Derivatives, Slope, Velocity, and Rate of Change

Lecture 1 Sept. 7, 2006 18.01 Fall 2006 Unit 1: Derivatives A. What is a derivative? • Geometric interpretation • Physical interpretation • Important for any measurement (economics, political science, finance, physics, etc.) B. How to differentiate any function you know. d � x arctan x � • For example: e . We will discuss what a derivative is today. Figuring out how to dx differentiate any function is the subject of the first two weeks of this course. Lecture 1: Derivatives, Slope, Velocity, and Rate of Change Geometric Viewpoint on Derivatives y Q Secant line Tangent line P f(x) x0 Figure 1: x0+∆x A function with secant and tangent lines The derivative is the slope of the line tangent to the graph of f (x). But what is a tangent line, exactly? 1 Lecture 1 Sept. 7, 2006 18.01 Fall 2006 • It is NOT just a line that meets the graph at one point. • It is the limit of the secant line (a line drawn between two points on the graph) as the distance between the two points goes to zero. Geometric definition of the derivative: Limit of slopes of secant lines P Q as Q → P (P fixed). The slope of P Q: Q (x0+∆x, f(x0+∆x)) Secant Line ∆f (x0, f(x0)) P ∆x Figure 2: lim Δx→0 Δf Δx = lim Δx→0 Geometric definition of the derivative f (x0 + Δx) − f (x0 ) Δx � �� � = “difference quotient” Example 1. f (x) = f � (x0 ) � �� � “derivative of f at x0 ” 1 x One thing to keep in mind when working with derivatives: it may be tempting to plug in Δx = 0 Δf 0 right away. If you do this, however, you will always end up with = . You will always need to Δx 0 do some cancellation to get at the answer. Δf = Δx 1 x0 +Δx − Δx 1 x0 � � � � 1 x0 − (x0 + Δx) 1 −Δx −1 = = = Δx (x0 + Δx)x0 Δx (x0 + Δx)x0 (x0 + Δx)x0 Taking the limit as Δx → 0, lim Δx→0 −1 −1 = 2 (x0 + Δx)x0 x0 2 Lecture 1 Sept. 7, 2006 18.01 Fall 2006 y x x0 Figure 3: Graph of Hence, f � (x0 ) = 1 x −1 x20 Notice that f � (x0 ) is negative — as is the slope of the tangent line on the graph above. Finding the tangent line. Write the equation for the tangent line at the point (x0 , y0 ) using the equation for a line, which you all learned in high school algebra: y − y0 = f � (x0 )(x − x0 ) Plug in y0 = f (x0 ) = 1 −1 and f � (x0 ) = 2 to get: x0 x0 y− 1 −1 = 2 (x − x0 ) x0 x0 3 Lecture 1 Sept. 7, 2006 18.01 Fall 2006 y x x0 Figure 4: Graph of 1 x Just for fun, let’s compute the area of the triangle that the tangent line forms with the x- and y-axes (see the shaded region in Fig. 4). First calculate the x-intercept of this tangent line. The x-intercept is where y = 0. Plug y = 0 into the equation for this tangent line to get: 1 x0 −1 x0 1 x x20 0− x = = = = −1 (x − x0 ) x20 −1 1 x+ x20 x0 2 x0 2 x20 ( ) = 2x0 x0 So, the x-intercept of this tangent line is at x = 2x0 . 1 1 and x = are identical equations, x y the graph is symmetric when x and y are exchanged. By symmetry, then, the y-intercept is at y = 2y0 . If you don’t trust reasoning with symmetry, you may follow the same chain of algebraic reasoning that we used in finding the x-intercept. (Remember, the y-intercept is where x = 0.) Next we claim that the y-intercept is at y = 2y0 . Since y = Finally, Area = 1 1 (2y0 )(2x0 ) = 2x0 y0 = 2x0 ( ) = 2 (see Fig. 5) 2 x0 Curiously, the area of the triangle is always 2, no matter where on the graph we draw the tangent line. 4 Lecture 1 Sept. 7, 2006 18.01 Fall 2006 y 2y0 x-1 y0 x0 2x0 Figure 5: Graph of x 1 x Notations Calculus, rather like English or any other language, was developed by several people. As a result, just as there are many ways to express the same thing, there are many notations for the derivative. Since y = f (x), it’s natural to write Δy = Δf = f (x) − f (x0 ) = f (x0 + Δx) − f (x0 ) We say “Delta y” or “Delta f ” or the “change in y”. If we divide both sides by Δx = x − x0 , we get two expressions for the difference quotient: Δy Δf = Δx Δx Taking the limit as Δx → 0, we get Δy Δx Δf Δx → dy (Leibniz’ notation) dx → f � (x0 ) (Newton’s notation) When you use Leibniz’ notation, you have to remember where you’re evaluating the derivative — in the example above, at x = x0 . Other, equally valid notations for the derivative of a function f include df � , f , and Df dx 5 Lecture 1 Sept. 7, 2006 18.01 Fall 2006 Example 2. f (x) = xn where n = 1, 2, 3... d n x ? dx To find it, plug y = f (x) into the definition of the difference quotient. What is Δy (x0 + Δx)n − xn0 (x + Δx)n − xn = = Δx Δx Δx (From here on, we replace x0 with x, so as to have less writing to do.) Since (x + Δx)n = (x + Δx)(x + Δx)...(x + Δx) n times We can rewrite this as � � xn + n(Δx)xn−1 + O (Δx)2 O(Δx)2 is shorthand for “all of the terms with (Δx)2 , (Δx)3 , and so on up to (Δx)n .” (This is part of what is known as the binomial theorem; see your textbook for details.) Δy (x + Δx)n − xn xn + n(Δx)(xn−1 ) + O(Δx)2 − xn = = = nxn−1 + O(Δx) Δx Δx Δx Take the limit: lim Δx→0 Δy = nxn−1 Δx Therefore, d n x = nxn−1 dx This result extends to polynomials. For example, d 2 (x + 3x10 ) = 2x + 30x9 dx Physical Interpretation of Derivatives You can think of the derivative as representing a rate of change (speed is one example of this). On Halloween, MIT students have a tradition of dropping pumpkins from the roof of this building, which is about 400 feet high. The equation of motion for objects near the earth’s surface (which we will just accept for now) implies that the height above the ground y of the pumpkin is: y = 400 − 16t2 The average speed of the pumpkin (difference quotient) = When the pumpkin hits the ground, y = 0, 400 − 16t2 = 0 6 Δy distance travelled = Δt time elapsed Lecture 1 Sept. 7, 2006 18.01 Fall 2006 Solve to find t = 5. Thus it takes 5 seconds for the pumpkin to reach the ground. Average speed = 400 ft = 80 ft/s 5 sec A spectator is probably more interested in how fast the pumpkin is going when it slams into the ground. To find the instantaneous velocity at t = 5, let’s evaluate y � : y � = −32t = (−32)(5) = −160 ft/s (about 110 mph) y � is negative because the pumpkin’s y-coordinate is decreasing: it is moving downward. 7