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Differentiation and Rates of Change. Lesson 3.2B. Rate of Change. Average rate of change What is the average rate of change from 5.6 hours to 11.35 hours?. Rate of Change. What is the instantaneous rate of change at time t = 9? This is the slope of the curve when t = 9.
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Differentiation and Rates of Change Lesson 3.2B
Rate of Change • Average rate of change • What is the average rate of change from 5.6 hours to 11.35 hours?
Rate of Change • What is the instantaneous rate of change at time t = 9? • This is the slope of the curve when t = 9 What will be the units of the answer?
Remember When? • "Two trains start at the same time heading in opposite directions. The first train … " • The basic relationship was • Also • Average velocity is
Average Velocity • Consider a pumpkin dropped from a height of 150 feet. • Then it's position will be a function of time • Find the average rate of change of distance from 0.5 sec. to 1.25 sec. • Remember rate of change of distance is called velocity
Instantaneous Velocity • Let the change in time (the denominator) get smaller and smaller • Then • That is, the derivative of the position function is the velocity function
Instantaneous Velocity • For any object dropped or thrown up into the air, we know • Where v0 = initial velocity • and s0 = initial position • Then instantaneous velocity = s'(t) • So what about our pumpkin at 2 sec?
Modeling Functions • You will use the regression feature of your calculator to do some of your homework • You are given data in a table • You find a modeling function • Then determine the rate of change of the modeling function
Data Modeling • From the table at the right • Enter data into data matrix ofcalculator • APPS, 6, Current, Clear contents
Using Regression On Calculator • Choose F5 for Calculations • Choose calculationtype (LinReg for this) • Specify columns where x and y values will come from
Using Regression On Calculator • It is possible to store the Regression EQuation to one of the Y= functions
Using Regression On Calculator • When all options areset, press ENTER andthe calculator comesup with an equation approximates your data Note both the original x-y values and the function which approximates the data
Using the Function • Resulting function: • Use function to find Caloriesfor 195 lbs. • C(195) = 5.24This is called extrapolation • Note: It is dangerous to extrapolate beyond the existing data • Consider C(1500) or C(-100) in the context of the problem • The function gives a value but it is not valid
Assignment • Lesson 3.2B • Page 138 • Exercises 89 – 103 odd