# 5.3: Applications of Derivatives

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

## Related Rates

The key to being successful in solving related rate problems is proper organization of given information at the start of the problem. By listing the given rate information, and the requested rate, labeled with the correct variable and differentials, the required equations will become clear and the process should be easier. Example:

ex. A spherical balloon is being inflated at a rate of \begin{align*}4\pi cm^3\end{align*}

First, identify the given and needed information. The rate that is given is a volume over time change, and the needed information is an area over time change. So:

\begin{align*}\text{given:} \ \frac{dV}{dt} = 4 \pi \qquad \text{needed:} \ \frac{dA}{dt} \qquad \text{when} \ t=9\end{align*}

These differentials indicate that we need the formulae for the volume and surface area of a sphere:

\begin{align*}V = \frac{4}{3} \pi r^3 \ \ A = 4 \pi r^2\end{align*}

We can take the derivatives of these equations to get the needed differentials:

\begin{align*}\frac{dV}{dt} = 4 \pi r^2 \frac{dr}{dt} \ \ \frac{dA}{dt} = 8 \pi r \frac{dr}{dt}\end{align*}

These derivatives tell us we need to things. First, we need the length of the radius of the sphere at \begin{align*}9\;\mathrm{seconds}\end{align*}

\begin{align*}4 \pi = 4 \pi(3)^3 \frac{dr}{dt} \rightarrow \frac{1}{9} = \frac{dr}{dt}\end{align*}

Substituting the radius and the change in radius over time into the second equations:

\begin{align*}\frac{dA}{dt} = 8 \pi(3) \left ( \frac{1}{9} \right ) \rightarrow \frac{dA}{dt} = \frac{8}{3} \pi\end{align*}

Therefore the rate the surface area is changing is \begin{align*}\frac{8}{3\pi} cm^2\end{align*}

The key item to notice is that by setting up the rates at the top, the next step was always dictated by what variables were in use and what needed to be found next.

## Extrema and the Mean Value Theorem

A useful principle related to the topics in this section is the racetrack principle:

*Suppose that* \begin{align*}g\end{align*}*and* \begin{align*}h\end{align*}*are continuous on* \begin{align*}[a,b]\end{align*}*and differentiable on* \begin{align*}(a,b)\end{align*}*and that* \begin{align*}g'(x) \le h'(x)\end{align*} *for* \begin{align*}a<x<b\end{align*}*. If* \begin{align*}g(a)=h(a)\end{align*}, *then* \begin{align*}g(x) \le h(x)\end{align*} *for* \begin{align*}a \le x \le b\end{align*}*. If* \begin{align*}g(b)=h(b)\end{align*}, *then* \begin{align*}g(x)\ge h(x)\end{align*} *for* \begin{align*}a \le x \le b\end{align*}.

An interpretation of this, and the origin of the name, is that there are two vehicles on a race track, and one vehicle, \begin{align*}h\end{align*}, is always moving faster. If they start at the same place, then \begin{align*}h\end{align*} will lead the entire time. Alternately if they end up in the same place, this means that \begin{align*}g\end{align*} will have to have been leading the whole time since its speed is slower. This is a handy principle to prove inequalities for two functions. A common application is: Show that \begin{align*}\sin (x) \le x\end{align*} for all \begin{align*}x \ge 0\end{align*}.

Since the idea here is to show that one function is greater than the other for the entire interval the racetrack principle should be helpful. When applying theorae or principles, it is always important to pay close attention to the conditions. The functions are both continuous and differentiable on the interval. We now need to decide if we need to show they start at the same point, or if they end at the same point. There is an intersection at the start of the interval \begin{align*}x=0\end{align*}, although it is worth noting, that there is nothing that says this is, or needs to be the only intersection. This lesser requirement is one of the useful aspects of the racetrack principle. Now differentiating both sides, we do see that \begin{align*}\cos (x) \le 1\end{align*} which is true. Therefore our original inequality does hold.

## The First Derivative Test

The first derivative tells much about the function. The temptation is for students who are raised in a graphing calculator environment to rely on the graphing or guess and check methods to answer questions that could easily be solved by testing using derivatives. Using a chart is a nice way to organize the information. Example:

Find all increasing and decreasing intervals for the function \begin{align*}f(x)=-x^3-4x^2+5x-1\end{align*}

First thing to do is to take the first derivative and set it equal to zero to find the critical points.

\begin{align*}f'(x)=-3x^2-8x+5 \rightarrow 0=-3x^2-8x+5\end{align*}

which is not factorable so applying the quadratic formula yields:

\begin{align*}x = \frac{8 \pm \sqrt{8^2 - 4 (-3)(5)}}{-6} \rightarrow x = \frac{8 \pm \sqrt{124}}{-6} \rightarrow x =.52,-3.19\end{align*}

Now set up a table with the critical points with some chosen values between each point:

\begin{align*}& x \text{value} && -5 && -3.19 && 0 && .52 && 1 \\ & \text{sign of derivative} && - && 0 && + && 0 && -\end{align*}

After substituting in the values to the derivative function. This means that the intervals where the function is decreasing is \begin{align*}(- \infty,-3.19)\cup(.52, \infty)\end{align*} and the function is increasing on the interval \begin{align*}(-3.19,.52)\end{align*}.

Setting up the table to dictate what values to choose is a key tool. I think of the critical points as being “partitions” for the real numbers. When the partitions are established then any values can be chosen inside those intervals. This is really important for some functions that may not be clear on the calculator, like functions that have critical points well outside the normal graphing window or functions that have critical points that are very close and do not appear correctly on a typical graphing window.

## The Second Derivative Test

The same tool that is used for finding first derivative information about increasing and decreasing functions is valuable for finding information about concavity, maxima and minima and inflection points. Here we'll look at an application of these techniques. Not only is it common to have optimization word problems where first and second derivatives will need to be evaluated, but analytic problems about functions can also be interesting. Example: Show that \begin{align*}x > 2\end{align*} ln \begin{align*}x\end{align*} for all \begin{align*}x > 0\end{align*}

Most students, having not seen problems like this before, will need to have a little guidance. What are you being asked to do with these two functions? Hopefully students will recognize that they are comparing the two functions, which can be evaluated by looking at the difference between the two. More accurately stated, is it true \begin{align*}x-2\end{align*} ln \begin{align*}x > 0\end{align*} for all values of \begin{align*}x\end{align*}? A good question is now, how do you find what the smallest value of the function \begin{align*}f(x) = x - 2\end{align*} ln \begin{align*}x\end{align*}? Smallest value should immediately trigger the “minimize/maximize” alarm that is growing in students' minds. Taking the derivative and setting equal to zero:

\begin{align*}f'(x)=1 - \frac{2}{x} \rightarrow 0 = 1 - \frac{2}{x} \rightarrow x = 2\end{align*}

There are a couple of ways to go about the next step, but it is important to understand that \begin{align*}x=2\end{align*} is *where* the minimum exists, not what the minimum is. Substituting back into the original functions shows:

\begin{align*}2-2 \text{ln}(2)=.614\end{align*}

Since the minimum value is greater than zero, then all values must be greater than zero, proving the original statement.

A couple of nice extensions on this question are: Is \begin{align*}e^x > x^2\end{align*} for all \begin{align*}x>0\end{align*}? This is actually just a corollary to the question above, and could be given as the first question asked to a strong student or class. Another good extension is the question: Is \begin{align*}x > 3\end{align*} ln \begin{align*}x\end{align*} for all \begin{align*}x > 0\end{align*}? This turns out to be false, showing how a simple number change can alter the problem.

## Limits at Infinity

l'Hopital's rule is fairly explicit in the instances which it can be used. This can sometimes cause trouble for students, as it is a really easy technique, and it is easy to try and apply it to situations when the required conditions are not met. It is such a powerful tool that it is worth trying to use in many circumstances. Therefore, the approach should be “Can I get this to fit the necessary conditions?” rather than “Does this meet the necessary conditions?” Here is an example of the subtle difference: Evaluate \begin{align*}\lim_{x \to 0+} x\end{align*} ln \begin{align*}x\end{align*} If you substitute zero into the expression you get \begin{align*}0 \;\mathrm{times}\end{align*} an undefined function. This is not one of the indeterminate forms that is accepted by l'Hopital's rule, but if you re-write the limit as:\begin{align*}\lim_{x \to 0+} \frac{\mathrm{ln} x}{\frac{1}{x}}\end{align*} now each function, top and bottom, has a defined right hand limit of \begin{align*} \pm \infty\end{align*} which is a form accepted by the rule. Now you can take the derivative of each and evaluate directly:

\begin{align*}\lim_{x \to 0+} \frac{\frac{1}{x}}{\frac{-1}{x^2}} = \lim_{x \to 0+} \frac{-x^2}{x} = \lim_{x \to 0+} - x = 0\end{align*}

Another tool is to use the property of logarithms to convert \begin{align*}\infty - \infty\end{align*} indeterminate forms to an expression that fits the rule:

\begin{align*}\lim_{x \to 0} x^{\sin x} = e^{\text{ln} \lim_{x \to 0} x^{\sin x}} = e^{\lim_{x \to 0} \sin x \text{ln} x} = e^{\lim_{x \to 0} \frac{\text{ln}x}{\frac{1}{\sin x}}}\end{align*}

Now the limit that is in the exponent is ∞ over ∞ meaning that l'Hopital's rule can be applied. Taking the derivative:

\begin{align*}e^{\lim_{x \to 0} \frac{\frac{1}{x}}{\frac{\cos x}{- \sin^2 x}}} = e^{\lim_{x \to 0} \frac{- \sin^2 x}{x \cos x}}\end{align*}

Which still results in \begin{align*}0\end{align*} over \begin{align*}0\end{align*}, so l'Hopital's rule can be applied again:

\begin{align*}e^{\lim_{x \to 0} \frac{-2 \sin x \cos x}{\cos x - x \sin x}}\end{align*}

Where the limit can be evaluated as going to \begin{align*}0\end{align*}, which means:

\begin{align*}\lim_{x \to 0}x^{\sin x} = 1\end{align*}

One thing to watch out for is the trap of using l'Hopital's rule in a circular manner. Sometimes now it may be tempting to find derivatives using the limit definition and applying l'Hopital's rule for \begin{align*}0\end{align*} over \begin{align*}0\end{align*} cases. This is circular, as a requirement for l'Hopital's rule is that the function has a derivative, and it is known. Therefore, l'Hopital's rule can't be used to find a derivative.

## Analyzing the Graphs of a Function

Often times tests require an interpretation of the derivatives of a graph without the function expressed in algebraic form. This can be made easier though using the same techniques used for algebraic functions, rather than simply try to sketch directly from the graph. Example: Sketch the first and second derivatives of the following function:

First up is the first derivative. Just like when given an analytic function, first find the places where this function is going to have a critical point. There are 3 critical points on this graph, with the sign of the slope in between each critical point:

It's possible at this point to sketch a good approximation, but it could be made better by looking for the inflection points, which will show up as maxima and minima for the first derivative:

Now indicate the concavity and sketch the second derivative:

The process is exactly the same, and can provide a good way to reinforce the conceptual parts of the derivative tests, as well as practice sketching graphs based on derivative information.

## Optimization

A very common question is asking for optimization of a path with different rates. Example:

A pipe needs to be laid from a well to a water treatment plant. The well is located along the shore of a river \begin{align*}5 \;\mathrm{km}\end{align*} from the treatment plant, which is on the other side of the river. The river is \begin{align*}250m\end{align*} wide, and the pipe costs \begin{align*}\$1.50\end{align*} per meter to lay under ground, but \begin{align*}\$4\end{align*} per meter to lay under the river. What is the cheapest way to lay the pipeline?

The first order of business for optimization problems is to know, and write down, the exact quantity to optimize. In some cases there will be a number of equations and rates, and it is easy to lose track of what exactly the question is asking for. In this case, we need to minimize the cost function for the pipeline. Taking into account the cost rates, the function is:\begin{align*} C=1.5g+4w\end{align*} where g is the meters of pipe in the ground and w is the meters of pipe under water.

The next thing to do is to draw an accurate diagram with all of the quantities labeled. Any variables that can be put in the diagram will help. In this case, students should be encouraged to think of what is likely to happen. If the cost of the pipe was equal, land or water, then a straight line between the two points is the least pipe, and therefore the cheapest. It is probably also not likely that the pipe runs perpendicular to the river as this would be the most amount of pipe possible. The standard diagram for this type of problem looks something like:

The next step is to try to develop a relationship between our two variables in our cost function. Put another way, there needs to only be a single variable to take a derivative and maximize, so one variable needs to be put in terms of the other. The diagram listed gives us a huge clue, in that the hypotenuse of the right triangle is going to be the distance traveled across the water, and it can be expressed in terms of the distance traveled along the shore using the Pythagorean theorem:

\begin{align*}w = \sqrt{(5000 - g)^2 + 250^2}\end{align*}

Substituting for the original function and taking the derivative:

\begin{align*}C = 1.5 g + 4 \sqrt{(5000 - g)^2 + 250^2} \rightarrow C' = 1.5 + \frac{- 4 (5000 - g)}{\sqrt{(5000 - g)^2 + 250^2}}\end{align*}

Now set the derivative equal to zero and solve for \begin{align*}g\end{align*}:

\begin{align*} 0 = \frac{1.5\sqrt{(5000 - g)^2 + 250^2} - 10000 + 4g}{\sqrt{(5000 - g)^2 + 250^2}}\end{align*}

Which will only be true when the numerator is equal to zero:

\begin{align*}0 & = 1.5 \sqrt{(5000 - g)^2 + 250^2} -10000 + 4g \rightarrow 666.67-2.67g = \sqrt{(5000 - g)^2 + 250^2} \\ 444448.89-3560.02g + 7.13g^2 & =24937500-10000g + g^2 \\ 0 & =24493051.11-6439.98g-6.13g^2\end{align*}

Applying the quadratic formula:

\begin{align*}g = \frac{6439.98 \pm \sqrt{642042955.62}}{- 12.26} \rightarrow g = 1541.48 - 259.05\end{align*}

The negative option does not fit with the context of the problem, so we know now that \begin{align*}1541.48m\end{align*} of pipe should be laid along the shore. Substituting back into the relationship between \begin{align*}w\end{align*} and \begin{align*}g\end{align*}:

\begin{align*} w = \sqrt{(5000-1541.48)^2+250^2} \rightarrow w = 3467.54m\end{align*}

Geometric relationships are the favorites of problem writers.In most circumstances for optimization problems the relationship between variables is going to come from an area, volume, or distance formula. It is useful then to have a couple of the more common ones memorized.

## Approximation Errors

While the example in the text shows that using a graphing calculator is the easiest method to find the interval where the approximation is within a certain error bound, sometimes all that is asked is to prove the existence of an interval of specific length . The first thing is to establish a definition for the error. If \begin{align*}f(x)\end{align*} has a known value, then the error will be reflected by:

\begin{align*}E(x)=f(x)-(f(c)+f'(c)(x-c))\end{align*}

The next thing we need to consider is a way to find an error bound, for which we need a guarantee that the function \begin{align*}f\end{align*} is differentiable. We need to use this for the following derivation. If we distribute the negative and then divide by the difference from \begin{align*}x\end{align*} to \begin{align*}c\end{align*}:

\begin{align*}\frac{E(x)}{(x-c)} = \frac{f(x)-f(c)-f'(c)(x-c)}{(x-c)} = \frac{f(x)-f(c)}{(x-c)}-f'(c)\end{align*}

If we now take the limit of each side of the equation as \begin{align*}x\end{align*} goes to \begin{align*}c\end{align*}, and using the definition of the derivative:

\begin{align*}\lim_{x \to c} \frac{E(x)}{x-c} = \lim_{x \to c} \frac{f(x) - f(c)}{x-c} - f'(c) = f'(c) - f'(c) = 0\end{align*}

We can use this to prove the existence of an interval about \begin{align*}x=0\end{align*} for the function \begin{align*}\sin(x)\end{align*} approximated by the linear function \begin{align*}x\end{align*}. \begin{align*}\sin(x) = x + E(x) \text{with} \lim_{x \to c} \frac{E(x)}{x} = 0\end{align*}

So if we need the error limit to be \begin{align*}.1\end{align*} then the strict definition of the limit states there exists a \begin{align*}\delta > 0\end{align*} such that \begin{align*}\left |\frac{E(x)}{x} \right | < .1\end{align*} for all \begin{align*}|x| < \delta\end{align*}. Therefore:

\begin{align*}|E(x)|<.1|x|\end{align*}

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