3.11: Related Rates – Calculus Study Notes

3.11: Related Rates

Introduction to Related Rates

Related rates problems involve finding the rate at which one quantity changes with respect to time, given information about the rate of change of another related quantity. These problems are a key application of derivatives in calculus, particularly in situations where multiple variables are related by an equation and each variable changes over time.

• Rate of Change: The rate of change of a function $f(t)$ with respect to time $t$ is given by its derivative $\frac{df(t)}{dt}$.

• Related Rates: When two or more quantities are related by an equation, their rates of change with respect to time are also related through differentiation.

General Strategy for Solving Related Rates Problems

1. Identify all given quantities and the rates at which they change.

2. Write an equation relating the variables involved.

3. Differentiate both sides of the equation with respect to time $t$ using the chain rule.

4. Substitute all known values and solve for the required rate.

Examples of Related Rates Problems

Example 1: Air Pumped into a Spherical Balloon

Problem: Air is being pumped into a spherical balloon so that its volume increases at a rate of 100 cm3/sec. How fast is the radius of the balloon increasing when the diameter is 50 cm?

• Given: $\frac{dV}{dt} = 100$ cm3/sec, diameter = 50 cm $\Rightarrow$ radius $R = 25$ cm

• Volume of a sphere: $V = \frac{4}{3}\pi R^3$

• Differentiating both sides with respect to $t$:

$\frac{dV}{dt} = \frac{4}{3}\pi 3R^2 \frac{dR}{dt} = 4\pi R^2 \frac{dR}{dt}$

• Substitute values:

$100 = 4\pi (25)^2 \frac{dR}{dt}$

• Solve for $\frac{dR}{dt}$:

$\frac{dR}{dt} = \frac{1}{25\pi}$ cm/sec

Interpretation: The radius increases at a rate of $\frac{1}{25\pi}$ cm/sec when the diameter is 50 cm.

Example 2: Melting Ice Cube

Problem: An ice cube that is 4 cm wide is melting at a rate of 3 cm3 per minute. How fast is the length of the side of the cube decreasing?

• Given: $\frac{dV}{dt} = -3$ cm3/min (negative because the volume is decreasing), $L = 4$ cm

• Volume of a cube: $V = L^3$

• Differentiating both sides with respect to $t$:

$\frac{dV}{dt} = 3L^2 \frac{dL}{dt}$

• Substitute values:

$-3 = 3(4)^2 \frac{dL}{dt}$ $-3 = 3 \cdot 16 \frac{dL}{dt}$ $\frac{dL}{dt} = -\frac{1}{16}$ cm/min

Interpretation: The side length of the cube is decreasing at a rate of $\frac{1}{16}$ cm/min.

Example 3: Expanding Circular Plate

Problem: When a circular plate of metal is heated in an oven, its radius increases at a rate of 3 cm/min. At what rate is the plate's area increasing when the radius is 10 cm?

• Given: $\frac{dR}{dt} = 3$ cm/min, $R = 10$ cm

• Area of a circle: $A = \pi R^2$

• Differentiating both sides with respect to $t$:

$\frac{dA}{dt} = \pi \cdot 2R \frac{dR}{dt}$

• Substitute values:

$\frac{dA}{dt} = \pi \cdot 2 \cdot 10 \cdot 3 = 60\pi$ cm2/min

Interpretation: The area is increasing at a rate of $60\pi$ cm2/min when the radius is 10 cm.

Example 4: Piston in a Cylindrical Chamber

Problem: A piston with a radius of 5 cm starts moving into the cylindrical chamber at a constant speed of 3 cm/s. What is the rate of change of the volume of the cylinder when the piston is 6 cm from the base?

• Given: $\frac{dH}{dt} = -3$ cm/s (height is decreasing), $R = 5$ cm, $H = 6$ cm

• Volume of a cylinder: $V = \pi R^2 H$

• Differentiating both sides with respect to $t$:

$\frac{dV}{dt} = \pi R^2 \frac{dH}{dt}$

• Substitute values:

$\frac{dV}{dt} = \pi (5)^2 (-3) = -75\pi$ cm3/s

Interpretation: The volume of the cylinder is decreasing at a rate of $75\pi$ cm3/s when the piston is 6 cm from the base.

Example 5: Sliding Ladder Problem

Problem: A 10 ft long ladder rests against a vertical wall. If the bottom of the ladder slides away from the wall at a rate of 1 ft/sec, how fast is the top of the ladder sliding down the wall when the bottom of the ladder is 6 ft from the wall?

• Given: $\frac{dx}{dt} = 1$ ft/sec, $x = 6$ ft, ladder length $= 10$ ft

• Relationship: $x^2 + y^2 = 10^2$

• Differentiating both sides with respect to $t$:

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

• At $x = 6$ ft, $y = \sqrt{100 - 36} = 8$ ft

• Substitute values:

$2 \cdot 6 \cdot 1 + 2 \cdot 8 \cdot \frac{dy}{dt} = 0$ $12 + 16 \frac{dy}{dt} = 0$ $\frac{dy}{dt} = -\frac{12}{16} = -\frac{3}{4}$ ft/sec

Interpretation: The top of the ladder is sliding down the wall at a rate of $\frac{3}{4}$ ft/sec when the bottom is 6 ft from the wall.

Summary Table: Common Related Rates Formulas

Shape/Scenario	Formula	Related Rate Equation
Sphere (radius $R$)	$V = \frac{4}{3}\pi R^3$	$\frac{dV}{dt} = 4\pi R^2 \frac{dR}{dt}$
Cube (side $L$)	$V = L^3$	$\frac{dV}{dt} = 3L^2 \frac{dL}{dt}$
Circle (radius $R$)	$A = \pi R^2$	$\frac{dA}{dt} = 2\pi R \frac{dR}{dt}$
Cylinder (radius $R$, height $H$)	$V = \pi R^2 H$	$\frac{dV}{dt} = \pi R^2 \frac{dH}{dt}$ (if $R$ constant)
Right Triangle (legs $x$, $y$; hypotenuse $c$)	$x^2 + y^2 = c^2$	$2x \frac{dx}{dt} + 2y \frac{dy}{dt} = 0$

Key Points to Remember

• Always relate all variables before differentiating.

• Use the chain rule when differentiating composite functions.

• Pay attention to the sign of the rates (increasing vs. decreasing quantities).

• Substitute all known values after differentiating, not before.