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Announcements Topics: - sections 6.4 (l’Hopital’s rule), 7.1 (differential equations), and 7.2 (antiderivatives) * Read these sections and study solved examples in your textbook! Work On: - Practice problems from the textbook and assignments from the coursepack as assigned on the course web page (under the link “SCHEDULE + HOMEWORK”) Extreme Value Theorem If f (x) is continuous for all x Î [a, b] , then there are points c1, c 2 Î [a, b] such that f (c1) is the global minimum and f (c 2 ) is the global maximum of f (x) on [a, b]. In words: If a function is continuous on a closed, finite interval, then it has a global maximum and a global minimum on that interval. Finding Absolute Extreme Values on a Closed Interval [a,b] 1. Find all critical numbers in the interval. 2. Make a table of values. The largest value of f(x) is the absolute maximum and the smallest value is the absolute minimum. Finding Absolute Extreme Values on a Closed Interval [a,b] Example: 1 2 3 Find the absolute extrema of g(x) = x (x - 2) on [-1, 1]. L’Hopital’s Rule Another application of derivatives is to help evaluate limits of the form where either or Idea: Instead of comparing the functions f(x) and g(x), compare their derivatives (rates) f’(x) and g’(x). L’Hopital’s Rule Suppose that f and g are differentiable functions such that is an indeterminate form of type 00 or ¥¥ . If g¢(x) ¹ 0 near a (could be 0 at a) then L’Hopital’s Rule Evaluate the following limits using L’Hopital’s Rule, if it applies. (a) (b) (c) (d) L’Hopital’s Rule Evaluate the following limits using L’Hopital’s Rule, if it applies. (a) (b) Differential Equations A differential equation is an equation that involves an unknown function and one or more of its derivatives. Examples: y'= 2 + y y"+2xy = x 2 y'= x + e 2 x Differential Equations A solution of a differential equation is a function that, along with its derivatives, satisfies the DE. Example: -x 3 Show that y = 2 + e is a solution of the differential equation y'+3x 2 y = 6x 2 and initial condition y(0) = 3. Pure-Time DEs A pure-time differential equation is obtained by measuring the rate of change of the unknown quantity and expressed as a function of time. Example: ds 2 = t - 3t + 5 dt Note that the formula for the rate of change depends purely on the time t. Example 1: Volume of a Cell Suppose we observe that 2.0 mm 3 of water enters a cell each second. Differential Equation: General Solution: V is called the ‘state variable’ Example 1: Volume of a Cell Suppose we observe that 2.0 mm 3 of water enters a cell each second. dV Differential Equation: = 2.0 ¬ pure - time DE dt General Solution: V is called the ‘state variable’ Example 1: Volume of a Cell Suppose we observe that 2.0 mm 3 of water enters a cell each second. dV Differential Equation: = 2.0 ¬ pure - time DE dt General Solution: V(t) = 2.0t +C V is called the ‘state variable’ Example 1: Volume of a Cell Suppose we are told that the initial volume of the cell is 150 mm 3 . General Solution: V(t) = 2.0t +C Initial Condition: V(0) =150 Particular Solution: Example 1: Volume of a Cell Suppose we are told that the initial volume of the cell is 150 mm 3 . General Solution: V(t) = 2.0t +C Initial Condition: V(0) =150 Particular Solution: V(t) = 2.0t +150 Autonomous DEs An autonomous differential equation is derived from a rule describing how a quantity changes and is expressed as a function of the unknown quantity. Example: Rule: The growth rate of a population is proportional to its size. db = k × b(t) or simply b'= k × b dt Example 2: Population Size Suppose we know that the growth rate of a population is half of its current population and the initial population is 10. Differential Equation: Initial Condition: Particular Solution: Example 2: Population Size Suppose we know that the growth rate of a population is half of its current population and the initial population is 10. Differential Equation: db = 0.5b ¬ autonomous DE dt Initial Condition: b(0) =10 Particular Solution: b(t) = ? Example 2: Population Size Suppose we know that the growth rate of a population is half of its current population and the initial population is 10. Differential Equation: db = 0.5b ¬ autonomous DE dt Initial Condition: b(0) =10 Particular Solution: b(t) =10e 0.5t Modelling: Verbal Descriptions IVPs Example: Write a differential equation and an initial condition to describe the following events. 1. The relative rate of change of the population of wild foxes in an ecosystem is 0.75 baby foxes per fox per month. Initially, the population is 74 thousand. 2. The population of an isolated island is 7500. Initially, 13 people are infected with a flu virus. The rate of change of the number of infected people is proportional to the product of the number who are infected and the number of people who are not yet infected. Solutions for General DEs Algebraic Solutions Geometric Solutions an explicit formula or algorithm for the solution (often, impossible to find) a sketch of the solution obtained from analyzing the DE Numeric Solutions an approximation of the solution using technology and and some estimation method, such as Euler’s method A Geometric Solution of a Pure-Time DEs Example: Sketch the graph of the solution to s'(t) = 3- t given the initial condition s(0) =1. Euler’s Method What information does an initial value problem tell us about the solution? Example: dy DE: =x+y dx IC: y(0) =1 slope of the solution curve y(x) an exact value of the solution Euler’s Method Euler’s Idea: First, using the initial condition as a base point, approximate the solution curve y(x) by its tangent line. First Euler approximation Euler’s Method Next, travel a short distance along this line, determine the slope at the new location (using the DE), and then proceed in that ‘corrected’ direction. Euler’s approximation with step size Euler’s Method Repeat, correcting your direction midcourse using the DE at regular intervals to obtain an approximate solution of the IVP. By increasing the number of midcourse corrections, we can improve our estimation of the solution. Euler approximation with step size Euler’s Method Summary: An approximate solution to the IVP dy = G(t, y), y(t 0 ) = y 0 dt is generated by choosing a step size and computing values according to the algorithm Euler’s Method Algorithm: Algorithm In Words: next time = current time + step size next approximation = current approximation + rate of change at current values x step size Example Consider the IVP y'= t + 3, y(0) = 2 Approximate the value of the solution at t=1 by applying Euler’s method and using a step size of 0.25. Example Calculations: Table of Approximate Values for the Solution y(t) of the IVP tn yn t0 = 0 y0 = 2 Example Graph of Approximate Solution: Plot points and connect with straight line segments. 6 5 4 tn yn t0 = 0 y0 = 2 t1 = 0.25 y1 = 2.75 2 t2 = 0.5 y2 = 3.5625 1 t3 = 0.75 y3 = 4.4375 t4 = 1 y4 = 5.375 3 0 0.25 0.5 0.75 1 Determining Properties of a Solution Example: A population P(t) of caribou is modeled by the autonomous DE æ P(t) ö P'(t) = 2P(t)ç1÷, P(t) > 0. è 2500 ø Analyze this equation to describe the behaviour of the population of caribou. Solving Pure-Time DEs The general form of a pure-time differential equation is dF = f (x) dx where F(x)is the unknown state variable and f (x) is the measured rate of change. Examples: (a) dF = 4x 3 + 1 dx (b) dy 1 x = 5e + 2 dx 1+ x Solving Pure-Time DEs Solve each by “guess and check”. Ask yourself: “What function has this as its derivative?” Examples: dF (a) = 4x 3 + 1 dx dy 1 x (b) = 5e + dx 1+ x 2 Antiderivatives/Indefinite Integrals An antiderivative (or indefinite integral) of a function f (x) is a function F(x) with derivative equal to f (x) . An antiderivative F(x) is a solution to the puretime differential equation dF = f (x). dx Initial Value Problems A differential equation has a whole family of solutions. Example: y'= 2x - 4 y = x 2 - 4x + C Initial Value Problems An initial value problem provides an initial condition so you can find a particular solution. Example: y'= 2x - 4, y(0) = 3 y = x 2 - 4x + 3 Initial Value Problems An initial value problem provides an initial condition so you can find a particular solution. Example: y'= 2x - 4, y(0) = 3 y = x 2 - 4x + 3 Antiderivatives/Indefinite Integrals Theorem 6.1: If F(x) is an antiderivative of f(x), then the most general antiderivative of f(x) is F(x)+C; i.e., ò f (x)dx = F(x) + C where C is a real number. If an initial value of the solution is given, then we can solve for C to find a specific or particular antiderivative of f(x). Rules for Antiderivatives The Power Rule for Integrals ò n +1 x x n dx = +C n +1 for n ¹ -1 Example: Integrate each. (a) ò x dx 7 (b) ò 1 dt 4 t (c) ò xdx Rules for Antiderivatives The Constant Product Rule for Integrals Suppose Then ò f (x)dx = F(x) + C. ò af (x)dx = aF(x) + C'. for any constant a. The Sum Rule for Integrals Suppose ò f (x)dx = F(x) + C Then ò [ f (x) + g(x)]dx = and ò g(x)dx = G(x) + C'. ò f (x)dx + ò g(x)dx = F(x) + G(x) + C''. Some More Examples Example 1: Integrate. (a) 3 ò (5x - x )dx 4 (b) Example 2: Solve the differential equation f '(x) = 2 + x with initial condition 1 2 x f (0) = 0. 2 4x (sec x + e )dx ò Application: A Differential Equation for AIDS During the early years of the AIDS epidemic, the number of new AIDS cases per year was » 523.8t 2, where t is measured in years since the beginning of 1981. rate at which new AIDS cases were reported: dA 2 = 523.8t dt Application: A Differential Equation for AIDS Surveys indicated that about 340 people had been infected at the beginning of 1981. Initial condition: A(0) = 340 Solution: A(t) represents the number of AIDS cases t years after 1981. A(t) =174.6t 3 + 340 Summary Of Some Basic Integration Formulas ò n +1 x x n dx = +C n +1 for n ¹ -1 ò cos xdx = sin x + C òx ò sin xdx = -cos x + C x x e dx = e +C ò ò sec 1 ò 1+ x 2 dx = arctan x + C 2 xdx = tan x + C -1 dx = ò 1 dx = ln x + C x