Page 47 - Focus SPM 2022 - Additional Mathematics

Additional Mathematics SPM Chapter 2 Differentiation

Try this HOTS Question F Interpreting and determining the
rates of change of related quantities
A factory intends to produce a cylindrical container
with an opening at one end with a volume of 1. A function y = f(x) expresses the relationship
27π cm . The factory estimates that the production between the variables y and x. The changes of
3
2
cost would be RM0.18 per cm . Determine the the variables y and x with respect to time t are
maximum number of containers that can be dy dx
produced if the production cost is RM3 500. written as dt and dt . Hence,
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Answer:
229 dy = dy × dx
dt dx dt

2. The above relationship exists between dy ,
dt
dy and dx follows the chain rule which helps
SPM Highlights dx dt
in solving many rate of change of variables
The diagram below shows the side view of a part of a problems. This relationship can also be written
roller coaster track in a park. as:
p dy dy 1 1 dt
= × 1 = 2
5 m dx dt dx dx dx
dt dt
Horizontal ground level dx 1 dy 1 dx
or = × 1 = 2
dt dy dt dy dt
The curve part of the roller coaster track is represented dt dx
by an equation y = 1 x – 3 x , with point p as the
2
3
64 16 dy dx
origin.Find the shortest vertical distance, in m, from the 3. dt . 0 or dt . 0 denotes the increase in the
track to ground level. dy
rate of change of the variable, while dt , 0
Solution or dx , 0 denotes the decrease in the rate of
y = 1 x – 3 x dt
3
2
64 16
dy 3 3 change of the variable.
2
= x – x
dx 64 8
At the turning point, 19
3
3 x – x = 0 Two variables x and y are related by the equation
2
64 8 14
4
3 x 3 1 x – 1 = 0 y = 2x + x . If y increases with a rate of 8.5 units per
8 8
x = 8, x = 0 second, find the rate of change of x when x = 2. State the
d y 3 3 type of rate of change of x at that particular moment.
2
Form 5
Also, = x –
dx 2 32 8 Solution
2
For x = 8, d y . 0 (y is minimum.) Given dy = 8.5 (Positive value because y is increasing.)
dx 2 dt
–1
2
For x = 0, d y , 0 (y is maximum (ignored)) and y = 2x + 14x
dx 2 dy
–2
1 3 = 2 – 14x
Substitute x = 8, y = (8) – (8) 2 dt
3
64 16 = 2 – 14
= –4 x 2
\ The shortest vertical distance is 5 – 4 = 1 m. When x = 2, dy = 2 – 14
dt (2) 2
= – 3
2
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