1. Rates of Change in the Natural and Social Sciences
Section 3.7

2. Physics and rate of change…
If s=f(t) and s is the position of a particle
Velocity
f’(t) = v(t) = ds/dt
Acceleration
f’’(t) = v’(t) = a(t) =dv/dt

3. Example 1; Physics The position of a particle is given by the equation
s = f (t) = t 3 – 6t 2 + 9t
where t is measured in seconds and s in meters.
(a) Find the velocity at time t.
Find derivative of the position function.
s = f (t) = t 3 – 6t 2 + 9t
v (t) = = 3t 2 – 12t + 9

4. (b) What is the velocity after 2 s? After 4 s?
Velocity after 2 s means the instantaneous velocity v(2)
v (2) = = 3(2)2 – 12(2) + 9
= –3 m/s
The velocity after 4 s is
v (4) = 3(4)2 – 12(4) + 9
= 9 m/s

5. c) When is the particle at rest?
At rest when v (t) = 0,
3t 2 – 12t + 9 = 3(t 2 – 4t + 3)
= 3(t – 1)(t – 3)
= 0
and this is true when t = 1 or t = 3.
Thus, particle is at rest after 1 s and after 3 s

6. (d) When is the particle moving forward (in the positive direction)?
positive direction when v (t) >0
3t 2 – 12t + 9 = 3(t – 1)(t – 3) > 0
Both factors are positive or both factors are negative (t > 3) or (t < 1)
Thus, particle moves in the positive direction on time intervals t < 1 and t > 3.
It moves backward (in the negative direction) when 1 < t < 3.

7. The total distance is 4 + 4 + 20 = 28 m.
(e) Find the total distance traveled by the particle during the first five seconds.
Calculate the distances traveled during pos/neg time intervals each separately
[0, 1] [1, 3] [3, 5]
distance in first second =| f (1) – f (0) | = | 4 – 0 | = 4 m
t = 1 to t = 3 distance =| f (3) – f (1) | = | 0 – 4 | = 4 m
t = 3 to t = 5 distance = | f (5) – f (3) | = | 20 – 0 | = 20 m
The total distance is = 28 m.

8. (g) Find the acceleration at time t and after 4 s.
Acceleration = derivative of the velocity
a (t) = =
a (t) = 6t – 12
a (4) = 6 (4) – 12
= 12 m/s2

9. Example 2; Physics I=current = rate at which charge flows
A current exists whenever electric charges move. Below is a wire and electrons moving through a plane surface, shaded red.
If Q is the net charge and t is the time period, then the average current during this time is
I=current = rate at which charge flows
Measured in; coulombs per second or amperes

10. Example 3; Chemistry
A chemical reaction results in the formation products from reactants.
For instance, the “equation”
2H2 + O2  2H2O
two molecules of hydrogen and one molecule of oxygen makes two molecules of water.
Look at the reaction
A + B  C

11. The concentration of a reactant A is the number of moles per liter
The concentration varies during a reaction, so A, B, and C all functions of time (t).
Average rate of reaction of the product C over a time interval t1 to t2
Chemists are look at the instantaneous rate of reaction,

12. Example 4; Thermodynamics
If a given substance is kept at a constant temperature, then its volume V depends on its pressure P.
The rate of change of V with respect to P —, dV/dP.
The compressibility is defined by introducing a minus sign and dividing this derivative by the V.

13. Example 4; Thermodynamics
β measures how fast, per unit volume, the volume decreases as the pressure increases (given a constant temperature)
For instance, the volume V (in cubic meters) of a sample of air at 25C was found to be related to the pressure P (in kilopascals) by equation

14. Find the rate of change of V with respect to P when P = 50 kPa is

15. Find the compressibility at that pressure is

16. Example 5, Biology
Let n = f (t) be the number of individuals in an animal or plant population at time t.
The average rate of change in the population size t1 and t2
The instantaneous rate of growth is obtained as follows:

17. The population function is n = n02t.
Consider a population of bacteria in a homogeneous nutrient medium.
Suppose the population doubles every hour.
If the initial population is n0 and the time t is (hours)
f (1) = 2f (0) = 2n0
f (2) = 2f (1) = 22n0
f (3) = 2f (2) = 23n0
The population function is n = n02t.

18. Find the rate of growth of the bacteria population at time t.
Find the rate of growth after 4 hours when the initial population of n0 = 100 bacteria.
= 1600 ln 2
 1109
After 4 hours, the bacteria population is growing at a rate of about 1109 bacteria per hour.

19. Example 6; Economics
Suppose C(x) is the total cost of producing x units
The function C is called a cost function.
The average rate of change of the cost is
The instantaneous rate of change of cost with respect to the number of items produced, is the marginal cost by economists

20. C (x) = a + bx + cx2 + dx3
a represents the overhead cost
other terms represent the cost of raw materials

21. Suppose a company has estimated that the cost (in dollars) of producing x items is
C (x) = 10, x x2
Find the marginal cost function.
C (x) = x
Find the marginal cost at the production level of 500 items.
C (500) = (500)
= $15/ item

22. This gives the rate at which costs are increasing with respect to the production level when x = 500 and predicts the cost of the 501st item.
Find the cost of producing the 501st item is
C (501) – C (500)
= [10, (501) (501)2]
– [10, (500) (500)2]
= $15.01

23. 3.7 Rates of Change in the Natural and Social Sciences
Summarize Notes
Read section 3.7
Homework
Pg.215 #1,9,11,17,23,25,27,35,49,51,53,76