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Μάθημα: Διαφορικός Λογισμός > Ενότητα 4

Μάθημα 1: 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

f^{'}

of a function

f

is that

f^{'} (k)

is the instantaneous rate of change of

f

x = 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_{1} (t) = \frac{2}{3} t

The slope of the function,

\frac{2}{3}

, represents its rate of change. In other words, the tank is being filled at a rate of

\frac{2}{3}

liters per second.

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_{2} (t) = 0.1 t^{2}

isn't linear.

Notice how the graph's growth is gradual at first and becomes more steep towards the end. The rate of change of

V_{2}

isn't constant.

If we want to analyze the rate of change of

V_{2}

, 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_{2}^{'} (t) = 0, 2 t

For example,

V_{2}^{'} (5) = 1

. Mathematically, this means that the slope of the line tangent to the graph of

V_{2}

when

x = 5

1

. What does this mean in the context of our water tank?

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_{2}^{'} (5) = 1

as follows:

At $t = 5$ ‍ seconds, the tank is being filled at a rate of $1$ ‍ 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.

t = 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 $D^{'} (t)$ ‍?

Πρόβλημα 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^{'} (5) = 3

in this context.

Can you match the teacher's comments to the interpretations?

1

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^{'}

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:

$M (t) = 20 \cdot e^{^{- 0, 8 t}}$ ‍

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^{'}

M^{'} (t) = - 16 \cdot e^{- 0, 8 t}

We are asked for the instantaneous rate of change after $1$ ‍ hour, which means we need to evaluate

M^{'}

t = 1

M^{'} (1) = - 16 \cdot e^{- 0, 8} \approx - 7, 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^{'}

are milligrams per hours.

In conclusion, the instantaneous rate of change of the remaining amount of medication after

1

hour is $- 7.2$ ‍ milligrams per hour.

Πρόβλημα 3

C

gives the cost, in dollars, to shred

w

pounds of confidential documents of a company.

C (w) = 0,001 w^{3} - 0, 15 w^{2} + 7, 5 w

What is the instantaneous rate of change of the costs when the weight of the documents is $10$ ‍ 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

f

, we want to look for the derivative

f^{'}

. Evaluating

f

at a point will not provide us any information about the rate of change of

f

at that point.

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