Applications of Derivatives

Related Rates

When you need to find the rate of change of two or more related variables, you can use chain rule to find these related rates.

Remember, you can think of derivatives as rates of change for the function you are deriving.

In the case of related rates, you are performing differentiation with respect to time, t.

When solving related rates, here are some steps you can take to aid solving the problem.

1. Identify all the variables you are given, and all of the variables to be determined.

2. Draw a picture of the scenario.

3. Find an equation related to all variables, known or unknown.

4. Use chain rule to differentiate implicitly with respect to time.

5. Substitute in all given information.

6. Solve for the unknown variable.

Example

Imagine a balloon that is having air pumped inside. If we knew the rate of change of the volume to be 8 inches per second, what would the rate of change be for the radius of the balloon when the radius is 4 inches?

These are related rates.

The first step is to determine all the known and unknown variables.

We know that the change in volume is 8 inches per second, and that the radius of the balloon will be 4 inches. We are trying to find the rate of change of the radius.

The second step is to draw a diagram that shows as air is being pumped into the balloon, its radius will be expanding.

Third, we need an equation relating these variables.

Since we are dealing with the volume of a sphere, we know this equation to be: $V={\frac {4}{3}}{\pi }r^{3}.$

Fourth, we need to differentiate with respect to time, this looks like:

${\frac {dV}{dt}}=4{\pi }r^{2}{\frac {dr}{dt}}.$

Fifth, we need to substitute in all known information, which is the rate of change of the volume, and the radius of the sphere.

$8=4{\pi }(4)^{2}{\frac {dr}{dt}}$

$8=64{\pi }{\frac {dr}{dt}}.$

Lastly, we need to solve for the rate of change of the radius. Since all the units were inches, no changes need to be made beforehand.

${\frac {8}{64{\pi }}}={\frac {dr}{dt}}$

${\frac {1}{8{\pi }}}={\frac {dr}{dt}}.$

The rate at which the radius is changing is ${\frac {1}{8{\pi }}}$ per second.

Sometimes related rates can be much more difficult and involve multiple unknown variables. In these cases you may need to use multiple equations to find unknown variables before finding the initial unknown you were looking for.

The situation of a related rate can be many things, from volumes, to airplanes and velocity, to basketballs, etc.

Something else to watch is the units of your variables. In some cases you may need to convert your variables to match units.

Here are some equations that are frequently used for related rates.

Circles/Spheres: Area = ${\pi }r^{2}$ Volume = ${\frac {4}{3}}{\pi }r^{3}$ Surface Area = $4{\pi }r^{2}$

Distance Formula: $d={\sqrt {x^{2}+y^{2}}}$

Cones: A = area of base = ${\pi }r^{2}$ Volume = ${\frac {Ah}{3}}$

Cylinder: Volume = ${\pi }r^{2}h$

Extrema and the Mean Value Theorem

The extrema, or extreme values, of a function are the minimum and/or maximum of a function. They are also known as absolute maximums, or absolute minimums.

Extreme Value Theorem

If a function $f$ is continuous on a closed interval $[a,b]$ then there exists both a maximum and minimum on the interval.

To find the relative extrema of a function, you first need to calculate the critical values of a function.

Do this by first finding the derivative of a function. Then set the derivative to equal 0, and solve for values of x.

Now, if the function is on a closed interval, $[a,b]$, then evaluate all the critical points, and endpoints, in your original function.

For absolute minimums/maximums on a closed interval, the highest of your evaluated points is the maximum; the smallest is the minimum.

Example

What are the absolute maximums/minimums, of $f(x)=x^{3}$ on the interval $[-2,2]$?

To begin, differentiate the function,

$f'(x)=3x^{2}.$

Then set the derivative to 0 and solve for critical points.

$f'(x)=3x^{2}=0$

$x=0.$

Now calculate the original function at your critical points and endpoints,

$f(-2)=-8,f(2)=8,f(0)=0$

Since $f(-2)=-8$ is the lowest, we know that the point, $x=-2$ is the absolute minimum, and 8 is the highest, thus the point, $x=2$ is the absolute maximum.

Rolle's Theorem

Let $f$ be a continuous function on a closed interval $[a,b]$, and differentiable on the open interval $(a,b).$

If $f(a)=f(b)$ then there is at least one number $c$ in the interval $(a,b)$ such that $f'(c)=0.$

Essentially, Rolle's theorem tells us that if a function starts and ends at the same point in a closed interval, then there exists some point $c$ where the derivative $f'(c)=0.$

You can informally think of this as at some point, the function must switch from increasing to decreasing, or vice versa, to get back to the y-coordinate where it began.

Mean Value Theorem

If $f$ is continuous on a closed interval $[a,b]$ and differentiable on the open interval $(a,b),$ then there is some number $c$ in $(a,b)$ that,

$f'(c)={\frac {f(b)-f(a)}{b-a}}$

*Note that the proof of the Mean Value Theorem uses Rolle's Theorem.

The Mean Value Theorem holds a couple different meanings. In Geometry, it tells us that if a secant line is drawn between our starting points, a and b, that there exists a tangent line parallel to the secant line, somewhere on the function.

The Mean Value Theorem also holds that somewhere on the open interval, $(a,b),$ that the instantaneous rate of change, the change at one point, is equal to the average rate of change over the whole interval.

When you evaluate with the Mean Value Theorem, it yields the average, or mean, rate of change over an interval.

Example

Determine if the Mean Value Theorem is applicable, and if it is, find all values of c in the open interval $(a,b)$ such that, $f'(c)={\frac {f(b)-f(a)}{b-a}}.$

Let $f(x)=x^{2},[-2,1].$

First, check if the function is continuous on the closed interval. Since we are dealing with $x^{2}$, we already know it is continuous for all x.

Next, check if the function is differentiable on the open interval, $(-2,1).$ We know our derivative is $f'(x)=2x,$ which exists everywhere on the open interval.

Now that we know both conditions for the theorem have been met, we can apply it to the problem.

Begin by evaluating the function at your endpoints. $f(-2)=4,f(1)=1.$

Next, plug all the information into the theorem, $f'(c)={\frac {1-4}{1-(-2)}}$

$={\frac {-3}{3}}$

$=-1.$

This tells us that the average rate of change over the interval is $-1.$

To find all the values of $c$, where $f'(c)=-1,$ replace $f'(c)$ with the derivative of $f(x).$

This looks like:

$2c=-1$

$c=-{\frac {1}{2}}$

The answer to the problem is $c=-{\frac {1}{2}}$

The First Derivative Test

The definition of a derivatives tells us that a derivative is the slope of the tangent line at a point on the function.

Derivatives can also tell us if a function is decreasing or increasing at a point.

A function $f(x)$ is increasing on an interval, if for two numbers $x_{1}$ and $x_{2}$ in the interval $x_{1}<x_{2},$ that $f(x_{1})<f(x_{2})$ is true.

A function $f(x)$ is decreasing on an interval, if for two numbers $x_{1}$ and $x_{2}$ in the interval $x_{1}<x_{2},$ that $f(x_{1})>f(x_{2})$ is true.

If a function $f(x)$ is continuous on a closed interval $[a,b],$ and differentiable on an open interval $(a,b),$ then the following applies:

1. If $f'(x)>0$ for all $x$ in $(a,b),$ then $f(x)$ is increasing on $[a,b].$

2. If $f'(x)<0$ for all $x$ in $(a,b),$ then $f(x)$ is decreasing on $[a,b].$

3. If $f'(x)=0$ for all $x$ in $(a,b),$ then $f(x)$ is constant on $[a,b].$

In the last section, we learned about absolute minimums/maximums. Inside a function, other extrema, known as relative extrema, can exist.

The relative extrema of a function are points on a function that are lower or higher than all of the points near them. Such points create "hills" or "valleys" within a given function.

Relative extrema occur at points on a function where the derivative at that point changes from increasing to decreasing, or decreasing to increasing.

If the derivative changes from increasing to decreasing, that point is known as a relative maximum.

If the derivative changes from decreasing to increasing, that point is known as a relative minimum.

By finding the relative extrema of a function, you can then calculate whether or not those extrema are relative minima or maxima using the derivative of the function at those points.

Relative extrema are always critical points of a function.

Example

Find the relative extrema of $f(x)=x^{3}-{\frac {3}{2}}x^{2}.$

First, check if the function is continuous for all $x.$

We can see the function exists for all $x$ therefore, it is continuous.

Second, find the critical numbers of $f(x)$ by using the derivative of the function.

Find the critical numbers by setting $f'(x)=0.$

$f'(x)=3x^{2}-3x$

$3x^{2}-3x=0$

$x(3x-3)=0$

$x=0,1.$

Third, create intervals with your critical numbers.

Since we have two critical numbers, we will have three intervals. They are:

$-\infty <x<0,0<x<1,1<x<\infty .$

Fourth, determine if $f'(x)$ is increasing or decreasing over each interval. Do this by evaluating a test number within each interval.

In most cases, it is beneficial to create a table to arrange the present data.

Interval	$-\infty <x<0$	$0<x<1$	$1<x<\infty$
Test Value	$x=-1$	$x={\frac {1}{2}}$	$x=2$
Sign of $f'(x)$	$f'(-1)=6$	$f'({\frac {1}{2}})={\frac {-3}{4}}$	$f'(2)=6$
Increasing/Decreasing	Increasing	Decreasing	Increasing

Lastly, determine if any relative maximums or minimums are present.

Since $f'(x)$ changes from increasing to decreasing to increasing, we can conclude that there is a relative maximum at $x=0,$ and a relative minimum at $x=1.$

Resources

• Limits & Derivative Formulas (Slides 28 & 29). PowerPoint file created by Professor Cynthia Roberts, UTSA.
• Exponential and Logarithms (Slides 24-26). PowerPoint file created by Professor Cynthia Roberts, UTSA.
• Product Rule and Quotient Rule. PowerPoint file created by Professor Cynthia Roberts, UTSA.
• Chain Rule. PowerPoint file created by Professor Cynthia Roberts, UTSA.
• Periodic Functions. PowerPoint file created by Professor Cynthia Roberts, UTSA.
• Local Max & Min. PowerPoint file created by Professor Cynthia Roberts, UTSA.
• Inflection Points. PowerPoint file created by Professor Cynthia Roberts, UTSA.
• Global Max & Min. PowerPoint file created by Professor Cynthia Roberts, UTSA.
• Logistic Growth & Surge Function. PowerPoint file created by Professor Cynthia Roberts, UTSA.

Licensing

Content obtained and/or adapted from:

• Related Rates, Wikibooks under a CC BY-SA license
• Extrema and the Mean Value Theorem, Wikibooks under a CC BY-SA license
• The First Derivative Test, Wikibooks under a CC BY-SA license