 ft and water height $h$

Step 3: $V=\pi r^{2}h=\pi (4{)}^{2}h=16\pi h$

Step 4: Differentiate: $\frac{dV}{dt}=16\pi \frac{dh}{dt}$

Step 5: Substitute: $8=16\pi \frac{dh}{dt}$

$\frac{dh}{dt}=\frac{8}{16\pi }=\frac{1}{2\pi }$ ft/min

Step 6: The water level is rising at $\frac{1}{2\pi }\approx 0.159$ feet per minute.

Common Mistakes and How to Avoid Them

Critical Error Prevention

Mistake 1: Substituting too early

Mistake 2: Sign errors

Mistake 3: Using similar triangles incorrectly

Mistake 4: Forgetting units

Mistake 5: Not finding needed values

Practice Section