Calculus I - Week 6

Related Rates

Related rates problems use implicit differentiation with respect to time to find how fast one quantity is changing when the rate of change of another is known. These appear constantly in physics and engineering: expanding circles, falling ladders, draining tanks, and more.

Learning Objectives

- Set up and solve related rates problems step by step. - Identify which quantity is changing and which rate is given. - Draw and label a diagram for each problem type. - Differentiate a geometric formula with respect to time. - Substitute known values only after differentiating. - Solve related rates problems involving circles, triangles, volume, and distance.

1. What Is a Related Rate?

A related rate problem gives you the rate of change of one variable and asks for the rate of change of another — both with respect to time. For example: a circular puddle is expanding. If you know how fast the radius is growing, how fast is the area growing? The key: radius and area are related by A = pi*r^2. Differentiating both sides with respect to time t gives the relationship between dA/dt and dr/dt.

2. The Step-by-Step Strategy

Step 1: Draw a diagram and label all quantities with variables. Step 2: Write an equation relating those variables. Step 3: Differentiate both sides with respect to t. Step 4: Substitute the known values. Step 5: Solve for the unknown rate. Do not substitute values before differentiating. This is the most common mistake.

3. Example: Expanding Circle

A stone is dropped in a pond. The radius of the ripple grows at 3 cm/s. How fast is the area growing when r = 5 cm? Relation: A = pi*r^2. Differentiate: dA/dt = 2*pi*r * (dr/dt). Substitute r = 5, dr/dt = 3: dA/dt = 2*pi*5*3 = 30*pi cm^2/s.

4. Example: Sliding Ladder

A 10-meter ladder leans against a wall. The bottom slides away at 1 m/s. How fast is the top sliding down when the bottom is 6 m from the wall? Pythagorean relation: x^2 + y^2 = 100. Differentiate: 2x(dx/dt) + 2y(dy/dt) = 0. When x = 6: y = sqrt(100 - 36) = 8. Substitute x = 6, y = 8, dx/dt = 1: 2(6)(1) + 2(8)(dy/dt) = 0 12 + 16(dy/dt) = 0 dy/dt = -3/4 m/s. The top slides down at 3/4 m/s.

5. Example: Draining Tank

Water drains from a conical tank (radius 5 m, height 10 m) at 2 m^3/min. How fast is the water level dropping when h = 4 m? At height h, the radius r satisfies r/h = 5/10 = 1/2, so r = h/2. Volume: V = (1/3)*pi*r^2*h = (1/3)*pi*(h/2)^2*h = pi*h^3/12. Differentiate: dV/dt = (pi/4)*h^2*(dh/dt). Substitute dV/dt = -2, h = 4: -2 = (pi/4)*16*(dh/dt) -2 = 4*pi*(dh/dt) dh/dt = -1/(2*pi) m/min.

6. Example: Two Cars

Car A drives north at 40 mph; Car B drives east at 30 mph. Both start from the same intersection. How fast is the distance between them increasing after 1 hour? After 1 hour: car A is 40 miles north, car B is 30 miles east. Distance: D = sqrt(x^2 + y^2) = sqrt(900 + 1600) = 50 miles. Relation: D^2 = x^2 + y^2. Differentiate: 2D(dD/dt) = 2x(dx/dt) + 2y(dy/dt). Substitute D=50, x=30, y=40, dx/dt=30, dy/dt=40: 2(50)(dD/dt) = 2(30)(30) + 2(40)(40) 100(dD/dt) = 1800 + 3200 = 5000 dD/dt = 50 mph.

7. Example: Shadow Length

A person 2 m tall walks away from a 6 m lamppost at 1.5 m/s. How fast is their shadow lengthening? Let x = distance from lamppost, s = shadow length. By similar triangles: 6/(x + s) = 2/s. 6s = 2(x + s) 4s = 2x s = x/2. Differentiate: ds/dt = (1/2)(dx/dt) = (1/2)(1.5) = 0.75 m/s.

8. Example: Balloon

Air is pumped into a spherical balloon at 100 cm^3/s. How fast is the radius growing when r = 5 cm? V = (4/3)*pi*r^3. dV/dt = 4*pi*r^2*(dr/dt). Substitute dV/dt = 100, r = 5: 100 = 4*pi*25*(dr/dt) 100 = 100*pi*(dr/dt) dr/dt = 1/pi cm/s.

9. Setting Up the Geometry

Many related rates problems require geometry: - Pythagorean theorem: x^2 + y^2 = c^2 (right triangles) - Circle: A = pi*r^2, C = 2*pi*r - Sphere: V = (4/3)*pi*r^3 - Cone: V = (1/3)*pi*r^2*h (use similar triangles to relate r and h) - Cylinder: V = pi*r^2*h Always simplify before differentiating when possible — for cones, substitute r = h*(R/H) first.

10. Common Mistakes

- Substituting values before differentiating with respect to t. - Forgetting the chain rule when r, x, or h are functions of t. - Missing a negative sign when a quantity is decreasing. - Mixing up which rate is given and which is asked for. - Not simplifying the geometric relation before differentiating (especially cones).