Chap2b

More on Differentiation Chapter 2B August 1999 Now that we have some basic rules for differentiation, we will expand our toolbox and look at some applications.

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Derivatives of trigonometric functions

We have already considered limits involving trigonometric functions.

1.1

Notes on trigonometric limits

Our first goal is to show that sin x = 1. x To aid in our discussion, consider the following graph. Actually, you’ll have to imagine this one for now; the graphic file seems to be missing, and I haven’t gotten around to recreating it. Sorry! A piece of a circle of radius 1 is shown, passing through points A and B. The line segments BD and BE intersect at a right angle. Note that, since the segment BD is a segment of a line that contains a radius of the circle, this means that the segment BE lies on the tangent line to the circle at the point B. The point C is chosen so that the line segment CB is perpindicular to the x− axis (So the point C has the same x− coordinate as the point B, but has y− coordinate 0). Y denotes the angle formed by the positive x− axis and the line passing through the origin and the points B and D (measured in radians). Recall that the segment of a circle of radius r determined by an angle θ (measured in radians) will have arc length rθ. So, in this case, the length of the segment of the circle determined by the points A and B will satisfy |arcAB)| = Y. First, we note that 0 < |CB| = sin Y < |arc (AB)| = Y . This tells us that (at least for Y > 0) lim

x→0

0<

sin Y <1 Y

(Why does this inequality also hold if Y < 0?) Also, applying the pinching or squeeze theorem to the first inequality tells us that lim+ (sin Y ) = 0. Y →0

On the other hand, Y = |arc(AB)| < |AE| + |EB| < |AD| = tan Y This tells us that Y <

sin Y cos Y

which is equivalent to cos Y <

sin Y Y

Combining the above results yields cos Y <

sin Y <1 Y 1

for Y 6= 0. Since lim cos Y = 1 and lim 1 = 1,

Y →0

Y →0

we can use the pinching theorem to deduce that lim

x→0

sin x = 1. x

A related result is the following:

1 − cos x = 0. x→0 x In fact this is a consequence of the previous result. To see why, note that lim

1 − cos x x

= = = =

1 − cos x 1 + cos x x 1 + cos x 1 − cos2 x x (1 + cos x) sin2 x x (1 + cos x)    sin x sin x x 1 + cos x

We can see that, as x approaches 0, this last expression also approaches 0. The limits we have just evaluated provide us with the slopes of the tangent lines to the graphs of the functions y = sin x and y = cos x at x = 0 (Be sure that you understand why!). But in fact the payoff for this work is much more. In the following excercise, we use the two limits we just calculated to determine the slope of the tangent line to the graph of either the sine or cosine at any point. We will need the identities: sin (A + B)

=

sin A cos B + cos A sin B

cos (A + B)

=

cos A cos B − sin A sin B

together with the limits computed above. Now, d (sin x) dx

=

sin (x + h) − sin x h sin x cos h + cos x sin h − sin x lim h→0 h sin x (cos h − 1) + cos x sin h lim h→0 h sin x (cos h − 1) cos x sin h lim + lim h→0 h→0 h h     cos h − 1 sin h (sin x) lim + (cos x) lim h→0 h→0 h h (sin x) (0) + (cos x) (1)

=

cos x.

= = = = =

lim

h→0

Similarly, we find d (cos x) dx

cos x (cos h − 1) − sin x sin h h = − sin x =

lim

h→0

2

Since other trigonometric functions can be expressed in terms of the sine and cosine, we can now compute the derivatives of those functions using the rules for differentiation we have already developed. For example, since  sin x  −1 = sin x (cos x) tan x = cos x we can use the quotient rule (or the product and chain rules) to learn that d (tan x) dx

= =

2 2.1

cos2 x + sin2 x 1 = cos2 x cos2 x sec2 x.

Applications and extensions Increasing and decreasing functions

Consider the cube function

Figure 1: Graph of y = x3 on [−3, 3]. As x increases (moves from left to right), we can see that f (x) also increases. That is, if s < t then f (s) < f (t). We say that the function f is increasing. Perhaps the simplest example of an increasing function is a line with positive slope. There is a connection between the notion of an increasing function and slopes of tangent lines. At a point on the graph of an increasing function where there is a tangent line, the slope of the tangent line will be positive (or at least non-negative; consider the derivative of f (x) = x3 at x = 0 in the example above). Since the slope of that tangent line to the graph at a point (x, f (x)) is the value of the derivative f 0 (x), we learn that an increasing function must have a non-negative derivative. We will be interested in understanding local as well as global behavior of functions, so we will also define a function to be increasing on an interval [a, b] if, for any pair of numbers s and t in the interval [a, b] with s < t, f (s) < f (t). In this case, we will have f 0 (x) ≥ 0 for all x ∈ [a, b]. In a similar fashion we define what it means for a function to be decreasing (on an interval). Consider the following example. Let g(x) = x3 + 2x2 − x + 1. Here is a plot of y = g(x) on the interval [−3, 2]. From the graph, we can see that the function is increasing until x is about −1.5 and again after x passes about 0.2. We can identify these values more accurately by graphing the derivative, g 0 (x) = 3x2 + 4x − 1. The points where the graph crosses the x-axis are the points where the derivative changes sign. You can zoom in on smaller intervals containing those crossings to get better and better approximations of the points where g changes from increasing to decreasing (or the reverse). (Actually, in this example, our first estimates look pretty good.) Alternatively, we can find the zeros of the derivative. √ √ −2 ± 7 −4 ± 16 + 12 0 2 = ≈ −1.549, .215 g (x) = 3x + 4x − 1 = 0 ⇐⇒ x = 6 3

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Figure 2:

Figure 3: Plot of the derivative of g. In this case, we could solve the equation explicitly. We will see that, in many cases, this is not possible, and we must approximate the zeros of the derivative. Note, too, that the zeros of the derivative in this case correspond to ”high” and ”low” points on the graph of g. These points are referred to as local maximums and local minimums. We will consider this issue in more detail later.

2.2

Higher order derivatives

Now that we know how to take the (first) derivative of a function, it is fairly straightforward to compute a second (or third, or fourth) derivative as well. For example, if f (x) = x2 , we already know that f 0 (x) = 2x. The second derivative of f will be f 00 (x) = 2, and the third derivative will be f 000 (x) = 0. We will soon learn what information about the original function is encoded in the second derivative.

2.3

Linear and quadratic approximation

The single most important thing to remember about the derivative is that it tells us how best to (locally) approximate a function by a straight line (the tangent line to the graph). The slope of this line is also referred to as the instantaneous rate of change of the function (at the point under consideration). We have already discussed the fact that the tangent line is the best local approximation to a function. The equation of the tangent line to the graph of a function f at a point (x0 , f (x0 )) is: y = f (x0 ) + f 0 (x0 ) (x − x0 ) . So, the idea is that, for x close to x0 , f (x) ≈ f (x0 ) + f 0 (x0 ) (x − x0 ) . This is called linear approximation; we are approximating a function with a line. 4

Let’s remind ourselves how this works by looking at an example. Suppose we consider the function √ f (x) = x. Then

1 f 0 (x) = √ . 2 x

We know the exact value of both f and f 0 at any perfect square (1, 4, 9, 16, ...), and can readily find the equation of the tangent line to the graph at any point corresponding to one of these. Let’s look at the tangent line at the point (16,4), together with the original function

Figure 4: We see that the tangent line remains close to the graph of the function (is a good approximation) close to the point (16,4). We can use the simpler equation of the tangent line to approximate the function. Here is a plot of the difference between the two:

Figure 5: Note that the approximation appears to be very good between 14 and 18: Note that, at the point (x0 , f (x0 )), the derivative of the function that defines the tangent line has the same value as the derivative of f at that point. That’s really obvious since we construct the tangent line with slope f 0 (x0 ). The point is that the tangent line matches the original funcion and its derivative at the point (x0 , f (x0 )). We can extend this notion of local approximation to higher orders. The next step is to consider quadratic approximations. For a function f, we wish to construct a quadratic polynomial which is the best approximation to f near a given point (x0 , f (x0 )). Following the reasoning above, we want to find a quadratic function which passes through (x0 , f (x0 )) and has the same first and second derivatives as f at that point. This is best done by writing a quadratic in the form: 2 q(x) = c + b (x − x0 ) + a (x − x0 ) .

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Figure 6: Remember that we want q and f to satisfy q (x0 )

=

f (x0 )

q 0 (x0 )

=

f 0 (x0 )

q 00 (x0 )

=

f 00 (x0 ) .

The first condition immediately yields c = f (x0 ). Since q 0 (x) = b + 2a(x − x0 ), the condition q 0 (x0 ) = f 0 (x0 ) requires that we choose b = f 0 (x0 ) . Note that the linear piece of our quadratic function is the same as the tangent line. Finally, since q 00 (x) = 2a, the condition q 00 (x0 ) = f 00 (x0 ) means that we must take a=

f 00 (x0 ) . 2

So, our quadratic approximation becomes q(x) = f (x0 ) + f 0 (x0 ) (x − x0 ) +

f 00 (x0 ) 2 (x − x0 ) . 2

In our example above, we looked at the linear approximation to the graph of f (x) = f 00 (x) = −

√

x at the point (16,4). Since

1 1 and f 00 (16) = − , 3/2 256 4x

the quadratic approximation at the same point is q(x) = 4 +

1 1 2 (x − 16) − (x − 16) . 8 512

Let’s look at a plot of this, together with a plot of f , to see how close the approximation is. This approximation looks very good near (16,4). Let’s look at the plot of the difference over a smaller interval. Comparing this approximation to the linear approximation, it appears that we have improved the accuracy by about a factor of 10. That’s pretty impressive, and fairly representative of what happens with other examples.

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Figure 7:

Figure 8:

2.4

Implicit differentiation

We have now developed several techniques for computing derivatives of functions defined by equations (y = f(x)). But many times, relationships between variables are not so simple. For example the equation of the circle centered at (a, b) with radius r is: (x − a)2 + (y − b)2 = r2 . While it is possible to ”solve” this equation for y (or x, for that matter), such a solution doesn’t tell the whole story, since for each value of x in the interval (−r, r) there are two possible values of y. Yet geometrically it is still clear that, at any point on the circle, we should be able to construct a tangent line. On your own In fact, for the circle described above, it should be clear from the geometry that the tangent line to a circle at any point (x0 , y0 ) on the circle is perpindicular to the line passing through the center (a, b) and (x0 , y0 ). Use this idea to construct the equation of the tangent line at an arbitrary point, and discuss the ”singularities” of your solution. In many cases, equations that define relationships between variables may not define one variable as a function of another, but may still define smooth curves that have tangent lines at any point (x0 , y0 ) on the curve. (We say that such equations define y as a function of x locally.) Assuming that, near a point (x0 , y0 ), y 0 exists, how can we determine its value? The answer is to differentiate both sides of the equation that defines the relationship between the two variables, and then ”plug in” the base point (x0 , y0 ). Let’s look at the example of the circle above: (x − a)2 + (y − b)2 = r2 . Differentiating both sides of this equation yields 2(x − a) + 2(y − b)y 0 = 0. Notice that, in the second term, we have used the chain rule, first differentiating the ”outside” (square) function, then the inside function, y − b. Since we don’t have an explicit expression for y 0 , we leave it symbolically. But, the resulting equation can be solved for y 0 : y 0 = −(x−a) (y−b) . So, at the point (x0 , y0 ), we will have 0 y = −(x0 − a)/(y0 − b) . Note that this won’t make sense if y0 = b. But what point does that correspond to? From the original equation (or the geometry of the circle, since the center is (a, b) ), if y0 = b, the point is on the horizontal diameter of the circle, and the tangent line is vertical, hence the slope is not defined. So, even in that case, the equation for y 0 makes sense.

7

On your own Verify that the equation for y’ just derived for the circle agrees with the result you obtained by constructing a line perpindicular to the radius. 2 2 Let’s look at another example. The equation x9 y4 = 1 defines the ellipse centered at the origin with major axis length 6 along the x-axis and minor axis length 4.

Figure 9: p At x = 2, there are two possible y values: ±2 ∗ (5)/3 . Let’s compute the slope of the tangent line at the point in the first quadrant. First we differentiate both sides of the equation defining the ellipse to obtain: 2

x y + 2 y 0 = 0. 9 4

Solving this equation for y 0 yields: y0 =

−4x . 9y

√

So, at the point (2,

2

(5) 3 ),

−4 . the slope of the tangent line will be y 0 = √ 3

(5)

We can now plot the ellipse together with the tangent line.

Figure 10:

On your own 1) For the previous example, compute the equation of the tangent line to the graph at the point ( 2,-2*sqrt(5)/3; ). 2) Consider the equation x2 ∗y−x∗y 2 = 2; .T hepoints(−1, 1)and(−1, −2)arebothonthegraphof thiscurve.Computetheslopeof

2.5

Applications in the sciences

E. P. Wigner wrote a now famous article, The unreasonable effectiveness of mathematics in the natural sciences, (Comm. Pure Appl. Math.13, Feb. 1960) in which he discussed a number of examples of the use of mathematics 8

to model real-world phenomena. The theme of that article, and in fact in much related scientific thought, is that the same mathematical concepts turn up in entirely unexpected connections. So, while we may be learning about the fundamental notions of the calculus from a strictly mathematical point of view, we wish to point out that there are many applications of these tools in diverse areas of science and engineering. Below are just a sampling. Note that there may be some missing things in the below, since the original version was interspersed with Maple code/commands as hints to the students. Problem 1: This problem allows you to look at some of the information that you can derive from a function. It also will introduce you to the Maple ”diff” function which, in this case, allows you to check your work. A ball is tossed into the air from a bridge. Its height (y) in feet, after t seconds is given by: y = t2 + 34t + 36 a. How high above the ground is the bridge? b. What is the average velocity of the ball for the first second? c. Approximate the velocity of the ball at t = 1 second (use Maple to get the derivative only to check your approximation. d. Graph the function. What is the maximum height the ball will reach? What is the velocity at that point? e. Look at the derivative of the function. What happens to the derivative as the ball crosses that point? Problem 2: Match the slopes with the points in the following graph. Do this visually without using Maple to differentiate. When you have made your estimates, check your answers using Maple. Slopes: -1.74, 0.54,-0.18,-0.94 Points: (1, y(1)), (2, y(2)), (3, y(3)), (4, y(4))

Figure 11: Problem 3: The removal of a drug by the kidneys usually proceeds at a rate that is a linear function of the concentration of the drug in the blood (i.e., the greater the concentration of the drug, the faster it is removed). Say for a specific drug:

Figure 12: Where concentration is on the x axis and removal rate per minute is on the y axis. 9

Say you started with an initial concentration of 20 micro-grams per 100 cc of blood. Conc := 20 The concentration after 1 minute would be given by the following equation. newConc:=Rate*1; newConc := 12.0 Now, take the new concentration and add it to the following list: C:=[20,: t:=[0,1,2,3,4,5,6] Next take the new concentration and plug it (the number) into ”Conc” and toggle down to get another ”newConc”. Repeat until you have the concentration of the blood as a function of time for the 6 minutes of the test. Next, zip the 2 lists and look at your plot (you’ll have to fill in the parameters for ”plot”). What kind of plot is this (what kind of a function governs drug removal)? As mentioned, this is the most common mechanism for ”clearing drugs” by the kidneys. It’s called a ”first order” reaction. Why? Hint: look at the slope of dC/dt as a function of C. See if you can fit a function to the data.

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