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Learning Log #6
1.6 HW
Chapter 2Limits & Derivatives
Yeah! We begin Calculus
EXAMPLE 1
If a ball is thrown into the air with an initial velocity of 60 ft/sec, its height (in feet) after seconds is given by
a) Find the average velocity for the time period beginning when and lasting
i) .5 s.ii) .1 s.iii) .05 s.iv) .01 s.
260 16y t t t
2t
distance traveledAverage velocity =
time elapsed
260 16y t t
Example 1: Numeric
2 2
) 2 2.5
60(2.5) 16(2.5) 60(2) 16(2)
2.5 2
i from t to t
v
50 56
.5v
6
.5v
12 / secv ft
Example 1 continued
2 2
) 2 2.1
60(2.1) 16(2.1) 60(2) 16(2)
2.1 2
ii from t to t
v
55.44 56
.1v
5.6 / secv ft
Example 1 continued
2 2
) 2 2.05
60(2.05) 16(2.05) 60(2) 16(2)
2.05 2
iii from t to t
v
.24
.05v
4.8 / secv ft
Example 1 continued
2 2
) 2 2.01
60(2.01) 16(2.01) 60(2) 16(2)
2.01 2
iv from t to t
v
.0416
.01v
4.16 / secv ft
Another way to calculate average velocity …
We can calculate the velocity from time t =2 to some time slightly later t = 2+h
2120 60 16(4 4 ) 56h h h
h
2 260(2 ) 16(2 ) 60(2) 16(2)h hv
h
continued …
2120 60 64 64 16 56h h h
h
216 4h h
h
( 16 4)h h
h
continued …4 16h ) .5 4 16(.5) 12 / sec
) .1 4 16(.1) 5.6 / sec
) .05 4 16(.05) 4.8 / sec
) .01 4 16(.01) 4.16 / sec
i h ft
ii h ft
iii h ft
iv h ft
Find the instantaneous velocity at t=2.
As h approaches zero, the limiting value is -4.
4 16velocity h
(2,56)(2.5,50)
Slope of secant line = Average velocity
Slope of Tangent Line= Instantaneous velocity
Example 2
The position of a car is given by the values in the table.
a)Find the average velocity for the time period beginning when t=2 lasting… i) 3 sec ii) 2 sec iii) 1 sec
t(seconds)
0 1 2 3 4 5
s(feet) 0 10
32
70
119
178
_178 32 146) 48.6 /
5 2 3i ft s
119 32 87) 43.5 /
4 2 2ii ft s
70 32 38) 38 /
3 2 1iii ft s
Guided practice#6 Large
whiteboards
2
58 / ,
58 .83
If anarrowis shot uponthemoonwith
avelocityof m s its height is given
by h t t
) 1,2 ) 1,1.1 ) 1,1.01
d)Find theinstantaneous velocityafter1second.
a b c
Homework
Pgs.89,90 (2,5)
Due tomorrow
