Page 58 - Hybrid PBD 2022 Tg 5 - Matematik Tambahan

Matematik Tambahan Tingkatan 5 Bab 2 Pembezaan

(ii) Luas permukaan bola kuprum
Surface area of the copper ball Tip Penting
A = 4pj 2 • Peratus perubahan dalam luas permukaannya ialah
dA = 8pj Percentage change in the surface area is defined by
dj
δA dA δj δA × 100%
Maka/Then ×100% = × × 100% A
A dj A
8pjdj

= × 100%
4pj 2
δj
= 2 ×100% = 2(2%) = 4%
j
Luas permukaan bola kuprum bertambah 4% apabila jejarinya bertambah 2%.
The surface area of the copper ball increases by 4% when the radius increases by 2%.
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(a) Sebuah kon tegak dengan bucunya ke bawah, sudut bucunya ialah 60°. Jika kon itu diisi dengan
3
minyak sehingga penuh dan minyak kemudiannya dibiar mengalir keluar dengan kadar malar 3 cm /s,
cari peratus perubahan tinggi minyak di dalam kon, dalam sebutan p ketika ketinggian minyak di
dalam kon ialah 6 cm.
A vertical cone with its tip at the bottom, the angle at the tip is 60°. If the cone is fully filled with oil and then the oil is allowed to drip
out from the tip at a constant rate of 3 cm /s, find the change of the percentage of the height of oil in the cone, in terms of π when
3
the height of the oil inside the cone is 6 cm.
Katakan jejari ialah j dan tinggi ialah h. / Let say the radius is j and the height is h.
tan 30 = j
0
h h
 3 = j
1 1 1 1

V = pj h = p  h h = p h 3 60°
2
2
3 3 3 9
dV = ph 2
1
dh 3
dV × 100% = dV × dh × 100%
dh dh V
1 dh
–3 = ph ×  3 × 100%
2
3 1 ph
9
dh
–3 = 3 × h × 100%
dh
× 100% = –1%
h
(b) Luas sebuah bulatan bertambah dengan kadar malar 4%. Tentukan
The area of a circle increases uniformly at 4%. Determine
(i) peratus perubahan jejarinya. (ii) peratus perubahan lilitannya.
the percentage change in the radius. the percentage change in the circumference.
δA Tafsirkan nilai peratus perubahan tersebut.
(i) Diberi/Given × 100% = 4% Interpret the value of change of the percentage.
A
Luas bulatan/Area of circle
A = pj 2 (ii) Lilitan bulatan/Circumference
dA = 2pj L = 2pj
dj dL = 2p
δA dA δj dj
Maka/Then × 100% = × × 100%
A dj A Maka/Then δL × 100% = dL × δj × 100%
2pj δj L dj L
4% = × 100%
pj 2 2p dj
δj 4% = 2pj × 100%
× 100% = = 2%
i 2 = 2 %
Jejari bulatan bertambah sebanyak 2% apabila Lilitan bulatan bertambah sebanyak 2%
luasnya bertambah sebanyak 4%. apabila jejarinya bertambah sebanyak 2%.
The radius increases by 2% when the area increases by 4 %. The circumference increases by 2% when the radius
increases by 2%

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