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Related Rates of Change

Key Terms

Chain rule for related rates
If y depends on x, and both depend on t, then dy/dt = (dy/dx) × (dx/dt).
Method
(1) Write an equation relating the relevant quantities. (2) Differentiate both sides with respect to time t using the chain rule. (3) Substitute known values and solve for the unknown rate.
Units matter
Always check that rates have consistent units (e.g. cm/s, m/s, km/h).
Implicit differentiation
For equations like x² + y² = r², differentiate both sides with respect to t, treating x and y as functions of t.
Key Formulas for Related Rates:
  • Sphere: V = 4πr³/3 ⇒ dV/dt = 4πr² × dr/dt
  • Cone: V = πr²h/3 (use similar triangles to relate r and h)
  • Circle: A = πr² ⇒ dA/dt = 2πr × dr/dt
  • Pythagoras: x² + y² = z² ⇒ 2x(dx/dt) + 2y(dy/dt) = 2z(dz/dt)
  • Angle: tanθ = h/d ⇒ sec²θ × dθ/dt = (1/d)(dh/dt) [with d constant]
Hot Tip Always draw a diagram and label all variables. Write the geometric relationship first (before differentiating). Be careful to evaluate the formula AT the given instant — substitute values only after differentiating, not before.

Worked Example — Expanding Balloon

Air is pumped into a spherical balloon at 100 cm³/s. Find the rate at which the radius is increasing when r = 10 cm.

Step 1: Volume of sphere: V = (4/3)πr³

Step 2: Differentiate with respect to t: dV/dt = 4πr² × dr/dt

Step 3: Substitute dV/dt = 100 and r = 10:

100 = 4π(100) × dr/dt ⇒ dr/dt = 100/(400π) = 1/(4π) cm/s ≈ 0.0796 cm/s.

What Are Related Rates?

In many real-world situations, several quantities all change simultaneously with time. Related rates problems involve finding the rate of change of one quantity given the rate of change of another, when the two quantities are connected by a geometric or physical relationship.

The key mathematical tool is the chain rule: if y = f(x) and x = g(t), then dy/dt = f′(x) × g′(t) = (dy/dx)(dx/dt).

The Four-Step Method

Step 1 — Draw and label: Sketch the situation and label all relevant quantities as functions of time. Identify what is given (the known rate) and what is required (the unknown rate).

Step 2 — Write the relationship: Find an equation connecting the relevant variables. This might be a geometric formula (volume, area, Pythagoras), a trigonometric identity, or a physical law.

Step 3 — Differentiate with respect to t: Differentiate both sides of the equation implicitly with respect to t, applying the chain rule to every variable that depends on t.

Step 4 — Substitute and solve: Substitute the given values (at the specified instant) and solve for the required rate.

Similar Triangles in Cone Problems

When water flows into or out of a conical vessel, we usually want dh/dt (rate of change of depth) given dV/dt. The cone formula V = πr²h/3 involves both r and h. We use the geometry of similar triangles to write r in terms of h (or vice versa), reducing the equation to one variable before differentiating.

For a cone of height H and base radius R: by similar triangles, r/h = R/H, so r = (R/H)h. Substituting: V = π(R/H)²h³/3 = πR²h³/(3H²). Then dV/dt = (πR²/H²)h² × dh/dt.

Ladder Problems

A ladder of length L leans against a wall. Let x = distance of foot from wall, y = height of top. Then x² + y² = L². Differentiating: 2x(dx/dt) + 2y(dy/dt) = 0, giving dy/dt = −(x/y)(dx/dt). The negative sign shows that as x increases (foot slides out), y decreases (top slides down).

Angle of Elevation Problems

For an observer at distance d from a vertical object of height h: tanθ = h/d. Differentiating (d constant): sec²θ × dθ/dt = (1/d)(dh/dt). Since sec²θ = 1 + tan²θ = 1 + h²/d², we can express dθ/dt in terms of known quantities.

Exam Tip: Never substitute numerical values for a variable BEFORE differentiating. The chain rule produces a relationship between rates; substituting too early gives an equation with zero derivatives (if you substitute a constant value for a changing quantity).
Exam Tip: For problems involving a right angle, Pythagoras is almost always the key relationship. For cone/cylinder filling problems, use the volume formula and eliminate one variable via similar triangles.

Mastery Practice

  1. A sphere’s volume is increasing at 10 cm³/s. Find dr/dt when r = 5 cm. Fluency

    Use V = 4πr³/3. Give your answer in exact form.

  2. Sliding ladder: a 5 m ladder leans against a wall. The base slides out at 0.5 m/s. Find the rate at which the top slides down when the base is 3 m from the wall. Fluency

  3. A conical tank (height 10 cm, radius 4 cm) fills with water at 2 cm³/s. Find dh/dt when h = 6 cm. Fluency

  4. Given x² + y² = 25 and dx/dt = 3, find dy/dt when x = 4. Fluency

  5. A kite flies at 60 m height; wind carries it horizontally at 4 m/s. Find the rate at which string length increases when 100 m of string is out. Understanding

  6. Two cars leave an intersection — one north at 60 km/h, one east at 80 km/h. How fast is the distance between them increasing 1 hour later? Understanding

  7. y = x³ − 3x. If x increases at 2 units/s, find the rate of change of y when x = 2. Understanding

  8. A rectangle has constant area 100 cm². Length increases at 3 cm/s. Find the rate of width decrease when length is 20 cm. Understanding

  9. Water drains from a hemispherical bowl of radius R at rate Q = k√h. Set up the differential equation for dh/dt. Problem Solving

    The volume of water in a hemispherical bowl of radius R to depth h is V = πh²(R − h/3). Use this to find dV/dh, then apply the chain rule.

  10. A rocket at height h(t) is observed from 1 km away. Angle of elevation θ satisfies tanθ = h/1000. If h increases at 200 m/s, find dθ/dt when θ = 30°. Problem Solving

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