If you're seeing this message, it means we're having trouble loading external resources on our website.

If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.

## AP®︎/College Calculus AB

### Unit 4: Lesson 1

Interpreting the meaning of the derivative in context

# Analyzing problems involving rates of change in applied contexts

AP.CALC:
CHA‑3 (EU)
,
CHA‑3.A (LO)
,
CHA‑3.A.1 (EK)
,
CHA‑3.A.2 (EK)
,
CHA‑3.A.3 (EK)
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 f, prime of a function f is that f, prime, left parenthesis, k, right parenthesis is the instantaneous rate of change of f at x, equals, k. 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 t seconds is given by the linear function V, start subscript, 1, end subscript, left parenthesis, t, right parenthesis, equals, start fraction, 2, divided by, 3, end fraction, t.
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, start fraction, 2, divided by, 3, end fraction, represents its rate of change. In other words, the tank is being filled at a rate of start fraction, 2, divided by, 3, end fraction 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 V, start subscript, 2, end subscript, left parenthesis, t, right parenthesis, equals, 0, point, 1, t, squared 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 V, start subscript, 2, end subscript isn't constant.
If we want to analyze the rate of change of V, start subscript, 2, end subscript, 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.
V, start subscript, 2, end subscript, prime, left parenthesis, t, right parenthesis, equals, 0, point, 2, t
For example, V, start subscript, 2, end subscript, prime, left parenthesis, 5, right parenthesis, equals, 1. Mathematically, this means that the slope of the line tangent to the graph of V, start subscript, 2, end subscript when x, equals, 5 is 1. 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 V, start subscript, 2, end subscript, prime, left parenthesis, 5, right parenthesis, equals, 1 as follows:
At t, equals, 5 seconds, the tank is being filled at a rate of 1 liter per second.
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. t, equals, 5 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.
Problem 1.A
In Problem set 1 we will analyze the following context:
Lindsay is walking home from school. Her distance from school, in meters, after t minutes is modeled by the differentiable function D.
What units should we use to measure D, prime, left parenthesis, t, right parenthesis?

Problem 2
H gives the height of a tree, in centimeters, t weeks after it was planted.
Four students were asked to interpret the meaning of H, prime, left parenthesis, 5, right parenthesis, equals, 3 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, H gets an input that is measured in weeks and gives an output that is measured in centimeters. Its derivative H, prime 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 t hours is given by the following function:
M, left parenthesis, t, right parenthesis, equals, 20, dot, e, start superscript, start superscript, minus, 0, point, 8, t, end superscript, end superscript
What is the instantaneous rate of change of the remaining amount of medication after 1 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 M, but let's make sure this is what we want: M 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 M, prime:
M, prime, left parenthesis, t, right parenthesis, equals, minus, 16, dot, e, start superscript, minus, 0, point, 8, t, end superscript
We are asked for the instantaneous rate of change after 1 hour, which means we need to evaluate M, prime at t, equals, 1:
M, prime, left parenthesis, 1, right parenthesis, equals, minus, 16, dot, e, start superscript, minus, 0, point, 8, end superscript, approximately equals, minus, 7, point, 2
Finally, we need to remember to use units. Since M gives an amount in milligrams for a given input in hours, the units in which we measure M, prime are milligrams per hours.
In conclusion, the instantaneous rate of change of the remaining amount of medication after 1 hour is minus, 7, point, 2 milligrams per hour.
Problem 3
C gives the cost, in dollars, to shred w pounds of confidential documents of a company.
C, left parenthesis, w, right parenthesis, equals, 0, point, 001, w, cubed, minus, 0, point, 15, w, squared, plus, 7, point, 5, w
What is the instantaneous rate of change of the costs when the weight of the documents is 10 pounds?