Page 60 - Hybrid PBD 2022 Tg 5 - Matematik Tambahan

Matematik Tambahan Tingkatan 5 Bab 2 Pembezaan
27. Selesaikan yang berikut. SP 2.4.9 TP5
Solve the following.

Mangkuk terbuka berbentuk hemisfera dengan Luas permukaan minyak,
jejari 13 cm. Minyak mengalir ke dalamnya dan Area of surface of oil
tinggi minyak dalam mangkuk berubah daripada A = pR 2
2
2
2
3 cm kepada 3.05 cm. Cari perubahan hampir luas Tetapi/But R = 13 – (13 – h) = 26h – h 2
2
permukaan minyak dalam mangkuk itu, dalam Maka/Hence A = p(26h – h )
sebutan p. dA = p(26 – 2h)
An open hemispherical bowl has a radius of 13 cm. Oil is flowing dh
into the bowl such that the height of the oil in the bowl increases δA ≈ dA .δh
from 3 cm to 3.05 cm. Find the approximate change of the surface dh
area of the oil in the bowl in terms of p. ≈ p(26 – 2h) δh
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Apabila/When δh = 0.05 dan h = 3
δA ≈ p(26 – 2(3)) (0.05)
≈ p
2
h R Jadi, luas permukaan bertambah sebanyak p cm .
Then, the surface area increases by p cm . 2

(a) Isi padu, V cm bagi suatu kon menyusut daripada 30 cm kepada 29.5 cm . Jika tinggi kon sentiasa
3.
3
3
tiga kali jejari tapak kon itu, cari
The volume, V cm of a cone decreases from 30 cm to 29.5 cm . If the height of the cone is always three times the length of the
3
3
3
base radius of the cone, find
(i) perubahan hampir tinggi apabila jejari tapak ialah 1.2 cm.
the approximate change in the height when the base radius is 1.2 cm.
(ii) perubahan hampir jejari jika tinggi kon ialah 6 cm.
the approximate change in the base radius when the height is 6 cm.
1 1 1
(i) h = 3j V = pj h = pj 3 (ii) V = pj h = ph 3
2
2
3 3 27
δV = 29.5 – 30 = –0.5 cm 3 dV 1
dV = 3 pj 2 dh = ph 2
9
dj dV
dV δV = dh .δh
δV = .δj
dj 1 2
–0.5 –0.5 = p(6) δh
2
–0.5 = 3 pj .δj δj = = –0.0368 9
3p(1.2) 2 –0.5
δh =
4p
Dari/From h = 3j
dh
dj = 3
dh
δh = dj .δj
–0.5 = 3 δj
4p
δj = –0.013

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