Rates of Change in Applied Contexts other than Motion Chapter Notes | Calculus AB - Grade 9 PDF Download

In our previous discussions, we explored how rates of change apply to motion problems, such as those involving position, velocity, and acceleration. Now, let’s dive into how rates of change work in other real-world scenarios that don’t involve motion.

Rates of Change in Everyday Scenarios

To understand the rate of change or derivative in non-motion problems, the key is to grasp what the given function represents. For instance, if a function f(t) describes the volume of water (in liters) left in a tank t minutes after opening the drain, it models volume as a function of time. The value f(15) would tell you the volume of water remaining after 15 minutes. Since the derivative represents the rate of change, f'(15) would give the rate at which water is draining (in liters per minute) at that specific moment.

Example Walkthrough: Pogo Stick Jumping

Let’s consider Karen, who’s bouncing on a pogo stick. Her height above the ground (in feet) after t seconds is modeled by the function:
H(t) = 3 sin(t/10) + 1/2
Question: What is the instantaneous rate of change of Karen’s height after 10 seconds?
Solution: To find the instantaneous rate of change, we need the derivative of H(t) at t = 10, which is H'(10).
First, compute the derivative:
H'(t) = (3/10) cos(t/10)
Evaluate at t = 10:
H'(10) = (3/10) cos(1) ≈ 0.162
Since H(t) measures height in feet with respect to time in seconds, the derivative’s units are feet per second. Thus, Karen’s height is changing at approximately 0.162 feet per second after 10 seconds.

Practice Problems to Test Your Skills

Problem 1: Social Media Likes
Thomas shares a post on Instagram, and the number of likes t days after posting is given by:
L(t) = 200 · e^(0.1t)
Question: What is the instantaneous rate of change of likes 5 days after the post?
Solution: We need L'(5), the derivative of L(t) at t = 5.
Compute the derivative:
L'(t) = 200 * 0.1 * e^(0.1t) = 20 e^(0.1t)
Evaluate at t = 5:
L'(5) = 20 e^(0.5) ≈ 32.97
Since L(t) represents likes with respect to days, the derivative’s units are likes per day. The answer is approximately 32.97 likes per day.

Problem 2: Filling a Gas Tank
Jen is filling her car’s gas tank, and the volume of gas (in liters) after t minutes is given by:
G(t) = 300 + 4t
Question: What is the instantaneous rate of change of the gas volume 4 minutes after she starts?
Solution: We need G'(4), the derivative of G(t) at t = 4.
Compute the derivative:
G'(t) = 4
Evaluate at t = 4:
G'(4) = 4
Since G(t) measures volume in liters with respect to time in minutes, the derivative’s units are liters per minute. The answer is 4 liters per minute.

Key Terms to Understand

• Derivative: A measure of how a function changes at a specific point, showing how one quantity responds to small changes in another.
• Negative: Refers to values below zero, often indicating a decreasing trend or opposite direction.
• Rates of Change: Describes how quickly a quantity changes over time or relative to another variable, capturing the speed of change.