# 4.3: Applications of Derivatives

**At Grade**Created by: CK-12

## Related Rates

CONTENT

This lesson is a kind of tour through implicit differentiation. It may be best to understand and teach these ideas based upon the idea of differentiation as a kind of operator. That is, if we are given some equality among any number of different variables like:

\begin{align*}a^2+b^2=c^2\end{align*}

then we can operate on both sides of this equation with the derivative operator \begin{align*}\frac{d}{da}\end{align*}

\begin{align*}2a + 2b \frac{db}{da} = 2c \frac{dc}{da} \Rightarrow \frac{dc}{da} = \frac{a}{c} + \frac{b}{c} \frac{db}{da} \end{align*}

This equality should be interpreted as explaining how the rates of change for the side lengths \begin{align*}c\end{align*}

Alternatively, we could have operated on both sides with the derivative operator \begin{align*}\frac{d}{db}\end{align*}

PROCESS

Related rates are readily applicable to real-world problems. For example, how does the radius of a balloon or a car tire change as air is pumped into it? How fast does the amount of carbon-dioxide in the atmosphere increase for a given rate human population increase? How much does a star become brighter as we increase its magnification? All of these questions are questions of related rates, and can be used to make this material more interesting.

A nice way to teach this is to have students actually experiment with the formulas. For example, the students could be split into groups of three to study the formula for a right triangle. One member would be the referee and would need some kind of a stopwatch. The other two students would stand in one corner and walk away from each other along the wall measuring the length to the next corner in steps. Ideally they should coordinate so that their step size is about the same.

The distances along the wall and the diagonal distance between the students at opposite corners should be recorded. Then the referee should count or clap off seconds and the other two students should walk along the wall back towards the corner. One should walk a rate of one step per second and the other at a rate of one step every two seconds. After \begin{align*}4 \mathrm {seconds}\end{align*}

\begin{align*}a \frac{da}{dt} + b \frac{db}{dt} = c \frac{dc}{dt}\end{align*}

where \begin{align*}a\end{align*}

PRODUCTS

Word problems are the classic approach to testing this material, but it should be recognized that some students are much better at these than others. For the students who struggle putting words into equations, it may be best to test this material purely using equations. Questions like the following are good for these students:

1. Suppose the variables \begin{align*}x, y\end{align*}

a. What is \begin{align*}\frac{dz}{dx}\end{align*}

b. When \begin{align*}x=1,y=0\end{align*}

2. The volume of a sphere \begin{align*}V\end{align*}

a. Find a formula for \begin{align*}A\end{align*}

b. If the radius of a unit sphere \begin{align*}(r=1)\end{align*}

## Extrema and the Mean Value Theorem

CONTENT

In order to teach effectively about extrema, Rolle’s Theorem, and the mean value theorem it is necessary to use geometric intuition and good pictures throughout. The language in the description of a **maximum** and a **minimum** is quite complicated but the concept could not be clearer. Forgetting endpoints for a moment, a function has a maximum at\begin{align*}x\end{align*}if its graph is peaked and a minimum if its graph has a trough. To get technical, maxima and minima can also occur at the boundaries where the graph could end at a high point or a low point. However this can only occur if the domain includes some boundary point, and this is undesirable anyway with everyday calculus functions since the function cannot be differentiable there. For example, the function \begin{align*}f(x)=\sqrt{x}\end{align*} may be defined with the domain \begin{align*}x \ge 0\end{align*} so that the boundary point \begin{align*}x=0\end{align*} is included. Then the point \begin{align*}x=0\end{align*} is easily seen to be a minimum of the function although the function fails to be differentiable there. Much more useful and interesting are *interior* maxima and minima.

The key to understanding everything in this lesson is that the function goes from increasing to decreasing at an interior maximum, and from decreasing to increasing at an interior minimum. This means that the derivative goes from positive to negative in the former, and from negative to positive in the latter. However, for a continuous derivative this is only possible if it is zero at the point in question. That is, a differentiable maximum/minimum must have horizontal tangent line with \begin{align*}\mathrm {slope} = 0\end{align*}. The proof of this “theorem” is given in the text but is intuitive in the following sense: If the derivative were positive at a max or min then the slope at the point would be positive and so the function would be increasing through the point in question. That is, the values would be slightly smaller to the left (so it couldn’t be a minimum) and slightly larger to the right (so it couldn’t be a maximum). The same argument with some elements reversed shows that if the derivative is negative the point again can neither be a maximum nor a minimum. So we are left with the fact that the derivative must be identically zero. This is a very simple geometric concept and students will have the best chance at understanding it in this context. The proof itself will make much more sense when it is clear that this picture is just being put into symbols.

Rolle’s Theorem is not proven in the book but the idea is extremely geometric and the opportunity to show a simple picture in class should not be missed. The point is that if a nice (differentiable) function over \begin{align*}[a,b]\end{align*} starts and ends at the same height \begin{align*}f(a)=f(b)\end{align*}, then it is either flat between or else changed directions somewhere in between. In both cases the derivative is zero somewhere in \begin{align*}[a,b]\end{align*}, either throughout if it is flat or at the point where it changes direction (a maximum or a minimum). The picture is:

The mean value theorem is proven by creating a function from \begin{align*}f\end{align*} that satisfies Rolle’s Theorem. Essentially we tilt the function \begin{align*}f(x)\end{align*} into a new function \begin{align*}g(x)\end{align*} that satisfies \begin{align*}g(a)=g(b)\end{align*}. The text describes the function \begin{align*}g(x)\end{align*} as the difference between \begin{align*}f(x)\end{align*} and the line joining \begin{align*}f(a)\end{align*} and \begin{align*}f(b)\end{align*}. An alternative way of explaining this function to students is to illustrate that our goal is to tilt \begin{align*}f\end{align*} so that we get a function whose values at \begin{align*}a\end{align*} and \begin{align*}b\end{align*} are equal. This amounts to subtracting off the *change* between \begin{align*}a\end{align*} and \begin{align*}b\end{align*} which can be done by subtracting any function which undergoes the same change between \begin{align*}a\end{align*} and \begin{align*}b\end{align*}. The simplest choice is to subtract any line with slope \begin{align*}m= \frac{f(b)-f(a)}{(b-a)}\end{align*} such as the line \begin{align*}y=m \ x\end{align*}. In the proof given the line connecting \begin{align*}f(a)\end{align*} and \begin{align*}f(b)\end{align*} is chosen instead, which clearly also has slope \begin{align*}m\end{align*}.

PROCESS

Teaching this material can be a little tricky since it is largely conceptual. The best technique may be to have students pair off so that one person will first explain how to prove that if a point is an interior extrema then its derivative is zero. The other person will then explain how to prove the mean value theorem. The partner who is explaining should have only paper and pencil whereas the other person can help him/her along with the help of the book. This will help students become more comfortable with the logical progression of these proofs.

PRODUCTS

Students can be tested on this material by finding maxima and minima and by sketching a great many plots. This will be particularly useful with the next few chapters in mind. By having the goal in mind that they are to use derivatives in order to plot functions, students will be far more likely to understand what follows.

## The First Derivative Test

CONTENT

The content of this chapter really is purely geometric and has the best chance at getting through to students in this way. It is quite simple, when the derivative is positive the function is increasing and when it is negative, it is decreasing. This is actually precisely what the derivative measures, whether a function is increasing or decreasing. Furthermore, it should be quite intuitive that a function increases to a maximum then decreases away from it, and that the opposite is true of a minimum.

PROCESS

This material can be conveyed to students by having them think before studying the formal ideas. They should consider some complicated looking function like

\begin{align*}f(x) = \frac{x}{\text{log}(x)}\end{align*}

We find that its derivative:

\begin{align*}f'(x) = \frac{\text{log}(x) - 1}{[\text{log}(x)]^2}\end{align*}

is zero whenever \begin{align*}\mathrm{log}(x) - 1 = 0\end{align*}. This is true only when \begin{align*}x=e\end{align*} so that means that\begin{align*}x=e\end{align*} is a critical point of the function \begin{align*}f(x)\end{align*}.

Students should then be asked how they can tell if\begin{align*}x=e\end{align*}is a maximum, a minimum, or a saddle point. After thinking about this they should either graph the function or discuss what is happening for \begin{align*}x\end{align*} close to \begin{align*}x=e\end{align*} in terms of the derivative. In the former case the graph of \begin{align*}f(x)=\frac{x}{\mathrm{log}(x)}\end{align*} should look like the following:

This plot makes it clear that\begin{align*}x=e\end{align*}is a minimum. In the latter case the students may notice that for \begin{align*}x < e, f'(x)\end{align*} has a negative numerator and positive denominator and is therefore negative. This means that \begin{align*}f\end{align*} is decreasing to the left of \begin{align*}x=e\end{align*}. On the other hand for \begin{align*}x>e, f'(x)\end{align*} has a positive numerator and a negative denominator and so is positive. This means \begin{align*}f\end{align*} is increasing to the right, so the point \begin{align*}x=e\end{align*} is a minimum. This is a great project for pairs. One student can plot the function to see that \begin{align*}x=e\end{align*} is a minimum and the other student can talk about the derivative.

PRODUCTS

A very nice little riddle is solved by the function above. Consider asking students to solve the problem:

** Using the following steps instead of your calculator, determine which is bigger:** \begin{align*}e^{\pi}\end{align*}

**\begin{align*}\pi^e\end{align*}**

*or*

*?*a. Start with the equation \begin{align*}e^{\pi}\neq \pi^e\end{align*}, take the log of both sides, and simplify:

i. Answer: \begin{align*}\pi \neq e \mathrm{log}(\pi)\end{align*}

b. Now manipulate this inequality so that one side involves only \begin{align*}\pi\end{align*} and the other side involves only \begin{align*}e\end{align*}:

i. Answer: \begin{align*}\frac{\pi}{\text{log}(\pi)}\neq e\end{align*}

c. Now notice that by replacing \begin{align*}\pi\end{align*} by the variable \begin{align*}x\end{align*} we can consider this side of the inequality to be a function of \begin{align*}x\end{align*}. What are the critical point/s of this function and are they maxima or minima?

i. Answer: We solved this above, the function \begin{align*}f(x) = \frac{x}{\text{log}(x)}\end{align*} has the critical point\begin{align*}x=e\end{align*} only. This, we determined, was a minimum!

d. Now for the punch line, what is the value of this function at its critical point/s?

i. Answer: At \begin{align*}x=e, f(x) = \frac{e}{\text{log}\ (e)} = e\end{align*}

e. How do these value/s compare with the function’s value at \begin{align*}x=\pi\end{align*}?

i. Answer: \begin{align*}x=e\end{align*}is a minimum, so \begin{align*}f(\pi)>f(e)\end{align*} or \begin{align*}\frac{\pi}{log(\pi)} >e\end{align*} or \begin{align*}\pi > e \mathrm{log}(\pi)\end{align*} or taking the exponential of both sides: \begin{align*}e^{\pi} > \pi^e\end{align*}

f. What are the numerical values of \begin{align*}e^{\pi}\end{align*} and \begin{align*}\pi^e\end{align*}?

i. Answer: \begin{align*}e^{\pi} \cong 23.1407, \pi^e \cong 22.4592\end{align*}

## The Second Derivative Test

CONTENT

Notice that in order to determine whether critical values were maxima, minima, or saddle points we appealed to how the derivative was changing. If the derivative was positive beforehand and then negative after, i.e. decreasing, then the point was a maximum. If on the other hand the derivative was negative beforehand and then positive after, i.e. increasing, then the point was a minimum. If on the other hand the sign did not change, i.e. the derivative was constant; the point was a saddle point. The quick way to check how something is changing is to take its derivative. So if we want to see how the derivative is changing, we should look at its derivative: The second derivative!

When the second derivative is positive the first derivative is increasing and this gives the function a *concave up shape*. If this is true at a critical point then the critical point must be a minimum. When the second derivative is negative the first derivative is decreasing and this gives the function a *concave down* shape. If this is true at a critical point then the critical point must be a maximum. If the second derivative is zero at a critical point then the function may have a maximum, minimum, or saddle point there.

PROCESS

Recall the complicated function we examined earlier:

\begin{align*}f(x) = \frac{x}{\text{log}(x)}\end{align*}

we used its first derivative:

\begin{align*}f'(x) = \frac{\text{log}(x) - 1}{[\text{log}(x)]^2}\end{align*}

to see that it had a critical point at \begin{align*}x=e\end{align*}. We can conclusively classify this critical point by looking at how the derivative is changing at \begin{align*}x=e\end{align*}. This is determined by finding \begin{align*}f''(e)\end{align*} and to get this we will need to calculate that:

\begin{align*}f''(x) = \frac{2 - \text{log}(x)}{x [ \text{log}(x)]^3}\end{align*}

From this we see easily that:

\begin{align*}f''(e) = \frac{2 - \text{log}(e)}{e[\text{log}(e)]^3} = \frac{2 - 1}{e 1^3} = \frac{1}{e} \cong 0.37\end{align*}

So the second derivative is positive at the critical point\begin{align*}x=e\end{align*}which means that \begin{align*}f'(x)\end{align*} is increasing there (going from negative to positive) and so\begin{align*}x=e\end{align*}is a minimum.

PRODUCTS

Students can be tested on this material with a large variety of word problems asking them to formulate a function and then classify its critical points. For students who do not work well with word problems it may be better to simply give functions directly and ask students to classify their critical points using the first and second derivatives. As a look ahead, again, it is useful to have students draw plots of the functions as well, using the information that they have determined.

## Limits at Infinity

CONTENT

In order to understand the behavior of functions, it is often important to look more closely near points they are not defined, and examine how they behave after they run off your graph paper. A function like \begin{align*}f(x)= \frac{x^2-1}{x-1}\end{align*} is undefined at \begin{align*}x=1\end{align*} however we see clearly that because \begin{align*}\lim_{x \to 1} f(x) = 2\end{align*} the function doesn’t do anything very exciting there. On the other hand, the simple function \begin{align*}f(x) = \frac{1}{x}\end{align*} is defined at \begin{align*}x=0\end{align*} and the limits:

\begin{align*}\lim_{x \to 0^+} f(x) & = \infty \\ \lim_{x \to 0^-} f(x) & = - \infty\end{align*}

tell us that to the left of the origin the function tends down towards negative infinity and to the right coming in towards zero it tends up to positive infinity.

We can also look at how functions behave for very very large values of\begin{align*}x\end{align*}by examining the limits: \begin{align*}\lim_{x \to \pm \infty} f(x)\end{align*}. The simplest case is if this limit is zero or some other constant number. Then we see that the function simply approaches this horizontal line. This is the case with the function \begin{align*}f(x) = \frac{1}{x}\end{align*} or with the function \begin{align*}f(x) = C+ \frac{1}{x^2}\end{align*} for example. However if the limit is \begin{align*}\pm \infty\end{align*} then we have to be a little more careful. For example, consider the function from above again:

\begin{align*}f(x)= \frac{x^2-1}{x-1}\end{align*}

Although we see clearly that \begin{align*}\lim_{x \to \infty} = \infty\end{align*}, there is more that can be said. For large values of \begin{align*}x\end{align*}, the constant values of \begin{align*}1\end{align*} in the numerator and in the denominator are irrelevant so the function begins to look more and more like:

\begin{align*}\lim_{x \to \infty} \frac{x^2 -1}{x-1} = \lim_{x \to \infty} \frac{x^2}{x} = \lim_{x \to \infty} x\end{align*}

That is to say that in this limit, for large \begin{align*}x\end{align*}, the function begins to look a lot like the straight-line function \begin{align*}g(x)=x\end{align*}. This is sometimes described by saying that \begin{align*}f(x)\end{align*} “behaves like\begin{align*}x\end{align*}or diverges like\begin{align*}x\end{align*}for large \begin{align*}x\end{align*}.” It is also sometimes said that \begin{align*}f(x)\end{align*} has a diagonal asymptote in the line \begin{align*}g(x)=x\end{align*}.

PROCESS

With an eye towards what follows, students may find it useful to practice this material by dividing into small groups and analyzing a function. Each group could be given a function like \begin{align*}f(x)= x + \mathrm {Sin}(x) + \frac{1}{x^2}\end{align*} and the group should be asked to thoroughly analyze their function. Find all of its critical points, classify these, examine all of its limits and points of discontinuity, and describe if there is any asymptotic behavior. For example, the function given has an infinite number of local maxima and minima corresponding to each half-integer multiple of \begin{align*}\pi\end{align*}, will diverge to plus \begin{align*}\infty\end{align*} from both sides at zero, and will asymptotically diverge like \begin{align*}g(x)=x\end{align*} for values of \begin{align*}x\end{align*} that are from zero.

PRODUCTS

This material is readily tested in a way that leads nicely towards using derivatives and limits to graph functions. Students should be given some function with a variety of critical points and discontinuities and asked to consider how the function behaves near these points. This should be done by taking limits if the derivative is not defined or one and two derivatives where it is defined and checking for their signs. Then the student should be asked to analyze how the function behaves for values of\begin{align*}x\end{align*}that are far from the origin by examining limits.

## Analyzing the Graph of a Function

CONTENT

This section is essentially a culmination of the ideas that students have been considering thus far. By putting together all of the tools they have been given, it should be much easier to get a feeling for how a function’s plot will look ahead of time.

PROCESS

This section should be taught by having students repeat the kinds of exercises they have done earlier. The class could be split into pairs and each team could thoroughly analyze some complicated function by finding all the critical points, classifying each, and examining the limits at discontinuities and with each tail. It would be nice if each pair could present their function to the class on the board so that everyone could see the variety of functions.

PRODUCTS

This material is best tested by having students thoroughly analyze at least one function with some interesting behavior. The best functions are usually rational functions.

## Optimization

CONTENT

Optimization problems are probably the best way to make calculus seem important. This content should be described as one of the most important things that you hope to convey to students in the course. The tools they have learned thus far will be consolidated in this section to provide an enormous application.

PROCESS

The best way to teach this material is to provide students with a simple question. Suppose you are CEO of Starbucks and are trying to decide how expensive a small coffee should cost to obtain the most profit. Clearly if the coffee is free the profit will be zero. On the other hand, if the price is too high nobody will buy it and the profit will also be zero. So there must be some *intermediate* price that is not too high and not too low so that the profit is the greatest. This is a problem of optimization, and it can be solved using calculus.

If students are interested, tell them that by analyzing years and years of data you have modeled the profit \begin{align*}y\end{align*} for different prices \begin{align*}x\end{align*} with the function:

\begin{align*}y = \frac{x^3}{1 + e^{x^2}}\end{align*}

It is clear that this function satisfies our expectations that profit \begin{align*}y\end{align*} should be zero for \begin{align*}x= \$0\end{align*} and that it should become zero for very large values of\begin{align*}x\end{align*}as well. This can be shown by using L’Hopital’s rule to find the limit at large \begin{align*}x\end{align*}.

In order to determine the optimal price, we must use calculus. We take a derivative of the function to see that:

\begin{align*}y'= \frac{3x^2+e^{x^2}(3x^2-2x^4)}{(1+e^{x^2})^2}\end{align*}

In order to find the maximum, we would like to look for critical points. That is, we want to look for prices \begin{align*}x\end{align*} that make the derivative vanish. Since the denominator is never zero, the fraction is zero whenever its numerator \begin{align*}3x^2+e^{x^2}(3x^2-2x^4)=0\end{align*}.

Now, this equation cannot be solved by algebra or by any other exact means. So we must resort to something like Newton’s Method. So we let:

\begin{align*}f(x)=3x^2+e^{x^2}(3x^2-2x^4)\end{align*}

and we calculate that:

\begin{align*}f'(x)=6x+e^{x^2}(6x-2x^3-4x^5)\end{align*}

If we guess that a good price for coffee might be close to \begin{align*}\$2\end{align*} then we might set \begin{align*}x_0=2\end{align*} and use Newton’s Method. We find that:

\begin{align*}x_0 & = \$2 \\ x_1 & = x_0 - \frac{f(x_0 )}{f'(x_0 )} = \$1.8499 \\ x_2 & = x_1- \frac{f(x_1 )}{f'(x_1 )} = \$1.70397 \\ & \qquad \qquad \ \ \ldots \\ \lim_{n \to \infty} x_n & = \$1.32607\end{align*}

So the derivative \begin{align*}y'\end{align*} is zero for \begin{align*}x \cong \$1.33\end{align*}. Furthermore, we see that the second derivative:

\begin{align*}y'' = \frac{2x (3 + e^{x^2}(6 - 7x^2 - 2x^4 + e^{x^2}(3 - 7x^2 + 2x^4)))}{(1 + e^{x^2})^3}\end{align*}

at \begin{align*}x=1.33\end{align*} is about \begin{align*}-1.47\end{align*} so the critical point is a maximum. The price for a small coffee that will maximize profit is \begin{align*}\$1.33\end{align*}.

PRODUCTS

There are some great problems in optimization for students who are good at word problems and for students who are not. Word problems can be avoided by asking for specific properties of an explicit function or by simply presenting a word problem with the equations explicitly written out.

## Approximation Errors

CONTENT

This lesson is best explained by first saying that essentially any function that we’ll use in calculus is exactly equal to its Taylor Polynomial as \begin{align*} n \rightarrow \infty\end{align*}. That is, we have that for any “nice” (infinitely differentiable) function we can pick a point \begin{align*}x_0\end{align*} and we have for all points that are in the same part of the domain as \begin{align*}x_0\end{align*} that:

\begin{align*}f(x)=f(x_0 )+f'(x_0 )(x-x_0 )+ \frac{f''(x_0 )}{2!} (x-x_0 )^2 + \frac{f'''(x_0 )}{3!} (x-x_0 )^3 + \ldots\end{align*}

where the dots are meant to imply that this summation goes on and on without ever ending. The fact is that the terms are guaranteed to eventually get smaller and smaller because of the factorial in the denominator, no matter how big \begin{align*}x-x_0\end{align*} may be.

An approximation just means that instead of taking *all* the infinite number of terms, we cut off the series somewhere. The remaining infinite terms added up are called the “tail” of the series and the approximation is good when this tail is small. That is because the tail represents the difference between the actual function and the approximate function. Thus, we make the approximation better and better when we take more terms or when \begin{align*}x-x_0\end{align*} is small since the tail of the series will be smaller in each case.

PROCESS

Students could be split into groups of three that are each given a function and a value for \begin{align*}x_0\end{align*}. For example, one group could be given \begin{align*}f(x)=\mathrm {Sin}(x)\end{align*} and \begin{align*}x_0 = \pi\end{align*}, and another could be given \begin{align*}f(x)= \sqrt{x}\end{align*} and \begin{align*}x_0=9\end{align*}. Then the groups should designate tasks to each member: One member will be the **artist**, one will be the **derivator**, and one will be the **calculator**.

The calculator will begin by making a table with a series of points near \begin{align*}x_0\end{align*} and the function’s values there, like the example below:

\begin{align*}& x && f(x)=\sqrt{x} && L(x)=f(x_0 ) + f'(x_0 )(x-x_0) && Q(x)=f(x_0 )+f'(x_0 )(x-x_0)+ \frac{(f''(x_0 )}{2} (x-x_0 )^2 \\ &&&&&=3 + \frac{1}{6}(x-3) && =3+ \frac{1}{6} (x-3)- \frac{1}{216}(x-3)^2\\ & 8.7 && 2.95 && 2.94958 && 2.94958 \\ & 8.8 && 2.96667 && 2.96648 && 2.96648 \\ & 8.9 && 2.98333 && 2.98329 && 2.98329\\ & 9 && 3 && 3 && 3 \\ & 9.1 && 3.01667 && 3.01662 && 3.01662 \\ & 9.2 && 3.03333 && 3.03315 && 3.03315 \\ & 9.3 && 3.05 && 3.04959 && 3.04959\end{align*}

Then the artist will work with this table to draw a careful graph of the function on graph paper near the point \begin{align*}x_0\end{align*} while the mathematician determines the equation for the linearization of the group’s function near the point \begin{align*}x_0\end{align*} and the quadratic approximation as in the table above.

The calculator will calculate and fill in the column of the table for the same \begin{align*}x-\end{align*}values using the linearization and quadratic approximation, and the artist will draw in a plot of the approximations. Enough points should be used with enough spacing to be able to see the approximations deviate from the function. This may require a very fine scale, a large spacing, or a great many points. This will be the task of the group as a whole to figure out what window to use and how many points to include.

PRODUCTS

Students can be tested on this material by having them create approximations for functions with various degrees of accuracy.

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