20.7 A Brief Introduction to Vectors 681 (c) How far has the object traveled horizontally when it hits the ground? (In other words, what is the horizontal component of its displacement vector?) (d) When the object hits the ground, how far is it from where it was launched? (In other words, what is the length of its displacement vector?) 10. Force A has a horizontal component of 3 pounds and a vertical component of 4 pounds. Force B has a horizontal component of 5 pounds and a vertical component of12 pounds. (a) What is the strength of force A? What angle does this force vector make with the horizontal? (Give a numerical approximation in degrees.) (b) What is the strength of force B? What angle does this force vector make with the horizontal? (Give a numerical approximation in degrees.) (c) What is the component of force A in the direction of force B? 11. There’s a 10-mile-per-hour wind, but a bicyclist calculates that its component in his direction of motion is only 9 miles per hour. What is the angle between the velocity vector of the wind and that of the bicyclist? 12. Due to the scampering of a goat, a rock has been dislodged from a mountain and is sliding down an incline making a 70 ◦ angle with level ground. The weight of the rock exerts a downward force of 3 pounds. What is the component of this force in the direction of motion of the rock? goat 3 lbs downward 70° rock 21 CHAPTER Differentiation of Trigonometric Functions 21.1 INVESTIGATING THE DERIVATIVE OF SIN X GRAPHICALLY, NUMERICALLY, AND USING PHYSICAL INTUITION Graphical Analysis Figure 21.1 presents the graphs of f(x)=sin x, g(x) = cos x, and sketches, qualitatively done, of their derivative functions. Where the function is increasing, its derivative is positive or zero; where the function is decreasing, its derivative is negative or zero. Where the function is concave up, its derivative is increasing; where the function is concave down, its derivative is decreasing. Sine and cosine are periodic with period 2π. Similarly, their derivatives must be periodic with period 2π . 1 Notice that no vertical scale is given on the graphs of the derivative functions. 1 A function with period k must have a derivative such that f (x + k) = f (x). In general, it is possible for the period to be some number s<ksuch that ns = k for some positive integer n. For example, if the graph of f is as shown below, then the period of f is half that of f . f x 2468 –1 –2 –3 683 684 CHAPTER 21 Differentiation of Trigonometric Functions sin x = f(x) cos x = g(x) x 1 –1 π 2 –π 2 3π 2 –3π 2 π–π 2π–2π f′(x) x π 2 –π 2 3π 2 –3π 2 π–π 2π–2π x π 2 –π 2 3π 2 –3π 2 π–π 2π–2π x 1 –1 π 2 –π 2 3π 2 –3π 2 π–π 2π–2π g′(x) Figure 21.1 Let’s focus on the derivative of sin x. Its graph looks seductively like cos x. However, simply from qualitative analysis we have no evidence leading us to believe, for instance, that the amplitude of its derivative is 1. In fact, the amplitude of the derivative will depend upon whether x is given in degrees or radians. Let’s stick to radians and gather some numerical data by approximating the derivative of sin x at x = 0. Numerical Investigation of the Slope of the Tangent Line to sin x at x = 0 The derivative of sin x at x = 0 is the slope of the tangent to sin x at (0, 0). We can approximate the slope of the tangent line by finding the slope of the secant line between (0, 0) and a nearby point on the sine curve. The slope of the secant line through (0, sin 0) and (0 + h, sin(0 + h))is sin h − sin(0) h − 0 = sin h h . f x (0, 0) (h, sin h) Figure 21.2 For h near 0 this is a good approximation to the slope of the tangent line. At h = 0.0001, the slope of this secant line is sin(0.0001) 0.0001 ≈ 0.9999999983. If we use h = 0.00001 to approximate d dx sin x| x=0 , we compute sin(0.00001) 0.00001 . A TI-83 calculator gives this quotient as 1 (although in fact it is not exactly 1). The conjecture that the derivative of sin x at x = 0 is 1 looks quite reasonable, 2 and following closely upon its heels is the delightfully appealing conjecture that d dx sin x = cos x. If we can prove this, it will make us happy. 2 Notice that it appears (although we have not yet proven it) that lim x→0 sin x x = 1. 21.1 Investigating the Derivative of sin x Graphically, Numerically, and Using Physical Intuition 685 EXERCISE 21.1 Graph j(x)= sin x x on a graphing calculator or a computer. j(x) is not defined at x = 0. What does the graph lead you to conjecture about lim x→0 sin x x ? Radians versus Degrees. The previous numerical analysis was done using radians. Let’s see what happens if we measure the angle x in degrees and look at the slope of the line tangent to sin x at x = 0. Again look at sin h h for h very small but this time measure h in degrees. According to a TI-83 calculator, for h = 0.001, 0.0001, and 0.00001, sin h ◦ h ◦ ≈ 0.0174532925 1 degrees . 3 This is a far cry from the value of approximately 1 when using radians. Actually, the number makes sense if it is indeed true that d dx sin x = cos x, when x is measured in radians. If w is an angle measured in degrees we can convert to radians and differentiate. d dw sin(w ◦ ) = d dw sin w ◦ · π radians 180 ◦ = d dw sin wπ 180 radians = cos wπ 180 · π 180 Using the hypothesis d dx sin x = cos x and the Chain Rule = cos w ◦ · π 180 ≈ 0.0174532925 cos w ◦ because π 180 ≈ 0.0174532925 Spinning Wheels: Learning from Physical Intuition We’ve been using the sine function to model the height of a marked point on a steadily spinning wheel, whether that point corresponds to position of a seat on a Ferris wheel, a mark on a gear, or gum on a suspended tire. Let’s see what we can learn about the derivative of sin t by looking at the real-world situation we’re modeling. Suppose a mark is made on the rim of a vertically oriented wheel of radius 3 meters. For simplicity’s sake, we’ll assume the wheel is spinning counterclockwise at a steady rate of 1 revolution every 2π minutes so that the height of the mark at time t is given by h(t) = 3 sin t + 3, where h is given in meters and t in minutes. Since the wheel is rotating steadily, we know the mark is moving at a constant speed. Its speed is given by circumference of wheel 2π seconds = 6π meters 2π seconds = 3 meters per second. 3 Radians are unitless. When we write sin x with no units, we mean radians. We have the awkwardness of 1/degree because degrees are not unitless. 686 CHAPTER 21 Differentiation of Trigonometric Functions Although the speed of the mark is constant, the direction it is moving is not. At each time t the velocity can be resolved into its vertical and horizontal components. 4 The vertical component is the rate at which the height is changing, and the horizontal component is the rate at which the horizontal coordinate of the point is changing. Q R P S Figure 21.3 For now we’ll focus on the vertical component of the velocity. h(t) = 3 sin t + 3gives the vertical position of the mark at time t,so dh dt must give the rate of change of height, or the vertical component of the velocity. dh dt = 3 · d dt [sin t] The point marked on the wheel has a vertical velocity of 3 m/sec when it is in position P (see Figure 21.4) because the entire velocity is directed upward. Its vertical velocity is decreasing but positive until it reaches position Q.AtQthe velocity is directed horizontally, so the vertical component is zero. t = 0 t = t = π v → → v → v → v → P Q R S π 2 t = 3π 2 vertical component of v is 0. → vertical component of v is 0. → vertical component of v is –3. → vertical component of v is 3. Figure 21.4 4 The velocity can be modeled as a vector, v of length 3 tangent to the wheel. We are resolving this vector into its vertical and horizontal components. 21.1 Investigating the Derivative of sin x Graphically, Numerically, and Using Physical Intuition 687 P Q 3 Figure 21.5 At position Q the point stops its ascent and is poised for its descent. Between Q and S (where the velocity again is entirely horizontal) the vertical component of velocity is negative. The most negative velocity is at position R where the entire velocity (3 m/sec) is directed downward. Since dh dt = 3 d dt sin t, this physical analysis indicates that the derivative of sin t has the following characteristics. It has a maximum value of 1 at t = 2πn, it has a minimum value of −1att=π+2πn, and it is zero at π/2 + πn. The results of our physical analysis support our graphical and numerical work and the conjecture that the derivative of sin x is cos x.We’ll prove this conjecture in the next section. PROBLEMS FOR SECTION 21.1 1. Estimate lim x→π sin x x both numerically and graphically. 2. Let f(x)=sin x. Use the difference quotient with h = 0.0001 to estimate the value of f (π), the slope of the tangent line to sin x at x = π . 3. Let f(x)=sin(x). (a) Using a calculator, tabulate f at x = 1.998, 1.999, 2.000, 2.001, 2.002 (make a table with values of x and f(x)). Round values of f to six decimal places. (b) Estimate f (1.999), f (2), and f (2.001) using the tabulated values. (c) Estimate f (2) using the results from part (b). 4. (a) What is the limit definition of d dx cos x | x=0 ? (b) Numerically approximate d dx cos x | x=0 . 5. Numerically approximate the derivative of cos x at x = π. 688 CHAPTER 21 Differentiation of Trigonometric Functions 21.2 DIFFERENTIATING SIN X AND COS X Proof that the Derivative of sin x is cos x To prove that d dx sin x = cos x we’ll go back to the limit definition of derivative. d dx sin x = lim h→0 sin(x + h) − sin(x) h We’ve already shown that sin(x + h) = sin x + sin h and asserted that sin(A + B) = sin A cos B + sin B cos A.We’ll use this fact below. d dx sin x = lim h→0 sin(x + h) − sin(x) h = lim h→0 sin(x) cos(h) + sin(h) cos(x) − sin(x) h = lim h→0 sin x[cos(h) − 1] + sin(h) cos x h (gathering the sin x terms) = lim h→0 sin x[cos(h) − 1] h + sin(h) cos x h = lim h→0 sin x cos(h) − 1 h + cos x sin(h) h = lim h→0 sin x cos(h) − 1 h + lim h→0 cos x sin(h) h (sin x and cos x are independent of h) = sin x lim h→0 cos(h) − 1 h + cos x lim h→0 sin(h) h There are two limits we need to evaluate: lim h→0 cos(h)−1 h and lim h→0 sin(h) h . Let’s take them on one by one. We already have evidence suggesting that lim h→0 sin(h) h = 1; we’ll prove this below. 5 Then we need only show the former limit is 0 to obtain the desired result. 6 Proof that lim x→0 sin x x = 1 First notice that sin x x is an even function: sin(−x) −x = − sin x −x = sin x x . Therefore, it is enough to show that lim x→0 + sin x x = 1, because the limit as we approach zero from the left will be the same as the limit as we approach zero from the right. 5 The proof given is the standard proof of this limit. When you first see this argument you may feel a bit bewildered. While each step follows logically from the previous one, chances are you will not see at the outset where you are being led. The proof is not laid out in the form in which someone thought of the argument. In contrast, the conclusion we want “drops out” at the end. Mathematicians spend a lotof time playing with ideas, poking, tugging, wrestling—and sometimes having a flash of inspiration. A mathematician may make the logical journey from point A to point B in a myriad of different ways. The proof he or she presents to the world may show a map of the journey or the proof may be polished and shined so that the marks of the hewing process are hidden from view. The aesthetics of a proof are very much part of the culture of mathematics. (A famous mathematician of our time, Paul Erd ¨ os (1913–1996), used to refer to what he called “The Book” of mathematical theorems in which each theorem exists along with its most beautiful and insightful proof.) 6 The proof that we are in the midst of (that the derivative of sin x is cos x) began in a very straightforward manner. We have come up against two sticking points, and will resolve these two as separate problems, importing the result back to our original argument. These building blocks of the puzzle are called “lemmas.” 21.2 Differentiating sin x and cos x 689 When looking at the limit as x goes to 0 from the right, we can restrict our discussion to 0 <x<π/2. That said, refer to Figure 21.6. B is the point (1, 0) and P = P(x)is a point on the unit circle in the first quadrant. Look at the area of the pie-slice shaped sector of the circle delineated by OBP.For positive x, the area of this sector is greater than the area of the small shaded triangle OAP and less than the area of the large triangle OBQ. Q v P x x ABO u cos x sin x 1 Figure 21.6 area of small triangle < area of sector < area of large triangle The area of the small triangle = 1 2 (length of AP ) · (length of OA) = 1 2 (sin x)(cos x) The area of the large triangle = 1 2 (length of BQ) · (length of OB) = 1 2 (tan x)(1) = 1 2 tan x To fi nd the area of the sector OBP look at the following ratio. area of the sector area of the circle = arclength circumference area of the sector π(1) 2 = x 2π area of the sector = x 2 x 1 Figure 21.7 690 CHAPTER 21 Differentiation of Trigonometric Functions It follows that 1 2 sin x cos x< x 2 < 1 2 tan x sin x cos x 2 < x 2 < 1 2 sin x cos x If A<B and k is any positive number, then kA < kB. 7 Multiply by 2. sin x cos x<x< sin x cos x Because sin x>0for x ∈ (0, π/2), we can divide by sin x. cos x< x sin x < 1 cos x This inequality holds for 0 <x<π/2. lim x→0 + cos x ≤ lim x→0 + x sin x ≤ lim x→0 + 1 cos x In the limit the strict inequality “<” becomes “≤”. 8 1 ≤ lim x→0 + x sin x ≤ 1 The limit is squeezed in a vise until completely determined. so lim x→0 + x sin x = 1 Therefore, lim x→0 + sin x x = 1. Because sin x x is an even function, we have shown that lim x→0 sin x x = 1. We now have that d dx sin x = sin x lim h→0 cos(h) − 1 h + cos x lim h→0 sin(h) h = sin x lim h→0 cos(h) − 1 h + cos x · 1 = sin x lim h→0 cos(h) − 1 h + cos x. If we can show lim h→0 cos(h)−1 h = 0, then our proof that d dx sin x = cos x is complete. WHAT IS lim h→0 cos(h)−1 h ? Looking at the graph of cos x−1 x for x near zero and numerically investigating lim h→0 cos(h)−1 h by evaluating cos(h)−1 h for h increasingly close to zero lead us to conjecture that lim h→0 cos(h)−1 h = 0. (Try this for yourself.) Alternatively, we can reason that lim h→0 cos(h)−1 h = lim h→0 cos(0+h)−cos 0 h = d dx cos x | x=0 . That is, lim h→0 cos(h)−1 h is just the derivative of cos x evaluated at x = 0. If we assume that cos x is differentiable at x = 0 and has a local maximum at x = 0, then it follows that lim h→0 cos(h)−1 h = d dx cos x | x=0 = 0. We can prove that lim h→0 cos(h)−1 h = 0 without these assumptions as follows. 7 If A<Band k is negative, then kA>kB.For instance, 1 < 2but(−5)(1)>(−5)(2). 8 For example, 1 x < 2 x for x positive but when we take the limit as x →∞we must write lim x→∞ 1 x ≤ lim x→∞ 2 x ; both these limits are zero. . area of this sector is greater than the area of the small shaded triangle OAP and less than the area of the large triangle OBQ. Q v P x x ABO u cos x sin x 1 Figure 21.6 area of small triangle < area. Investigation of the Slope of the Tangent Line to sin x at x = 0 The derivative of sin x at x = 0 is the slope of the tangent to sin x at (0, 0). We can approximate the slope of the tangent line. velocity vector of the wind and that of the bicyclist? 12. Due to the scampering of a goat, a rock has been dislodged from a mountain and is sliding down an incline making a 70 ◦ angle with level