More Differentiation Problems: Going Off on a Tangent - Differentiation - Calculus For Dummies

Calculus For Dummies, 2nd Edition (2014)

Part IV. Differentiation

Chapter 13. More Differentiation Problems: Going Off on a Tangent


Tangling with tangents

Negotiating normals

Lining up for linear approximations

Profiting from business and economics problems

In this chapter, you see three more applications of differentiation: tangent and normal line problems, linear approximation problems, and economics problems. The common thread tying these problems together is the idea of a line tangent to a curve — which should come as no surprise since the meaning of the derivative of a curve is the slope of the tangent line.

Tangents and Normals: Joined at the Hip

By now you know what a line tangent to a curve looks like — if not, one or both of us has definitely dropped the ball. A normal line is simply a line perpendicular to a tangent line at the point of tangency. Problems involving tangents and normals are common applications of differentiation.

The tangent line problem

I bet there have been several times, just in the last month, when you’ve wanted to determine the location of a line through a given point that’s tangent to a given curve. Here’s how you do it.

Example: Determine the points of tangency of the lines through the point images that are tangent to the parabola images.

Solution: If you graph the parabola and plot the point, you can see that there are two ways to draw a tangent line from images: up to the right and up to the left. See Figure 13-1.


FIGURE 13-1: The parabola images and two tangent lines through images.

The key to this problem is in the meaning of the derivative: Don’t forget — The derivative of a function at a given point is the slope of the tangent line at that point. So, all you have to do is set the derivative of the parabola equal to the slope of the tangent line and solve:

1. Because the equation of the parabola is images, you can take a general point on the parabola, images, and substitute images for y.

So, label the two points of tangency images.

2. Take the derivative of the parabola.


3. Using the slope formula, images, set the slope of each tangent line from images to images equal to the derivative at images which is 2x, and solve for x.

(By the way, the math you do in this step may make more sense to you if you think of it as applying to just one of the tangent lines — say the one going up to the right — but the math actually applies to both tangent lines simultaneously.)




So, the x-coordinates of the points of tangency are images and images.

4. Plug each of these x-coordinates into images to obtain the y-coordinates.



5. Thus, the two points of tangency are images and images, or about images and images.

The normal line problem

Here’s the companion problem to the tangent line problem in the previous section. Find the points of perpendicularity for all normal lines to the parabola, images, that pass through the point images.

remember Definition of normal line: A line normal to a curve at a given point is the line perpendicular to the line that’s tangent to the curve at that same point.

Graph the parabola and plot the point images. Now, before you do the math, try to estimate the locations of all normal lines. How many can you see? It’s fairly easy to see that, starting at images, one normal line goes down and to the right and another goes down to the left. But did you see that there’s actually a second normal line that goes down to the left? No worries if you didn’t see it, because when you do the math, you get all three solutions.

tip Making commonsense estimates enhances mathematical understanding. When doing calculus, or any math for that matter, come up with a common sense, ballpark estimate of the solution to a problem before doing the math (when possible and time permitting). This deepens your understanding of the concepts involved and provides a check to the mathematical solution. (This is a powerful math strategy — take my word for it — despite the fact that in this particular problem, most people will find at most two of the three normal lines using an eyeball estimate.)

Figure 13-2 shows the parabola and the three normal lines.


FIGURE 13-2: The parabola images and three normal lines through images.

Looking at the figure, you can appreciate how practical this problem is. It’ll really come in handy if you happen to find yourself standing inside the curve of a parabolic wall, and you want to know the precise location of the three points on the wall where you could throw a ball and have it bounce straight back to you.

The solution is very similar to the solution of the tangent line problem, except that in this problem you use the rule for perpendicular lines:

remember Slopes of perpendicular lines. The slopes of perpendicular lines are opposite reciprocals.

Each normal line in Figure 13-2 is perpendicular to the tangent line drawn at the point where the normal meets the curve. So the slope of each normal line is the opposite reciprocal of the slope of the corresponding tangent line — which, of course, is given by the derivative. So here goes:

1. Take a general point, images, on the parabola images, and substitute images for y.

So, label each point of perpendicularity images.

2. Take the derivative of the parabola.


3. Using the slope formula, images, set the slope of each normal line from images to image equal to the opposite reciprocal of the derivative at images, and solve for x.


Now, there’s no easy way to get exact solutions to this cubic (3rd degree) equation like the way the quadratic formula gives you the exact solutions to a 2nd degree equation. Instead, you can graph images, and the x-intercepts give you the solutions, but with this method, there’s no guarantee you’ll get exact solutions. (If you don’t, the decimal approximations you get will be good enough.) Here, however, you luck out — actually, I had something to do with it — and get the exact solutions of imagesimages and 12. (You should graph the cubic function so you see how this works.)

4. Plug each of these x-coordinates into image to obtain the y-coordinates.


Thus, the three points of normalcy are images, images, and images — play ball!

Straight Shooting with Linear Approximations

Because ordinary functions are locally linear (that’s straight) — and the further you zoom in on them, the straighter they look — a line tangent to a function is a good approximation of the function near the point of tangency. Figure 13-3 shows the graph of images and a line tangent to the function at the point images. You can see that near images, the curve and the tangent line are virtually indistinguishable.


FIGURE 13-3: The graph of images and a line tangent to the curve at images

Determining the equation of this tangent line is a breeze. You’ve got a point, images, and the slope is given by the derivative of f at 9:


Now just take this slope (this derivative) of images and the point images, and plug them into point-slope form:


That’s the equation of the line tangent to images at images. I suppose you may be wondering why I wrote the equation as images. It might seem more natural to put the 3 to the right of images, which, of course, would also be correct. And I could have simplified the equation further, writing it in images form. I explain later in this section why I wrote it the way I did.

(If you have your graphing calculator handy, graph images and the tangent line. Zoom in on the point images a couple times. You’ll see that — as you zoom in — the curve gets straighter and straighter and the curve and tangent line get closer and closer to each other.)

Now, say you want to approximate the square root of 10. Because 10 is pretty close to 9, and because you can see from Figure 13-3 that images and its tangent line are close to each other at images, the y-coordinate of the line at images is a good approximation of the function value at images, namely images.

Just plug 10 into the line equation for your approximation:


Thus, the square root of 10 is about images. This is only about 0.004 more than the exact answer of 3.1623… . The error is roughly a tenth of a percent.

Now I can explain why I wrote the equation for the tangent line the way I did. This form makes it easier to do the computation and easier to understand what’s going on when you compute an approximation. Here’s why. You know that the line goes through the point images. right? And you know the slope of the line is images. So, you can start at images and go to the right (or left) along the line in the stair-step fashion, as shown in Figure 13-4: over 1, up images; over 1, up images; and so on. (Note that since images, when the run is 1 (as shown in Figure 13-4), the rise equals the slope.)


FIGURE 13-4: The linear approximation line and several of its points.

So, when you’re doing an approximation, you start at a y-value of 3 and go up images for each 1 you go to the right. Or if you go to the left, you go down images for each 1 you go to the left. When the line equation is written in the above form, the computation of an approximation parallels this stair-step scheme.

Figure 13-4 shows the approximate values for the square roots of 7, 8, 10, 11, and 12. Here’s how you come up with these values. To get to 8, for example, from images, you go 1 to the left, so you go down images to images; or to get to 11 from images, you go two to the right, so you go up two-sixths to images. (If you go to the right one half to images, you go up half of a sixth, that’s a twelfth, to images, the approximate square root of images.)

The following list shows the size of the errors for the approximations shown in Figure 13-4. Note that the errors grow as you get further from the point of tangency images. Also, the errors grow faster going down from 9 to 8 then 7, etc. than going up from 9 to 10 then 11, etc.; errors often grow faster in one direction than the other with linear approximations because of the shape of the curve.


mathrules Linear approximation equation: Here’s the general form for the equation of the tangent line that you use for a linear approximation. The values of a function images can be approximated by the values of the tangent line images near the point of tangency, image, where


This is less complicated than it looks. It’s just the gussied-up calculus version of the point-slope equation of a line you’ve known since Algebra I, images, with the images moved to the right side:


This equation and the equation for images differ only in the symbols used; the meaning of both equations — term for term — is identical. And notice how they both resemble the equation of the tangent line in Figure 13-4.

tip Look for algebra-calculus and geometry-calculus connections. Whenever possible, try to see the basic algebra or geometry concepts at the heart of fancy-looking calculus concepts.

Business and Economics Problems

Believe it or not, calculus is actually used in the real world of business and economics — learn calculus and increase your profits! Tell me: When you’re driving around an upscale part of town and you pass by a huge home, what’s the first thing that comes to your mind? I bet it’s “Just look at that home! That guy (gal) must know calculus.”

Managing marginals in economics

Look again at Figures 13-3 and 13-4 in the previous section. Recall that the derivative and thus the slope of images at images is images, and that the tangent line at this point can be used to approximate the function near the point of tangency. So, as you go over 1 from 9 to 10 along the function itself, you go up about images. And, thus, images is about images more than images. The mathematics of marginals works exactly the same way.

mathrules Marginal cost, marginal revenue, and marginal profit work a lot like linear approximation. Marginal cost, marginal revenue, and marginal profit all involve how much a function goes up (or down) as you go over 1 to the right — just like with linear approximation.

Say you’ve got a cost function that gives you the total cost, images of producing x items. See Figure 13-5.


FIGURE 13-5: The graph of a cost function images

Look at the blown-up square on the right in the figure. The derivative of images at the point of tangency gives you the slope of the tangent line and thus the amount you go up as you go 1 to the right along the tangent line. (This amount is labeled in the figure as marginal cost.) Going 1 to the right along the cost function itself shows you the increase in cost of producing one more item. (This is labeled as the extra cost.) Because the tangent line is a good approximation of the cost function, the derivative of C — called the marginal cost — is the approximate increase in cost of producing one more item. Marginal revenue and marginal profit work the same way.

mathrules Definition of marginal cost, marginal revenue, and marginal profit:

· Marginal cost equals the derivative of the cost function.

· Marginal revenue equals the derivative of the revenue function.

· Marginal profit equals the derivative of the profit function.

Before doing an example involving marginals, there’s one more piece of business to take care of. A demand function tells you how many items will be purchased (what the demand will be) given the price. The lower the price, of course, the higher the demand; and the higher the price, the lower the demand. You’d think that the number purchased should be a function of the price — input a price and find out how many items people will buy at that price — but traditionally, a demand function is written the other way around. The price is expressed as a function of the number demanded. I know that seems a bit odd, but don’t sweat it — the function works either way. Think of it like this: If a retailer wants to sell a given number of items, the demand function tells the retailer what he or she should set the selling price at.

Okay, so here’s the example. A widget manufacturer determines that the demand function for his widgets is


where p is the price of a widget and x is the number of widgets demanded. The cost of producing x widgets is given by the following cost function:


Determine the marginal cost, marginal revenue, and marginal profit at images widgets. Also, how many widgets should be manufactured, and what should they be sold for to produce the maximum profit, and what is that maximum profit? (If you get through all this, I’ll nominate you for the Nobel Prize in Economics.)

Marginal cost

Marginal cost is the derivative of the cost function, so take the derivative and evaluate it at images:


Thus, the marginal cost at images is $15 — this is the approximate cost of producing the 101st widget.

Marginal revenue

Revenue, images, equals the number of items sold, x, times the price, p:


Marginal revenue is the derivative of the revenue function, so take the derivative of images and evaluate it at images:


Thus, the approximate revenue from selling the 101st widget is $50.

Marginal profit

Profit, images, equals revenue minus cost. So,


Marginal profit is the derivative of the profit function, so take the derivative of images, and evaluate it at images:


Selling the 101st widget brings in an approximate profit of $35.

tip Marginal profit short cuts: Did you notice either of the two shortcuts you could have taken here? First, you can use the fact that


to determine images directly, without first determining images. Then, after getting images, you just plug 100 into x for your answer.

And, if all you want to know is images, you can use the following really short shortcut:


This is common sense. If it costs you about $15 to produce the 101st widget and you sell it for about $50, then your profit is about $35.

I did it the long way because you need both the profit function, images, and the marginal profit function, images, for the problems below.

Maximum profit

To determine maximum profit, set the derivative of profit — that’s marginal profit — equal to zero, solve for x, and then plug the result into the profit function:


So, maximum profit occurs when 2,025 widgets are sold. Plug this into images:


Thus, the maximum profit is $10,250. (Extra credit: Did you see where I got a bit lazy here? The derivative of the profit function is zero at x = 2,025, but that doesn’t guarantee that there’s a max at that x-value. There could instead be a min or an inflection point there. You could use either the first or second derivative test (see Chapter 11) to show that it’s actually a max. But I just peaked at a graph of the profit function and saw that it’s sort of an upside-down cup shape, so I knew that there was a max at the top of the cup at x = 2,025.)

Finally, plug the number sold into the demand function to determine the profit-maximizing price:


So, the maximum profit of $10,250 occurs when the price is set at $22.22. At this price, 2,025 widgets will be sold. Figure 13-6 sums up these results. Note that because profit equals revenue minus cost, the vertical distance or gap between the revenue and cost functions at a given x-value gives the profit at that x-value. Maximum profit occurs where the gap is greatest.


FIGURE 13-6: The revenue and cost functions. The vertical distance between them, at a given x-value, represents the profit.
(Note that although the scale of this graph makes images look like a straight line, its middle term of images means that it is not exactly straight.)

And here’s another thing. Because maximum profit occurs where images, and because images, it follows that the profit will be greatest where images — in other words, where images. And when images the slopes of the functions’ tangent lines are equal. So, if you were to draw tangent lines to images and images where the gap between the two is greatest, these tangents would be parallel. Right about now you’re probably thinking something like — Such symmetry, such simple elegance, such beauty! Verily, the mathematics muse seduces the heart as much as the mind. Yeah, it’s nice all right, but let’s not get carried away.