Lesson 1: Related Rates
Estimated time: 40-50 minutes
Learning Objectives
By the end of this lesson, you will be able to:
- Identify related rates problems and set up the appropriate equation
- Differentiate both sides of an equation with respect to time
- Solve for an unknown rate given known rates
- Apply related rates to geometric and physical problems
What Are Related Rates?
In a related rates problem, two or more quantities change with time, and they are connected by an equation. We differentiate that equation with respect to t (time) to relate the rates of change.
Related Rates Strategy
- Draw a picture and label all variables.
- Write an equation connecting the variables.
- Differentiate both sides with respect to t (using the Chain Rule).
- Substitute known values and solve for the unknown rate.
Classic Examples
Example 1: Expanding Circle
A stone is dropped into a pond, creating a circular ripple. The radius increases at 3 ft/s. How fast is the area increasing when the radius is 10 ft?
Step 1: Variables: r = radius, A = area. Given: dr/dt = 3 ft/s. Find: dA/dt when r = 10.
Step 2: Equation: A = πr².
Step 3: Differentiate: dA/dt = 2πr (dr/dt).
Step 4: Substitute: dA/dt = 2π(10)(3) = 60π ft²/s ≈ 188.5 ft²/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?
Step 1: Let x = distance of base from wall, y = height on wall. Given: dx/dt = 2. Find: dy/dt when x = 5.
Step 2: Pythagorean theorem: x² + y² = 169.
Step 3: Differentiate: 2x(dx/dt) + 2y(dy/dt) = 0.
Step 4: When x = 5: y = √(169 − 25) = 12. Substitute: 2(5)(2) + 2(12)(dy/dt) = 0.
20 + 24(dy/dt) = 0, so dy/dt = −5/6 ft/s. The top slides down at 5/6 ft/s.
Example 3: Filling a Cone
Water flows into a conical tank at 2 m³/min. The cone has height 10 m and top radius 5 m. How fast is the water level rising when the depth is 4 m?
Step 1: By similar triangles, r/h = 5/10, so r = h/2.
Step 2: Volume: V = (1/3)πr²h = (1/3)π(h/2)²h = πh³/12.
Step 3: Differentiate: dV/dt = (π/4)h² (dh/dt).
Step 4: 2 = (π/4)(16)(dh/dt), so dh/dt = 2/(4π) = 1/(2π) m/min ≈ 0.159 m/min.
Key Takeaways
- Related rates problems use implicit differentiation with respect to time.
- Always eliminate extra variables before differentiating (e.g., use similar triangles).
- Do NOT substitute specific numbers until after differentiating.
- A negative rate means the quantity is decreasing.
Check Your Understanding
1. A sphere's radius grows at 2 cm/s. How fast is its volume increasing when r = 3 cm? (V = 4πr³/3)
2. A 10-ft ladder: the base is pulled at 1 ft/s. How fast does the top slide when the base is 6 ft out?
3. In Example 3, what changes if the flow rate increases to 5 m³/min?
4. Two cars leave an intersection: one east at 30 mph, one north at 40 mph. How fast is the distance between them increasing after 2 hours?