BAB Matematik Tambahan Tingkatan 5 Bab 1 Sukatan Membulat 1 1 CONTOH NOTA IMBASAN NOTA IMBASAN NOTA IMBASAN NOTA IMBASAN BAB 1 Sukatan Membulat Circular Measure 1.1 Radian Radian 1. Sudut boleh diukur dalam darjah dan minit atau radian (rad). Angle can be measured in degrees and minutes or in radians (rad). 2. Sudut yang dicangkum pada pusat bulatan oleh lengkok yang sama panjang dengan jejarinya ditakrifkan sebagai 1 radian. Lihat contoh dalam rajah. The angle subtended at the centre of the circle by the arc length which is the same length as the radius is defined as 1 radian. 3. Apabila panjang lilitan ialah 2πj, maka sudut yang tercangkum ialah 2π radian. When the circumference is 2pj, then the subtended angle is 2p radian. 4. Diketahui bahawa sudut bulatan ialah 360°, maka 2π radian adalah setara dengan 360°. It is known that the angle of a circle is 360° , then 2π radian is equivalent to 360°. 5. Secara am, kita boleh tulis In general, we can write q 360° = q rad 2π rad 1. Tukarkan setiap yang berikut kepada radian. Convert each of the following to radians. TP 1 (i) 65°42’ (ii) 279°24’ Penyelesaian: (i) Gunakan/Use θ 360° = θ rad 2π rad 65°42’ 360° = θ rad 2π rad θ = 65°42’ × 2π 360° = 1.147 rad (ii) θ = 279°24’ × 2π 360° = 4.876 rad (a) 43°32’ Gunakan/Use 43°32’ 360° = θ rad 2π rad θ = 43°32’ × 2π 360° = 0.76 rad 1 rad j j O j 2 rad 2j j j O
Matematik Tambahan Tingkatan 5 Bab 1 Sukatan Membulat 2 BAB 1 CONTOH (b) 157°16’ Gunakan/Use 157°16’ 360° = θ rad 2π rad θ = 157°16’ × 2π 360° = 2.745 rad (c) 247°54’ Gunakan/Use 247°54’ 360° = θ rad 2π rad θ = 247°54’ × 2π 360° = 4.327 rad (d) 303.68° Gunakan/Use 303.68° 360° = θ rad 2π rad θ = 303.68° × 2π 360° = 5.3 rad (e) 354.74° Gunakan/Use 354.74° 360° = θ rad 2π rad θ = 354.74° × 2π 360° = 6.191 rad 2. Tukarkan setiap yang berikut kepada darjah dan minit. Convert each of the following into degree and minutes. TP 1 (a) 2 3 π rad θ° 360° = 2 3 π rad 2π rad θ° = 360° × 2 3 π 2π = 120° (b) 4 9 rad θ° 360° = 4 9 rad 2π rad θ° = 360° × 4 9 2π = 25.46° atau/or 25°28’ (c) 4.65 rad θ° 360° = 4.65 rad 2π rad θ° = 360° × 4.65 2π = 266.43° atau/or 266°26’ (d) 1.67π rad θ° 360° = 1.67π rad 2π rad θ° = 360° × 1.67π 2π = 300.6° atau/or 300°36’ (e) 0.72π rad θ° 360° = 0.72π rad 2π rad θ° = 360° × 0.72π 2π = 129.6° atau/or 129°36’ 3.2 rad Penyelesaian: Gunakan/Use θ 360° = θ rad 2π rad θ 360° = 3.2 rad 2π rad θ = 360° × 3.2 2π = 183.21’ atau/or 183.35°
BAB Matematik Tambahan Tingkatan 5 Bab 1 Sukatan Membulat 3 1 CONTOH NOTA IMBASAN NOTA IMBASAN NOTA IMBASAN NOTA IMBASAN 1. Panjang lengkok, s, suatu bulatan berkadaran dengan sudut yang tercangkum di pusat bulatan. Maka, semakin besar sudut yang dicangkum, semakin besar panjang lengkok. The arc length, s, of a circle is directly proportional to the size of the angle subtended at the centre of the circle. Hence, the bigger the angle, the longer the arc length. 2. Secara am, kita boleh tulis seperti yang berikut. In general, we can write as follows. q° 360° = q rad 2π rad = panjang lengkok (arc length) panjang lilitan (circumference) dengan panjang lilitan = 2πj dan jejari ialah j. where the circumference = 2πj and the radius is j. 3. Pasangan yang digunakan untuk menyelesaikan masalah bergantung kepada sudut yang diberi dalam darjah atau radian. The pair to solve a problem will depend on whether the angle given is in degree or radian. 4. Jika sudut θ ialah dalam radian, panjang lengkok boleh diberi oleh s = jθ. If the angle θ is in radian, the arc length can be given by s = jθ. 5. Panjang lengkok major adalah lebih panjang daripada panjang lengkok minor. The major arc length is longer than the minor arc length. 6. Kawasan berlorek yang dibatasi oleh lengkok AB dan perentas AB dinamakan tembereng bulatan. The region bounded by the arc length AB and the chord AB is called a segment. 7. Perentas AB dapat diperolehi dengan petua kosinus, AB = j 2 + j2 − 2j 2 kosθ dengan θ dalam darjah atau AB = 2j sin θ 2 The chord AB can be obtained by using cosine rule, AB = j 2 + j2 − 2j2 cosθ with the angle θ is in degree or AB = 2j sin θ 2 . 3. Tentukan panjang lengkok, s bagi setiap bulatan yang diberi. Determine the arc length, s for each of the following given circles. TP 1 Penyelesaian: Oleh kerana sudut ialah dalam radian, kita pilih Since the angle is in radian, we choose θ rad 2π rad = panjang lengkok, s/ arc length, s 2πj 1.2 rad 2π rad = panjang lengkok, s/ arc length, s 2πj Panjang lengkok, s Arc length, s = 1.2 rad × 2π(3.2) 2π rad = 3.84 cm (a) 0.8 rad 2π rad = panjang lengkok, AB Arc length AB 2πj Panjang lengkok, s Arc length, s = 0.8 rad × 2π(7.4) 2π rad = 5.92 cm 1.2 Panjang Lengkok Suatu Bulatan Arc Length of a Circle O j θ A B s O A B 1.2 rad 3.2 cm O A B s 0.8 rad 7.4 cm Kaedah Alternatif Guna/Use AB = jθ = 3.2(1.2) = 3.84 cm
Matematik Tambahan Tingkatan 5 Bab 1 Sukatan Membulat 4 BAB 1 (b) 0.5π rad 2π rad = panjang lengkok, s 2πj Panjang lengkok, s Arc length, s = 0.5π rad × 2π(4.5) 2π rad = 7.07 cm (c) Sudut yang dicangkum = 360° – 125° = 235° Angle subtended θ 360° = panjang lengkok, s 2πj Panjang lengkok, s Arc length, s = 235° × 2π(10.2) 360° = 41.84 cm (d) Sudut yang dicangkum = (2π − 1.3) rad Angle subtended (2π – 1.3) rad 2π rad = panjang lengkok, s 2πj Panjang lengkok, s Arc length, s = (2π – 1.3) rad × 2π(20.5) 2π rad = 102.16 cm (e) Sudut yang dicangkum = (2π − 4.5) rad Angle subtended (2π – 4.5) rad 2π rad = panjang lengkok, s 2πj Panjang lengkok, s Arc length, s = (2π – 4.5) rad × 2π(8.4) 2π rad = 15 cm O s 4.5 cm 0.5fi rad B A O A B s 125fi 10.2 cm 1.3 rad 20.5 cm O A B s 4.5 rad 8.4 cm O A B s
BAB Matematik Tambahan Tingkatan 5 Bab 1 Sukatan Membulat 5 1 CONTOH 4. Tentukan jejari bulatan, j dengan diberikan panjang lengkok, s dan sudut bagi setiap bulatan yang berikut. Determine the radius, j of the given the arc length, s and the angle of each of the following circles. TP 2 Penyelesaian: Oleh kerana sudut ialah dalam radian, kita pilih Since the angle is in radian, we choose θ rad 2π rad = panjang lengkok, s / arc length, s 2πj Apabila diringkaskan, kita dapat When simplified, we get j = s θ = 8.4 1.4 = 6 cm (a) 204° 360° = panjang lengkok, s 2πj j = 20.3 × 360° 204° × 2π = 5.7 cm (b) j = 14.5 2π – 1.6 = 3.1 cm (c) j = 11.5 2π – 0.8π = 3.05 cm (d) 360° – 105° 360° = panjang lengkok, s 2πj j = 14.2 × 360° 255° × 2π = 3.19 cm (e) j = 3 2π – 3 4 π = 0.76 cm 1.4 rad 8.4 cm O j A B O j 204fi 20.3 cm 1.6 rad 14.5 cm O j 0.8fi rad 11.5 cm O j 105fi 14.2 cm O j 3 cm fi rad 3 4 O j
Matematik Tambahan Tingkatan 5 Bab 1 Sukatan Membulat 6 BAB 1 CONTOH 5. Tentukan sudut tercangkum, θ dalam radian, di pusat bulatan dengan diberikan jejari bulatan, j dan panjang lengkok, s bagi setiap bulatan yang berikut. Determine the angle subtended, θ in radian, at the centre of the circle given the radius, j and the arc length, s for each of the following circles. TP 3 Penyelesaian: θ rad 2π rad = panjang lengkok, s / arc length, s 2πj Ini boleh dringkaskan kepada This can be simplified into θ = s j (2π – θ)= 9.8 4.4 θ = 4.06 rad (a) (2π – θ) = 7.2 6.1 θ = 5.1 rad (b) (2π – θ) = 4.8 3.2 θ = 4.78 rad (c) (2π – 2θ) = 2.8 1.8 θ = 2.36 rad (d) (2π – θ) 3 = 3.2 2.5 θ = 2.44 rad (e) 2πj = 6.3 + 8.2 + 11.4 j = 4.12 cm θ = 11.4 4.12 = 2.77 rad Q O fi 4.4 cm 9.8 cm P fi 6.1 cm 7.2 cm O 4.8 cm O 3.2 cm fi = = fi 1.8 cm 2.8 cm O = = = O 2.5 cm 3.2 cm fi fi 8.2 cm 11.4 cm 6.3 cm O
BAB Matematik Tambahan Tingkatan 5 Bab 1 Sukatan Membulat 7 1 CONTOH 6. Tentukan perimeter tembereng bagi setiap bulatan berpusat O yang berikut Determine the perimeter of the segment of each of the following circles with centre O. TP 4 Penyelesaian: Perentas PQ dapat diperoleh dengan petua kosinus, iaitu PQ = j 2 + j 2 − 2j 2 kos θ , dengan θ dalam darjah. The length of the chord PQ can be obtained by using the cosine rule, that is PQ = j2 + j2 − 2j2 cos θ , such that θ is in degree. 2 rad = 2 × 360° 2π = 114.59° Maka/Hence, PQ = 6.22 + 6.22 − 2(6.2)2 kos 114.59° = 10.43 cm Panjang lengkok PQ = jθ = 6.2(2) Arc length PQ = 12.4 cm Perimeter = (12.4 + 10.43) cm = 22.83 cm (a) 0.4π rad = 0.4π × 360° 2π = 72° Maka/Hence, PQ = 9.62 + 9.62 − 2(9.6)2 kos72° = 11.28 cm Panjang lengkok PQ = jθ Arc length PQ = 9.6(0.4π) = 12.06 cm Perimeter = (12.06 + 11.28) cm = 23.34 cm (b) Panjang lengkok minor PQ = 2π(8) – 30 Minor arc length PQ = 20.27 cm ∠POQ = 20.27 8 = 2.53 rad = 145.14° Maka, PQ = 82 + 82 −2(8)2 kos145.14° = 15.27 cm Perimeter = 15.27 + 20.27 Hence = 35.54 cm (c) Panjang lengkok minor PQ = 2π(3) – 4.5(3) Minor arc length PQ = 5.35 cm ∠POQ = 2π − 4.5 = 1.78 rad = 102.17° Maka, PQ = 32 + 32 − 2(3)2 kos102.17° Hence = 4.67 cm Perimeter = 5.35 cm + 4.67 cm = 10.02 cm 2 rad O Q P 6.2 cm Q P O 0.4fi rad 9.6 cm O P Q 8 cm 30 cm 4.5 rad P 3 cm Q O
Matematik Tambahan Tingkatan 5 Bab 1 Sukatan Membulat 8 BAB 1 CONTOH (d) ∠POQ = 150° Panjang jejari/Arc length, j = (7.3 × 360°) (210 × 2π) = 1.99 cm Maka/Hence, PQ = 1.992 + 1.992 − 2(1.99)2 kos150° = 3.84 cm Panjang lengkok = 150° 360° × 2π(1.99) Arc length = 5.2 cm Perimeter = 5.2 cm + 3.84 cm = 9.04 cm (e) Panjang lengkok PQR = 2π(5.3) − 5.3(1.4π) Arc length PQR = 10 cm ∠POQ = (2π − 1.4π) ÷ 2 = 0.3π = 54° Panjang perentas PQ Length of chord PQ = 5.32 + 5.32 − 2(5.3)2 kos 54° = 4.81 cm Perimeter = (4.81 × 2 + 10) cm = 19.62 cm 7. Selesaikan masalah yang melibatkan panjang lengkok. Solve the problems involving the arc lengths. TP 4 Rajah di atas menunjukkan dua bulatan berpusat O dan masing-masing mempunyai jejari 8 cm dan 14 cm. Cari perimeter rantau berlorek itu. The diagram above shows two circles with centre O and with radii 8 cm and 14 cm respectively. Find the perimeter of the shaded region. Penyelesaian: ∠POQ = s j = 4.7 14 = 0.336 rad Panjang lengkok AB Arc length AB = 8(0.336) = 2.69 cm Perimeter = 2.69 + 6(2) + 4.7 = 19.39 cm (a) Rajah di atas menunjukkan sebuah sektor AOB berpusat O dan mempunyai jejari 9.2 cm. Cari perimeter rantau berlorek itu. The diagram above shows a sector AOB with centre O and radius 9.2 cm. Find the perimeter of the shaded region. 0.7 rad = 0.7 × 360° 2π = 40.1° sin 40.1° = AD 9.2 AD = 5.93 cm OD = 9.22 − 5.932 = 7.03cm Panjang lengkok = 9.2 × 0.7 Arc length = 6.44 cm Perimeter = 5.93 + 6.44 + (9.2 – 7.03) = 14.54 cm O P Q fi 12 rad 7.3 cm P O R Q = = 5.3 cm 1.4fi rad P Q O A B 4.7 cm O 0.7 rad D B A 9.2 cm
BAB Matematik Tambahan Tingkatan 5 Bab 1 Sukatan Membulat 9 1 (b) Rajah di atas menunjukkan dua sektor, POQ dan MON. M ialah titik tengah OP. Ungkapkan α dalam sebutan θ jika perimeter MPQR dan ORN adalah sama. The diagram above shows two sectors, POQ and MON. M is the midpoint of OP. Express α in terms of θ if the perimeter of MPQR and ORN are the same. Katakan OM = j cm Let Maka, jθ + 2j + 2jθ = 2j + jα Hence, α = 3θ (c) Rajah di atas menunjukkan dua sektor, AOB dan APB. Diberi nisbah jejari OA kepada PA ialah 3 : 1, AP = 2 cm dan perimeter rantau berlorek ialah 10 cm. The diagram above shows two sectors, AOB and APB. Given that the ratio of radius OA to PA is 3 : 1, AP = 2 cm and the perimeter of the shaded region is 10 cm. (i) Tunjukkan bahawa θ = (5 − 3a) rad. Show that θ = (5 − 3a) rad. (ii) Jika a = 30°, cari nilai θ, dalam radian. If a = 30°, find the value of θ, in radian. (i) Diberi AP = 2 cm, maka OA = 6 cm Given hence 6a + 2θ = 10 θ = 5 − 3a (ii) a = 30° = π 6 rad θ = 5 − 3 π 6 = 3.43 rad (d) Rajah di atas menunjukkan satu semibulatan berpusat O dan satu sektor berpusat P dan ∠APB = π 3 . Cari jejari PA jika perimeter rantau berlorek ialah 35π 6 cm. The diagram above shows a semicircle with centre O and a sector with centre P and ∠APB = π 3. Find the radius PA if the perimeter of the shaded region is 35π 6 cm. π 3 = 60° Maka, AP = AB = j. Hence (e) Rajah di atas menunjukkan sektor KOL berpusat O dan MLN berpusat L. Diberi bahawa OM = ML = 5 cm, dan ∠MON = π 4 rad, cari perimeter rantau berlorek itu. The diagram above shows sector KOL with centre O and sector MLN with centre L. Given that OM = ML = 5 cm, and ∠MON = π 4 rad, find the perimeter of the shaded region. Perimeter = OM + ON + panjang lengkok/arc length MN + MK + panjang lengkok/ arc length KL + ML = 5 + (5 2 − 5) + 5 π 4 + (5 2 − 5) + 5 2 π 4 + 5 = 23.62 cm fi ff R Q O M P N P A B O fi ff - fi 3 O A B P C O N L K M fi 4 = = OA = j 2 = 35π 6 = πj 3 + πj 2 j = 7 cm
Matematik Tambahan Tingkatan 5 Bab 1 Sukatan Membulat 10 BAB 1 CONTOH NOTA IMBASAN NOTA IMBASAN NOTA IMBASAN NOTA IMBASAN 1. Didapati bahawa luas sektor AOB bagi suatu bulatan adalah berkadaran dengan saiz sudut, θ rad yang tercangkum di pusat. It is known that the area of the sector AOB of a circle is proportional to the size of the angle, θ rad which is subtended at the centre. Maka, luas sektor Hence, the area of a sector = θ 2π × πj 2 = θ 2 j 2 2. Jika diungkapkan dalam nisbah, kita dapat luas sektor luas bulatan = θ 2π = q° 360° If expressed in the ratio form, we get area of sector area of circle = θ 2π = θ° 360° 3. Luas tembereng/Area of a segment = luas sektor – luas segi tiga area of sector − area of triangle = θ 2 j 2 – 1 2 j 2 sin θ 4. Dari apa yang telah dipelajari, hubungan secara am antara sudut, panjang lengkok dan luas sektor adalah seperti berikut. From the earlier studies, the general relationship between the angles, arc length and the area of sector are as follows. q° 360° = q rad 2π rad = panjang lengkok, s 2πj = luas sektor luas bulatan θ° 360° = θ rad 2π rad = arc length, s 2πj = area of sector area of circle 1.3 Luas Sektor Suatu Bulatan Area of Sector of a Circle 8. Tentukan luas sektor bagi setiap bulatan yang berikut. Berikan jawapan anda kepada dua tempat perpuluhan. Determine the area of the sector for each of the following circles. Give your answer to two decimal places. TP 3 Penyelesaian: Sudut yang tercangkum = (2π – 2.3) rad Angle subtended Maka, luas = 2π – 2.3 2 (6)2 = 71.70 cm2 Hence, the area (a) Sudut yang tercangkum = 5 6 π rad Angle subtended Maka, luas = 5 6 π 2 (3.5)2 = 16.04 cm2 Hence, the area O A j θ B O A B j θ O Q P 2.3 rad 6 cm O P Q 3.5 cm 5 6 fi rad
BAB Matematik Tambahan Tingkatan 5 Bab 1 Sukatan Membulat 11 1 (b) AOB ialah diameter dan panjangnya ialah 10.4 cm, ∠AOC = 2∠BOC. AOB is the diameter and the length is 10.4 cm, ∠AOC = 2∠BOC. Sudut yang tercangkum = 2 3 π rad Angle subtended Maka, luas = 2 3 π 2 ( 5.2)2 Hence, the area = 28.32 cm2 (c) Sudut yang tercangkum = (360 – 215)° Angle subtended = 145° Maka, luas = 145° 360° × π × ( 7.8)2 Hence, the area = 76.98 cm2 (d) Sudut yang tercangkum = 6 2.5 = 2.4 rad Angle subtended Maka, luas = 2.4 2 × (2.5)2 = 7.50 cm2 Hence, the area (e) Jejari sektor = 10 3π 5 Radius of the sector = 50 3π cm Maka, luas = 3 5 π 2 50 3π 2 Hence, the area = 26.53 cm2 O A B C P Q O 7.8 cm 215fi O B A 2.5 cm 6 cm 10 cm 3 5 fi rad B O A
Matematik Tambahan Tingkatan 5 Bab 1 Sukatan Membulat 12 BAB 1 CONTOH 9. Tentukan jejari bagi sektor bulatan yang berikut. Beri jawapan kepada dua tempat perpuluhan. Determine the radius of the following sectors of circles. Give the answers to two decimal places. TP 4 Diberi luas sektor ialah 20.5 cm2 . Given the area of the sector is 20.5 cm2. Penyelesaian: Gunakan luas sektor luas bulatan = θ 2π Use area of sector area of circle = θ 2p 20.5 πj 2 = 1.6 2π j 2 = 20.5 × 2 1.6 j = 5.06 cm (a) Diberi luas sektor ialah 10.21 cm2 . Given the area of the sector is 10.21 cm2. Gunakan /Use luas sektor luas bulatan = θ 2p area of sector area of circle = θ 2p 10.21 πj 2 = 2π – 3.5 2π j 2 = 10.21 × 2 2p – 3.5 j = 2.71 cm (b) Diberi luas sektor ialah 42 cm2 dan panjang lengkok PQ ialah 12 cm. Given that the area of the sector is 42 cm2 and the arc length of PQ is 12 cm. Gunakan/Use panjang lengkok 2πj = luas sektor luas bulatan arc length 2πj = area of sector area of circle 12 2πj = 42 πj 2 j = 42 × 2 12 = 7 cm (c) Diberi luas sektor ialah 85 cm2 dan panjang lengkok PQ ialah 38 cm. Given that the area of the sector is 85 cm2 and the arc length of PQ is 38 cm. Gunakan/Use 38 2πj = πj2 – 85 πj 2 j = 2(πj2 – 85) 38 2(πj2 – 85) = 38j πj2 – 19j – 85 = 0 j = 19 ± (–19)2 – 4π(–85) 2π j = 9.04 cm 1.6 rad 3.5 rad 12 cm P Q O 38 cm O P Q
BAB Matematik Tambahan Tingkatan 5 Bab 1 Sukatan Membulat 13 1 CONTOH (d) Diberi luas sektor ialah 7.5 cm2 . Given the area of the sector is 7.5 cm2. Gunakan/ Use luas sektor/area of sector luas bulatan/area of circle = θ 360° 7.5 πj 2 = 88 360° j 2 = 7.5 × 360° 88π j = 3.13 cm (e) Diberi luas sektor ialah 30 cm2 . Given the area of the sector is 30 cm2. Gunakan/ Use 30 πj 2 = 12.5 2πj j = 30 × 2 12.5 = 4.8 cm 10. Tentukan sudut tercangkum, θ, dalam radian di pusat bulatan bagi yang berikut. Berikan jawapan anda kepada dua tempat perpuluhan. Determine the angle subtended, θ, in radian at the centre of the circle for each of the following. Give your answer to two decimal places. TP 4 Diberi luas sektor ialah 18 cm2 Given the area of the sector is 18 cm2. Penyelesaian: Gunakan luas sektor luas bulatan = θ 2π Use area of sector area of circle = θ 2p 18 π(9.3)2 = θ 2π θ = 18 × 2 (9.3)2 = 0.42 radian (a) Diberi luas sektor ialah 33 cm2 . Given the area of the sector is 33 cm2. Gunakan/ Use luas sektor luas bulatan = θ 2p area of sector area of circle = θ 2p 33 π(4.8)2 = 2π – θ 2π 66π 23.04π = 2π – θ 2.865 = 2π – θ θ = 2p − 2.865 = 3.42 radian 88fi 12.5 cm O 9.3 cm fi O fi 4.8 cm O
Matematik Tambahan Tingkatan 5 Bab 1 Sukatan Membulat 14 BAB 1 (b) Diberi luas sektor ialah 46 cm2 . Given the area of the sector is 46 cm2. Gunakan/Use luas sektor luas bulatan = panjang lengkok 2πj area of sector area of circle = arc length 2πj 46 πj 2 = 12 2πj j = 46 × 2 12 = 7.67 cm s = jθ θ = 12 7.67 = 1.56 radian (c) Diberi luas sektor ialah 30 cm2 dan perimeter ialah 15 cm. Ungkapkan θ dalam sebutan jejari, j. Given the area of the sector is 30 cm2 and the perimeter is 15 cm. Express θ in terms of the radius, j. 2j + jθ = 15 …… ➀ 1 2 j 2 θ = 30 …… ➁ ➁ ÷ ➀ 1 2 j 2 q 2j + jq = 2 jq 2(2 + q) = 2 jθ = 8 + 4θ θ = 8 j – 4 (d) Diberi luas sektor ialah 43 cm2 . Given the area of the sector is 43 cm2. 1 2 × 82 × θ = 43 θ = 1.34 radian (e) Diberi luas sektor ialah 54 cm2 . Given the area of the sector is 54 cm2. Gunakan/Use luas sektor luas bulatan = panjang lengkok 2πj = θ 2π area of sector area of circle = arc length 2πj = θ 2π 54 πj 2 = 21 2πj = θ 2π j = 54 × 2 21 = 5.14 cm 2p – θ = 21 5.14 θ = 2.2 radian 12 cm j fi 8 cm fi fi 21 cm O
BAB Matematik Tambahan Tingkatan 5 Bab 1 Sukatan Membulat 15 CONTOH 1 11. Tentukan luas tembereng yang berlorek bagi suatu bulatan yang berikut. Berikan jawapan anda kepada dua tempat perpuluhan. Determine the area of the shaded segment for each of the following circles. Give your answer to two decimal places. TP 4 Penyelesaian: Luas berlorek = luas sektor POQ – luas segi tiga POQ Shaded area = area of sector POQ − area of triangle POQ Luas sektor/Area of sector = 1 2 j 2 θ = 1 2 (5.3)2 (2.3) = 32.3 cm2 2.3 rad = 2.3 2π × 360° = 131.78° Luas segi tiga/Area of triangle = 1 2 ×(5.3)2 × sin131.78° = 10.47 cm2 Maka, luas tembereng Hence, the area of the segment = 32.3 − 10.47 = 21.83 cm2 (a) Luas berlorek/Shaded area = luas sektor POQ – luas segi tiga POQ area of sector POQ – area of triangle POQ Luas sektor = 1 2 j 2 2π − 4π 3 Area of sector = 1 2 (8.1)2 2π 3 2 = 68.71 cm2 2π 3 rad = 120° Luas segi tiga = 1 2 × (8.1)2 × sin 120° Area of triangle = 28.41 cm2 Maka, luas tembereng = 68.71 − 28.41 Hence, the area of segment = 40.3 cm2 (b) Diberi POR dan SOQ adalah garis lurus yang melalui pusat O. Given that POR and SOQ are straight lines passing through the centre O. Luas sektor POS = luas sektor QOR area of sector POS = area of sector QOR 1 2 j2 θ = 1 2 (3.5)2 (π − 1.4) = 10.67 cm2 1.4 rad= 1.4 2π × 360° = 80.21° Luas segi tiga = 1 2 × (3.5)2 × sin80.21° Area of triangle = 6.04 cm2 Maka, luas tembereng = 2(10.67 − 6.04) Hence, the area of segment = 9.26 cm2 (c) Perentas AB = 5.52 + 5.52 − 2(5.5)2 kos50° Chord AB = 4.65 cm Kos/ cos ∠AOB = 4.32 + 4.32 − 4.652 2(4.3)2 ∠AOB = 65.46° = 1.14 rad 50° = 0.87 rad Luas tembereng = 1 2 (5.5)2 (0.87) − 1 2 (5.5)2 Area of segment sin50° + 1 2 (4.3)2 (1.14)− 1 2 (4.3)2 sin65.46° = 3.7 cm2 O Q P 5.3 cm 2.3 rad P Q 8.1 cm 4 3 fi rad O P Q S R O 3.5 cm 1.4 rad 4.3 cm 5.5 cm B A P O 50fi
Matematik Tambahan Tingkatan 5 Bab 1 Sukatan Membulat 16 BAB 1 CONTOH 1.4 Aplikasi Sukatan Membulat Application of Circular Measures 12. Selesaikan masalah yang melibatkan sukatan membulat yang berikut. Solve the following problems involving the circular measures. TP 5 Rajah di atas menunjukkan sekeping kek yang telah dipotong kepada 8 keping yang bersaiz sama daripada satu kek yang bulat berpusat O dan berjejari 18 cm. Cari The diagram above shows a piece of cake that is cut into 8 pieces of the same size of a round cake at centre O and with a radius of 18 cm. Find (i) sudut θ dalam radian. the angle θ in radian. (ii) panjang lengkok PQ. the arc length PQ. (iii)jumlah luas permukaan kepingan kek itu jika tebal kek ialah 5 cm. the total surface area of the cake if the thickness is 5 cm. Penyelesaian: (i) 8θ = 2p. Maka/Hence, θ = 1 4 p (ii) Panjang lengkok PQ = jθ Arc length PQ = 18 1 4 p = 14.14 cm (iii) Luas sektor/Area of sector = 1 2 × 182 × 1 4 p = 127.23 cm2 Luas permukaan/Total surface area = 2(18 × 5) + 14.14 × 5 + 2(127.23) = 505.16 cm2 (a) Rajah di atas menunjukkan satu corak yang terdiri daripada satu sektor major OPRQ berpusat O dan sebuah semibulatan POQ berpusat S dengan jejari 8 cm. Tunjukkan bahawa luas rantau berlorek ialah 32(π + 2) cm2 . The diagram above shows a pattern which is made up of a major sector OPRQ with centre O and a semicircle POQ of centre S with a radius of 8 cm. Show that the shaded area is 32(π + 2) cm2. OP = 8 2 cm Luas sektor major = 270° 360° × π(8 2)2 Area of major sector = 96π cm2 Luas segi tiga = 1 2 × (8 2)2 Area of triangle = 64 cm2 Jumlah luas = 96π + 64 Total area Luas semibulatan = π(8)2 = 64π Area of semiclrcle Luas rantau berlorek = 96π + 64 − 64π Area of shaded region = 32π + 64 = 32(π + 2) cm2 P O Q fi R P Q O S
BAB Matematik Tambahan Tingkatan 5 Bab 1 Sukatan Membulat 17 1 (b) Rajah di atas menunjukkan sebuah kipas yang dibina daripada dua keping kain, P dan Q yang berlainan. Cari beza luas kain P dan Q. The diagram above shows a fan made up of two different pieces of clothes, P and Q. Find the difference of the area of the clothes P and Q. Luas kain Q Area of cloth Q = 1 2 × (20)2 × 1.5 − 1 2 × (15)2 × 1.5 = 300 cm2 – 168.75 cm2 = 131.25 cm2 Luas kain P Area of cloth P = 1 2 × (15)2 × 1.5 − 1 2 × (10)2 × 1.5 = 168.75 cm2 – 75 cm2 = 93.75 cm2 Beza luas Difference of area = 131.25 cm2 − 93.75 cm2 = 37.5 cm2 (c) Rajah di atas menunjukkan dua sayap yang serupa bagi sebuah kipas. OPQ, OTR dan QSR adalah tiga semibulatan dengan diameter yang ditunjukkan. Cari The diagram above shows two similar wings of a fan. OPQ, OTR and QSR are three semicircles with the diameters as shown. Find (i) perimeter keseluruhan dalam sebutan π. the total perimeter of the wings in terms of π. (ii) luas keseluruhan dalam sebutan π. the total area in terms of π. (i) Lilitan/ Circumference OPQ = 9π Lilitan/ Circumference QSR = 5π Lilitan/ Circumference OTR = 4π Perimeter = 2(9π + 5π + 4π) = 36π cm (ii) Luas semibulatan OPQ = 1 2 × (9)2 × π Area of semicircle OPQ Luas semibulatan QSR = 1 2 × (5)2 × π Area of semicircle QSR Luas semibulatan OTR = 1 2 × (4)2 × π Area of semicircle OTR Luas keseluruhan = 2 × 1 2 π × (81 + 25 – 16) Total area = 90π cm2 1.5 rad 5 cm 10 cm 5 cm Q P O R 8 cm 10 cm T S Q P O
Matematik Tambahan Tingkatan 5 Bab 1 Sukatan Membulat 18 BAB 1 PRAKTIS PRAKTIS SPM SPM 11 Kertas 1 1. Rajah menunjukkan pandangan hadapan sebahagian mural berbentuk segi empat tepat 10 m × 8 m. ABP dan DCP adalah dua sektor serupa berpusat P dan BC adalah selari dengan FE. Bahagian berlorek perlu dicatkan. The diagram shows a front view of a rectangular mural of 10 m × 8 m. ABP and DCP are two similar sectors with centre P and BC is parallel to FE. The shaded region needs to be painted. Cari luas bahagian berlorek itu. Find the area of the shaded region. 42 = 52 + 52 – 2(5)(5) kos θ kos θ = 52 + 52 – 42 2(5)(5) θ = 47.16° ∠APB= ∠CPD = 180° − 47.16° 2 = 66.42° Luas sektor APB = 66.42° 360° × π(5)2 = 14.49 cm2 Area of sector APB Luas ∆BCP = 1 2 (5)(5)sin 47.16° = 9.17 cm2 Luas rantau berlorek = 10 × 8 − 9.17 – 2(14.49) Area of shaded region = 41.85 m2 2. Rajah menunjukkan sebuah sektor berpusat O. The diagram shows a sector with centre O. Diberi bahawa panjang perentas ialah 8 cm, cari Given the length of the chord is 8 cm, find (a) sudut θ, dalam radian, the angle θ, in radian, (b) luas rantau berlorek itu. the area of the shaded region. (a) 82 = 152 + 152 – 2(15)2 kos θ kos θ = 152 + 152 – 82 2(15)(15) θ = 30.93° 30.93° 360° × 2π = 0.54 rad (b) Luas tembereng Area of segments = 1 2 (15)2 (0.54)− 1 2 (15)2 sin30.93° = 2.93 cm2 3. Rajah menunjukkan satu bulatan dengan pusat O dan jejari 7 cm. Diberi bahawa panjang lengkok minor AB ialah 10 cm, cari The diagram shows a circle with centre O and radius 7 cm. Given that the minor arc length AB is 10 cm, find (a) sudut θ, dalam radian, the angle θ, in radian, (b) luas sektor minor. the area of the minor sector. (a) 10 = 7(2π − θ) 7θ = 14p − 10 θ = 4.85 rad (b) Luas = 1 2 j 2 θ = 1 2 (7)2 (2p − θ) Area = 1 2 (7)2 10 7 = 35 cm2 2014 2015 2016 A P D B C F E 4 m 10 m 8 m A P D B C F E 4 m 10 m 8 m 5 m fi P O Q fi 15 cm O B A fi
BAB Matematik Tambahan Tingkatan 5 Bab 1 Sukatan Membulat 19 1 Kertas 2 1. Rajah menunjukkan sebuah semibulatan ABC berpusat O dan berjejari 12 cm. BAD ialah satu sektor berpusat A dan BE adalah berserenjang dengan AC dengan keadaan E adalah titik tengah OC. The diagram shows a semicircle with centre O and radius 12 cm. BAD is a sector with centre A and BE is perpendicular to AC such that E is the midpoint of OC. Cari/Find (a) sudut BOD dalam radian, the angle of BOD in radian, (b) panjang lengkok BD, the arc length BD, (c) luas berlorek rantau dalam cm2 . the shaded area in cm2. (a) kos θ = 6 12 = 1 cos θ 2 θ = 60° = π 3 rad (b) kos 30° = 18 AB cos 30° AB = 20.78 cm Panjang lengkok BD = 20.78 π 6 Arc length BD = 10.88 cm (c) Luas sektor BAD = 1 2 (20.78)2 π 6 Area of sector BAD = 113.05 cm2 Luas ∆ABE = 1 2 (18)(20.78) sin30° Area = 93.51 cm2 Luas rantau berlorek = 19.54 cm2 Area of shaded region 2. Rajah (a) menunjukkan topi lampu dan rajah (b) menunjukkan bentangannya dalam bentuk dua sektor berpusat O yang dilukis pada sekeping kad berbentuk segi empat tepat. Diagram (a) shows the cover of a lamp and diagram (b) shows the net of the cover drawn on a piece of a rectangular cardboard. (a) Hitung nilai minimum, dalam cm, bagi panjang dan lebar kad kepada integer yang terdekat. Calculate the minimum value, in cm, of length and breadth, of the card to the nearest integer. (b) Seterusnya, cari luas bagi kad yang tidak digunakan. Hence, find the area of the unused card. (a) PQ2 = 152 + 52 = 250 PQ = 5 10 cm Panjang lengkok AB = 2π(7.5) Arc length AB = 15π Panjang lengkok CD = 2π(12.5) Arc length CD = 25π Maka, 15π = jq ….. ➀ 25π = (j + 5 10 )q ….. ➁ ➁ ÷ ➀ 5 3 = j + 5 10 j 2j = 15 10 j = 15 10 2 cm q = 15π × 2 15 10 = 1.99 rad = 113.84° kos ∠EOC = BO OC 15 10 2 + 5 10 kos(180° − 113.84°) = EO EO = 15.98 cm Panjang kad = 15.98 + 15 10 2 + 5 10 Length of card = 55.5 cm ≈ 56 cm Lebar kad / Breadth of card = j + 5 10 = 15 10 2 + 5 10 = 39.53 cm ≈ 40 cm (b) Luas kad tidak digunakan Area of unused card = 56 × 40 − 39.532 2 × 1.99 + 1 2 15 10 2 2 (1.99) = 1244.88 cm2 2014 B A O E D C B A 12 cm O E D C 6 cm fi fi 2 2015 15 cm 25 cm 15 cm O (a) (b) 15 cm 12.5 cm 7.5 cm P Q O fi E j B D A C 5 10 √
Matematik Tambahan Tingkatan 5 Bab 1 Sukatan Membulat 20 BAB 1 3. Rajah menunjukkan sebuah logo syarikat. ABC ialah segi tiga sama sisi dalam satu bulatan yang dilukis dengan keadaan AB, BC dan AC adalah tangennya. Terdapat tiga tembereng, P, Q dan R serupa dilukis dengan keadaan A, B dan C masingmasing adalah pusat sektornya. The diagram shows a logo of a company. ABC is an equilateral triangle in a circle that is drawn such that AB, BC and AC as tangents. The three similar segments, P, Q and R are drawn such that A, B and C are the centres respectively. Jika jejari bulatan ialah 3 cm, cari If the radius of the circle is 3 cm, find (a) sudut ABC dalam radian, the angle ABC in radians, (b) panjang lengkok AC, the arc length AC, (c) luas rantau berlorek itu. the area of the shaded region. (a) 60° = π 3 rad (b) AC = 2 3 tan 30° = 10.39 cm (c) Luas satu tembereng Area of a segment = 1 2 (10.39)2 π 3 − 1 2 (10.39)2 sin60° = 9.78 cm2 Luas bulatan = π(3)2 Area of circle Jumlah luas = 9π + 3(9.78) Total area = 57.61 cm2 4. Rajah menunjukkan satu sektor SRQ berpusat R dan TS ialah tangen pada S. The diagram shows a sector SRQ with centre R and TS is a tangent at S. Jika panjang lengkok PQ ialah 5 cm, cari If the arc length PQ is 5 cm, find (a) sudut PRQ dalam radian, the angle PRQ in radian, (b) luas rantau berlorek itu. the area of the shaded region. (a) 5 = 11q q = 5 11 rad = 0.45 rad (b) Luas sektor = 1 2 (11)2 (0.45) Area of sector = 27.2 cm2 tan 50° = TS 11 TS = 13.11 cm Luas/Area ∆TSR = 1 2 (11)(13.11) = 72.1 cm2 50° = 50° 360° × 2p = 0.87 rad Luas sektor SRP = 1 2 (11)2 (0.87) Area of sector SRP = 52.6 cm2 Luas berlorek = 72.1 − 52.6 + 27.2 Shaded area = 46.7cm2 2018 A B C O Q R P A B C O P Q 60fi 30fi 2019 S R Q P T 50fi 11 cm S R Q P T 50fi 11 cm 5 cm ff Praktis SPM Ekstra
BAB Matematik Tambahan Tingkatan 5 Bab 1 Sukatan Membulat 21 Rajah menunjukkan sebuah bulatan berpusat O dan 1 berjejari 10 cm. PQ dan SR adalah dua perentas yang selari dengan PQ = 10 cm dan ∠SOR = 120°. The diagram shows a circle with centre O and radius 10 cm. PQ and SR are two parallel chords such that PQ = 10 cm and ∠SOR = 120°. P S Q R O P S Q R O 10 cm 120fi 10 cm Sudut Sudut KBAT KBAT KBAT Ekstra Quiz 1 (a) Cari sudut SOP dalam radian. Find the angle of SOP in radian. (b) Tentukan luas rantau berlorek itu. Determine the area of the shaded region. (c) Tunjukkan bahawa perimeter rantau berlorek ialah 101 + 3 + π 3 cm. Show that the perimeter of the shaded region is 101 + 3 + π 3 cm. (a) PQ2 = 102 + 102 − 2(10)2 kos POQ kos POQ = 102 + 102 − 102 2(100) ∠POQ = 60° ∠SOP = 120°−60° 2 = 30° = π 2 rad (b) Luas tembereng PQ /Area of segment PQ = 1 2 (10)2 π 3 − 1 2 (10)2 sin 60° = 9.06 cm2 Luas tembereng SR / Area of segment SR = 1 2 (10)2 2π 3 − 1 2 (10)2 sin 120° = 61.42 cm2 Maka, luas berlorek Hence, the shaded area = 61.42 – 9.06 = 52.36 cm2 (c) Panjang lengkok SP = QR Arc length SP = QR 10 π 6 = 5π 3 SR2 = 102 + 102 − 2(10)2 kos 120° SR = 10 3 Perimeter = 10 + 2 5π 3 + 10 3 = 101+ π 3 + 3
Matematik Tambahan Tingkatan 5 Bab 2 Pembezaan 22 BAB 2 NOTA IMBASAN NOTA IMBASAN NOTA IMBASAN NOTA IMBASAN BAB 2 Pembezaan Differentiation 2.1 Had dan Hubungannya dengan Pembezaan Limit and its Relation to Differentiation 1. Apabila x dalam ungkapan f(x) = 1 x + 2 menghampiri 2, nilai 1 x + 2 menghampiri 1 4 , iaitu had 1 x + 2 = 1 4 . x → 2 When x in the expression f(x) = 1 x + 2 approaches 2, the value of 1 x + 2 approaches 1 4 , that is lim 1 x + 2 = 1 4 . x → 2 2. Apabila x menghampiri a, dengan keadaan x ≠ a, maka had bagi f(x) ialah L dan ditulis sebagai had f(x) = L. x → a When x approaches to a, where x ≠ a, then the limit for f(x) is L and written as lim f(x) = L. x → a 3. Langkah-langkah menentukan had f(x), dengan x → a keadaan a ∈ R adalah seperti berikut: Steps to determine lim f(x), where a ∈ R are as follows: x → a (a) Jika f(a) ≠ 0 0 , maka nilai had f(x) = f(a). Contoh: x → a had (x + 2) = 1 + 2 = 3. x → 1 If f(a) ≠ 0 0 , then the value of lim f(x) = f(a). For example: x → a lim (x + 2) = 1 + 2 = 3. x → 1 (b) Jika f(a) = 0 0 , maka f(x) perlu diringkaskan dengan pemfaktoran, atau merasionalkan pengangka atau penyebut fungsi itu. Contoh: had x2 – 16 x – 4 = x → 4 had (x – 4)(x + 4) (x – 4) = had(x + 4) = 8 x → 4 x → 4 If f(a) = 0 0 , then f(x) needs to be simplified by factorisation, or rationalising the numerator or the denominator of the function. For example: lim x2 – 16 x – 4 = lim (x – 4)(x + 4) (x – 4) = lim (x + 4) = 8 x → 4 x→ 4 x → 4 (c) Jika f(a) = ∞ ∞, maka setiap sebutan dalam f(x) perlu dibahagikan dengan kuasa tertinggi x. Contoh: had x2 – 16 x – 4 = had 1 + x2 2 – x2 = had 1 x2 + x2 x2 2 x2 – x2 x2 = x → ∞ x → ∞ x → ∞ 0 + 1 0 – 1 = −1 If f(a) = ∞ ∞ , then each term in f(x) needs to be divided by the highest power of x. For example: lim x2 – 16 x – 4 = x → ∞ lim 1 + x2 2 – x2 = lim 1 x2 + x2 x2 2 x2 – x2 x2 = 0 + 1 0 – 1 = −1 x → ∞ x → ∞ 4. Jika y = f(x), dy dx ditakrifkan sebagai dy dx = had δy δx , dx → 0 dengan δx ialah tokokan kecil dalam x dan δy adalah tokokan kecil dalam y yang sepadan. If y = f(x), dy dx is defined as dy dx = lim δy δx , where δx is the small dx → 0 change in x and δy is the small change in the corresponding y. 5. Juga, dy dx = had dy dx . δx → 0 Also, dy dx = lim dy dx . δx → 0 6. dy dx disebut sebagai terbitan pertama y terhadap x. dy dx is called the first derivative of y with respect to x. 7. dy dx juga boleh ditulis sebagai f’(x) apabila y = f(x). dy dx can be written as f’(x) when y = f(x).
BAB Matematik Tambahan Tingkatan 5 Bab 2 Pembezaan 23 2 CONTOH 1. Cari had bagi setiap fungsi berikut apabila x → 0. Find the limit for each of the following functions when x → 0. TP 1 x2 +3x −5 Penyelesaian: had x2 + 3x – 5 x → 0 lim x2 + 3x – 5 x → 0 = 02 + 3(0) – 5 = −5 (a) k 2k + x had = k x → 0 2k + x limx → 0 = 1 2 (b) 4 – 2x + x2 2. Cari nilai bagi setiap yang berikut. Find the value for each of the following. TP 2 (a) had (6 – 2x) x → 0 limx → 0 (6 – 2x) = 6 (b) had 1 3 x x → ∞ lim x → ∞ 1 3 x = 0 (c) had 2 x2 x → 0 – x + 3 lim x → 0 2 x2 – x + 3 = 2 3 (d) had 2x2 – 8 x → 2 x – 2 limx → 2 = had 2(x2 – 4) x → 2 x – 2 limx → 2 = had 2(x + 2)(x – 2) x → 2 x – 2 limx → 2 = 8 (e) had 2x2 – 6x – 8 x2 x → 2 – x + 3 limx → 2 = had 2(x2 – 3x – 4) x → 2 x + 1 limx → 2 = had 2(x – 4)(x + 1) x → 2 x + 1 limx → 2 = –4 (f) had 1 + 2x x → ∞ 4 + x limx → ∞ = had 1 x + 2x x 4 x + x x x → ∞ limx → ∞ = 0 + 2 0 + 1 = 2 CONTOH had= 4 – 2x + x2 x → 0 limx → 0 = 4 = 2 (i) had (2 – x) x → 1 lim (2 – x) x → 1 Penyelesaian: (i) had (2 – x) = 2 – 1 x → 1 limx → 1 = 1 (iii) had (0.02)x = 0 x → ∞ limx → ∞ (ii) had x2 – 2x – 3 x → 3 x – 3 lim x2 – 2x – 3 x → 3 x – 3 (ii) had x2 – 2x – 3 x – 3 = had (x – 3)(x + 1) x → 3 x → 3 x – 3 limx → 3 limx → 3 = had (x + 1) = 4 x → 3 limx → 3 (iii) had (0.02)x x → ∞ lim (0.02)x x → ∞
Matematik Tambahan Tingkatan 5 Bab 2 Pembezaan 24 BAB 2 3. Cari dy dx dengan menggunakan prinsip pertama bagi setiap fungsi y = f(x) yang berikut. Find dy dx by using the first principle for each of the following functions y = f(x). TP 2 (a) y = f(x) = 4 − x2 δy = f(x + δx) = 4 − x2 = 4 – (x + δx)2 – 4 + x2 = 4 – x2 – 2xδx – (δx)2 – 4 + x2 = −(2x + δx)δx δy δx = –(2x + δx) had δy δx = dy dx = –2x δx → 0 (b) y = f(x) = (x – 2)(x + 1) x2 – x – 2 δy = (x + δx)2 – (x + δx) – 2 – x2 + x + 2 = x2 + 2x·δx + (δx)2 – x − δx – x2 + x = 2xδx – δx + (δx)2 δy = [(2x – 1) + δx]δx δy δx = (2x – 1) + δx had δy δx = dy dx = 2x – 1 δx → 0 (c) y = f(x) = 1 x2 δy = 1 (x + δx)2 = 1 x2 = x2 – (x + δx)2 x2 (x + δx)2 = –2xδx – (δx)2 x2 (x + δx)2 δy δx = –2x – δx x2 (x + δx)2 had δy δx = dy dx = –2x x4 = –2 x3 δx → 0 (d) y = f(x) = 1 x + 1 δy = 1 (x + δx + 1) – 1 (x + 1) = (x + 1) – x – 1 – δx (x + 1)(x + 1 + δx) = –δx (x + 1)(x + 1 + δx) δy δx = –1 (x + 1)(x + 1 + δx) had δy δx = –1 (x + 1)(x + 1) = – 1 (x + 1)2 δx → 0 CONTOH (i) y = 2x + 3 Penyelesaian: (i) y + δy = 2(x + δx) + 3 2x + 3 + δy = 2x + 2δx + 3 δy = 2δx δy δx = 2 Maka/Hence dy dx = had δy δx / dy dx = lim δy δx → 0 δx → 0 δx = had 2 δx → 0 = 2 (ii) y + δy = 2 x + δx (y + δy)(x + δx) = 2 yx + yδx + xδy + δx·δy = 2 xδy + δx·δy = –yδx (x + δx)δy = –yδx δy δx = –2 x(x + δx) Maka/Hence dy dx = had δy δx / dy dx = lim δy δx δx → 0 δx → 0 = had = –2 x(x + δx) = –2 x(x) = –2 x2 δx → 0 (ii) y = 2 x
BAB Matematik Tambahan Tingkatan 5 Bab 2 Pembezaan 25 NOTA IMBASAN NOTA IMBASAN 2 NOTA IMBASAN NOTA IMBASAN 2.2 Pembezaan Peringkat Pertama The First Derivative 1. JIka y = k, dengan keadaan k ialah pemalar, maka dy dx = 0. If y = k, where k is a constant, then dy dx = 0. 2. Jika y = axn , dengan keadaan a dan n adalah pemalar, maka dy dx = anxn − 1. If y = axn , where a and n are constants, then dy dx = anxn − 1. 3. Jika f(x) and g(x) adalah fungsi x dan y = f(x) ± g(x), maka dy dx = d dx [f(x) ± g(x)] = df(x) dx ± dg(x) dx . Fungsi ini dibezakan secara sebutan demi sebutan. If f(x) and g(x) are functions of x and y = f(x) ± g(x), then dy dx = d dx [f(x) ± g(x)] = df(x) dx ± dg(x) dx . The function is differentiated term by term. 4. Jika y = g(u) dan u = h(x), maka pembezaan y terhadap x diberi oleh f'(x) = g' (u) × h'(x), iaitu dy dx = dy du × du dx . If y = g(u) and u = h(x), then differentiate y with respect to x is given by f‘(x) = g'(u) × h'(x) , that is dy dx = dy du × du dx . 5. Jika u dan v adalah fungsi x dan y = uv, maka dy dx = d dx (uv) = u dv dx + v du dx . If u and v are functions of x and y = uv, then dy dx = d dx (uv) = u dv dx + v du dx . 6. Jika u dan v adalah fungsi x dan y = u v , maka dy dx = dy dx u v = v du dx – u dv dx v2 . If u and v are functions in x and y = u v , then dy dx = dy dx u v = v du dx – u dv dx v2 . CONTOH (i) 5 Penyelesaian: (i) d(5) dx = 0 (ii) 4x2 (ii) d(4x2 ) dx = 4(2)x2 – 1 = 8x (iii) 3 x3 = 3x – 3 d(3x –3) dx = 3(–3)x–3 – 1 = –9x–4 (iii) 3 x3 4. Bezakan yang berikut terhadap x. Differentiate the following with respect to x. TP 2 Tukarkan kepada bentuk axn . Change to the form axn .
Matematik Tambahan Tingkatan 5 Bab 2 Pembezaan 26 BAB 2 CONTOH (a) 1 2 d dx 1 2 = 0 (b) x d dx ( x ) = d dx x 1 2 = 1 2 x (c) 8x4 5 d dx 8x4 5 = 32 5 x3 (d) –6 (2x)3 d dx –6 (2x)3 = d dx –6 8x3 = d dx –3 4 x–3 = 9 4x4 (e) –5x 3x 4 d dx –5x 3x4 = d dx –5 3 x –3 = 5 x 4 (f) 2x x d dx 2x x = d dx ( 2x – 1 2 ) = – 2 2 x – 3 2 = – 2 2x x 5. Cari f’(x) bagi setiap fungsi yang berikut. Find f' (x) for each of the following functions. TP 2 (a) f(x) = 3 x (b) f(x) = 10 3x –3 (c) f(x) = –5 4x –2 f(x) = 1 2 x5 Penyelesaian: f’(x) = 1 2 (5)x4 = 5 2 x4 f(x) = 3 x = 3x 1 – 2 f’(x) = 3 – 2 x 3 – 2 = –3 2x x f(x)= –5 4x–2 = –5 4 x3 f’(x) = –5 2 x f(x) = 10 3x –3 = 10 3 x3 f’(x) = 10x2 Tip df(x) dx = f'(x)
BAB Matematik Tambahan Tingkatan 5 Bab 2 Pembezaan 27 2 6. Cari nilai dy dx bagi setiap fungsi yang berikut. Find the value of dy dx for each of the following functions. TP 2 (a) y = 3 x3 , x = 4 (b) y = 7x 2 x , x = 9 (c) y = –8 5x –2 , x = 1 – 8 7. Diberi fungsi polinomial yang berikut, cari terbitan pertama terhadap x bagi setiap yang berikut. Given the following polynomial functions, find the first derivative with respect to x of each of the following. TP 3 CONTOH y = x5 2x2 , x = 2 Penyelesaian: y = x5 2x2 = 1 2 x3 dy dx = 1 2 (3)x2 = 3 2 x2 Apabila/When x = 2 dy dx = 3 2 (2)2 = 6 Tip Ungkapkan kepada bentuk y = axn . Express to the form of y = axn . y = 3x 3 2 dy dx = 9 2 x 1 2 = 9 2 x Apabila x = 4, dy dx = 9 2 (2) When = 9 y = –8 5x–2 = 8 – 5 x2 dy dx = 16 – 5 x Apabila x = 1 – 8 , dy dx = 16 – 5 1 – 8 When = 2 5 y = 7x 2 x = 7 2 x 1 2 dy dx = 7 4 x 1 – 2 = 7 4 x Apabila x = 9, dy dx = 7 When 4 9 = 7 12 CONTOH (i) y = 2x3 – 5x2 + 6 (ii) y = (3x – 7)(7 – x) x Penyelesaian: (i) dy dx = dy dx (2x3 ) – dy dx (5x2 ) + dy dx (6) = 6x2 – 10x + 0 = 6x2 – 10x (ii) y = (3x – 7)(7 – x) x = 21x – 3x2 – 49 + 7x x = 28x – 3x2 – 49 x = 28 – 3x – 49x−1 dy dx = –3 + 49x –2 = –3 + 49 x2 Beza setiap sebutan secara berasingan. Differentiate for each term separately. Ringkaskan supaya menjadi sebutan berasingan dan dalam bentuk axn . Simplify to become separate terms and in the form of axn .
Matematik Tambahan Tingkatan 5 Bab 2 Pembezaan 28 BAB 2 (a) y = x5 – 1 2 x3 + 1 x (b) y = – 2 x5 – 1 x2 + 2 (c) y = x2 2 – 3 x (d) y = x ( x + 4x2 ) (e) y = (2x + 3)(1 – x3 ) 3x 2 (f) y = 2(x + 1)2 3x (g) y = x3 (3 – 1 2 x)2 (h) y = 4x2 2 – 3 x 7x y = 2x2 + 3x dy dx = 4x – 3 y = x3 9 – 3x + 1 4 x2 = 9x3 – 3x4 + 1 4 x5 dy dx = 27x2 – 12x3 + 5 4 x4 = x2 27 – 12x + 5 4 x2 y = 4 7 x2 – 3 x = 8 7 x – 12 7 dy dx = 8 7 y = 2x – 2x4 + 3 – 3x3 3x2 = 2 3 x –1 – 2 3 x2 + x –2 – x dy dx = –2 3x2 – 4 3 x – 1 x3 – 1 y = x + 4x 5 2 dy dx = 1 + 10x 3 2 = 1 + 10x x y = x5 – 1 2 x3 + x –1 dy dx = 5x4 – 3 2 x2 – 1 x2 y = –2x – 5 – x – 2 + 2 dy dx = 10 x6 + 2 x3 y = 2(x2 + 2x + 1) 3x = 2 3 (x + 2 + x –1) dy dx = 2 3 1 – 1 x2
BAB Matematik Tambahan Tingkatan 5 Bab 2 Pembezaan 29 2 8. Cari nilai terbitan pertama untuk fungsi berikut bagi nilai x yang diberikan. Find the value of the first derivative for each of the following functions for the given value of x. TP 3 (a) y = 8 3x – x2 + x , x = 4 (b) y = 5x – 1 x (2x )2 , x = –1 9. Tentukan terbitan pertama bagi fungsi gubahan yang berikut. Determine the first derivative for the following composite functions. TP 3 CONTOH y = 2 – x 3 2 , x = –1 Penyelesaian: y = 2 – x 3 2 = 4 – 4x + x2 9 dy dx = – 4 9 + 2 9 x = 2 9 (x – 2) Apabila/When x = −1, dy dx = 2 9 (–1 – 2) = – 2 3 y = 8 3 x–1 – x2 + x 1 2 dy dx = –8 3x2 – 2x + 1 2 x Apabila/When x = 4, dy dx = –8 3(4)2 – 8 + 1 4 = –711 12 y = 5x – x–1 4x2 = 5 4 x–1 – 1 4 x–3 dy dx = – 5 4 x–2 + 3 4 x–4 Apabila/When x = –1, dy dx = – 5 4 + 3 4 = – 1 2 CONTOH (i) (3 – 4x)4 (ii) y = 3 (2x – x2 )3 Penyelesaian: (i) Katakan/Let y = (3 − 4x)4 dan/and u = 3 – 4x Maka/Hence y = u4 dan du dx = −4 dy dx = 4u3 dy dx = dy du × du dx = 4u3 (−4) = −16u3 = −16(3 – 4x)3 (ii) Katakan/Let y = 3 (2x – x2 )3 dan/and u = 2x – x2 Maka/Hence y = 3u–3 dy du = –9u–4 dan/and du dx = 2 – 2x dy dx = dy du × du dx = –9u–4(2 – 2x) = – 9 u4 (2 – 2x) = –18(1 – x) (2x – x2 )4 Kaedah Alternatif (i) y = (3 − 4x) 4 dy dx = 4(3 − 4x)4 − 1· d dx (3 − 4x) = 4(3 − 4x)3 (−4) = −16(3 – 4x)3 (ii) y = 3(2x – x2 )−3 dy dx = 3(−3)(2x −x2 )−3 − 1· d dx (2x − x2 ) = −9(2x – x2 )−4(2 – 2x) = –18(1 – x) (2x – x2 )4
Matematik Tambahan Tingkatan 5 Bab 2 Pembezaan 30 BAB 2 (a) (1 – 4x2 )2 3 (b) 2 – 1 x 5 (c) 1 2 (3x3 – 2 5 x2 )4 (d) 2 5 x2 + x 3 10. Cari nilai dy dx bagi setiap nilai x yang diberi berikut. Find the value of dy dx for each of the following given value of x. TP 4 (a) y = 5 2 + x2 4 , x = 4 y = (1 – 4x2 )2 3 = 1 3 (1 – 4x2 )2 dy dx = 2 3 (1 – 4x2 )· d dx (1 – 4x2 ) = 2 3 (1 – 4x2 )(–8x) = – 16 3 x(1 – 4x2 ) y = 1 2 3x3 – 2 5 x2 4 dy dx = 1 2 (4)3x3 – 2 5 x2 3 · d dx 3x3 – 2 5 x2 = 23x3 – 2 5 x2 3 ·9x2 – 4 5 x y = 2 – 1 x 5 dy dx = 52 – 1 x 4 · d dx (2 – x–1 ) = 52 – 1 x 4 · 1 x2 = 5 x2 2 – 1 x 4 y = 2 5 x2 + x 3 = 2 5 x2 + x –3 dy dx = 2(–3) 5 x2 + x –4– 10 x3 + 1 = –61 – 10 x3 5 x2 + x 4 = 6(10 – x3 ) x3 5 x2 + x 4 CONTOH y = 2 3 4 – x , x = 3 Penyelesaian: y = 2 3 4 – x = 2 3 (4 – x) – 1 2 Jika/If u = 4 – x, du dx = –1, y = 2 3 u – 1 2 dy du = – 1 3 u – 3 2 dy dx = dy du × du dx = – 1 3u 3 2 ·(–1) = 1 3(4 – x) 4 – x Apabila/When x = 3, dy dx = 1 3(1) 1 = 1 3 Tip u 3 2 = u·u 1 2 y = 5 4 (2 + x2 ) 1 2 dy dx = 5 4 1 2 (2 + x2 ) 1 – 2 (2x) = 5x 4 2 + x2 Apabila x = 4, dy dx = 5(4) 4 18 When = 5 3 2 = 5 2 6
BAB Matematik Tambahan Tingkatan 5 Bab 2 Pembezaan 31 2 (b) y = (x4 − 3x2 − 2)3 , x = −1 (c) y = 6 (x + x 2 ) , x = –2 11. Tentukan terbitan pertama bagi suatu fungsi yang melibatkan hasil darab ungkapan algebra yang berikut. Determine the first derivative for the following functions which involve the multiplication of algebraic expressions of the following. TP 4 (a) y = x3 (1 − 2x)2 (b) y = (x + 4)2 (1 −3x)2 (c) y = (2x − 1)(1 + x2 )4 y = (x4 – 3x2 – 2)3 dy dx = 3(x4 – 3x2 – 2)2 (4x3 – 6x) = 6(2x3 – 3x)(x4 – 3x2 – 2)2 Apabila x = –1, dy dx = 6(–2 + 3)(1 – 3 – 2)2 When = 96 y = 6(x + x2 )–1 dy dx = –6(x + x2 )–2(1 + 2x) = –6(1 + 2x) (x + x2 )2 Apabila x = –2, dy dx = –6(–3) (–2 + 4)2 When = 9 2 y = x3 (1 – 2x)2 u = x3 v = (1 – 2x)2 du dx = 3x2 dv dx = 2(1 – 2x)(–2) dy dx = x3 (2)(–2)(1 – 2x) + (1 – 2x)2 (3x2 ) = x2 (1 – 2x)[–4x + 3(1 – 2x)] = x2 (1 – 2x)(3 – 10x) y = (2x –1)(1 + x2 )4 dy dx = (2x – 1)(4)(2x)(1 + x2 )3 + (2)(1 + x2 )4 = 2(1 + x2 )3 [4(2x – 1)(x) + 1 + x2 ] = 2(1 – x2 )3 [9x2 – 4x + 1] y = (x + 4)2 (1 – 3x)2 dy dx = (x + 4)2 (2)(–3)(1 – 3x) + (1 – 3x)2 (2)(x + 4) = 2(1 – 3x)(x + 4)[–3(x + 4) + 1 – 3x] = 2(1 – 3x)(x + 4)[–6x – 11] CONTOH y = (x – 2)2 (3 – x2 )3 Penyelesaian: y = (x – 2)2 (3 – x2 )3 Kata/Let u = (x – 2)2 v = (3 – x2 )3 du dx = 2(x – 2) dv dx = 3(–2x)(3 – x2 )2 = –6x(3 – x2 )2 dy dx = u dv dx + v du dx = (x – 2)2 (–6x)(3 – x2 )2 + (3 – x2 )3 (2)(x – 2) = 2(x – 2)(3 – x2 )2 [–3x(x – 2) + 3 – x2 ] = 2(x – 2)(3 – x2 )2 [6x + 3 – 4x2 ] Ambil faktor yang sama. Take the same factor.
Matematik Tambahan Tingkatan 5 Bab 2 Pembezaan 32 BAB 2 (d) y = x2 x – 1 (e) y = 2(x − 2)(1 − 3x)3 12. Tentukan terbitan pertama bagi suatu fungsi yang melibatkan hasil bahagi ungkapan algebra yang berikut. Determine the first derivative for the following functions which involve the division of algebraic expressions. TP 4 (a) y = (x – 3)3 (1 + 2x)2 y = x2 (x – 1) 1 2 dy dx = x2 1 2 (x – 1) 1 – 2 + x – 1(2x) = x2 2 x – 1 + 2x x – 1 = x2 + 4x(x – 1) 2 x – 1 = 5x2 – 4x 2 x – 1 y = 2(x – 2)(1–3x)3 dy dx = 2(x – 2)(3)(–3)(1 – 3x)2 + 2(1 – 3x)3 = 2(1 – 3x)2 [–9(x – 2) + 1 – 3x] = 2(1 – 3x)2 [19 – 12x] CONTOH y = (1 + 2x)2 (1 – x)3 Penyelesaian: y = (1 + 2x)2 (1 – x)3 Katakan u = (1 + 2x)2 , v = (1 – x)3 Let du dx = 4(1 + 2x) dv dx = –3(1 – x)2 Gunakan dy dx = v du dx – u dv dx v2 Use dy dx = (1 – x)3 (4)(1 + 2x) – (1 + 2x)2 (–3)(1 – x)2 (1 – x)6 = (1 – x)2 (1 + 2x)[4(1 – x) + 3(1 + 2x)] (1 – x)6 = (1 + 2x)[4 – 4x + 3 + 6x] (1 – x)4 = (1 + 2x)[7 + 2x] (1 – x)4 y = (x – 3)3 (1 + 2x)2 dy dx = (1 + 2x)2 (3)(x – 3)2 – (x – 3)3 (2)(2)(1 + 2x) (1 + 2x)4 = (1 + 2x)(x – 3)2 [3(1 + 2x) – 4(x – 3) (1 + 2x)4 = (x – 3)2 [15 + 2x) (1 + 2x)3 3
BAB Matematik Tambahan Tingkatan 5 Bab 2 Pembezaan 33 2 (b) y = x2 (1 – 2x)3 (c) y = 1 + 2x2 2 + x3 (d) y = 2x3 – 3 (x – 1) (e) y = x x + 1 2.3 Pembezaan Peringkat Kedua The Second Derivative NOTA IMBASAN NOTA IMBASAN NOTA IMBASAN NOTA IMBASAN 1. Jika y = f(x), dy dx = f'(x) ialah pembezaan peringkat pertama. If y = f(x), dy dx = f'(x) is the first order differentiation. d dx dy dx = d2 y dx2 = f''(x) ialah pembezaan peringkat kedua d dx dy dx = d2 y dx2 = f''(x) is the second order differentiation. 2. Perhatikan bahawa d2 y dx2 ≠ dy dx 2 . Note that d2 y dx2 ≠ dy dx 2 . y = x2 (1 – 2x)3 dy dx = (1 – 2x)3 (2x) – x2 (3)(–2)(1 – 2x)2 (1 – 2x)6 = 2x(1 – 2x)2 [1 – 2x + 3x] (1 – 2x)6 = 2x(1 + x) (1 – 2x)4 y = 2x3 – 3 (x – 1) dy dx = (x – 1)(6x2 ) – (2x3 – 3) (x – 1)2 = 4x3 – 6x2 + 3 (x – 1)2 y = 1 + 2x2 2 + x3 dy dx = (2 + x 3 )(4x) – (1 + 2x2 )(3x2 ) (2 + x3 )2 = x[4(2 + x3 ) – 3(1 + 2x2 )x] (2 + x3 )2 = x(8 + 4x3 – 3x – 6x3 (2 + x3 )2 = x[8 – 3x – 2x3 ] (2 + x3 )2 y = x x + 1 = x 1 2 x + 1 dy dx = (x + 1) 2 x – x (x + 1)2 = x + 1 – 2x 2 x (x + 1)2 = 1 – x 2 x (x + 1)2 4
Matematik Tambahan Tingkatan 5 Bab 2 Pembezaan 34 BAB 2 13. Cari terbitan pertama dan kedua bagi fungsi berikut yang berkenaan dengan x. Seterusnya, bandingkan bentuk d2 y dx2 dan dy dx 2 2 Find the first and the second derivatives for the following functions with respect to x. Then, compare the expressions of d2y dx2 and dy dx 2 . TP 3 (a) y = (x − 3)2 (4 − x2 ) (b) y = 1 x – 1 x2 + 2x2 (c) y = x 2x + 1 CONTOH y = 4x3 – 2x2 + 1 2 x Penyelesaian: dy dx = 12x2 – 4x + 1 2 d2 y dx2 = 24x – 4 dy dx 2 = 12x2 – 4x + 1 2 2 Maka/Hence d2 y dx2 ≠ dy dx 2 dy dx = (x – 3)2 (–2x) + (4 – x2 )(2)(x – 3) = 2(x – 3)[–x(x – 3) + 4 – x2 ] = 2(x – 3)(3x + 4 – 2x2 ] d2 y dx2 = (2x – 6)(–4x + 3) + (– 2x2 + 3x + 4)(2) = –8x2 + 6x + 24x – 18 – 4x2 + 6x + 8 = –12x2 + 36x – 10 dy dx 2 = 4(x – 3)2 (3x + 4 – 2x2 ]2 ∴ d2 y dx2 ≠ dy dx 2 y = x –1 – x–2 + 2x2 dy dx = – x–2 + 2x –3 + 4x d2 y dx2 = 2x–3 – 6x–4 + 4 = 2 x3 – 6 x4 + 4 dy dx 2 = 4x – 1 x2 + 2 x3 2 ∴ d2 y dx2 ≠ dy dx 2 dy dx = (2x + 1) – 2x (2x + 1)2 = 1 (2x + 1)2 d2 y dx2 = –4 (2x + 1)3 dy dx 2 = 1 (2x + 1)4 ∴ d2 y dx2 ≠ dy dx 2
BAB Matematik Tambahan Tingkatan 5 Bab 2 Pembezaan 35 2 14. Jawab semua soalan yang berikut. Answer all the following questions. TP 4 (a) Diberi y = x2 + 2x. Tunjukkan bahawa 2x2 d2 y dx2 − dy dx 2 + 4 dy dx = 4. Given y = x2 + 2x. Show that 2x2 d2y dx2 − dy dx 2 + 4 dy dx = 4. (b) Diberi y = x3 – 2x2 . Tunjukkan bahawa x2 d2 y dx2 – 6y = 8x2 . Given y = x3 – 2x2. Show that x2 d2y dx2 – 6y = 8x2. (c) Diberi y = x + 2 x . Tunjukkan bahawa d2 y dx2 + 2 x dy dx + x = y. Given y = x + 2 x . Show that d2y dx2 + 2 x dy dx + x = y. CONTOH y = x2 + 2x y’ = 2x + 2 y’’ = 2 Sebelah kiri = 2x2 [2] – (2x + 2)2 + 4(2x + 2) Left side = 4x2 – [4x2 + 8x + 4] + 8x + 8 = 4 sebelah kanan right side Diberi y = 2x – 1 x . Tunjukkan bahawa x2 d2 y dx2 + 4x dy dx = 12x − 2y Given y = 2x – 1 x . Show that x2 d2y dx2 + 4x dy dx = 12x − 2y. Penyelesaian: y = 2x – x –1 dy dx = 2 + 1 x2 d2 y dx2 = –2 x3 Sebelah kiri = x2 d2 y dx + 4x dy dx Left side = x2 – 2 x3 + 4x2 + 1 x2 = – 2 x + 8x + 4 x = 2 x + 8x Sebelah kanan= 12x – 2y Right side = 12x – 22x – 1 x = 12x – 4x + 2 x = 8x + 2 x ∴ x2 d2 y dx2 + 4x dy dx = 12x – 2y y = x3 – 2x2 y’ = 3x2 – 4x y’’ = 6x – 4 Sebelah kiri = x2 [6x – 4] – 6[x3 – 2x2 ] Left side = –4x2 + 12x2 = 8x2 sebelah kanan right side y = x + 2x –1 y' = 1 – 2x –2 y'' = 4x –3 Sebelah kiri = 4 x3 + 2 x 1 – 2 x2 + x Left side = 2 x + x sebelah kanan/right side
Matematik Tambahan Tingkatan 5 Bab 2 Pembezaan 36 BAB 2 NOTA IMBASAN NOTA IMBASAN 2.4 Aplikasi Pembezaan Application of Differentiation NOTA IMBASAN NOTA IMBASAN 1. Tafsiran dy dx sebagai kecerunan lengkung y = f(x). Defining dy dx as the gradient of a curve y = f(x). (a) Katakan P(x, y) dan titik terdekat Q(x + δx, y + δy) adalah dua titik pada lengkung y = f(x), maka kecerunan PQ = δy δx . Let P(x, y) and the nearest point Q(x + δx, y + δy) are two points on the curve y = f(x), then the gradient PQ = δy δx . (b) Jika kecerunan tangen pada P ialah m, maka m = δy δx , iaitu had δy δx = dy dx = m δx → 0 If the gradient of the tangent at P is m, then m = δy δx , that is lim δy δx = dy dx = m δx → 0 2. Kecerunan tangen, m1 × kecerunan normal, m2 = −1. The gradient of the tangent, m1 × the gradient of the normal, m2 = −1. 3. Persamaan tangen atau normal boleh ditentukan dengan menggunakan rumus (y – y1 ) = m(x − x1 ), dengan m ialah kecerunan tangen atau normal. The equation of the tangent or normal can be obtained by using the formula (y – y1 ) = m(x − x1 ), where m is the gradient of the tangent or the normal. 4. Satu titik pada lengkung y = f(x) yang mempunyai dy dx = 0 dinamakan titik pusingan. Pada titik ini, tangen adalah selari dengan paksi-x. A point on the curve y = f(x) whose dy dx = 0 is known as the turning point. At this point, the tangent is parallel to the x-axis. 5. Pada rajah (i), titik P merupakan titik maksimum. Pada rajah (ii), titik Q merupakan titik minimum. Perhatikan tanda dy dx di sebelah kiri dan sebelah kanan masing-masing bagi titik P dan Q. In diagram (i), P is the maximum point. In diagram (ii), Q is the minimum point. Note the signs of dy dx on the left and right of the turning points P and Q. 6. Langkah berikut boleh digunakan untuk menentukan titik maksimum atau minimum bagi suatu lengkung y = f(x). The following steps can be used to determine the maximum or minimum point for a curve y = f(x). x y fix fiy P (x, y) Q (x+δx, y+δy) 0 Tangen pada P Tangent at P x y δy δx = 0 δy δx < 0 δy δx > 0 P x1 x < x1 x > x1 0 x y δy δx = 0 δy δx < 0 δy δx > 0 Q x1 x < x1 x > x1 0 (i) (ii)
BAB Matematik Tambahan Tingkatan 5 Bab 2 Pembezaan 37 2 NOTA IMBASAN NOTA IMBASAN NOTA IMBASAN NOTA IMBASAN NOTA IMBASAN NOTA IMBASAN (i) Cari/Find dy dx (ii) Cari nilai x apabila dy dx = 0. Find the value of x when dy dx = 0. (iii) Tentukan sama ada titik pusingan (x, y) adalah titik maksimum atau minimum dengan menggunakan Determine whether the turning point (x, y) is a maximum point or minimum point by using the (a) jadual/table Nilai x Value of x Pilih satu nilai x lebih kecil daripada nilai x pada titik pusingan. Choose a value of x smaller than the value of x at the turning point. Nilai x pada titik pusingan. The value of x at the turning point. Pilih satu nilai x lebih besar daripada nilai x pada titik pusingan. Choose a value of x larger than the value of x at the turning point. Titik pusingan Turning point Nilai dy dx Value of dy dx Jika/If dy dx 0, x x1 dy dx = 0 Jika/If dy dx 0, x x1 Minimum Maksimum/maximum Titik refleks Reflex point (b) terbitan kedua second derivatives Pada titik pusingan (x1 , y1 ), jika d2 y dx2 0 pada x = x1 , maka titik tersebut ialah titik maksimum. At the turning point (x1 , y1 ), if d2 y dx2 0 at the x = x1 , then the point is a maximum point. Pada titik pusingan (x1 , y1 ), jika d2 y dx2 0 pada x = x1 , maka titik tersebut ialah titik minimum At the turning point (x1 , y1 ), if d2 y dx2 0 at the x = x1 , then the point is a minimum point Pada titik pusingan (x1 , y1 ), jika d2 y dx2 = 0 pada x = x1 , maka titik tersebut ialah titik reflek. At the turning point (x1 , y1 ), if d2 y dx2 = 0 at the x = x1 , then the point is a reflex point. 7. Kadar perubahan yang terhubung Rate of related change Jika y = f(x) dan x = g(t), maka kadar perubahan y terhadap masa, t ialah dy dt yang diberi oleh petua rantai dy dt = dy dx × dx dt . If y = f(x) and x = g(t), then the rate of change of y with respect to time, t is dy dt which is obtained by the chain rule dy dt = dy dx × dx dt . 8. Tokokan kecil dan penghampiran Small changes and approximation Kita diketahui bahawa had δy δx = dy dx . δx → 0 We know that lim δy δx = dy dx . dx → 0 Jika δy dan δx adalah sangat kecil, maka δy δx adalah satu hampiran bagi dy dx , iaitu δy δx ≈ dy dx → δy ≈ dy dx × δx. If δy and δx are small changes, then δy δx is an approximation for dy dx , that is δy δx ≈ dy dx → δy ≈ dy dx × δx.
Matematik Tambahan Tingkatan 5 Bab 2 Pembezaan 38 BAB 2 15. Cari fungsi kecerunan bagi setiap lengkung yang diberi dan seterusnya cari nilai kecerunan pada titik x yang diberi dan nilai koordinat y yang sepadannya. Find the gradient function for the following curves and then find the value of the gradient at the given value of point x with its corresponding coordinates of y. TP 3 (a) y = x(5 − 3x) pada/at x = −2. (b) f(x) = 4x2 – 6x + 1 pada/at x = –1. (c) f(x) = −2x + 4x2 pada/at x = 1. CONTOH y = x3 − 2x pada x = 2. y = x3 − 2x at x = 2. Penyelesaian: Fungsi kecerunan, dy dx = 3x2 − 2 The gradient function, dy dx = 3x2 − 2 Apabila/When x = 2, y = 23 − 2(2) = 4 dy dx = 3(2)2 – 2 = 10 Maka, kecerunan lengkung pada titik (2, 4) ialah 10. Hence, the gradient of the curve at point (2, 4) is 10. y = 5(–2) – 3(–2)2 x = –10 – 12 = –22 y' = 5 – 6x Pada/At x = –2 dy dx = 5 – 6(–2) = 17 f(1) = –2 + 4 = 2 f'(x) = –2 + 8x f'(1) = –2 + 8(1) = 6 ∴ Pada (1, 2), f'(x) = 6 At f(–1) = 4(–1)2 – 6(–1) + 1 = 11 f'(x) = 8x – 6 f'(–1) = 8(–1) – 6 = –14 ∴ Pada (–1, 11), f'(x) = –14 At
BAB Matematik Tambahan Tingkatan 5 Bab 2 Pembezaan 39 2 16. Cari nilai p dan q bagi yang berikut. Find the value of p and of q for the following. 4 (a) Kecerunan lengkung y = p x + qx pada (2, 4) ialah −8. The gradient for the curve y = p x + qx at (2, 4) is −8. (b) Kecerunan lengkung y = px3 + qx pada (−1, 1) ialah 5. The gradient for the curve y = px3 + qx at (−1, 1) is 5. (c) Kecerunan lengkung y = px2 + qx + 1 pada (2, 5) ialah 8. The gradient for the curve y = px2 + qx + 1 at (2, 5) is 8. CONTOH Kecerunan lengkung y = px + qx2 pada (1, 1) ialah −5. The gradient for the curve y = px + qx2 at (1, 1) is −5. Penyelesaian: Pada/At (1, 1), 1 = p(1) + q(1)2 ... ➀ dy dx = p + 2qx Maka/Hence p + 2q(1) = −5 ... ➁ ➁ − ➀, q = −6 Gantikan dalam ➀, Substitute into ➀, p = 1 + 6 = 7 y = px–1 + qx 4 = p 2 + 2q y' = –p x2 + q Apabila x = 2 –p 4 + q = –8 When –p + 4q = –32 ...➁ ➁ + ➂ p + 4q = 8 ...➂ 8q = –24 q = –3 p = (4 + 6)2 = 20 y = px2 + qx + 1 (2, 5) 5 = 4p + 2q + 1 2 = 2p + q ...➀ y' = 2px + q, Apabila x = 2 y' = 4p + q = 8 ...➁ When ➁ – ➀ 2p = 6 p = 3, q = –4 y = px3 + qx (–1, 1) 1 = –p – q ...➀ y' = 3px2 + q Apabila x = –1 y' = 3p + q = 5 ...➁ When ➁ + ➀ 2p = 6 p = 3, q = –4
Matematik Tambahan Tingkatan 5 Bab 2 Pembezaan 40 BAB 2 17. Cari persamaan tangen dan normal kepada lengkung pada titik dengan koordinat-x yang diberi. Find the equation of the tangent and the normal to the curve at the point with the given coordinates of x. TP 4 (a) y = 1 x pada/at x = 2 (b) y = x3 – 4x pada/at x = –1 (c) y = x pada/at x = 4 CONTOH y = 3 – 2x – x2 pada/at x = −2. Penyelesaian: y = 3 – 2(−2) – (−2)2 = 3 dy dx = −2 − 2x Maka, kecerunan tangen pada x = −2 ialah Hence, the gradient of tangent at x = −2 is −2 – 2(−2) = 2 Persamaan tangen pada (−2, 3) The equation of the tangent at (−2, 3) y – 3 = 2[x – (−2)] y – 3 = 2x + 4 y = 2x + 7 Kecerunan normal/Gradient of the normal = – 1 2 Persamaan normal pada (−2, 3). Equation of the normal at (−2, 3). y – 3 = – 1 2 (x + 2) y – 3 = – 1 2 x – 1 y = – 1 2 x + 2 m1 × m2 = –1 y = 1 x x = 2, y = 1 2 dy dx = – 1 x2 Pada x = 2, dy dx = – 1 4 Persamaan tangen/Equation of tangent y – 1 2 = – 1 4 (x – 2) y = – 1 4 x + 1 Persamaan normal//Equation of normal y – 1 2 = 4(x – 2) y = 4x – 15 2 y = x3 – 4x x = –1 y = –1 – 4(–1) = 3 dy dx = 3x2 – 4 pada x = –1 dy dx = 3(–1)2 – 4 = –1 Persamaan tangen/Equation of tangent y – 3 = –(x + 1) y = –x + 2 Persamaan normal/Equation of normal y – 3 = x + 1 y = x + 4 y = x = x 1 2 x = 4, y = 2 dy dx = 1 2 x Pada x = 4, dy dx = 1 4 Persamaan tangen/Equation of tangent y – 2 = 1 4 (x – 4) y = 1 4 x + 1 Persamaan normal//Equation of normal y – 2 = –4(x – 4) y = –4x + 18 Kaedah Alternatif Persamaan tangen y = 2x + c pada titik (–2, 3) 3 = 2(–2) + c c = 7 Persamaan normal y = 1 – 2 x + c pada titik (–2, 3) y = 1 – 2 (–2) + c1 c1 = +2 ∴y = 1 – 2 x + 2
BAB Matematik Tambahan Tingkatan 5 Bab 2 Pembezaan 41 2 CONTOH 18. Selesaikan masalah yang melibatkan tangen dan normal. Solve the problems involving tangents and normals. TP 4 (a) Kecerunan titik P yang terletak pada lengkung y = 3x2 − 6x adalah selari dengan garis lurus y = 12x − 3. The gradient of point P on the curve y = 3x2 − 6x is parallel to the line y = 12x − 3. Cari/Find (i) koordinat P, the coordinates of P, (ii) persamaan normal pada P. the equation of the normal at P. (b) Cari koordinat bagi titik-titik pada lengkung y = 2x + 8 x , dengan tangennya selari dengan paksi-x. Find the coordinates of the points on the curve y = 2x + 8 x , such that the tangents are parallel to the x-axis. y = 2x + 8 x dy dx = 2 – 8 x2 = 0 2x2 = 8 x = ±2 y = 2(2) + 8 2 , 2(–2) – 8 2 = 8, –8 ∴(2, 8), (–2, –8) (i) y = 3x2 – 6x dy dx = 6x – 6 = 12 6x = 18 x = 3 y = 3(3)2 – 6(3) = 9 ∴P[3, 9] Diberi f(x) = x − 1 x . Cari koordinat bagi titik jika f'(x) = 2. Seterusnya, cari persamaan tangen pada titik dengan x 0. Given that f(x) = x − 1 x . Find the coordinates of the point if f'(x) = 2. Then, find the equation of the tangent at the point whose x 0. Penyelesaian: f(x) = x – 1 x f'(x) = 1 + 1 x2 = 2 1 x2 = 1 x = ±1 Apabila /When x = 1 f(1) = 1 – 1 = 0 x = –1 f(–1) = –1 + 1 = 0 (1, 0) dan (–1, 0) Persamaan tangen pada (1, 0) Equation of the tangent at (1, 0) y – 0 = 2(x – 1) y = 2x – 2 Persamaan normal pada (1, 0) Equation of the normal at (1, 0) y – 0 = – 1 2 (x – 1) y = – 1 2 x + 1 2 (ii) Persamaan normal pada (3, 9) Equation of the normal at (3, 9) y – 9 = – 1 12(x – 3) y = – 1 12x + 37 4
Matematik Tambahan Tingkatan 5 Bab 2 Pembezaan 42 BAB 2 (c) Tangen kepada lengkung y = ax2 + bx + 2 pada titik 1, 1 2 adalah selari dengan normal kepada lengkung y = x2 + 6x + 4 pada (−2, −4). Cari nilai a dan nilai b. The tangent to the curve y = ax2 + bx + 2 at the point 1, 1 2 is parallel to the normal for the curve y = x2 + 6x + 4 at the point (−2, −4). Find the value of a and of b. 19. Cari titik pusingan bagi lengkung dan tentukan sifat-sifat bagi titik tersebut. Find the turning point(s) of the curves and determine the characteristics of these points. TP 4 y = ax2 + bx + 2 dy dx = 2ax + b pada (1, 1 2 ), dy dx = 2a + b y = x2 + 6x + 4 dy dx = 2x + 6 pada (–2, –4), dy dx = 2(–2) + 6 = 2 ∴2a + b = – 1 2 ...➀ 1 2 = a + b + 2 a + b = – 3 2 ...➁ ➀ – ➁ a = 1, b = –1 = – 5 2 CONTOH y = 6x – x2 Penyelesaian: dy dx = 6 − 2x Untuk titik pusingan, kecerunan dy dx = 0 For the turning point, gradient dy dx = 0 Maka/Then 6 − 2x = 0 x = 3 y = 18 − 9 = 9 Titik pusingan ialah (3, 9). The turning point is (3, 9). Dari jadual berikut. From the following table. x Sedikit kurang 3 Less than 3 3 Sedikit lebih 3 More than 3 dy dx + 0 − Lakaran tangen Sketch of the tangent Kecerunan dy dx menukar tanda dari positif ke negatif semasa melalui x = 3, jadi titik pusingan itu ialah maksimum. The gradient dy dx changes from positive to negative as it passes through x = 3, hence the turning point is a maximum point. Kaedah Alternatif Gunakan terbitan kedua, jadi d2 y dx2 = −2 Use second derivative, so d2 y dx2 = −2 Apabila d2 y dx2 0, titik pusingan (3, 9) ialah maksimum. When d2 y dx2 < 0, the turning point (3, 9) is maximum
BAB Matematik Tambahan Tingkatan 5 Bab 2 Pembezaan 43 2 (a) y = 2x2 − 4x − 1 (b) y = x3 − 3x2 (c) y = 1 3 x3 − x2 − 3x + 1 dy dx = x2 − 2x – 3 Untuk titik pusingan, kecerunan dy dx = 0 Maka x2 − 2x – 3 = 0 (x – 3)(x + 1) = 0 x = 3 dan –1 y = –8 dan 8 3 Titik pusingan ialah (3, –8) dan −1, 8 3 Dari jadual berikut. x Sedikit kurang 3 0 Sedikit lebih 3 dy dx − 0 + Lakaran tangen x Sedikit kurang –1 –1 Sedikit lebih –1 dy dx + 0 – Lakaran tangen (3, −8) ialah titik minimum dan −1, 8 3 ialah titik maksimum. (3, –8) is a minimum point and (–1, 8 3 ) is a maximum point. x Sedikit kurang 1 1 Sedikit lebih 1 dy dx − 0 + Lakaran tangen Kecerunan dy dx menukar tanda dari negaif ke positif semasa melalui x = 1, jadi titik pusingan itu ialah minimum. The gradient dy dx changes from negative to positive as it passes through x = 1, hence the turning point is minimum. dy dx = 4x − 4 Untuk titik pusingan, kecerunan dy dx = 0 For the turning point, gradient dy dx = 0 Maka, 4x − 4 = 0 Then x = 1 dan y = 2 − 4 − 1 = −3 Titik pusingan ialah (1, –3) The turning point is (1, –3) Dari jadual berikut. From the following table. dy dx = 3x2 − 6x Untuk titik pusingan, kecerunan dy dx = 0 For the turning point, gradient dy dx = 0 Maka 3x2 − 6x = 0 x = 0 dan 2 y = 0 dan −4 Titik pusingan ialah (0, 0) dan (2, −4) The turning point is (0, 0) and (2, −4) Dari jadual berikut./From the following table. x Sedikit kurang 0 0 Sedikit lebih 0 dy dx − 0 + Lakaran tangen x Sedikit kurang 2 2 Sedikit lebih 2 dy dx + 0 – Lakaran tangen (0, 0) ialah titik minimum dan (2, −4) ialah titik maksimum. (0, 0) is a minimum point and (2, –4) is a maximum point.
Matematik Tambahan Tingkatan 5 Bab 2 Pembezaan 44 BAB 2 CONTOH 20. Selesaikan masalah yang melibatkan nilai maksimum dan nilai minimum yang berikut. Solve the following problems involving the maximum and minimum values. TP 5 (a) Rajah menunjukkan satu kad jemputan. Ia terdiri daripada satu segi empat tepat dengan panjang y cm dan lebar x cm. Pada kedua-dua hujung segi empat tepat itu, terdapat satu segi tiga sama sisi dengan panjang sisi x cm. The diagram shows an invitation card. It is made up of a rectangle with a length of y cm and width of x cm. At the both ends of the rectangle, there are equilateral triangles with sides of x cm. y cm x cm 60fi Jika perimeter keseluruhan rajah ialah 26 cm, buktikan bahawa luasnya adalah satu nilai maksimum apabila x bernilai 13 4 – 3 cm. If the perimeter of the whole diagram is 26 cm, prove that the area is maximum when the x value is 13 4 – 3 cm. 2x m x m h m Sebuah tangki segi empat yang tertutup hendak dibuat supaya dapat mengisi 9 m3 air. Panjang tangki mesti dua kali lebarnya. Cari dimensi tangki itu jika jumlah luas permukaan ialah minimum. A closed rectangular tank is made to fill 9 m3 of water. The length of the tank is twice its breadth. Find the dimensions of the tank if the total surface area is minimum. Penyelesaian: Isi padu/Volume = 9 m3 = 2x2 h Jumlah luas permukaan/Total surface area A = 2(2xh) + 2xh + 4x2 = 6xh + 4x2 = 6x 9 2x2 + 4x2 = 27 x + 4x2 dA dx = − 27 x2 + 8x = 0 Untuk minimum/For minimum 8x = 27 x2 x3 = 27 8 x = 3 2 d2 A dx2 = 54 x3 + 8 Apabila/When x = 3 2 , d2 A dx2 0 (minimum) h = 9 2 9 4 = 2 Panjang/Length = 3 m, Lebar/Breadth = 3 2 m, Tinggi/Height = 2 m Perimeter = 26 = 2y + 4x y = 13 – 2x Luas A = xy + 1 2 x2sin 60°2 Area A = x[13 – 2x] + x2sin60° Untuk A maksimum, dA dx = 0 For a maximum A dA dx = 13 – 4x + 2xsin60° = 0 13 = x4 – 2 3 2 ∴x = 13 4 – 3
BAB Matematik Tambahan Tingkatan 5 Bab 2 Pembezaan 45 2 (b) Seutas dawai 360 cm panjang digunakan untuk membuat 12 sisi sebuah kotak yang berbentuk segi empat tepat dengan panjangnya adalah dua kali lebarnya. Jika lebar kotak ialah x cm, A wire of length 360 cm is used to make 12 sides of a rectangular box whose length is twice its breadth. If the breadth is x cm. (i) buktikan bahawa isi padu kotak, V diberi oleh V = 180x2 – 6x3 . prove that the volume of the box, V is given by V = 180x2 – 6x3. (ii) seterusnya tentukan dimensi kotak supaya isi padunya bernilai maksimum. then determine the dimensions of the box so that its volume is maximum. (c) Rajah menunjukkan sekeping kadbod ABCD, dengan BC = 6 cm, AB = 10 cm dan AD = 26 cm. Ahmad ingin menggunting sebuah segi empat AEFG dari kadbod supaya luasnya adalah maksimum. The diagram shows a piece of cardboard ABCD, where BC = 6 cm, AB = 10 cm and AD = 26 cm. Ahmad wants to cut out a rectangle AEFG from the cardboard so that the area of the rectangle is maximum. Jika FE ialah x cm, If FE is x cm, (i) nyatakan ED, AE dan luas AEFG dalam sebutan x. state ED, AE and the area of AEFG in terms of x. (ii) Seterusnya, cari luas maksimum segi empat itu. Then, find the maximum area of the rectangle. 2x x t A E D F B C G x cm (i) 4t + 8x + 4x = 360 t + 3x = 90 I = 2x2 t = 2x2 [90 – 3x] = 180x2 – 6x3 (ii) dI dx = 360x – 18x2 = 0 x[360 – 18x] = 0 x = 0 x = 20 Dimensi 20 × 40 × 30 cm Dimension (i) x 10 = ED 20 ED = 2x ∴AE = 26 – 2x Luas AEFG = x[26 – 2x] Area AEFG (ii) dA dx = 26 – 4x = 0 x = 13 2 cm Luas maksimum = 13 2 26 – 2 13 2 Maximum area = 84 1 2 cm2
Matematik Tambahan Tingkatan 5 Bab 2 Pembezaan 46 BAB 2 21. Bagi setiap persamaan yang berhubung dengan x dan y berikut, jika kadar perubahan x ialah 0.5 unit s−1, cari kadar perubahan y pada ketika yang diberi. Tafsirkan jawapan anda. For each of the following equations relating x and y, if the rate of change of x is 0.5 unit s−1, find the rate of change of y at the given instant. Interpret your answers. TP 3 (a) y = (3x − 6)2 , x = 1 (b) y = x x – 1 , x = 4 Diberi dx dt = 0.5 unit s−1 apabila x = 1. Given When dy dx = 6(3x − 6) Maka dy dt = dy dx × dx dt Hence = 6(3x − 6) dx dt Apabila x = 1 dan dx dt = 0.5 When and Maka dy dt = 6(−3)(0.5) Hence = −9 unit s−1 Kadar perubahan y ialah −9 unit s−1. y menyusut. The rate of change of y is –9 unit s–1. y decreases. Diberi dx dt = 0.5 unit s−1 apabila x = 4. Given when dy dx = (x – 1) – x (x – 1)2 = –1 (x – 1)2 Maka dy dt = dy dx × dx dt Hence = –1 (x – 1)2 dx dt Apabila x = 4 dan dx dt = 0.5 When and Maka dy dt = –1 (4 – 1)2(0.5) Hence = – 1 18 unit s−1 Kadar penyusutan y ialah 1 18 unit s−1 The rate of y decreasing is 1 18 unit s–1 CONTOH y = 2x2 + 1, x = 1.2 Penyelesaian: Diberi/Given dx dt = 0.5 unit s−1 apabila/when x = 1.2 dy dx = 4x Maka/Hence dy dt = dy dx × dx dt = 4x dx dt Apabila/when x = 1.2 dan/and dx dt = 0.5 Maka/Hence dy dt = 4(1.2)(0.5) = 2.4 unit s−1 Kadar perubahan y ialah 2.4 unit s−1. The rate of change of y is 2.4 unit s−1. Tip Oleh sebab dy dt 0, perubahan y dikatakan bertambah. Since dy dt 0, the change of y is increasing.
BAB Matematik Tambahan Tingkatan 5 Bab 2 Pembezaan 47 2 22. Bagi setiap persamaan yang berhubung x dan y berikut, jika kadar perubahan y ialah 8 unit s–1, cari kadar perubahan x pada ketika yang diberi. Tafsirkan jawapan anda. For each of the following equations relating x and y, if the rate of change of y is 8 unit s–1, find the rate of change of x at the instant given. Interpret your answers. TP 3 (a) y = 4x2 x – 1 , x = 3 (b) y = 2x – 3, x = 2 Diberi dy dt = 8 unit s−1 apabila x = 2 Given when dy dx = 1 2x – 3 Maka dy dt = dy dx × dx dt Hence = 1 2x – 3 dx dt Apabila x = 2 dan dx dt = 8 When and Maka 8 = 1 dx dt Hence dx dt = 8 unit s−1 Kadar perubahan x ialah 8 unit s−1. x bertambah. The rate of change of x is 8 unit s–1. x increases. CONTOH Tip Oleh sebab dx dt < 0, perubahan x dikatakan menyusut. Since dx dt 0, the change of x is decreasing. y = x2 − 6x, x = −1 Penyelesaian: Diberi/Given dy dt = 8 unit s−1 apabila/when x = −1 dy dx = 2x – 6 Maka/Hence dy dt = dy dx × dx dt = (2x – 6) × dx dt Apabila/When x = −1 dan/and dy dt = 8 Maka/Hence 8 = [2(–1) – 6] dx dt dx dt = 8 –8 unit s−1 = –1 unit s−1 Kadar perubahan x ialah –1 unit s−1 The rate of change of x is –1 unit s−1 Diberi dx dt = 8 unit s−1 apabila x = 3 Given when dy dx = (x – 1)(8x) – 4x2 (x – 1)2 Maka dy dt = dy dx × dx dt Hence = (x – 1)(8x) – 4x2 (x – 1)2 dx dt Apabila x = 3 dan dy dt = 8 When and Maka 8 = (3 – 1)(8)(3) – 4(3)2 (3 – 1)2 dx dt Hence dx dt = 8 3 unit s−1 Kadar perubahan x ialah 8 3 unit s−1. x bertambah. The rate of change of x is 8 3 unit s–1. x increases.