1. Derivatives and Rates of Change
Section 2.7

3. Example 1, equation of tangent line
Given parabola y = x2 at the point P(1, 1).
Solution:
a = 1 and f (x) = x2, so the slope is
Using the point-slope form of the equation the equation of the tangent line at (1, 1)
y – 1 = 2(x – 1)
y = 2x – 1
= = 2

4. Example 1, equation of tangent line
Solution:
a = 3 and f (x) = 3/x, so the slope is

5. The average velocity over this time interval is

6. Example 3, velocity of ball
Suppose that a ball is dropped from the upper observation deck of the CN Tower, 450 m above the ground.
(a) What is the velocity of the ball after 5 seconds?
Velocity after 5 sec
v(5) = (9.8)(5)
= 49 m/s.

7. Example 3, velocity of ball
(b) How fast is the ball traveling when it hits the ground?
Ball must travel 450 m
the ball will hit the ground at the time t
s(t) = 450
4.9t2 = 450
Velocity of the ball as it hits the ground is
v(t) = 9.8t
t2 =
v( ) = 9.8
t =  9.6 sec
= 94 m/s

9. Example 4, find the derivative
f (x) = x2 – 8x + 9
Solution:

10. Example 5, find tangent line
f (x) = x2 – 8x + 9 at point (3, -6)
Solution:

11. Rates of Change
In other words, f (a) is the velocity of the particle at time t = a.
The speed of the particle is the absolute value of the velocity, that is, | f (a) |.

12. Example 5
A manufacturer produces bolts of a fabric with a fixed width. The cost of producing x yards of this fabric is C = f (x) dollars.
(a) What is the meaning of the derivative f (x)?
What are its units?
Instantaneous rate of change of the production cost with respect to the number of yards produced.
Since C is measured in dollars and x in yards, it follows that the units for f (x) are dollars per yard.

13. Example 6(b) – Solution
(b) In practical terms, what does it mean to say
f (1000) = 9?
After 1000 yards of fabric have been manufactured, the rate production cost is increasing is $9/yard.

14. Example 6 f(t) = number of centimeters of rainfall since midnight
t = time in hours
(a) What is the meaning of f -1(x)? Domain? Range?
- Inverse is time where a certain #of cm has fallen
- Domain; 0 to max. total rainfall
- Range; Midnight to the end of the storm
(b) Interpret, f(5)=2
- After 5 hours 2 cm of rain has fallen

15. f(t) = number of centimeters of rainfall since midnight
t = time in hours
(c) Interpret, f -1(5)= 2
- 5 cm of rain has fallen after 2 hours
(d) Interpret, f ‘(5)= ½
- After 5 hours the rain is falling at a rate of ½ cm/hr
(e) Interpret, (f -1) ‘ (5)= ½
- After 5 cm of rain has fallen, time is passing at a rate of ½ hr/cm

16. 2.7 Derivatives and Rates of Change
Summarize Notes
Read Section 2.7
Homework
Pg. 150
#5,7,9,13,23,27,29,31,39,43,47