warm up
DESCRIPTION
Warm UP. Find the derivative of. 3.4 A – Velocity and other rates of change. Want some extra practice? This website will randomly generate practice derivative problems for you. It's not perfect but it will give you some good basic practice. http://www.bluffton.edu/~nesterd/java/derivs.html - PowerPoint PPT PresentationTRANSCRIPT
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Warm UP
Find the derivative of
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3.4 A – Velocity and other rates of change
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Want some extra practice?This website will randomly generate practice derivative problems for you. It's not perfect but it will give you some good basic practice.
http://www.bluffton.edu/~nesterd/java/derivs.html
Set your page up as below:
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Goal•I will understand the relationship between a function and its derivative as it relates to motion and finances.
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Example
•Consider the volume of a cube given by the formula
Find the volume if the side length is 3.
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You try!
• What if the side length is 5?
• Answer= 125
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Instantaneous Rate of Change (Iroc)
• The Iroc at a point x=a is simply .
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Continued
•Now, find the Instantaneous rate of change at x=5
•What does this mean?
•When the side length is 5, the volume is changing at a rate of 75
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You Try!
• The area of a circle is given by .
1. Find the area when the radius is 8.
2. Find the Iroc when r is 2.
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Key terms for motion
1. Position function is s(t) where t is time. 2. Displacement – Change in position over an interval.
(neg=left, pos=right)3. Average Velocity= 4. Instantaneous velocity is the derivative of the
position function at a point.
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Continued
5. Speed is the absolute value of velocity
6. Acceleration is the derivative of the velocity function.
7. Free fall constants on earth:
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Example
Given the position of an object can be given by the function s(t) = -t2 + 10t where t is measured in seconds and s is measured in feet.
a. Graph this functionb. What is the displacement from 0-5 seconds? 0-10? c. What is the average velocity from 0-5 sec? 0-10?d. When does the object reach it’s highest point?
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Solutions
a. Graph this functionb. What is the displacement from 0-5 seconds? 0-10? 25-0 = 25 ft -100 + 100 = 0 ftc. What is the average velocity from 0-5 sec? 0-10?25/5 = 5 ft/sec 0/10 = 0 ft/sec d. When does the object reach it’s highest point?When the slope=0. So, let’s find the derivative and set it equal to 0.
At t=5 sec the item reaches the highest point.
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You try!
• A particle moves along a line so its position at any time is given by the function , where s is measured in meters and t is in seconds.
a. Fins the displacement of the particle during the first 2 secondsb. Find the average velocity during the first 4 seconds.c. Find the instantaneous velocity when t=4.d. Find the acceleration of the particle when t=4.
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Answers
a. s(2)-s(0) = -4 (so it is 4 meters to the left from where it started)
b. =0/4 = 0 m/sec
c. d.
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Homework
• 3.4: Do NOT do 20,27,28 (we will do those Friday)
• Stuck on #13? See example on page 130-131