Related Rates in Calculus: Study Notes and Examples

Related Rates Problems in Calculus

Related rates problems are a key application of differential calculus, where two or more quantities are related and change with respect to time. These problems often require the use of the chain rule to relate the rates of change of different variables.

Key Concepts

• Related Rates: Problems involving two or more variables that are functions of time, where the rate of change of one variable is related to the rate of change of another.

• Chain Rule: A fundamental rule in calculus used to differentiate composite functions. In related rates, it allows us to relate the derivatives of different variables with respect to time.

• Implicit Differentiation: Often used when variables are related by an equation, but not explicitly solved for one variable in terms of the other.

General Steps for Solving Related Rates Problems

1. Draw a diagram if possible, and assign variables to all quantities that change over time.

2. Write an equation relating the variables.

3. Differentiate both sides of the equation with respect to time (t), using the chain rule as needed.

4. Substitute all known values and rates at the instant of interest.

5. Solve for the required rate.

Examples and Applications

Pebble Causing Ripples

When a pebble is dropped into a pond, it creates circular ripples. The area of the circle and the radius both change over time.

• Formula: The area of a circle is $A = \pi r^2$.

• Variables: Both area (A) and radius (r) are functions of time (t).

• Related Rates Equation: $\frac{dA}{dt} = 2\pi r \frac{dr}{dt}$

• Application: If the radius increases at a known rate, this formula allows us to find how fast the area is increasing at a particular instant.

Sand Falling Into An Hourglass

• Formula: The volume of sand in the lower part of the hourglass can be modeled as a function of time.

• Variables: Volume (V) and height (h) are changing over time.

• Application: If the rate at which sand falls is known, we can relate it to the rate at which the height of sand increases.

Sliding Ladder

• Formula: For a ladder of length L leaning against a wall, $x^2 + y^2 = L^2$, where x is the distance from the wall to the base, and y is the height on the wall.

• Variables: Both x and y change over time.

• Related Rates Equation: $2x \frac{dx}{dt} + 2y \frac{dy}{dt} = 0$

• Application: If the base slides away at a known rate, we can find how fast the top is sliding down.

Worked Examples

Example 2: Implicit Differentiation with Related Rates

If $x^2y^2 = 8$ and $\frac{dx}{dt} = \frac{1}{2}$ and $\frac{dy}{dt} = \frac{1}{3}$, what is $\frac{dz}{dt}$ when $x = 2$ and $y = \frac{1}{2}$?

• Step 1: Differentiate both sides with respect to t using the product and chain rules.

• Step 2: Substitute the given values to solve for the unknown rate.

Example 3: Expanding Circular Plate

When a circular plate of metal is heated, its radius increases at a rate of 0.01 cm/min. At what rate is the plate's area changing when the radius is 50 cm?

• Given: $\frac{dr}{dt} = 0.01$ cm/min, $r = 50$ cm

• Formula: $A = \pi r^2$

• Related Rate: $\frac{dA}{dt} = 2\pi r \frac{dr}{dt}$

• Calculation: Substitute values to find $\frac{dA}{dt}$.

Example 4: Sliding Ladder Problem

A 25-foot ladder is leaning against a house. When the base is 20 feet from the house and moving away at 4 ft/sec, how fast is the top sliding down?

• Given: $L = 25$ ft, $x = 20$ ft, $\frac{dx}{dt} = 4$ ft/sec

• Equation: $x^2 + y^2 = 25^2$

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

• Solve for: $\frac{dy}{dt}$

Example 6: Cylinder Volume Change

The radius of a 6-inch deep cylinder increases at 0.1 inch every 3 minutes. How rapidly is the cylinder volume increasing when the bore diameter is 3.8 inches?

• Formula: $V = \pi r^2 h$ (h is constant at 6 inches)

• Given: $\frac{dr}{dt} = 0.1/3$ in/min, $d = 3.8$ in so $r = 1.9$ in

• Related Rate: $\frac{dV}{dt} = 2\pi r h \frac{dr}{dt}$

Example 7: Water Flowing from a Conical Reservoir

Water is flowing out of a conical reservoir at 500 cm3/min. When the water level is 5 cm deep, the radius is 10 cm and the radius is changing at 4 cm/min. How fast is the water level falling?

• Formula: $V = \frac{1}{3}\pi r^2 h$

• Given: $\frac{dV}{dt} = -500$ cm3/min, $r = 10$ cm, $h = 5$ cm, $\frac{dr}{dt} = 4$ cm/min

• Related Rate: $\frac{dV}{dt} = \frac{1}{3}\pi (2r h \frac{dr}{dt} + r^2 \frac{dh}{dt})$

• Solve for: $\frac{dh}{dt}$

Summary Table: Common Related Rates Formulas

Shape/Scenario	Formula	Related Rate Equation
Circle (Area)	$A = \pi r^2$	$\frac{dA}{dt} = 2\pi r \frac{dr}{dt}$
Sphere (Volume)	$V = \frac{4}{3}\pi r^3$	$\frac{dV}{dt} = 4\pi r^2 \frac{dr}{dt}$
Cylinder (Volume)	$V = \pi r^2 h$	$\frac{dV}{dt} = 2\pi r h \frac{dr}{dt} + \pi r^2 \frac{dh}{dt}$
Right Triangle (Pythagorean)	$x^2 + y^2 = L^2$	$2x \frac{dx}{dt} + 2y \frac{dy}{dt} = 0$
Cone (Volume)	$V = \frac{1}{3}\pi r^2 h$	$\frac{dV}{dt} = \frac{1}{3}\pi (2r h \frac{dr}{dt} + r^2 \frac{dh}{dt})$

Additional info: The original file references videos and interactive applets for visualizing related rates problems, which are common in calculus courses to help students understand dynamic changes in geometric quantities.