ApplicationsDerivativesII.nb 1 Applications of Derivatives Now that we have covered the basic concepts and mechanics of derivatives it's time to look at some applications. We will start with a summary of what we have learned thus far. The Connection between Derivatives of a Function and the Function Itself If f ' > 0 on an interval, then f is increasing over that interval. If f ' < 0 on an interval, then f is decreasing over that interval. If f " > 0 on an interval, then f ' is increasing, and the graph of f is concave upward on that interval. If f ' < 0 on an interval, then f ' is decreasing, and the graph of f is concave downward on that interval. Critical Points For any function, f, a point p in the domain of f where f '(p) = 0 or where f '(p) is undefined is called a critical point of the function. We sometimes call the coordinates of the point, (p, f(p)) a critical point as well. A critical value of f is the value, f(p), of the function at a critical point, p. The First Derivative Test for Local Maxima and Minima Suppose p is a critical point of a continuous function f
ApplicationsDerivativesII.nb 2 If f ' changes signs from negative to positive at p, then f has a local minimum at p (note that the graphic interpretation is that the function changes from decreasing to increasing at p). If f ' changes signs from positive to negative at p, then f has a local maximum at p (note that the graphic interpretation is that the function changes from increasing to decreasing at p). The Second Derivative Test for Local Maxima and Minima If f ' (p) = 0 and f "(p) >0, then f has a local minimum at p (note that the graphical interpretation here is that we have a horizontal tangent line to the graph of f at p and the graph is concave upward at point p). If f '(p) = 0 and f "(p) < 0, then f has a local maximum at p (in this case we have a horizontal tangent at p and the graph is concave downward at p). Inflection Points A function f with a continuous derivative has an inflection point at p if either of the following conditions hold: f ' has a local minimum or local maximum at p. f " changes signs at p.
ApplicationsDerivativesII.nb 3 Global Maxima and Minima The single greatest (or least) value of a function f over a specified domain is called the global (or absolute) maximum (or minimum) of f. Recall (from above) that the local maxima and minima tell us where a function is locally largest or smallest. Now we are interested in where the function is absolutely largest or smallest in a given domain. Look at the definitions below. 1. f has a global minimum at p if f(p) is less than or equal to all values of f. 2. f has a global maximum at p if f(p) is greater than or equal to all values of f. The procedure for finding the global maximum and global minimum of a continuous function on a closed interval is to compare the values of the function at all critical points in the interval and at its endpoints. The largest value among these choices is the global maximum and the smallest value is the global minimum. To find the global maximum and minimum of a continuous function on an open interval or on all real numbers find the value of the function at all critical points and investigate the behavior of the function when x approaches the endpoints of the interval, or approaches ±, as appropriate. Now let's look at some applications. Rates of Change (Motion on a Line) Since we define the derivative of a function at a as the instantaneous rate of change of f at a, one application of derivatives is the study of motion on a line (known as rectilinear motion).
ApplicationsDerivativesII.nb 4 If we let x(t) define the position of a particle on a line then the rate of change of the position with respect to time can be defined to be the velocity, v(t), of the particle. Likewise, the rate of change of the velocity with respect to time can be defined as the acceleration, a(t), of the particle with respect to time. We can summarize these relationships as floows: Let x(t) = the position function Then x '(t) = v(t), the velcocity function and x '(t) = v '(t) = a(t), the accelration function. Although most AP Exam questions involving rectilinear motion involve both derivatives and integrals we are not ready at this time of the review to handle both so we will confine ourselves to just derivatives. There will be a section in the third chapter under "Applications of Integration" that will deal with both derivatives and integrals. Let's look at an example. Example 1: A particle moves along the x-axis so that its position at any time t 0 is given by x(t) = t 3 - t 2 - t + 3. (a) Write a polynomial expression for the velocity and acceleration of the particle for any time t 0. (b) On what interval(s) is the particle moving to the right? To the left? (c) On what interval(s) is velocity increasing? Solution: (a) Since the position of the particle is given by x(t) = t 3 - t 2 - t + 3 we can find the velocity and acceleration functions using differentiation. v(t) = x '(t) = 3 t 2-2 t - 1
ApplicationsDerivativesII.nb 5 a(t) = v '(t) = x "(t) = 6t - 2 (b) For this question we use the fact that a particle is moving to the right if its velocity is positive and moving to the left if its velocity is negative. So we need to set the velocity function equal to 0, find the critical points (in this case the critical points would indicate where the particle is not moving), and set up a sign chart to indicate where the velocity function is positive and neagtive on the indicated interval. v(t) = 3 t 2-2 t - 1 = 0 (3t + 1)(t - 1) = 0 ï that t = -1/3 or 1 Note that t = -1/3 is not in the domain of our function (t 0) so we may ignore it. v(t) - 0 + + + + + + (3t + 1)+ + + + + + + (t - 1) - 0 + + + + + 0 1 2 3 4 5 So, on the interval indicated (t 0) the particle is moving to the right on (1, ) and moving to the left on [0, 1). (c) If we want to find the intervals where a function is increasing we look at its derivative. So, if we want to know where the velocity is increasing we look at its derivative, the acceleration function. We will set the acceleration equal to zero and set up a sign chart as we did in (b). a(t) = 6t - 2 = 0 ï 2(3t - 1) = 0 ï t = 1/3. a(t) - 0 + + + + + (3t - 1) - 0 + + + + +
ApplicationsDerivativesII.nb 6 0 1 3 1 2 3 4 5 So we can see that the acceleration of the particle is negative on the interval [0, 1/3) and positive on the interval (1/3, ). This means that the velcoity of the particle is decreasing on the interval [0, 1/3) and increasing on the interval (1/3, ). Optimization The concept of finding either local (relative) or global (absolute) extrema is known as optimization. You might be asked to find the local maxima or minima of a function or the gloabl maxima or minima of a function. You might have to actually write a function and determine the domain yourself. Let's look at a couple of examples. You can review your homework and worksheets from class for more examples. Example 2: Find the global maximum and minimum of the function f(x) = x 3 + x 2-8 x + 5on the closed interval [0, 4]. Solution: First we find the critical points. f ' (x) = 3 x 2 + 2 x - 8 = 0 (3x + 4)(x - 2) = 0 ï x = -4/3 and 2 Note that -4/3 is not in the indicated interval so we ignore it. We now evaluate the function at the remaining critical number and at the endpoints. f(0) = 5 f(2) = 1 f(4) = 53
ApplicationsDerivativesII.nb 7 Since 1 is the least of these values it is the global minimum. Since 53 is the largest of these values it is the global maximum. Note that the y-values are the global maximum and global minimum. The x-vales are where they occur. Example 3: A rectangle is to be inscribed under one arch of the cosine curve. What is the largest area the rectangle can have, and what dimensions give that area? Solution: Look at the graph below. Since the cosine graph is periodic, each arch is the same so we chose to graph the arch between x = -p/2 and x = p/2. Obviously, there are an infinite numbers of rectangles that can be inscribed under this arch. We are trying to find the one that has the greatest area. Notice that since the cosine curve is symetric with respect to the y-axis that the wih of the rectangle is 2x. Ü Graphics Ü y=cos x - p ÅÅÅÅÅ 2 p ÅÅÅÅÅ 2 x The height of this rectangle is cos x. Using the area formula for a rectangle we have Area = A(x) = 2x cos x So we are asked to maximize A(x). Notice that the probelm, as written, determines the domain of this function; x must lie between 0 and p/2. To summarize, then, we are asked to find the global maximum value of A(x) = 2x cos x on the closed interval [0, p/2].
ApplicationsDerivativesII.nb 8 Let's find the critical points. A'(x) = 2x(-sin x) +2 cos x = 0. We will use our graphing calculator to solve this equation. If we enter the function 2x(-sin x) + 2 cos x as Y 1 we can find its zeros. We get x = 0.860 to three decimal places. Note that at the endpoints that the area is 0; A(0) = 2(0)cos(0) = 0(1) = 0 and A(p/2) = 2(p/2)cos(p/2) = p(0) = 0. So the only point that can result in the largest area is x = 0.866. The area of the rectangle at this value is A(0.860) = 2(0.860)cos(0.860) = 1.122. The dimensions that give us this area are 1.720 0.652. Let's check our answer for reasonableness. The area under one arch of the cosine curve is 2 (check this on your calculator or using the Fundamental Theorem). So an answer of 1.122 is certainly reasonable. Related Rates We have already seen how the Chain Rule can be used to find dy/dx implicitly. Another important use of the Chain Rule is to find rates of change of two or more related variable that are changing with repsect to time. For instance, suppose we are inflating a spherical balloon. As the volume of the ballon increases so does the radius. Both the radius and the volume are related and both are changing with respect to time. This concept is best illustrated with examples. Example 3: Air is being pumped into a spherical ballon at a rate of 4.5 cubic feet per minute. Find the rate of change of the radius when the radius is 3 feet. Solution: The volume of a sphere and its radius are related by the volume formula; V = 4 ÅÅÅÅ 3 pr3 Now we assume that both V and r are functions of t and differntiate both sides of the equation with respect to t. d ÅÅÅÅÅ HVL = d ÅÅÅÅÅ H ÅÅÅÅ 4 3 pr3 L
ApplicationsDerivativesII.nb 9 dv ÅÅÅÅÅÅÅÅ = 4 pr 2 dr ÅÅÅÅÅÅ Since we know ÅÅÅÅÅÅÅÅ dv (the rate at which the ballon is being inflated; 4.5 cubic feet per minute) and we know the radius; 3 feet) we can solve for the rate at which the radius is increasing, ÅÅÅÅÅÅ dr. 4.5 ft. 3 ê min= 4 ph3ft.l 2 dr ÅÅÅÅÅÅ ï ÅÅÅÅÅÅ dr = 4.5 ft.3 êmin ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ = 0.0397 ft. ê min 36 p ft. 2 Example 4: A 13 foot long ladder leans against the wall of a building. If the ladder is sliding down the wall at a rate of 4 feet/sec. how fast is the bottom of the ladder sliding away from the wall when the bottom of the ladder is 5 feet from the base of the wall? Solution: Look at the diagram below. y 13 ft. x We can see that the height, y, of the ladder and the distance of the bottom of the ladder from the wall, x, are related to the length of the ladder by the Pythagorean Theorem. x 2 + y 2 = 13 2 Differentiating both sides of the equation with respect to t gives us
ApplicationsDerivativesII.nb 10 d ÅÅÅÅÅ Hx2 L + d ÅÅÅÅÅ Hy2 L = d ÅÅÅÅÅ H132 L 2x ÅÅÅÅÅÅ dx + 2 y dy ÅÅÅÅÅÅ = 0 We know from the problem that the base of the ladder is 5 feet from the wall, so this is x. We are also told that the top of the ladder is sliding down the wall at 4 feet per second so this is ÅÅÅÅÅÅ dy dx (we will use -4ft./sec for the motion down). We are trying to find ÅÅÅÅÅÅ so all we need is y. We can find y using the Pythagorean Theorem. In fact if we recognize the fact that we have a Pythagorean Triple (A 5-12-13 triangle) w can shorten our work. Substituting in the known quantities and solving for the unknown quantity we get 2(5 ft.)( ÅÅÅÅÅÅ dx ) + 2(12 ft.)(-4 ft./sec) = 0 dx ÅÅÅÅÅÅ = 9.6 ft. ê sec So the base of the base of the ladder is sliding away from the wall at a rate of 9.6 feet/sec. Again, look at your worksheets and notes for more examples of related rates. The Tangent Line Approximation Once we know the process for finding the equation of the tangent line it's time to put it to good use. In calculus we use the tangent line to the graph of a function at a point to approximate the y-values of the function. Since the graph of a function and its tangent line have the same slope at the point of tangency (namely, the derivative at that point), the tangent line stays close to the graph of a function near that point. This is known as local linearization. Look at the three graphs below. These are the graphs of a parabola and its tangent line. As we "zoom" in notice that the graph of the function begins to "straighten" out. It is becoming more linear. We use this phenomenon to use the tangent line to approximate y-values of the curve. As long as we stay close to the point of tangency, the y-values on the tangent line are very close to the y-values on the curve.
ApplicationsDerivativesII.nb 11 Thus, if we find the equation of the tangent line we can "sub" in values of x near the point of tangency and the y-values that result will be very near the y-values on the curve. So why do we do this? A tangent line is a linear function. "Subbing" into a linear function is much easier than subbing into a complicated function of a curve. If we are trying to find approximations to the curve at many points then finding the derivative function once (and writing the equation of the tangent line) allows us to approximate y-values on the curve much easier. Since the slope of the tangent line to the graph of y = f(x) at x = a is f ' (a), then the equation of the tangent line is
ApplicationsDerivativesII.nb 12 y = f(a) + f ' (a)(x - a) This gives us the following relationship: The Tangent Line Approximation For values of x near a, f(x) ª f(a) + f ' (a)(x - a) Since a is "fixed" (a constant) so are f(a) and f ' (a). Let's look at an example. Example 5 :Use the tangent line to approximate the y-value of the curve defined by f(x) = 5 + 5x -x 2 at x near 2. Solution: First we find the derivative function. f ' (x) = 5-2x Since the slope of the tangent line, m, is given by f ' (2) we evaluate the derivative at x = 2. f ' (2) = 5-2(2) = 1, so m = 1 Now we need the y-value associated with x = 2. We get this value by substituting 2 in for x into the original function. f(2) = 5 + 5(2) - 2 2 = 5 +10-4 = 11, so when x= 2, y = 11.
ApplicationsDerivativesII.nb 13 We now write the equation of the tangent line using the point-slope form of a linear equation. y - 11 = 1(x - 2) Solving for y gives y = 1(x - 2) + 11 or y = x + 9. So, f(x) º x + 9 at x near 2. Let's approximate the function value near x = 2, say at x = 2.1. According to our tangent line equation, f(2.1) º 2.1 + 9 = 11.1. The actual value of f(2) can be found by "subbing" 2.1 into the original equation. f(2.1) = 5 + 5(2.1) - 2.1 2 = 11.09. So our approximation is very close. Notice how easy it was to approximate this value with the tangent line. Let's try another. Approximate the function value at x = 1.99. According to our tangent line f(1.99) º 1.99 + 9 = 10.99 The actual value of f(x) at x = 1.99 is 10.9899 so our approximation is even closer. The fact is, the closer our x-value is to 2 the closer our approximation will be to the actual value. Worksheet 10a: Critical Points and the First derivative Test Worksheet 10b: Using the Sceond Derivative Test
ApplicationsDerivativesII.nb 14 Worksheet 10c: Inflection Points and Concavity Worksheet 10d: Finding Global Maxima and Minima Worksheet 10e: Rectilinear Motion Worksheet 10f: Optimization Worksheet 10g: Related Rates Worksheet 10h: Tangent Line Approximations