Principles of Calculus

Talk for Learning, Think-pair-share, Experiential Learning, and Group Work/Collaborative Learning.
Activity 1: Derivative Applications In pairs or groups, task learners to discuss how to find a derivative of a relation that is not a function and implement it on a given function. 2𝑥 " + 3𝑦 " = 9𝑥
Solution: <A <A 6𝑥 ) + 3𝑦 ) <0 = 9(𝑥 <0 + 𝑦) 𝑑 𝑑 6𝑥 ) + 3𝑦 ) = 9𝑥 + 9𝑦 𝑑 𝑑 <A 6𝑥 ) − 9𝑦 = <0 (9𝑥 − 3𝑦 ) )
<A /0 ! *2A = 20*"A ! <0 )0 ! *"A <A = "0*A ! <0
Activity 2: Implicit Differentiation Learners will be working in convenient groups (ability, mixed-ability and mixed gender) to solve problems on implicit differentiation and share it across groups. E.g. In convenient groups, learners find the derivative of: i. 𝑥 ) 𝑦 + 2 = 5𝑦 , − 𝑥 ii. 𝑒 )A 𝑥 + 2 𝑠 𝑠 (𝑦) = 5𝑥 + 2 − 3𝑥
Activity 3: Everyday Problems Involving Derivatives Learners in specific groups use chain notes to solve everyday problems involving derivatives of relations that are not functions. Learners discuss and demonstrate their understanding of the problem and the solutions arrived at.
Example 1: Two cars start out 500 miles apart. Car A is to the west of Car B and starts driving to the east (i.e., towards Car B) at 35 mph, and at the same time, Car B starts driving south at 50 mph. After 3 hours of driving, at what rate is the distance between the two cars changing? Is it increasing or decreasing?
Solution:

In this figure, y represents the distance driven by Car B, x represents the distance separating Car A from the initial position of Car B, and z represents the distance separating the two cars. After 3 hours of driving time, we have the following values of x and y:
𝑥 = 500 − 35(3) = 395 𝑦 = 50(3) = 150 By Pythagoras theorem, 𝑧 ) = 𝑥 ) + 𝑦 ) = 395 ) + 150 ) = 178525 𝑧 = √178525 = 422.5222 Now to determine 𝑧 - , we apply implicit differentiation as all variables change with time. Given 𝑥 - = −35 𝑎 𝑦 - = 50 𝑧 ) = 𝑥 ) + 𝑦 ) ⟹ 2𝑧 = 2𝑥 + 2𝑦 - - - 𝑧 - (422.5222) = (395)(−35) + (150)(50)
- /"). 𝑧 - = = −14.9696 ,))..)))
So, after three hours, the distance between them is decreasing at a rate of 14.9696 mph.
Example 2: Air is being pumped into a spherical balloon at a rate of 5 cm 3 /min. Determine the rate at which the radius of the balloon is increasing when the diameter of the balloon is 20 cm.
Solution: The volume 𝑉(𝑡) and radius 𝑟(𝑡) are varying with time. 𝑑 𝑉 - (𝑡) = 5, 𝑟 - (𝑡) =? 𝑤ℎ𝑒 𝑟(𝑡) = = 10𝑐 2 Volume of a sphere is given by; 𝑉(𝑡) = " 𝜋[𝑟(𝑡)] " ,
By implicit differentiation; 𝑉 - = 4𝜋𝑟 ) 𝑟 - 5 = 4𝜋(10 ) )𝑟 - ! ⟹ 𝑟 - = 1( 𝑐/𝑚
Example 3: A trough of water is 8 meters in length, and its ends are in the shape of isosceles triangles whose width is 5 meters and height is 2 meters. If water is pumped in at a constant rate of 6m 3 /sec, at what rate is the height of the water changing when the water has a height of 120 cm? At what rate is the width of the water changing when the water has a height of 120cm?
Solution: 

/# " 𝑉 - = ℎ - =? 𝑎 ℎ = 1.2𝑚 ON; 𝑉 = (𝐴 𝑜 𝐸)(𝑑ℎ) = i ) 𝑏 × ℎ𝑒ℎ𝑡j (𝑑ℎ) !
! = ) ℎ𝑤(8) = 4ℎ𝑤
h W . = ) ⟹ 𝑤 = ) ℎ .
. 𝑉 = 4ℎ𝑤 = 4ℎ i ℎj = 10ℎ ) ) By implicit differentiation; 𝑉 - = 20ℎℎ - 6 = 20(1.2)ℎ - ⟹ ℎ - = 0.25 𝑚/𝑠
So, the height of the water is rising at a rate of 0.25 m/sec
. 𝑤 = ) ℎ . ⟹ 𝑤 - = ) ℎ -
𝑤 - = (0.25) = 0.625 𝑚/𝑠 . ) Therefore, the width is increasing at a rate of 0.625 m/sec.
Teaching and Learning Resources:
- Reading resource
- Colour pens
- Notebook
- Graph sheets
- Mathematical sets
- Technological tools
Assessment (2.3.1.AS.3). The document marks these depth-of-knowledge levels for this indicator: Level 2 Skills of conceptual understanding; Level 4 Extended critical thinking and reasoning.