| 17 | Introduction to Derivatives |
The relationship between a quantity and its rate of change can provide shortcuts in calculating. These rates, called derivatives, are sometimes the only way to find an accurate result. In this chapter, you’ll learn how rates of change can be determined by a technique called differentiation.
RATES OF CHANGE
To start understanding the basic principle behind differentiation, consider a car traveling along a highway. The rate at which it moves along the highway is its speed. If the direction is also specified, or if the car moves in a straight line, speed can be called velocity. In contrast, acceleration is the rate at which velocity changes. When the speed increases, the acceleration is positive; when the speed decreases, the acceleration is negative (this is also called deceleration).
Before radar speed guns were invented, police timed the movement of a car between two points. If the time was less than what traveling at the legal speed would require, the driver got a ticket. A smart driver would see the first cop and slow down to a speed well below the limit before she got to the second cop. The time check would show that she wasn’t speeding, even though she was when she saw the first cop. Radar speed guns stopped this practice by reading speed at an instant, instead of averaging it over a distance.
The weight-and-spring system, described earlier in this book, shows how velocity and acceleration change during its movement. How do you check those facts? Differential calculus (sometimes called infinitesimal calculus), which is straightforward despite its imposing name, helps you study these types of problems.
INFINITESIMAL CHANGES
Any relationship can be plotted as a graph. Any graph can have an algebraic equation to express the function that is plotted. Some equations are simple, and others are complicated. Those that relate to the real world have two types of variables: independent and dependent. In an equation such as y = xn, x is the independent variable on which the value of y depends. That means x is the independent variable, and y is the dependent variable.
A tiny change in y, divided by the tiny change in x that causes it, gives you the slope of a graph at a point. Use the symbols dx and dy to represent infinitesimals (vanishingly small changes) of x and y. Compared to x and y, these infinitesimals are too small to be measurable. You can make them as minuscule as you want, either positively or negatively, as long as you don’t make them equal to zero. This makes interesting things happen! When the technique of infinitesimals was first worked out by Isaac Newton and Gottfried Leibniz in the 17th century, the results, which we now call calculus, seemed like mathematical magic!
Figure 17-1 shows the general solution for the slope of a graph for y = xn. The binomial expansion is used on (x + dx)n. Then y = xn is subtracted from the expression for y + dy = (x + dx)n, leaving you with an equation for dy in terms of dx. The smaller you make a change, the smaller the higher powers become. As dx/x approaches zero, so does the value of (dx)2/x2, but faster! Using your imagination, you can make these infinitesimals so small that you can treat them as zeros, even though they really aren’t. They converge toward zero, but never quite get all the way there.
Figure 17-1
Infinitesimal changes and the principle of binomial expansion can be used to find the derivative dy/dx of the function y = xn.
Now apply these principles to a few cases and see how they work. Start with the simplest possible equation, y = x, as shown in Fig. 17-2. Dividing dy by dx, after subtracting y = x, we have dy/dx = 1. This statement is easily seen to be correct, because for every change of 1 in x, the value of y also changes by 1.
Figure 17-2
The derivative of the function y = x is dy/dx = 1.
Now take y = x2, as shown in Fig. 17-3. From the formula already derived for the derivative of a variable raised to a power, dy/dx = 2x. Plotting the curve for y = x2, for integer values of x from −5 to 5, draw a smooth curve through the points. If your curve represents the true curve for y = x2, you can draw right triangles under the curve with a base length of 1, and a slanting hypotenuse whose top just touches the curve at the same slope as the curve. In each case, the vertical height of the triangle is 2x for the particular point. Values of dy/dx can be plotted as another graph (this time a straight line), which shows the slope of the first curve at every point.
Figure 17-3
The derivative of the function y = x2 is dy/dx = 2x.
Now try the graph of y = x3, as shown in Fig. 17-4. From the general formula, you can easily find that dy/dx = 3x2. For higher powers of x, the scale for y has to be compressed in order to fit the graph on the page. Draw tangent triangles for the unit horizontal base, just as you did in the previous graph, and measure the height to verify the formula. Then plot a graph for dy/dx. You’ll get a parabola for this, with a shape similar to that of the graph of y = x2 (but with a different vertical aspect). Notice that the cubic curve, unlike the square curve or parabola, curves upward for positive values of x and downward for negative values of x. Its slope is always upward, except when x = 0, where it is momentarily horizontal, the zero slope. The slope is equally positive for the same values of x, positive and negative.
Figure 17-4
The derivative of the function y = x3 is dy/dx = 3x2.
Take one more power of x for the time being: y = x4, as shown in Fig. 17-5. Compressing the y scale again to get the graph on the page, plot from x = −5 to x = 5. This curve is similar to that for y = x2, in that values of y are positive for both positive and negative values of x. But the curvature is different. Correspondingly, the curve for dy/dx is negative when x is negative, because the y = x4 slope is downward (from left to right). Once again, the triangle constructions verify the formula result.
Figure 17-5
The derivative of the function y = x4 is dy/dx = 4x3.
SUCCESSIVE DIFFERENTIATION
In the preceding examples, the slopes of the curves were found by a single differentiation of y with respect to x, denoted dy/dx. Some problems require successive differentiation (as mathematicians call it). For example, velocity or speed is a rate of change of position, also called displacement or distance. Acceleration is a rate of change of velocity or speed. You can even talk about the rate of change of acceleration, sometimes called jerk. Velocity is the first derivative of position, acceleration is the second derivative of position, and jerk is the third derivative of position.
Starting with y = x4, you can repeatedly differentiate to find the successive derivatives. Previously the simplest equation, y = xn, was used. As things work out, a constant will transfer to the derivative. The work in Fig. 17-6 is an example of this. The general statement is
Figure 17-6
The first, second, third, and fourth derivatives of y = x4.
DIFFERENTIATING A POLYNOMIAL
Suppose an equation for y in terms of x takes the form of a polynomial with several terms that involve different powers of x, all added up. In general form, this might be
It could continue to any number of terms, although only two are taken here.
Increasing x by dx, substitute x + dx for x each time it occurs in the original equation. Similarly, increase y by dy to make y + dy. Using the same method as that shown in Fig. 17-1, you get
If you subtract the parts that correspond to y, the parts that correspond with dy are left:
Dividing through by dx gives an expression showing that dy/dx is equal to the sum of the derivatives of the individual terms of the polynomial:
This rule holds true for all expressions made up of terms added together. An easy way to remember it is this: “The derivative of a sum is equal to the sum of the derivatives.” Here is an example:
The derivative is
Notice that where the original term has a minus sign (as does 50x3), the term in the derivative also has a minus sign, as long as the original power is positive.
Now work through this with tabulated values of y and dy/dx, term by term, for integer values of x from x = − 6 to x = 6. Study the curves as they are shown in Fig. 17-7. Where the curve for y reaches a maximum or minimum, the curve for dy/dx passes through a zero value. Momentarily, y is neither increasing nor decreasing. Where the curve for y crosses the zero line (except at the ends where both curves go off scale), the dy/dx curve is at a maximum positive or negative. At those points, the slope of y reaches a maximum, either up or down.
Figure 17-7
Table and graphs for y = x5 − 50x3 + 520x and its derivative.
SUCCESSIVE DIFFERENTIATION OF MOVEMENT
Successive differentiation can be applied to movement (for example, a traveling car). Time is the independent variable because whatever happens, time continues. Distance is the dependent variable because it depends on time. The first derivative of distance is velocity, which is expressed in terms of distance traveled per unit time. The second derivative of distance is acceleration, the rate of change of velocity, measured as distance per unit time squared. Jerk, or rate of change of acceleration, is expressed as distance per unit time cubed. These relationships can be demonstrated during five successive time intervals for a hypothetical moving car, as shown in Fig. 17-8.
Figure 17-8
Here is how distance, velocity, acceleration, and rate of acceleration change (jerk) might vary with time for a moving car.
• In the first interval, the car is stationary. The distance (from any other point) is fixed. Velocity and acceleration are both zero.
• For the second interval, acceleration increases. Assume a steady rate, shown by the straight slanting line. Velocity increases with a quadratic curve. Distance begins increasing with a cubic curve.
• For the third interval, acceleration holds constant, so velocity increases steadily in a straight line. That part of the distance curve is quadratic.
• For the fourth interval, acceleration decreases back to zero. Velocity follows an inverted quadratic curve, and distance approaches a straight line.
• For the fifth interval, acceleration is again zero. Velocity is constant (a horizontal straight line), and distance is a steeply sloping straight line.
This is only one example. Of course, there are infinitely many other ways in which a car can move with respect to time.
CIRCULAR MEASURE OF ANGLES
Angles can be measured in a variety of ways. Degree measure, shown at A in Fig. 17-9, divides a complete circle (rotation or revolution) into 360 equal degrees, symbolized deg or °. The number of degrees tells you how much of a complete circle the angle represents. Advocates of the metric system originally wanted to divide the circle into 100 equal parts, which they called grades. This measure, shown in Fig. 17-9B, is rarely used nowadays, but the principle is the same as for the others. Dividing a circle into quadrants is quite basic, as shown at C.
Figure 17-9
Measures of angles. At A, by degrees. At B, by equal parts with 100 to a complete circle. At C, by quadrants, with four parts to a circle. At D, by radians, defined as arc length divided by radius.
The last method of measuring angles considers a ratio of distances. In many fields of mathematics, physics, and engineering, an angle is defined as the ratio of the arc length around the circumference of a circle, divided by the radius of that same circle. This unit, called the radian and symbolized rad, is illustrated at D in Fig. 17-9.
The circular measure of angles works well with the definitions of trigonometric ratios. Sine, cosine, and tangent are all ratios that identify an angle. None of them is conveniently proportional to all angles. Sine and tangent begin at small angles, in direct proportion to the angle, but this proportionality breaks down long before the first right angle. The cosine begins at 1 and decreases, slowly at first, reaching zero at the first right angle.
Arc length, measured along the circumference of a circle and divided by the radius, is always proportional to the angle that the ratio identifies. Measuring distance around the circumference, the first semicircle (180°) accommodates a little over 3 radii. A whole circle accommodates twice as many, a little over 6 radii. The ratio of the length of a semicircular arc to its radius has the universal symbol π (the Greek lowercase letter pi). For the present, you can use its approximate value of 3.14.
On this basis, a right angle is exactly π/2 rad. From this, you can derive radian values for angles of 60°, 45°, 30°, and 15° as shown in Fig. 17-10. As you tabulate the radian equivalents, sines, and tangents, notice how close the angle in radians, the sine of the angle, and the tangent of the angle are at 15°. For very small angles, the figures are even closer. As the angle approaches zero, the radian measure, the sine, and the tangent all come together. You can verify this with a calculator. Likewise, as the angle increases toward π/2 rad, or 90°, the differences between the values becomes greater.
Figure 17-10
Comparison of circle and triangle definitions for the basic trig ratios, along with tabulations for several angles.
DERIVATIVES OF SINE AND COSINE
Circular measure allows you to apply the principles of differential calculus concerning infinitesimal changes. As already pointed out, for very small angles, the ratio for the sine is almost the same as its circular measure in radians. So for a nearly zero angle, you can consider that sin dx = dx. Because the adjacent equals the hypotenuse at zero angle and 1/1 = 1, it follows that cos dx = 1. These values are true regardless of the value of x, because they concern only a vanishingly small angle dx.
Figure 17-11 shows two examples of how the sine and cosine functions can be differentiated. At A in the figure, y = sin x. Apply the sum formula to the right-hand part of this equation:
Figure 17-11
At A, differentiation of the sine function. At B, differentiation of the cosine function. Note the use of the infinitesimal angle dx, where sin dx = dx and cos dx = 1.
Using the formula you learned earlier for the sine of a sum, and substituting sin dx = dx and cos dx = 1, you get
Now take away the original part, y = sin x. This leaves
Dividing both sides by dx, you obtain the derivative you’re looking for:
Similarly, at B in the figure, if y = cos x, you can again use the sum formula and substitute for cos dx and sin dx, so you get dy/dx = −sin x.
SUCCESSIVE DERIVATIVES OF SINE
These facts about trigonometric ratios can help you draw curves representing the sine and the cosine. The curves provide an excellent way of learning how these functions are related. In Fig. 17-12, the first, second, third, and fourth derivatives of the sine function are tabulated for values of x between 0 rad and 2π rad (0° and 360°). Note that
Figure 17-12
Tabulated values for successive derivatives of the sine function for selected angles in radians.
The notation dn/dxn means that you should take the nth derivative, with respect to x, of whatever comes after it.
Differentiating the sine yields the cosine, differentiating the cosine yields minus the sine, differentiating minus the sine yields minus the cosine, and differentiating minus the cosine yields the sine. If you keep on taking derivatives after that, you will cycle through these same four functions over and over. Figure 17-13 shows graphs of these functions based on the tabulated values, with both degree measure and radian measure. Note that all the curves have the same basic shape. This shape is called a sine wave or sinusoid. Also note that the waves all have identical heights and identical lengths. The only difference is their horizontal position, called the phase. Each time you take a derivative of one of these functions, you shift the phase to the left by 90°, or π/2 rad.
Figure 17-13
Curves showing successive derivatives of the sine function for angles in degrees and radians.
QUESTIONS AND PROBLEMS
This is an open-book quiz. You may refer to the text in this chapter (and earlier ones, too, if you want) when figuring out the answers. Take your time! Consider all values given as exact, so you don’t have to worry about significant figures. The correct answers are in the back of the book.
1. Find dy/dx for the following functions:
(a) y = 5x2
(b) y = x2 + 3x − 5
(c) y = (x + 5)(x − 2)
(d) y = x2 + cos x
(e) y = 4x3 − 4x2 − 4x − 4
(f) y = sin x + 2 cos x
(g) y = 5x4 + 2x2
(h) y = x2 + 4x4
2. Suppose a car starts from a stationary position and accelerates at a rate of 5 miles per hour per second [5 (mi/h)/s]. How far will the car travel in the first 10 s after it starts?
3. In the situation of the previous problem, how fast will the car be moving, in miles per hour, after the first 10 s has elapsed?
4. Suppose the car in the scenario of Probs. 2 and 3 stops accelerating (the acceleration becomes zero) at the end of the first 10 s, and then the car continues traveling indefinitely at a zero acceleration rate. How far will the car have gone from its initial position 20 s after it started moving?
5. An object in free fall, in the gravitational field of the earth, accelerates downward at a rate of 32 feet per second per second (32 ft/s2), assuming there is no effect from air resistance. Suppose you drop a lead ball (heavy enough to overcome all air resistance) off the top of a building 20 stories tall, where each story is 10 ft high. How long will it be, in seconds, from the time the ball is dropped until it hits the ground?
6. In the situation of Prob. 5, how fast will the ball be traveling, in feet per second, when it hits the ground?
7. Refer to the graph of acceleration versus time shown by Fig. 17-14. Suppose a car starts out traveling at a speed of 44 ft/s. Write an equation for the speed ν of the car in feet per second as a function of the elapsed time t in seconds, based on the information given in this graph.
Figure 17-14
Illustration for problems 7 through 10.
8. Write an equation, based on the information in Fig. 17-14, for the displacement s of the car in feet as a function of the elapsed time t in seconds. Assume that when t = 0, s = 0.
9. Based on the information in Fig. 17-14 and the derived equations, how fast will the car be traveling, in feet per second, after the following lengths of time? Express your answers in decimal form to four significant digits.
(a) 1.000 s
(b) 2.000 s
(c) 5.000 s
(d) 10.00 s
10. Based on the information in Fig. 17-14 and the derived equations, how far will the car have traveled, in feet, from its starting point after the following lengths of time? Express your answers in decimal form to four significant digits.
(a) 1.000 s
(b) 2.000 s
(c) 5.000 s
(d) 10.00 s
11. Refer to the graph of voltage versus time in Fig. 17-15. The curve is a sine wave. The maximum voltage is +8.00, and the minimum voltage is −8.00. Write down an equation for the voltage V (at any given instant) in volts as a function of time t in seconds.
Figure 17-15
Illustration for problems 11 and 12.
12. How rapidly, in volts per second (V/s), is the voltage in Fig. 17-15 changing at the following times t? Use a calculator if you need one, and express your answers to four significant digits.
(a) t = 0.000 s
(b) t = 2.000 s
(c) t = 2.500 s
(d) t = 5.000 s
(e) t = 6.000 s
(f) t = 7.500 s
13. What are the measures, in radians, of the following angles? Use a calculator and specify to four significant digits.
(a) 10.00°
(b) 30.00°
(c) 75.00°
(d) 145.0°
(e) 220.0°
(f) 300.0°
14. What are the measures, in degrees, of the following angles? Use a calculator and specify to four significant digits.
(a) 0.2000 rad
(b) 0.5000 rad
(c) 1.000 rad
(d) 1.700 rad
(e) 2.200 ra
(f) 3.500 rad
15. Assume that the earth makes one complete revolution around the sun, relative to the distant stars, in exactly 365.25 days. Also assume that the earth revolves at a constant speed at all times in its orbit, and that the orbit of the earth is a perfect circle with the sun at the center. (These assumptions are not quite true in reality, but assuming them simplifies this problem tremendously!) Given this information:
(a) How many degrees of arc does the earth advance around the sun in one day? Express your answer to five significant digits.
(b) How many radians of arc does the earth advance around the sun during the month of April, which has exactly 30 days? Express your answer to five significant digits.
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