Page 49 - Focus SPM 2022 - Additional Mathematics

Additional Mathematics SPM Chapter 2 Differentiation
SPM Highlights 4. The smaller the value of dx, the more accurate
the approximate value of dy.
The surface area of a cube increase at a constant rate 5. dx . 0 or dy . 0 denotes the small increase in x
of 15 cm s . Find the rate of change of side length, in or y, dx , 0 or dy , 0 denotes the small decrease
2 –1
3
–1
cm s , when the volume of the cube is 125 cm .
in x or y.
Solution
x 22
2
x Given y = 3x + 6x, find the small change in y when x
x changes from 4 units to 4.02 units.
Let the length of the side of the cube = x cm
2
Therefore the area of the cube, A = 6x Solution
2
3
When the volume of the cube = 125 cm , From y = 3x + 6x
x = 125 dy = 6x + 6
3
x = 5 cm dx
dA When x = 4,
\ dx = 12x dy
= 12(5) dx = 6(4) + 6
= 60 = 30
and dA = 15 cm s dx = 4.02 – 4
2 –1
dt = 0.02
dA dA dx dy
From = × Therefore, dy = × dx
dt dx dt dx
dx
= 30 × 0.02

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15 = 60 ×
dt = 0.6 unit
dx 15
=
dt 60 \ y experiences small changes for 0.6 unit.
= 1 cm s
–1
4 23
\ The length of side increase with a rate of 1 cm s –1 Estimate the approximate value of √63.98 by using
3
4
when the volume of the cube is 125 cm . differentiation.
3
Solution
3
Try Questions 15 – 16 in ‘Try This! 2.4’ Let y = √x
1
= x
3
H Interpreting and determining small \ dy = 3x
1
Penerbitan ≈ dy when dx → 0. Choose a value of x close to 63.98, such that its cube
changes and approximations of
2
dx
3
lim dy
3
1. From dx → 0 dx = dy , where dx is the small Let 3 √63.98 = √64 + dy
dx
SPM Tips
changes in x and dy is the small changes in y, it is
Form 5
dy
found out that
dx
dx
dy
dy
dy
≈
2. formula is used to estimate the approximate root value can be obtained easily. dy is the change or
can be written as dy ≈
× dx. This
the approximate difference between √63.98 and
3
dx
dx
dx
3
3
√64 , that is dy = √63.98 – √64 .
3
change or the small change in y when there is a
small change dx in x. Small changes in y,
3. Observe that the value of dy obtained from dy ≈ dy × dx
dx
dy ≈ dy × dx is not the exact change in y. The ≈ 1 × dx
dx
2
value is just an approximation to the real value 3x
3
of change.
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