determining rates of change from an equation recall that we were able to determine the average rate...
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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 y-value (Temperature) for any x-value (time).
Temperature Example revisited
Example 2 continued• The formula for the Temperature-time
graph is given by:
.800,20t
400120tT(t)
t
•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 ?
20
400120)(
t
ttT
20)20(
400)20(120)20(
T
70)20(40
2800)20(
T
T
20)0(
400)0(120)0(
T
20
400)0( T
20)0( T s
C
t
T
020
2070
sCt
T/5.2
The average rate of increase in temperature is 2.5 degrees per second.
How did we
do this yesterday?
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
2036
400)36(120)36(
T
285714.84)36( T
sC /649351.0t
T
3536
636363.83285714.84
t
T
The instantaneous rate of
change at t=35s is approximately 0.649351 degrees per second.
Compare that to 0.67 using a 5 second interval.
56
4720)36( T
2035
400)35(120)35(
T
55
4600)35( T
636363.83)35( T
•Use a 0.1s following interval to estimate the instantaneous rate of change at t=35s.
You try!
201.35
400)1.35(120)1.35(
T
sC /65996.0t
T
351.35
636363.83702359.83
t
T
702359.83)1.35( T
2035
400)35(120)35(
T
55
4200)35( T
636363.83)35( 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 xfy 32)( xxf
2)1(2)1( 3 f
0)0(2)0( 3 f
201
02
x
y
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
32)( xxf
82.15.06.0
)5.0(2)6.0(2 33
x
y
Use a 0.1s following interval.
The instantaneous rate of change
at x=0.5 is 1.82.
Difference Quotient
h
xfhxf
x
y
xhx
xfhxf
x
y
)()(
)()(
11
11
11
Why is called the difference quotient?
You try• Given the function:
xxf )(
Estimate the instantaneous rate of change of y with respect to x at x=6.