Transcript Determining Rates of Change from Data

Determining Rates of Change
from an Equation
•Recall that we were able to determine the average rate of change of a
quantity by calculating the slope of the secant line joining two points on the
curve.
•We were also able to estimate the instantaneous rate of change in four ways:
•Drawing a tangent line and then using two points on this tangent line,
calculate the slope of the tangent line.
•Estimate the slope of the tangent line by calculating the slope of the secant
line using a small preceding interval and the given table of values.
•Estimate the slope of the tangent line by calculating the slope of the secant
line using a small following interval and the given table of values.
•Estimate the slope of the tangent line by calculating the slope of the secant
line using a small centered interval and the given table of values.
Using an Equation
• We can do all of these things again today
but without having the graph or the table of
values.
• Instead we will use a formula for our
calculations. This formula gives us more
flexibility because we can calculate the yvalue (Temperature) for any x-value (time).
Temperature Example revisited
Example 2 continued
• The formula for the Temperature-time graph
is given by:
T (t)
120t 400
,0  t  80.
t  20
•This is written using function notation.
T(t) is read as T as a function of t, or T of t.
• Later we will see examples such as
f(x)=3x. This is read as f of x equals 3x.
To find the value of the function when x is
4, we write f(4)=3(4). We say f of 4 =12.
You just substitute 4 for the x.
•In this case y=f(x).
Average rate of change
• What was the average rate of change of the
temperature with respect to time from t=0s
to=20s ?
120(0)  400
T (0) 
120 t  400
T (t ) 
t  20
120(20)  400
(20)  20
2800
T (20) 
40
T (20)  70
T (20) 
(0)  20
400
T (0) 
20
How did we
T (0)  20
do this

T 70  20 C yesterday?

20  0s
t
T
 2.5 C / s
t
The average rate of increase in temperature is 2.5
degrees per second.
Estimating instantaneous rate of change
• Estimate the instantaneous rate of increase
in temperature at t=35s.
• Yesterday we used a 5s interval because
those were the only values we had in the
table.
• With the formula we can calculate any value
so we can use a smaller interval and get
more accurate estimate.
Use a 1 second interval.
• Use a 1-s following interval.
• Use t=35s and t=35+1=36s
120 (36)  400
T (36 ) 
36  20
4720
T (36 ) 
56
T (36)  84.285714
120 (35)  400
35  20
4600
T (35) 
55
T (35) 
T (35)  83.636363
The instantaneous rate of
T 84.285714 83.636363
change at t=35s is

36  35
t
approximately 0.649351
T
degrees per second.
 0.649351 C / s
t
Compare that to 0.67 using a 5
second interval.
You try!
•Use a 0.1s following interval to estimate
the instantaneous rate of change at t=35s.
120 (35.1)  400
T (35.1) 
35.1  20
T (35.1)  83.702359
120 (35)  400
35  20
4200
T (35) 
55
T (35) 
T (35)  83.636363
T 83.702359 83.636363

35.1  35
t
T
 0.65996 C / s
t
The instantaneous rate of
change at t=35s is
approximately 0.65996
degrees per second.
Example 2
• Given the function y=2x3
• a) Find the average rate of change from x=0
to x=1.
Let y  f ( x)
f ( x)  2 x3
f (1)  2(1)3  2
f (0)  2(0)3  0
y 2  0

2
x 1  0
The average rate of change from
x=0 to x=1 is 2.
• Given the function y=2x3
• b) Find the instantaneous rate of change at
x=0.5
f ( x)  2 x3
Use a 0.1s following
interval.
y 2(0.6)3  2(0.5)3

 1.82
x
0.6  0.5
The instantaneous rate of change
at x=0.5 is 1.82.
Difference Quotient
y f ( x1  h)  f ( x1 )

x
x1  h  x1
y f ( x1  h)  f ( x1 )

x
h
Why is called the difference quotient?
You try
• Given the function:
f ( x)  x
Estimate the instantaneous rate of change of y
with respect to x at x=6.