copyright © 2013 all rights reserved, government of newfoundland and labrador calculus 3208...
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Copyright © 2013 All rights reserved, Government of Newfoundland and Labrador
Calculus 3208Derivative (18)
Unit 4: Chapter # 2 – Section 2.1 (Essential Calculus)Average and Instantaneous Slope
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Average and Instantaneous Slope
• A Concrete Example of a Rate of Change• Average Rate of Change• Instantaneous Rate of Change
Copyright © 2013 All rights reserved, Government of Newfoundland and Labrador
An Apple Falls From a TreeSynopsis:
An apple falls from atop a 20m tree and is in free-fall as it drops toward the ground. It’s height above the ground (in meters) is a decreasing function of time (in seconds).
A Table of Heights and Times:
Time (s) 0 0.5 1 1.5 2Height (m) 20 18.775 15.1 8.975 0.4
A Graph of Height vs. Time:
(0,20)(0.5, 18.775)
(1. 15.1)
(1.5, 8.975)
(2, 0.4)
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Average Rate of ChangeQuestion: At what average speed is the apple falling between 0.5s and 1.5s after it starts to drop?
12
12
tt
hhvavg
(0,20)(0.5, 18.775)
(1. 15.1)
(1.5, 8.975)
(2, 0.4)
sm /8.91
8.95.05.1
775.18975.8
Avg. Rate of Change = Slope of Secant =x
y
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Instantaneous Rate of ChangeQuestion: How fast is the apple falling, exactly 1 second after it starts to drop?
12
12
tt
hhvinst
Inst. Rate of Change = Slope of Tangent
(0,20)(0.5, 18.775)
(1. 15.1)
(1.5, 8.975)
(2, 0.4)
We need to estimate:
),( 11 htWhere: and ),( 22 htShould be as close as possible to the point of tangency!
Examples:
sm
vinst
/8.91
8.95.05.1
775.18975.8
sm
vinst
/25.125.0
125.615.1
1.15975.8
OR
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Over the span of a month, the depth of the water in a city reservoir varies with time. The table shown gives selected values for the depth of the water (in ft.) as a function of time (in days since the beginning of March).
Time (days) 2 10 15 23 30
Depth (ft.) 10 12 20 21 27
(A) What is the average rate of change (in ft. per day) of the depth of the reservoir between the 10th and 30th of March?(B) What is the approximate instantaneous rate of change of the depth of the reservoir (in ft.
per day) on March 23rd ?
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1. Let D(t) be the US national debt at time t. The table shown gives approximate values of this function by providing end of year estimates, in billions of dollars, from 1990 to 2010. Approximately how fast was the US national debt rising at the end of 2005?
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The graph at the right represents the population of bacteria in a laboratory culture as a function of time (in minutes). (A) Determine the average growth rate of the bacteria culture over the first three minutes. (B) Estimate the instantaneous growth rate of the bacteria 6 minutes after the culture was started.
popu
latio
n
time (minutes)
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The graph at the right represents the population of bacteria in a laboratory culture as a function of time (in minutes). (A) Determine the average growth rate of the bacteria culture over the first three minutes. (B) Estimate the instantaneous growth rate of the bacteria 6 minutes after the culture was started.
popu
latio
n
Time (minutes)
- Solution
(A) The average rate of change can be calculated by finding the slope of the secant joining (0, 100), and (3, 200). (Shown in Red)
3.3303
100200
t
Pbacteria/min
(B) The exact instantaneous rate of change can be calculated by finding the slope of the tangent touching the curve at (6, 400). (Shown in Green) We can estimate using the secant joining (6, 400), and (9, 800). Other reasonable estimates are acceptable.
3.1333
400
69
400800
instm bacteria/min
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• The average rate of change of a function on an interval , can be calculated by finding the slope of the secant line joining , and :
• The instantaneous rate of change of a function at the point , is the slope of the tangent that touches at that point.
• To estimate the instantaneous rate of change of a function at the point , we can chose a second point , where c is reasonably close to a. We then calculate:
)(xf ba,
)(, afa )(, bfb
ab
afbfmavg
)(xf )(, afa)(xf
)(xf )(, afa
)(, cfc
ac
afcfmisnt