SHS3 Additional Mathematics · Semester 2, Week 6
Application of Calculus
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Curriculum details
- Strand
- Calculus (Strand 3)
- Sub-strand
- Application of Calculus (3.2)
- Content standard
- 3.3.2.CS.1 - Demonstrate conceptual Find the area under a constant function, curves and distinguish between total, net and understanding of integration to find distances, areas under curve, volumes and other related real life problems. 3.3.2.LO.1 Determine distance, area under curve and solid of volumes formed under revolution.
- Indicator
- 3.3.2.LI.1 - signed area.
- Suggested placement
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Semester 2, Week 6
(Week 26 of the year)
Our suggestion, laid out in curriculum order across three terms of twelve weeks. NaCCA does not fix the week, so follow your school's scheme of learning.
- Source document note
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The official curriculum document has a fault in this entry, so the curriculum text above is transcribed exactly as printed:
- exemplars - p.521: adjacent duplicated CambriaMath characters were collapsed, but this PDF also maps some equation glyphs to the wrong letter or operator; verify every mathematical expression in this row against the rendered page
- Curriculum reference
- NaCCA curriculum document, p. 521
Exemplars (from the NaCCA curriculum)
Talk for Learning, Think-pair-share, Experiential Learning; and Group Work/Collaborative Learning. Activity 1: Area under constant function Using a real life instances or scenario involving area under curve, Learners in small convenient groups (for instance, mixed gender, mixed-ability, etc.) investigate the total are, net area and signed area under a given function. Example 1 - Compute the area under the curve 𝑦 = 𝑥 ) − 4𝑥 + 5 𝑓 𝑥 = −1 𝑡 𝑥 = 3 Solution:
" 𝑥 " î (𝑥 ) − 4𝑥 + 5 ) 𝑑 = ) − 2𝑥 ) + 5𝑥* 3 − 1 3 - ! " " (*!) " = É " − 2(3) ) + 5(3)Ê − É − 2(−1) ) + 5(−1)Ê " ,( = " NB: the function, 𝑓 > 0, hence the integral represent the area. determine - An object moving along a straight line has velocity function 𝑣(𝑡) = 𝑠 (𝑡). If the object starts at position 0, the total distance travelled and the object's position at time "\ 𝑡 = . ) Solution:
The total area bounded by the curve is "5 \ 𝐴 = ∫ ( 𝑠 𝑠 𝑡 𝑑 − ∫ \ ! 𝑠 𝑠 𝑡 𝑑 = 2 + 1 = 3 This means that the total distance travelled is 3 units. The overall or total change in position of the object is given by "5 \ 𝐴 = ∫ ( 𝑠 𝑠 𝑡 𝑑 + ∫ \ ! 𝑠 𝑠 𝑡 𝑑 = 2 + (−1) = 1 So, if the object starts at position 0, it ends up at position 0 + 1 = 1 Activity 2: Solving problems on Collaborative learning/Group work. Learners are to work in pairs to debate the expected solution(s) provided to a question and provide reasons for the approaches used. Question: Suppose an object moves so that its speed, or more properly velocity, is given by v(t) = −t ) + 5t in cm/sec. Assume that at time t=0, the position is 0. Find the position at t=6 Solution: Let us examine the motion of this object carefully. Let 𝑠(𝑡) be the position of the particle at time t. Then,
Approach 1 G " .G ! 𝑠(𝑡) = ∫ ( 𝑣(𝑡) 𝑑 = ∫ ( −𝑡 ) + 5𝑡 𝑑 = (− " + ) + 𝐶)| G4/ / / G4( =18 Approach 2 𝑠(𝑡) = ∫ ( 𝑣(𝑡) 𝑑 = ∫ ( −𝑡 ) + 5𝑡 𝑑 = / / . / 7! ∫ ( −𝑡 ) + 5𝑡 𝑑 − ∫ . −𝑡 ) + 5𝑡 𝑑 = " NB: Learners should be aware of the concepts of "total" and "net." Teaching and Learning Resources: - Reading resource colour pens - Notebook - Graph sheets - Mathematical sets, - Technological tools. - Curriculum - Cardboard Assessment (3.3.2.AS.1). The document marks these depth-of-knowledge levels for this indicator: Level 1 Recall; Level 2 Skills of conceptual understanding; Level 3 Strategic reasoning; Level 4 Extended critical thinking and reasoning.