Fall 2025: Chapter 3.9 - Related Rates

Welcome to a recap of Professor Baker's Math Class! This post focuses on Chapter 3.9, which covers related rates. Related rates problems involve finding the rate at which a quantity changes by relating it to other quantities whose rates of change are known. We'll explore both geometric and trigonometric examples to solidify your understanding.

Key Concepts and Problem Solving Strategies

Here's a breakdown of the core concepts and a suggested problem-solving strategy for tackling related rates problems:

  1. Read the problem carefully. Understanding the problem is the first and most crucial step.
  2. Draw a diagram (if possible). Visualizing the problem can significantly aid in understanding the relationships between variables.
  3. Introduce notation. Assign symbols to all quantities that are functions of time. For instance, let $V$ represent volume, $r$ represent radius, and $t$ represent time.
  4. Express the given information and the required rate in terms of derivatives. For example, if the volume is increasing at a rate of 10 $cm^3/s$, then $\frac{dV}{dt} = 10$.
  5. Write an equation that relates the various quantities of the problem. This is often the most challenging step. Use geometry, trigonometry, or other relevant formulas to establish the relationship.
  6. Use the Chain Rule to differentiate both sides of the equation with respect to $t$. Remember that each variable is a function of time.
  7. Substitute the given information into the resulting equation and solve for the unknown rate.

Examples

Example 1: Inflating a Balloon

Air is being pumped into a spherical balloon so that its volume increases at a rate of 100 $cm^3/s$. How fast is the radius of the balloon increasing when the diameter is 50 cm?

We are given $\frac{dV}{dt} = 100$ $cm^3/s$ and $d = 50$ cm (so $r = 25$ cm). We want to find $\frac{dr}{dt}$.

The volume of a sphere is $V = \frac{4}{3}\pi r^3$. Differentiating with respect to $t$ using the chain rule, we get:

$\frac{dV}{dt} = 4\pi r^2 \frac{dr}{dt}$

Solving for $\frac{dr}{dt}$:

$\frac{dr}{dt} = \frac{1}{4\pi r^2} \frac{dV}{dt} = \frac{1}{4\pi (25)^2} (100) \approx 0.0127$ $cm/sec$

Example 2: Sliding Ladder

A ladder 10 ft long rests against a vertical wall. If the bottom of the ladder slides away from the wall at a rate of 4 ft/s, how fast is the top of the ladder sliding down the wall when the bottom of the ladder is 6 ft from the wall?

Let $x$ be the distance from the wall to the bottom of the ladder and $y$ be the distance from the ground to the top of the ladder. We have $x^2 + y^2 = 10^2 = 100$. We are given $\frac{dx}{dt} = 4$ ft/s and want to find $\frac{dy}{dt}$ when $x = 6$ ft. When $x = 6$, $y = \sqrt{100 - 6^2} = 8$ ft.

Differentiating $x^2 + y^2 = 100$ with respect to $t$, we get:

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

Substituting the known values:

$2(6)(4) + 2(8) \frac{dy}{dt} = 0$

Solving for $\frac{dy}{dt}$:

$\frac{dy}{dt} = \frac{-48}{16} = -3$ ft/s

The negative sign indicates that the top of the ladder is sliding down the wall.

Keep practicing, and you'll master related rates in no time!