4.3 Rates of Change in Applied Contexts other than Motion

4.3 Rates of Change in Applied Contexts Other than Motion

Previously, you learned how to apply rates of change to solve rectilinear motion problems involving position, speed, velocity, and acceleration. In this section, you’ll learn about rates of change in other contexts (non-motion problems).

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📉Rates of Change in Applied Contexts Other than Motion

When trying to figure out the meaning of a rate of change or derivative in a context other than motion, you simply have to understand what the given function models.

If a problem states that “$f(x)$ gives the volume, in liters, of the water remaining in a tank $t$ minutes after the drain is opened”, this indicates $f(x)$ models volume with respect to time in minutes.

For example, $f(15)$ would model the volume of water in the tank $15$ minutes after opening the drain. Because the derivative is the rate of change, the derivative of this function is volume (of water drained) per minute. Therefore, if asked for the volume of water drained per minute at a certain point in time, again say $15$ minutes after first opening the drain, you simply need to find $f'(15)$.

✏️ Rates of Change: Walkthrough

Karen is pogo stick jumping. The following function gives her height above ground, in feet, $t$ seconds after jumping:

What is the instantaneous rate of change of Karen’s height after $10$ seconds?

Because the problem is asking what the instantaneous rate of change at a certain point in time is, this indicates that we should calculate the derivative of the function at that point, which in this case is $10$ seconds. Thus we want to calculate $H'(10)$.

$H'(10)=H'(t)$ evaluated at $t=10$.

Since the function $H(t)$ models the height above ground in feet with respect to time in seconds, the unit for the derivative is feet per second, so the answer is $0.162$ feet per second.

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📝Practice Problems

Give it a try yourself!

❓Problems

Question 1:

Thomas posted on Instagram, which rapidly gained likes over time. The following function gives the number of likes $t$ days after posting:

What is the instantaneous rate of change of the number of likes $5$ days after the post was uploaded?

Question 2:

Jen is filling up her car’s gas tank. The following function gives the volume, in liters, of gas in the tank $t$ minutes after she starts pumping:

What is the instantaneous rate of change of the volume of gas $4$ minutes after she started filling up the tank?

✅Answers and Solutions

Question 1:

Because the problem is asking what the instantaneous rate of change at a certain point in time is, this indicates that you should calculate the derivative of the function at that point, which in this case is $5$ days.

$L'(5)=L'(t)$ evaluated at $t=5.$

Since the function $L(t)$ models the number of likes with respect to time in days, the unit for the derivative is likes per day, so the answer is $32.97$ likes per day.

Question 2:

Because the problem is asking what the instantaneous rate of change at a certain point in time is, this suggests you should calculate the derivative of the function at that point, which in this case is $4$ minutes.

$G'(4)=G'(t)$ evaluated at $t=4.$

$G'(t)=4$

$G'(4)=4$

Since the function $G(t)$ models the volume of gas in liters with respect to time in minutes, the unit for the derivative is liters per minute, so the answer is $4$ liters per minute.

You did it again! 🥳