Applications of Calculus

Collaborative learning, Experiential Learning, and initiate Talk for Learning.
Learning Experience: Learners working in mixed-ability groups explore ways of finding the rate of change of a given function or phenomenon at certain points.
Activity 1: Think-pair share: Learners in a mixed-ability group use their previous knowledge of finding the slope between two points to find the rate of change at a point.
Example 1: A dropped ball has height (m) h(t)=100-4.9t 2 , t seconds after it is released. How fast is the ball going at time t=2?
Solution: At time t=2 height is 100-4.9*(2 2 )= 80.4. A second later, that is, at t=3, what will be the height?
Solution: The height, h (3) =100-4·9*(9)=55.9, so in that second, the ball has travelled 80.4-55.9=24.5 meters. This means that the average speed during that time was 24.5 meters per second.
Note: If ∆t is some tiny amount of time, what we want to know is what happens to the average speed (h(2)-h(2+∆t))/∆t as ∆t gets smaller and smaller. Doing a bit of algebra: (ℎ(2) − ℎ(2 + ∆𝑡)) (80.4 − (100 − 4.9(2 + ∆𝑡) ) ) = ∆𝑡 ∆𝑡 (80.4 − 100 + 19.6 + 19.6∆𝑡 + 4.9∆𝑡 ) = ∆𝑡 (19.6∆𝑡 + 4.9∆𝑡 ) = ∆𝑡 = 19.6 + 4.9∆𝑡
When ∆t is very small, this is very close to 19.6, and indeed, it seems clear that as ∆t goes to zero, the average speed goes to 19.6, so the exact speed at t=2 is 19.6 meters per second.
When the COVID-19 hit Accra, Public Health Officials estimated that the number of persons having the COVID-19 at time t (measured in days from the beginning of the COVID-19) is approximated by P(t)=45t 2 - t 3 , provided that 0≤t≤50. At what rate is the COVID-19 spreading when t=10? When is the COVID-19 spreading at the rate of 600 per day?
Solution: The rate at which the COVID-19 is spreading is given by the derivative; P'(t)=90t-3t 2 (0≤t≤50) Since P(t) is measured in people and time is measured in days, the rate P'(t) is measured in people per day. When t=10 P' (10)=90(10)-3(10) 2 =600 Thus, 10 days after the beginning of COVID-19, it spreads at the rate of 600 people per day.
In this case, we are given the rate of change of P(t), and we must find the time corresponding to that rate. We set the expression for P'(t) equal to 600 and solve for t; 90t-3t 2 = 600 90t-3t 2 -600 = 0 Dividing by -3 and factoring, we have t 2 -30t+200=0 (t-10)(t-20)=0 Then t=10 or t=20. At both times COVID-19 is spreading at the rate of 600 people per day.
Example 2: Suppose that f(x)=x 2. Calculate the average rate of change of f(x) over the intervals 1 to 2, 1 to 1.1, and 1 to 1.01 Determine the (instantaneous) rate of change of f(x) when x=1
Solution: The intervals are of the form 1 to 1+∆x for ∆x=1,0.1, and 0.01. The average rate of change is given by the ratio Δy/Δx=(f(1+∆x)-f(1))/∆x=(〖(1+∆x) 2 -1)/∆x For the three given values of ∆x, this expression has the following respective values
∆x=1: Δy/Δx= (2 2 -1 2 )/1=(4-1)/1=3 Δy/Δx= (〖1.1〗^2-1^2)/0.1=(1.21-1)/0.1=2.1 ∆x=0.1: ∆x=0.01: Δy/Δx= (〖1.01〗^2-1^2)/0.01=(1.0201-1)/0.01=2.01 Thus, the average rate of change for ∆x=1,0.1, and 0.01 is 3,2.1,2.01 units per unit change in x, respectively
The instantaneous rate of change of f(x) at x=1 is equal to f'(1). We have f^' (x)=2x f^' (1)=2·1=2 That is, the instantaneous rate of change is 2 units per unit change in x.
Example 3: Let the production function p(x) give the number of units of goods produced when employing x units of labour. Supposed 4000 units of labour are currently employed, p(4000)=200, and p'(4000)=4. Estimate the number of additional units of goods produced when employing: One additional unit of labour An additional 1/4 unit of labour One less unit of labour.
Teaching and Learning Resources:
- GeoGebra and PhET
- Technology tools, mathematical sets and calculators
- Learners' textbooks and graph sheets curriculum
Assessment (1.3.2.AS.2). The document marks these depth-of-knowledge levels for this indicator: Level 3 Strategic reasoning; Level 4 Extended critical thinking and reasoning.