Page 57 - Hybrid PBD 2022 Tg 5 - Matematik Tambahan

Matematik Tambahan Tingkatan 5 Bab 2 Pembezaan
24. Bagi setiap persamaan yang berhubung x dan y berikut, jika kadar perubahan y ialah 0.08 unit s , cari kadar
–1
perubahan x pada ketika nilai x yang diberi. Tafsirkan jawapan anda. SP 2.4.6 TP4
For each of the following equations relating x and y, if the rate of change of y is 0.08 unit s , find the rate of change of x at the instant given
–1
value of x. Interpret your answers.

1 (b) y = 5x + 6, x = 2
x 2 (a) y = x – 4x , x = –1
2
y = , x = 4 2
(x – 1) d y
Diberi / Given Diberi d y = 0.08 unit/s Diberi dt = 0.08 unit/s
dt
d y = 0.08 unit/s apabila/ when apabila x = –1 apabila x = 2
dt d y = 5 
5x + 6
x = 4 d y = x – 4 dx 2
d y = [(x –1)(2x) – x ] dx
2
dx (x –1) 2 d y d y d x Maka/Then d y = d y × d x
dx
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dt
dt
Guna petua rantai/Use chain rule Maka/Then dt = dx × dt
d x
d x
5x + 6
d y = d y × d x = (x – 4)   = 5   
dt
2
dt dx dt dt
2
[(x –1)(2x) – x ] d x Apabila/When x = –1 dan Apabila x = 2 dan d y = 0.08
=   dt
(x –1) 2 dt d y
= 0.08
dt
 
Apabila/When x = 4 dan/and Maka/Then Maka/Then 0.08 = 5 d x
dy 8 dt
d x
dt = 0.08 0.08 = (–1– 4)   d x = 0.128 unit/s
Maka/Then dt dt
[(4 – 1)(2(4) –(4) ] d x d x 0.08
2
0.08 =   = unit/s x bertambah dengan kadar
(4 – 1) 2 dt dt –5 0.128 unit/s.
= –0.016 unit/s
9
dx = 0.08 × unit/s x menyusut dengan kadar x increases at a rate of 0.128 unit/s.

dt 8
= 0.09 unit/s 0.016 unit/s.
x decreases at a rate of 0.016 unit/s.
x bertambah dengan kadar
0.09 unit/s.
x increases at a rate of 0.09 unit/s.

25. Selesaikan masalah yang berikut. SP 2.4.7 TP6
Solve the following problems.
Semasa dipanaskan, panjang jejari, j cm bagi sebiji bola kuprum bertambah dengan kadar malar 2%. Cari
While heating, the radius, j cm of a copper ball increases uniformly at 2%. Find
(i) peratus perubahan dalam isi padu.
the percentage change in the volume.
(ii) peratus perubahan dalam luas permukaan bola kuprum.
the percentage change in the surface area of the copper ball.
Tafsirkan nilai peratus perubahan tersebut.
Interpret the value of change of the percentage.
(i) Diberi/Given δj × 100% = 2 %
j
4
Isi padu bola kuprum V = pj Tip Penting
3
Volume of copper ball 3 • Jejari, j cm bagi sebuah bola kuprum bertambah
dV = 4pj 2 dengan malar 2% ialah δj j × 100% = 2%.
dj
δV dV δj The radius, j cm of a copper ball increases uniformly
Maka/Then × 100% = × × 100% δj
V dj V at 2% is j × 100% = 2%.
δj
= 4pj × × 100% • Peratus perubahan dalam isi padunya ialah
2
4pj 3 Percentage change in the volume is defined by
3 δV × 100%
δj v
= 3 × 100% = 3(2%) = 6%
j
Isi padu bola kuprum bertambah 6% apabila jejarinya bertambah 2%
The volume of the copper ball increases by 6% when the radius increases by 2%.

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