# 4.1: Prelude to Applications of Derivatives

Summary: A rocket launch involves two related quantities that change over time. In addition, we examine how derivatives are used to evaluate complicated limits, to approximate roots of f

A rocket is being launched from the ground and cameras are recording the event. A video camera is located on the ground a certain distance from the launch pad. At what rate should the angle of inclination (the angle the camera makes with the ground) change to allow the camera to record the flight of the rocket as it heads upward? (See [link].)

Figure (PageIndex{1}): As a rocket is being launched, at what rate should the angle of a video camera change to continue viewing the rocket? (credit: modification of work by Steve Jurvetson, Wikimedia Commons)

A rocket launch involves two related quantities that change over time. In addition, we examine how derivatives are used to evaluate complicated limits, to approximate roots of functions, and to provide accurate graphs of functions.

## 4.1: Prelude to Applications of Derivatives

In the final section of this chapter let’s take a look at some applications of derivatives in the business world. For the most part these are really applications that we’ve already looked at, but they are now going to be approached with an eye towards the business world.

Let’s start things out with a couple of optimization problems. We’ve already looked at more than a few of these in previous sections so there really isn’t anything all that new here except for the fact that they are coming out of the business world.

How many apartments should they rent in order to maximize their profit?

All that we’re really being asked to do here is to maximize the profit subject to the constraint that (x) must be in the range (0 le x le 250).

First, we’ll need the derivative and the critical point(s) that fall in the range (0 le x le 250).

[P'left( x ight) = - 16x + 3200hspace <0.5in>Rightarrow hspace <0.25in>3200 - 16x = 0hspace<0.25in>, Rightarrow hspace <0.25in>x = frac<<3200>><<16>> = 200]

Since the profit function is continuous and we have an interval with finite bounds we can find the maximum value by simply plugging in the only critical point that we have (which nicely enough in the range of acceptable answers) and the end points of the range.

[Pleft( 0 ight) = - 80,000hspace<0.5in>Pleft( <200> ight) = 240,000hspace<0.5in>Pleft( <250> ight) = 220,000]

So, it looks like they will generate the most profit if they only rent out 200 of the apartments instead of all 250 of them.

Note that with these problems you shouldn’t just assume that renting all the apartments will generate the most profit. Do not forget that there are all sorts of maintenance costs and that the more tenants renting apartments the more the maintenance costs will be. With this analysis we can see that, for this complex at least, something probably needs to be done to get the maximum profit more towards full capacity. This kind of analysis can help them determine just what they need to do to move towards that goal whether it be raising rent or finding a way to reduce maintenance costs.

Note as well that because most apartment complexes have at least a few units empty after a tenant moves out and the like that it’s possible that they would actually like the maximum profit to fall slightly under full capacity to take this into account. Again, another reason to not just assume that maximum profit will always be at the upper limit of the range.

Let’s take a quick look at another problem along these lines.

How many widgets per day should they produce in order to minimize production costs?

Here we need to minimize the cost subject to the constraint that (x) must be in the range (0 le x le 60,000). Note that in this case the cost function is not continuous at the left endpoint and so we won’t be able to just plug critical points and endpoints into the cost function to find the minimum value.

Let’s get the first couple of derivatives of the cost function.

The critical points of the cost function are,

Now, clearly the negative value doesn’t make any sense in this setting and so we have a single critical point in the range of possible solutions : 50,000.

Now, as long as (x > 0) the second derivative is positive and so, in the range of possible solutions the function is always concave up and so producing 50,000 widgets will yield the absolute minimum production cost.

Recall from the Optimization section we discussed how we can use the second derivative to identity the absolute extrema even though all we really get from it is relative extrema.

Now, we shouldn’t walk out of the previous two examples with the idea that the only applications to business are just applications we’ve already looked at but with a business “twist” to them.

There are some very real applications to calculus that are in the business world and at some level that is the point of this section. Note that to really learn these applications and all of their intricacies you’ll need to take a business course or two or three. In this section we’re just going to scratch the surface and get a feel for some of the actual applications of calculus from the business world and some of the main “buzz” words in the applications.

Let’s start off by looking at the following example.

Answer each of the following questions.

1. What is the cost to produce the 301 st widget?
2. What is the rate of change of the cost at (x = 300)?

We can’t just compute (Cleft( <301> ight)) as that is the cost of producing 301 widgets while we are looking for the actual cost of producing the 301 st widget. In other words, what we’re looking for here is,

[Cleft( <301> ight) - Cleft( <300> ight) = 97,695.91 - 97,400.00 = 295.91]

So, the cost of producing the 301 st widget is $295.91. In this part all we need to do is get the derivative and then compute (C'left( <300> ight)). [C'left( x ight) = 350 - 0.18xhspace <0.5in>Rightarrow hspace<0.5in>C'left( <300> ight) = 296.00] Okay, so just what did we learn in this example? The cost to produce an additional item is called the marginal cost and as we’ve seen in the above example the marginal cost is approximated by the rate of change of the cost function, (Cleft( x ight)). So, we define the marginal cost function to be the derivative of the cost function or, (C'left( x ight)). Let’s work a quick example of this. What is the marginal cost when (x = 200), (x = 300) and (x = 400)? So, we need the derivative and then we’ll need to compute some values of the derivative. [eginC'left( x ight) & = - 10 - 0.02x + 0.0006 C'left( <200> ight) & = 10hspace<0.5in>C'left( <300> ight) = 38hspace<0.5in>C'left( <400> ight) = 78end] So, in order to produce the 201 st widget it will cost approximately$10. To produce the 301 st widget will cost around $38. Finally, to product the 401 st widget it will cost approximately$78.

Note that it is important to note that (C'left( n ight)) is the approximate cost of producing the ( ight)^<>>>) item and NOT the n th item as it may seem to imply!

Let’s now turn our attention to the average cost function. If (Cleft( x ight)) is the cost function for some item then the average cost function is,

Here is the sketch of the average cost function from Example 4 above.

We can see from this that the average cost function has an absolute minimum. We can also see that this absolute minimum will occur at a critical point when (overline C'left( x ight) = 0) since it clearly will have a horizontal tangent there.

Now, we could get the average cost function, differentiate that and then find the critical point. However, this average cost function is fairly typical for average cost functions so let’s instead differentiate the general formula above using the quotient rule and see what we have.

Now, as we noted above the absolute minimum will occur when (overline C'left( x ight) = 0) and this will in turn occur when,

[x,C'left( x ight) - Cleft( x ight) = 0hspace <0.5in>Rightarrow hspace<0.5in>C'left( x ight) = frac<> = overline Cleft( x ight)]

So, we can see that it looks like for a typical average cost function we will get the minimum average cost when the marginal cost is equal to the average cost.

We should note however that not all average cost functions will look like this and so you shouldn’t assume that this will always be the case.

Let’s now move onto the revenue and profit functions. First, let’s suppose that the price that some item can be sold at if there is a demand for (x) units is given by (pleft( x ight)). This function is typically called either the demand function or the price function.

The revenue function is then how much money is made by selling (x) items and is,

[Rleft( x ight) = x,pleft( x ight)]

The profit function is then,

[Pleft( x ight) = Rleft( x ight) - Cleft( x ight) = x,pleft( x ight) - Cleft( x ight)]

Be careful to not confuse the demand function, (pleft( x ight)) - lower case (p), and the profit function, (Pleft( x ight)) - upper case (P). Bad notation maybe, but there it is.

Finally, the marginal revenue function is (R'left( x ight)) and the marginal profit function is (P'left( x ight)) and these represent the revenue and profit respectively if one more unit is sold.

Let’s take a quick look at an example of using these.

and the demand function for the widgets is given by,

[pleft( x ight) = 200 - 0.005xhspace<0.5in>0 le x le 10000]

Determine the marginal cost, marginal revenue and marginal profit when 2500 widgets are sold and when 7500 widgets are sold. Assume that the company sells exactly what they produce.

Okay, the first thing we need to do is get all the various functions that we’ll need. Here are the revenue and profit functions.

Now, all the marginal functions are,

[eginC'left( x ight) & = 100 - 0.06x + 0.000012 R'left( x ight) & = 200 - 0.01x P'left( x ight) & = 100 + 0.05x - 0.000012end]

The marginal functions when 2500 widgets are sold are,

[C'left( <2500> ight) = 25hspace<0.5in>R'left( <2500> ight) = 175hspace<0.5in>P'left( <2500> ight) = 150]

The marginal functions when 7500 are sold are,

[C'left( <7500> ight) = 325hspace<0.5in>R'left( <7500> ight) = 125hspace<0.5in>P'left( <7500> ight) = - 200]

So, upon producing and selling the 2501 st widget it will cost the company approximately $25 to produce the widget and they will see an added$175 in revenue and $150 in profit. On the other hand, when they produce and sell the 7501 st widget it will cost an additional$325 and they will receive an extra $125 in revenue, but lose$200 in profit.

We’ll close this section out with a brief discussion on maximizing the profit. If we assume that the maximum profit will occur at a critical point such that (P'left( x ight) = 0) we can then say the following,

[P'left( x ight) = R'left( x ight) - C'left( x ight) = 0hspace <0.5in>Rightarrow hspace<0.5in>R'left( x ight) = C'left( x ight)]

We then will know that this will be a maximum we also were to know that the profit was always concave down or,

[P''left( x ight) = R''left( x ight) - C''left( x ight) < 0hspace <0.5in>Rightarrow hspace<0.5in>R''left( x ight) < C''left( x ight)]

So, if we know that (R''left( x ight) < C''left( x ight)) then we will maximize the profit if (R'left( x ight) = C'left( x ight)) or if the marginal cost equals the marginal revenue.

In this section we took a brief look at some of the ideas in the business world that involve calculus. Again, it needs to be stressed however that there is a lot more going on here and to really see how these applications are done you should really take some business courses. The point of this section was to just give a few ideas on how calculus is used in a field other than the sciences.

## 2.1: Prelude to Applications of Derivatives

A rocket is being launched from the ground and cameras are recording the event. A video camera is located on the ground a certain distance from the launch pad. At what rate should the angle of inclination (the angle the camera makes with the ground) change to allow the camera to record the flight of the rocket as it heads upward?

Figure (PageIndex<1>): As a rocket is being launched, at what rate should the angle of a video camera change to continue viewing the rocket? (credit: modification of work by Steve Jurvetson, Wikimedia Commons)

A rocket launch involves two related quantities that change over time. Being able to solve this type of problem is just one application of derivatives introduced in this chapter. We also look at how derivatives are used to find maximum and minimum values of functions. As a result, we will be able to solve applied optimization problems, such as maximizing revenue and minimizing surface area. In addition, we examine how derivatives are used to evaluate complicated limits, to approximate roots of functions, and to provide accurate graphs of functions.

## 2.1: Prelude to Applications of Derivatives

A rocket is being launched from the ground and cameras are recording the event. A video camera is located on the ground a certain distance from the launch pad. At what rate should the angle of inclination (the angle the camera makes with the ground) change to allow the camera to record the flight of the rocket as it heads upward?

Figure (PageIndex<1>): As a rocket is being launched, at what rate should the angle of a video camera change to continue viewing the rocket? (credit: modification of work by Steve Jurvetson, Wikimedia Commons)

A rocket launch involves two related quantities that change over time. Being able to solve this type of problem is just one application of derivatives introduced in this chapter. We also look at how derivatives are used to find maximum and minimum values of functions. As a result, we will be able to solve applied optimization problems, such as maximizing revenue and minimizing surface area. In addition, we examine how derivatives are used to evaluate complicated limits, to approximate roots of functions, and to provide accurate graphs of functions.

## Teaching Calculus

Unit 4 covers rates of change in motion problems and other contexts, related rate problems, linear approximation, and L’Hospital’s Rule. (CED – 2019 p. 82 – 90). These topics account for about 10 – 15% of questions on the AB exam and 6 – 9% of the BC questions.

You may want to consider teaching Unit 5 (Analytical Applications of Differentiation) before Unit 4. Notes on Unit 5 will be posted next Tuesday September 29, 2020

### Topics 4.1 – 4.6

Topic 4.1 Interpreting the Meaning of the Derivative in Context Students learn the meaning of the derivative in situations involving rates of change.

Topic 4.2 Linear Motion The connections between position, velocity, speed, and acceleration. This topic may work better after the graphing problems in Unit 5, since many of the ideas are the same. See Motion Problems: Same Thing, Different Context

Topic 4.3 Rates of Change in Contexts Other Than Motion Other applications

Topic 4.4 Introduction to Related Rates Using the Chain Rule

Topic 4.5 Solving Related Rate Problems

Topic 4.6 Approximating Values of a Function Using Local Linearity and Linearization The tangent line approximation

Topic 4.7 Using L’Hospital’s Rule for Determining Limits of Indeterminate Forms. I ndeterminate Forms of the type and . (Other forms may be included, but only these two are tested on the AP exams.)

Topic 4.1 and 4.3 are included in the other topics, topic 4.2 may take a few days, Topics 4.4 – 4.5 are challenging for many students and may take 4 – 5 classes, 4.6 and 4.7 two classes each. The suggested time is 10 -11 classes for AB and 6 -7 for BC. of 40 – 50-minute class periods, this includes time for testing etc.

This is a re-post and update of the third in a series of posts from last year. It contains links to posts on this blog about the differentiation of composite, implicit, and inverse functions for your reference in planning. Other updated post on the 2019 CED will come throughout the year, hopefully, a few weeks before you get to the topic.

## Exercises 4.1

### Terms and Concepts

T/F: Implicit differentiation is often used when solving “related rates” type problems.

T/F: A study of related rates is part of the standard police officer training.

### Problems

The area of a square is increasing at a rate of 42 ft 2 /min. How fast is the side length increasing when the length is 7 ft?

Consider the traffic situation introduced in Example 4.1.3 . How fast is the “other car” traveling if the officer and the other car are each 1/2 mile from the intersection, the other car is traveling due west , the officer is traveling north at 50mph, and the radar reading is - 80 mph?

An F-22 aircraft is flying at 500mph with an elevation of 10,000ft on a straight-line path that will take it directly over an anti-aircraft gun.

## Motion along a Line

Another use for the derivative is to analyze motion along a line. We have described velocity as the rate of change of position. If we take the derivative of the velocity, we can find the acceleration, or the rate of change of velocity. It is also important to introduce the idea of speed, which is the magnitude of velocity. Thus, we can state the following mathematical definitions.

### Definition

Let be a function giving the position of an object at time .

The velocity of the object at time is given by .

The speed of the object at time is given by .

The acceleration of the object at is given by .

### Comparing Instantaneous Velocity and Average Velocity

A ball is dropped from a height of 64 feet. Its height above ground (in feet) seconds later is given by .

1. What is the instantaneous velocity of the ball when it hits the ground?
2. What is the average velocity during its fall?

#### Solution

The first thing to do is determine how long it takes the ball to reach the ground. To do this, set . Solving , we get , so it takes 2 seconds for the ball to reach the ground.

1. The instantaneous velocity of the ball as it strikes the ground is . Since m we obtain ft/s.
2. The average velocity of the ball during its fall is

ft/s.

### Interpreting the Relationship between and

A particle moves along a coordinate axis in the positive direction to the right. Its position at time is given by . Find and and use these values to answer the following questions.

1. Is the particle moving from left to right or from right to left at time />?
2. Is the particle speeding up or slowing down at time />?

#### Solution

Begin by finding and .

and .

Evaluating these functions at , we obtain and .

1. Because , the particle is moving from right to left.
2. Because and , velocity and acceleration are acting in opposite directions. In other words, the particle is being accelerated in the direction opposite the direction in which it is traveling, causing to decrease. The particle is slowing down.

### Position and Velocity

The position of a particle moving along a coordinate axis is given by .

1. Find .
2. At what time(s) is the particle at rest?
3. On what time intervals is the particle moving from left to right? From right to left?
4. Use the information obtained to sketch the path of the particle along a coordinate axis.

#### Solution

.

The path of the particle can be determined by analyzing .

A particle moves along a coordinate axis. Its position at time is given by . Is the particle moving from right to left or from left to right at time ?

#### Solution

Find and look at the sign.

## 4.1: Prelude to Applications of Derivatives

The purpose of this section is to remind us of one of the more important applications of derivatives. That is the fact that (f'left( x ight)) represents the rate of change of (fleft( x ight)). This is an application that we repeatedly saw in the previous chapter. Almost every section in the previous chapter contained at least one problem dealing with this application of derivatives. While this application will arise occasionally in this chapter we are going to focus more on other applications in this chapter.

So, to make sure that we don’t forget about this application here is a brief set of examples concentrating on the rate of change application of derivatives. Note that the point of these examples is to remind you of material covered in the previous chapter and not to teach you how to do these kinds of problems. If you don’t recall how to do these kinds of examples you’ll need to go back and review the previous chapter.

First, we’ll need to take the derivative of the function.

[g'left( x ight) = - 6 + 20sin left( <2x> ight)]

Now, the function will not be changing if the rate of change is zero and so to answer this question we need to determine where the derivative is zero. So, let’s set this equal to zero and solve.

[ - 6 + 20sin left( <2x> ight) = 0hspace <0.5in>Rightarrow hspace<0.5in>sin left( <2x> ight) = frac<6><<20>> = 0.3]

The solution to this is then,

[egin<4>2x = & 0.3047 + 2pi n & & hspace<0.5in>,,,,>hspace<0.5in>,,,, & 2x = & 2.8369 + 2pi n & & hspace<0.25in>n = 0, pm 1, pm 2, ldots x = & 0.1524 + pi n & & hspace<0.5in>,,,,>hspace<0.5in>,,, & x = & 1.4185 + pi n & & hspace<0.25in>n = 0, pm 1, pm 2, ldots end]

If you don’t recall how to solve trig equations check out the Solving Trig Equations sections in the Review Chapter.

As with the first problem we first need to take the derivative of the function.

[A'left( t ight) = 135 - 180 - 390 = 15left( <9- 12t - 26> ight)]

Next, we need to determine where the function isn’t changing. This is at,

So, the function is not changing at three values of (t). Finally, to determine where the function is increasing or decreasing we need to determine where the derivative is positive or negative. Recall that if the derivative is positive then the function must be increasing and if the derivative is negative then the function must be decreasing. The following number line gives this information.

So, from this number line we can see that we have the following increasing and decreasing information.

If you don’t remember how to solve polynomial and rational inequalities then you should check out the appropriate sections in the Review Chapter.

Finally, we can’t forget about Related Rates problems.

The first thing to do here is to get sketch a figure showing the situation.

In this figure (y) represents the distance driven by Car B and (x) represents the distance separating Car A from Car B’s initial position and (z) represents the distance separating the two cars. After 3 hours driving time with have the following values of (x) and (y).

[x = 500 - 35left( 3 ight) = 395hspace<0.5in>hspace<0.25in>y = 50left( 3 ight) = 150]

We can use the Pythagorean theorem to find (z) at this time as follows,

Now, to answer this question we will need to determine (z') given that (x' = - 35) and (y' = 50). Do you agree with the signs on the two given rates? Remember that a rate is negative if the quantity is decreasing and positive if the quantity is increasing.

We can again use the Pythagorean theorem here. First, write it down and the remember that (x), (y), and (z) are all changing with time and so differentiate the equation using Implicit Differentiation.

Finally, all we need to do is cancel a two from everything, plug in for the known quantities and solve for (z').

[z'left( <422.5222> ight) = left( <395> ight)left( < - 35> ight) + left( <150> ight)left( <50> ight)hspace <0.25in>Rightarrow hspace<0.5in>z' = frac<< - 6325>><<422.5222>> = - 14.9696]

So, after three hours the distance between them is decreasing at a rate of 14.9696 mph.

So, in this section we covered three “standard” problems using the idea that the derivative of a function gives the rate of change of the function. As mentioned earlier, this chapter will be focusing more on other applications than the idea of rate of change, however, we can’t forget this application as it is a very important one.

## Applications of Derivatives Class 12 Maths

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## Applications of derivatives

Applications of derivatives: rate of change of bodies, increasing/decreasing functions, tangents and normal, use of derivatives in approximation, maxima and minima (first derivative test motivated geometrically and second derivative test is given as a provable tool)

To solve practical problems such as engineering optimization, the greatest challenge is often to convert the word problem into a mathematical optimization problem—by setting up the function that is to be maximized or minimized. Similarly, we can also find the rate of change of a body using differentiation.

Recall that dy/dx is positive if y increases as x increases and is negative if y decreases as x increases.Also, the slope of the tangent of a curve f(x) can be found out by deriving the function of the curve at the given point. Because the normal line is perpendicular to the curve f(x) at a given point, the gradient of the normal is -1/f’(x).

Example: A ship can reach its top speed in 5 minutes.During that time its distance from the start can be calculated using the formula D = t + 50t 2 where t is the time in seconds and D is measured in metres. How fast is it accelerating?

Speed, v m/s, is the rate of change of distance with respect to time.

Acceleration, a m/s 2 , is the rate of change of speed with respect to time or second derivative of distance with respect to time.

Now consider the graph below:

The signs indicate where the gradient of the curve is =, – or 0. In each case:

A function is strictly increasing in a region where f´(x) > 0. A function is strictly increasing in a region where f´(x) < 0. A function is stationary where f´(x) = 0

Identify where the function is (i) increasing (ii) decreasing (iii) stationery

Solution: f’(x) = 6x 2 – 6x – 12

A sketch of the derivative shows us that

When a function is defined on a closed interval, axb, then it must have a maximum and a minimum value in that interval.

These values can be found either at

• a stationary point [where f´(x) = 0]

• an end-point of the closed interval. [f(a) and f(b)]

All you need do is find these values and pick out the greatest and least values.

Example: A manufacturer is making a can for holding 250 ml of juice.The cost of the can is dependent on its radius, x cm.For practical reasons, the radius must be between 2.5 cm and 4.5 cm.The cost can be calculated from the formula

Calculate the maximum and minimum values of the cost function.

which equals zero at stationary points.

• (3x– 1)(x – 3) = 0
• x = 1 /3 or x = 3
• Working to 1 d.p.
• f( 1 /3) = 15.5
• f(3) = 6
• f(2.5) = 6.9
• f(4·5) = 18.4
• By inspection fmax = 18.4 (when x = 4.5) and fmin = 6 (when x = 3).

Let f : D → R, D ⊂R, be a given function and let y = f (x). Let ∆x denote a small increment in x. Recall that the increment in y is corresponding to the increase in x, denoted by ∆y, is given by ∆y = f (x + ∆x) – f (x). We define the following:

• The differential of x, denoted by dx, is defined by dx = ∆x.
• The differential of y, denoted by dy, is defined by dy = f′( x) dx

In case dx = ∆x is relatively small when compared with x, and for y, we denote dy ≈ ∆y.