⚠️ IMPORTANT: Don't substitute values until AFTER differentiating!

Common Rate Notation

In related rates, we use derivatives with respect to time:

• $\frac{dr}{dt}$ = rate of change of radius
• $\frac{dV}{dt}$ = rate of change of volume
• $\frac{dA}{dt}$ = rate of change of area
• $\frac{dh}{dt}$ = rate of change of height
• $\frac{dx}{dt}$ = rate of change of position

Units matter! If $\frac{dx}{dt} = 5$ ft/sec, that means the position is increasing at 5 feet per second.

Example 1: Expanding Circle

Problem: The radius of a circle is increasing at 3 cm/sec. How fast is the area increasing when the radius is 10 cm?

Step 1: Identify information

Given: $\frac{dr}{dt} = 3$ cm/sec (radius increasing)

Find: $\frac{dA}{dt}$ when $r = 10$ cm

Step 2: Write equation relating variables

$A = \pi r^2$

Step 3: Differentiate with respect to time

$\frac{dA}{dt} = \frac{d}{dt}[\pi r^2]$

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

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

Step 4: Substitute known values

When $r = 10$ and $\frac{dr}{dt} = 3$:

$\frac{dA}{dt} = 2\pi(10)(3) = 60\pi \text{ cm}^2/\text{sec}$

Answer: The area is increasing at $60\pi \approx 188.5$ square cm per second.

Example 2: Ladder Sliding Down Wall

Problem: A 13-foot ladder is leaning against a wall. The bottom is being pulled away from the wall at 2 ft/sec. How fast is the top sliding down when the bottom is 5 feet from the wall?

Step 1: Draw and identify

Let $x$ = distance from wall to bottom of ladder

Let $y$ = height of top of ladder on wall

Given: $\frac{dx}{dt} = 2$ ft/sec (bottom moving away)

Find: $\frac{dy}{dt}$ when $x = 5$ ft

Step 2: Write equation

By Pythagorean theorem:

$x^2 + y^2 = 13^2 = 169$

Step 3: Find $y$ when $x = 5$

$5^2 + y^2 = 169$ $y^2 = 144$ $y = 12$ feet

Step 4: Differentiate with respect to time

$\frac{d}{dt}[x^2 + y^2] = \frac{d}{dt}[169]$

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

Step 5: Substitute and solve

When $x = 5$, $y = 12$, and $\frac{dx}{dt} = 2$:

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

$20 + 24\frac{dy}{dt} = 0$

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

Answer: The top is sliding down at $\frac{5}{6}$ ft/sec (negative means decreasing).

Common Formulas for Related Rates

Geometry Formulas

Circle:

• Area: $A = \pi r^2$
• Circumference: $C = 2\pi r$

Sphere:

• Volume: $V = \frac{4}{3}\pi r^3$
• Surface Area: $S = 4\pi r^2$

Cone:

• Volume: $V = \frac{1}{3}\pi r^2 h$

Cylinder:

• Volume: $V = \pi r^2 h$

Triangle:

• Area: $A = \frac{1}{2}bh$

Pythagorean Theorem: $a^2 + b^2 = c^2$

Similar Triangles: $\frac{a_1}{a_2} = \frac{b_1}{b_2}$

⚠️ Common Mistakes

Mistake 1: Substituting Too Early

❌ Substitute values before differentiating ✅ Differentiate first, then substitute

Mistake 2: Wrong Signs

If a quantity is decreasing, its rate is negative!

Mistake 3: Forgetting Chain Rule

When differentiating $r^2$ with respect to $t$: $\frac{d}{dt}[r^2] = 2r\frac{dr}{dt}$

Mistake 4: Units

Always include units in your answer! (ft/sec, cm³/min, etc.)

Mistake 5: Missing the Relationship

Make sure your equation actually relates the variables in the problem.

Problem Types

Type 1: Area and Volume

Expanding/shrinking circles, spheres, cones, etc.

Key: Use geometry formulas and differentiate with respect to time.

Type 2: Pythagorean Relationships

Ladders, shadows, moving vehicles, etc.

Key: Set up right triangle and use $a^2 + b^2 = c^2$

Type 3: Similar Triangles

Shadows, conical tanks with changing water levels, etc.

Key: Set up proportion using similar triangles, then differentiate.

Type 4: Distance and Position

Two objects moving, finding rate of change of distance between them.

Key: Use distance formula or Pythagorean theorem.

Tips for Success

Before Differentiating:

1. Draw a clear diagram
2. Label variables that change with time
3. Label constants (like ladder length)
4. Write the equation relating variables

When Differentiating:

1. Differentiate the entire equation with respect to $t$
2. Use Chain Rule for every variable
3. Don't substitute until after differentiating

After Differentiating:

1. Substitute known values
2. Solve for the unknown rate
3. Check the sign (increasing or decreasing?)
4. Include proper units

Word Problem Clues

Certain phrases tell you what rate is given or asked:

• "increasing at..." → positive rate
• "decreasing at..." → negative rate
• "how fast is..." → find a derivative
• "at the instant when..." → specific value to substitute
• "at what rate..." → find a derivative
• "constant rate" → the rate doesn't change

📝 Practice Strategy

1. Read carefully and identify what's given and what's asked
2. Draw and label everything
3. Find the equation relating your variables
4. Differentiate with respect to time (use implicit differentiation)
5. Substitute values at the specific instant
6. Solve for the unknown rate
7. Check sign and units

Find: $\frac{dr}{dt}$ when $r = 5$ inches

Step 2: Write volume formula for a sphere

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

Step 3: Differentiate with respect to time

$\frac{dV}{dt} = \frac{d}{dt}\left[\frac{4}{3}\pi r^3\right]$

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

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

Step 4: Substitute known values

When $r = 5$ and $\frac{dV}{dt} = 100$:

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

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

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

Step 5: Solve for $\frac{dr}{dt}$

$\frac{dr}{dt} = \frac{100}{100\pi} = \frac{1}{\pi}$

Answer: The radius is increasing at $\frac{1}{\pi} \approx 0.318$ inches per minute.

Part (a): Find $\frac{dV}{dt}$ when $r = 10$.

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

Differentiate with respect to time:

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

Substitute $r = 10$ and $\frac{dr}{dt} = 2$:

$\frac{dV}{dt} = 4\pi (10)^2 (2) = 4\pi (100)(2) = 800\pi$ cm³/s

Part (b): Find $\frac{dS}{dt}$ when $r = 10$.

$S = 4\pi r^2$

$\frac{dS}{dt} = 4\pi \cdot 2r \cdot \frac{dr}{dt} = 8\pi r \frac{dr}{dt}$

Substitute:

$\frac{dS}{dt} = 8\pi (10)(2) = 160\pi$ cm²/s