Understanding Change: Average and Instantaneous Rates, Derivatives, Study notes of Mathematics

Download Understanding Change: Average and Instantaneous Rates, Derivatives and more Study notes Mathematics in PDF only on Docsity! MA123, Chapter 2: Change, and the idea of the derivative (pp. 17-45) Date: Chapter’s Goal: • Understand average rates of change; • understand the ideas leading to instantaneous rates of change; • understand the connection between instantaneous rates of change and the derivative; • know the definition of the derivative at a point; • use the definition of the derivative to calculate derivatives; • understand the connection between a position function, a velocity function, and the derivative; • understand the connection between the derivative and the slope of a tangent line. Assignments: Assignment 02 (Due date: Sept. 11) Assignment 03 (Due date: Sept. 15) Roughly speaking, Calculus describes how quantities change, and uses this description of change to give us extra information about the quantities themselves. ◮ Average rates of change: We are all familiar with the concept of velocity (speed): If you drive a distance of 120 miles in two hours, then your average velocity, or rate of travel, is 120/2 = 60 miles per hour. In other words, the average velocity is equal to the ratio of the distance traveled over the time elapsed: average velocity = distance traveled time elapsed = ∆s ∆t . In general, the quantity y2 − y1 x2 − x1 = ∆y ∆x is called the average rate of change of y with respect to x. Note: Often, a change in a quantity q is expressed by the symbol ∆q (you should not think of this as ∆ times q, but rather as one quantity!). Note: Finding average rates of change is important in many contexts. For instance, we may be interested in knowing how quickly the air temperature is dropping as a storm approaches, or how fast revenues are increasing from the sale of a new product. Note: In this course we use the terms “speed” and “velocity” for the same concept. This is not the case in some other courses. Thus “instantaneous speed” and “instantaneous velocity” have the same meaning, and “average speed” and “average velocity” have the same meaning. Example 1: A train travels from city A to city B. It leaves A at 10:00 am and arrives at B at 2:30 pm. The distance between the cities is 150 miles. What was the average velocity of the train in miles per hour (mph)? Do you think the train was always traveling at the same speed? 7 Example 2: A train leaves station A at 8:00 am and arrives at station B at 10:00 am. The train stops at station B for 1 hour and then continues to station C. It arrives at station C at 3:00 pm. The average velocity from A to B was 40 mph and the average velocity from B to C was 50 mph. What was the average velocity from A to C (including stopping time)? Generally, in computing average rates of change of a quantity y with respect to a quantity x, there is a function that shows how the values of x and y are related. ◮ Average rates of change of a function: The average rate of change of the function y = f(x) between x = x1 and x = x2 is average rate of change = change in y change in x = f(x2) − f(x1) x2 − x1 The average rate of change is the slope of the secant line between x = x1 and x = x2 on the graph of f , that is, the line that passes through (x1, f(x1)) and (x2, f(x2)). x y 0 f(x1) f(x2) x1 x2 • • y = f(x) f(x2) − f(x1) x2 − x1 Example 3: Find the average rate of change of g(x) = 2 + 4(x − 1) with respect to x as x changes from −2 to 5. Could you have predicted your answer using your knowledge of linear equations? Example 4: Find the average rate of change of k(t) = √ 3t + 1 with respect to t as t changes from 1 to 5. 8 The approach we will see now has the tremendous advantage that it yields a formula for the instantaneous velocity of this object as a function of time t. Example 10: A particle is traveling along a straight line. Its position at time t is given by s(t) = t2 + 3. Find the velocity of the particle as a function of t. Note: Even if you have a formula for a quantity, knowing how the quantity is changing can give you extra information that is not obvious from the formula. For example, if s(t) denotes the position of a ball being thrown up into the air, how high does the ball go? Observe that when the ball is going up it has positive velocity, because its height is increasing in time, whereas when the ball is going down it has negative velocity, because its height is decreasing in time. Thus, the instant at which the ball reaches its highest point is exactly the one when its velocity is 0. Example 11: A particle is traveling along a straight line. Its position at time t is given by s(t) = t2 + 3. 1. Find the velocity of the particle when t = 4 seconds; 2. Find the velocity of the particle when t = 2 seconds; 3. Find the velocity of the particle when t = 7 seconds; 4. When is the velocity of the particle equal to 5 feet per second? 5. When is the velocity of the particle equal to 0 feet per second? 11 Example 12: Find the instantaneous rate of change of g(k) = 2k2 + k − 1 at k = 3. Example 13: Find the instantaneous rate of change of g(k) = 2k2 + k − 1 as a function of k. ◮ The derivative: The derivative of f with respect to x, at x = x1, is the instantaneous rate of change of f with respect to x, at x = x1, and is thus given by the formula lim h→0 f(x1 + h) − f(x1) h Now just drop the subscript ‘1’ from the x in the above formula, and you obtain the instantaneous rate of change of f with respect to x at a general point x. This is called the derivative of f at x and is denoted with f ′(x): f ′(x) = lim h→0 f(x + h) − f(x) h Alternative notations: As we remarked earlier, a change in a quantity q is often expressed by the symbol ∆q (you should not think of this as ∆ times q, but rather as one quantity!). Thus the above formulas are often rewritten as f ′(x) = lim ∆x→0 ∆f ∆x = df dx Given a general function f(x), it is often common to think in terms of y = f(x) so that the above formulas are often rewritten as y′ = lim ∆x→0 ∆y ∆x = dy dx A preview: Often, the information you have about a quantity is not about the quantity itself, but about its rate of change. This means that you know the derivative of a function, and want to find the function. This occurs frequently in Physics. Newton’s formula gives information about the acceleration of an object, that is, about the rate of change of velocity with respect to time. From this, one can often get information about the velocity, the rate of change of position with respect to time, and then information about the position itself of the object. 12 Example 14: Let f(x) = mx + b be an arbitrary linear function (here m and b are constants). Prove that f ′(x) = m. Example 15: Let f(x) = ax2 + bx + c be an arbitrary quadratic function (here a, b and c are constants). Prove that f ′(x) = 2ax + b. Example 16: Let g(x) = 2x2 + x − 1. Find a value c between 1 and 4 such that the average rate of change of g(x) from x = 1 to x = 4 is equal to the instantaneous rate of change of g(x) at x = c. 13