ESTIMATE INSTANTANEOUS RATE OF CHANGE USING AN EQUATION

Problem 1 :

The population of a small town appears to be growing exponentially. Town planners think that the equation 

P(t) = 35000 (1.05)^t

where P(t) is the number of people in the town and t is the number of years after 2000, models the size of the population. Estimate the instantaneous rate of change in the population in 2015.

Solution :

P(t) = 35000 (1.05)^t

We find the instantaneous rate of change of population in the year 2015, since this is 5 years after 2000, we find the rate of change in the interval 14 ≤ t ≤ 15

Average rate of change = P(15) - P(14) / (15 - 14)

P(15) = 35000 (1.05)¹⁵

= 35000 (2.0789)

= 72762.48

P(14) = 35000 (1.05)¹⁴

= 35000 (1.979)

= 69297.60

Average rate of change

= (72762.48 - 69297.60) / 1

= 3464.88

Approximately 3464 people per year.

Using preceding intervals, 14.5 ≤ t ≤ 15

P(14.5) = 35000 (1.05)^14.5

= 35000 (2.0288)

= 71008

Average rate of change

= P(15) - P(14.5) / (15 - 14.5)

= (72762.48 - 71008) / 0.5

= 1754.48/0.5

= 3508.96

Approximately 3509 people per year.

Using preceding intervals, 15.5 ≤ t ≤ 16

P(15.5) = 35000 (1.05)^15.5

= 35000 (2.130)

= 74559.36

P(16) = 35000 (1.05)¹⁶

= 35000 (2.182)

= 76400

Average rate of change

= P(16) - P(15.5) / (16 - 15.5)

= (76400 - 74559.36) / 0.5

= 1840.64/0.5

= 3681.28

Approximately 3681 people per year.

Finding average = (3464 + 3509)/2

= 3486.5

= 3487 approximately.

Problem 2 :

A soccer ball is kicked into the air. The following table of values shows the height of the ball above the ground at various times during its flight

a) Estimate the instantaneous rate of change in the height of the ball at exactly t =  2.0 s using the preceding and following interval method. 

b) Estimate the instantaneous rate of change in the height of the ball at exactly t = 2.0 s using the centered interval method.

c) Which estimation method do you prefer? Explain

Solution :

a) Preceding interval method :

Using preceding intervals, 1.5 ≤ t ≤ 2

= (30.9 - 26.98)/(2 - 1.5)

= 3.92/0.5

= 7.84

Using preceding intervals, 2 ≤ t ≤ 2.5

= (32.38 - 30.9)/(2.5 - 2)

= 1.48/0.5

= 2.96

Average of instantaneous rate of change = (7.84 + 2.96)/2

= 10.8/2

= 5.4 m/s

b) Instantaneous rate of change using centered interval method. :

At t = 2, then the centered interval will be 1.5 and 2.5 where h = 0.5

= f(2.5)-  f(1.5) / (2.5 - 1.5)

= (32.38 - 26.98)/1

= 5.4 m/s

c) I prefer the centred interval method. Fewer calculations are required, and it takes into account points on each side 

Problem 3 :

A population of raccoons moves into a wooded area. At t months, the number of raccoons P(t) can be modelled using the equation 

P(t) = 100 + 30t + 4t²

a) Determine the population of raccoons at 2.5 months.

b) Determine the average rate of change in the raccoon population over the interval from 0 months to 2.5 months.

c) Estimate the rate of change in the raccoon population at exactly 2.5 months.

d) Explain why your answers for parts a), b), and c) are different.

Solution :

P(t) = 100 + 30t + 4t²

a) When t = 2.5

P(2.5) = 100 + 30(2.5) + 4(2.5)²

= 100 + 75 + 4(6.25)

= 100 + 75 + 25

= 200

b) 0 ≤ t ≤ 2.5

= [P(2.5) - P(0)] / (2.5 - 0)

= (200 - 100)/2.5

= 40 raccoons per month.

c) Rate of change at t = 2.5

Using centered interval method,

2 ≤ t ≤ 3

P(2) = 100 + 30(2) + 4(2)²

= 100 + 60 + 16

= 176

P(3) = 100 + 30(3) + 4(3)²

= 100 + 90 + 36

= 226

Rate of change = (226 - 176)/(3 - 2)

= 50

d)  The three answers represent different things: the population at a particular time, the average rate of change prior to that time, and the instantaneous rate of change at that time