AP®︎/College Calculus AB
Course: AP®︎/College Calculus AB > Unit 4Lesson 1: Interpreting the meaning of the derivative in context
Analyzing problems involving rates of change in applied contexts
Differential calculus is all about instantaneous rate of change. Let's see how this can be used to solve real-world word problems.
One way to interpret the derivative of a function is that is the instantaneous rate of change of at . Let's see how this interpretation can be used to solve word problems.
Say a water tank is being filled and the volume, in liters, of the water in the tank after seconds is given by the linear function .
Function V sub 1 is graphed. The x-axis, labeled time in seconds, goes from negative 1 to 10. The y-axis is labeled volume in liters. The graph is a line. The line starts at (0, 0), moves upward through (6, 4).
The slope of the function, , represents its rate of change. In other words, the tank is being filled at a rate of liters per second.
The graph of function V sub 1 starts at (0, 0), moves upward through points (3, 2) and (6, 4), and ends in quadrant 1. An arrow representing + 3 seconds moves rightward from point (3, 2) to (6, 2). An arrow representing + 2 liters moves upward from (6, 2) to point (6, 4).
The rate of change of a linear function is always constant, which makes it relatively easy to reason about.
Now say a different tank is being filled, and this time the volume function isn't linear.
Function V sub 2 is graphed. The x-axis, labeled time in seconds, goes from negative 1 to 10. The y-axis is labeled volume in liters. The graph is a curve. The curve starts at (0, 0), moves upward through about (3, 0.9) and (7, 4.9), and ends in quadrant 1.
Notice how the graph's growth is gradual at first and becomes more steep towards the end. The rate of change of isn't constant.
If we want to analyze the rate of change of , we can talk about its instantaneous rate of change at any given point in time. The instantaneous rate of change of a function is given by the function's derivative.
For example, . Mathematically, this means that the slope of the line tangent to the graph of when is . What does this mean in the context of our water tank?
The graph of function V sub 2 has a tangent line starting in quadrant 4, moving upward, touching the curve at (5, 2.5), and ending in quadrant 1.
The tangent line's slope indicates the curve's slope at that particular point in time. Since we already saw how slope gives us the rate of change, we can interpret as follows:
At seconds, the tank is being filled at a rate of liter per second.
Notice a couple things about this interpretation:
First, the rate is given in liters per second. The units of a derivative are always a ratio of the dependent quantity (e.g. liters) over the independent quantity (e.g. seconds).
Second, the rate is given for a specific point in time (e.g. seconds). This is because it's instantaneous. Take another point in time, and the rate might be different. Look at an interval of time, and the rate isn't constant.
In Problem set 1 we will analyze the following context:
Lindsay is walking home from school. Her distance from school, in meters, after minutes is modeled by the differentiable function .
What units should we use to measure ?
gives the height of a tree, in centimeters, weeks after it was planted.
Four students were asked to interpret the meaning of in this context.
Can you match the teacher's comments to the interpretations?
Common mistake: Forgetting to include units, or using incorrect units
Remember: When analyzing problems in applied contexts, remember we should always use units.
For example, in Problem 2, gets an input that is measured in weeks and gives an output that is measured in centimeters. Its derivative also gets an input that is measured in weeks, but its output is the rate centimeters per week.
Another common mistake: Using phrases that refer to “over an interval of time” rather than “at a point in time”
Derivatives are all about instantaneous rate of change. Therefore, when we interpret the rate of a function given the value of its derivative, we should always refer to the specific point when that rate applies.
Solving problems that involve instantaneous rate of change
Consider the following problem:
Carlos has taken an initial dose of a prescription medication. The amount of medication, in milligrams, in Carlos's bloodstream after hours is given by the following function:
What is the instantaneous rate of change of the remaining amount of medication after hour?
The first thing that should pop up to us as we read this problem is that we are asked for the instantaneous rate of change of a quantity. This means we need to use derivatives.
The only function whose derivative we can use is , but let's make sure this is what we want: gives the amount of medication in Carlos's bloodstream over time, and we are asked for the instantaneous rate of change of that amount. So yes, we need :
We are asked for the instantaneous rate of change after hour, which means we need to evaluate at :
Finally, we need to remember to use units. Since gives an amount in milligrams for a given input in hours, the units in which we measure are milligrams per hours.
In conclusion, the instantaneous rate of change of the remaining amount of medication after hour is milligrams per hour.
gives the cost, in dollars, to shred pounds of confidential documents of a company.
What is the instantaneous rate of change of the costs when the weight of the documents is pounds?
Want more practice? Try this exercise.
Common mistake: Evaluating the original function rather than the derivative
Remember: When we are asked about the rate of change of a function , we want to look for the derivative . Evaluating at a point will not provide us any information about the rate of change of at that point.
Want to join the conversation?
- About this problem: "Lindsay is walking home from school. Her distance from school, in meters, after t minutes is modeled by the differentiable function D. What is the meaning of D′(3) = 80?"
The answer you identified as correct was "After 3 minutes, Lindsay is walking at a rate of 80 meters per minute."
Isn't it more accurate to say, "After 3 minutes, Lindsay's distance from school is increasing at a rate of 80 meters per minute"?
For example, she could be walking extremely fast but doing so on a trajectory that's not a straight line away from the school, and therefore the net rate of distance change is only 80m/min.(16 votes)
- Yep, I think you are right. The answer can be "After 3 minutes, Lindsay is walking at a rate of 80 meters per minute" if it is assumed her route to her home is a straight line.(9 votes)
- what rule did you use in the prime of M(t)=20(e)^-0.8t(5 votes)
- It's the derivative of e^x which is e^x itself. But you have to multiply it with x's coefficient (chain rule) which is -0.8. So you just multiply 20.e^-0.8t with -0.8t which gives us -16.e^-0.8t(11 votes)
- I have a question that's really bothering me.
Ok so in problem 3, the function:
C(w) = 0.001w^3 −0.15w^2 + 7.5w
Well it tells us the cost to shred "w" pounds of paper.
The function C' however is supposed to tell us the dollars per pound with weight w.
Here's my question: By using the original function, we can find the cost in dollars to shred 10 pounds of paper right? After doing the math we get 61 dollars. So what's the cost per pound of paper? Well that would be 61/10 which is 6.1 dollars per pound.
The derivative of C tells us the instantaneous rate of dollars per pound at w. When we plug 10 into the derivative function, we get 4.8 dollars per pound.
So is the cost per pound at w=10 6.1 dollars per pound or 4.8 dollars per pound? Obviously both should be true yet they conflict each other.(5 votes)
- That is a great question Andrew Escobedo. It all has to deal with approximating slope and instantaneous slope.
We find that indeed, C(10) = 61, meaning that it costs $61 to shred 10 pounds of paper. When you divided by 10, you obtained the approximate rate of change, which is $6.1 dollars per pound.
C'(W) is the derivative of the function C and gives the instantaneous rate of change, or in this case cost per pound for any W > 0. C'(10) = 4.8, meaning that when at the moment the weight is 10 pounds, it cost exactly $4.8 per pound.
Think of it like this. You plot the function C(w) from [0,10]. When you calculated C(10)/10, you calculated the average slope, right?
m = [C(10)-C(0)]/(10 - 0) = C(10)/10
That is the average rate of change.
When you calculate C'(10), that is the instantaneous rate of change at W = 10.
Hope that helps! Best of luck!(12 votes)
- In the practice questions there is an appearance of an e in the equations. What is that?(4 votes)
- e is like pi, a number that they gave a symbol.(5 votes)
- Is there any way that you could find the rate of change for a nonlinear equation without taking the instantaneous or average rate of change? Like, is it possible for there to be some slope that is defined for all the points on a parabola without having an average slope?(3 votes)
- That's what derivatives are. Finding the derivative of a function allows you to know its rate of change at any point.(4 votes)
- I understand the derivative is the instantaneous rate of change, but this doesn't seems helpful because it doesn't suggest the tendency of other outputs
How could be really useful in the real world?(2 votes)
- Derivative does have more real world applications since it is describing how fast a certain function is changing. For example, velocity is the derivative of the displacement respect to time, and acceleration is the derivative of the velocity respect to time. Think about physics applications, most of them are using calculus.(3 votes)
- In problem 2, the correct answer for "At 5 weeks, the rate at which the tree is growing is 3." is that you didn't use the correct units. I'm confused as to why you can have incorrect units if there are no units at all?(1 vote)
- You can have incorrect units if there are no units at all, since the only units they were using were centimeters it's incorrect.(2 votes)
- (F^-1)’ (5)=1.5? how would u interpret this in a sentence(1 vote)
- Literally, it says:
"Evaluated at 𝑥 = 5, the derivative of the inverse function of 𝐹(𝑥) is equal to 1.5"
– – –
Let 𝐹(𝑥) be the number of ants in an anthill of mass 𝑥 (in grams).
Then 𝐹 ⁻¹(𝑥) is the mass of an anthill with 𝑥 ants.
And (𝐹 ⁻¹)'(𝑥) is the instantaneous rate of change of the mass of an anthill that houses 𝑥 ants.
So, (𝐹 ⁻¹)'(5) = 1.5 means that if there are only 5 ants, then the instantaneous rate of change of the anthill's mass is 1.5 grams/ant.(1 vote)
- In one of the answer choices to a problem dealing with time the answer choice is "After 5 seconds, Herman's acceleration is 1.5 meters per second squared." & "After 5 seconds, Herman's velocity is 1.5 meters per second." How do we know when the answer is velocity or acceleration and when the second is squared?(1 vote)
- what actually is a derivitive im still confused on that(1 vote)
- The derivative is simply the rate of change of the dependent variable, but for a very, very small change in the independent variable.
Think of a simple example. Say you went from your home to the market. The market is 4 km away and you arrived there in 2 hours. Now, your average speed would be given by distance/time = 4/2 = 2km/hr. This isn't the speed throughout the journey but as the name says, is the average speed.
Now, say I want the exact speed at t = 1 hour. In other words, when you finish travelling for exactly 1 hour, what is your speed at that instant? Clearly, averages won't help. So, this is what we do.
Firstly, we assume the distance to be a function of time (i.e. as time changes, distance does as well). Let the function be distance = s(t). Now, we take the distance covered between two instants of time. One being t = 1 hour and the other being t = (1+h) hours (where "h" is a very, very small interval of time). Now, when t goes from 1 to 1+h, your distance goes from s(1) to s(1+h). See that we now have a change in the independent variable (time) and a subsequent change in the dependent variable (distance). So, we can find the average speed between these two time intervals, just like we did in the earlier problem. So, we get speed = [s(1+h) - s(1)]/[1+h-1]. This is just coming from (change in distance)/(change in time). Simplifying this, we get speed = [s(1+h) - s(1)]/h. Now, this is the average speed between times t = 1 and t = 1+h. But, we don't want that. We want the exact speed at t = 1. So, we want t=1+h to coincide with t=1. How? Well, make h = 0...right? Well, see that our formula has an "h" in the denominator. if h = 0, the speed becomes undefined. So, we need another approach.
Well, what about limits? Let's take the limit as h approaches 0. That shouldn't cause problems, and it'll give us the speed at that instant. So, we have speed = lim(h-->0)[s(1+h) - s(1)]/h. And this beautiful thing right here is the very definition of a derivative. As mentioned at the start, the derivative is the rate of change of the dependent variable, but for a very, very small change in the independent variable. See how this formula gives us exactly that!!(1 vote)