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MAT1613 ASSIGNMENT 1 2021
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This document contains MAT1613 ASSIGNMENT 1 2021 solutions. All workings are shown clearly step by step and explanations are also provided.
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MAT1613ASSIGNMENT 1 2021
QUESTION 1
𝐴 = 4𝜋𝑟 2 (𝑎𝑟𝑒𝑎 𝑜𝑓 𝑎 𝑠𝑝ℎ𝑒𝑟𝑒)
4
𝑉 = 𝜋𝑟 3 (𝑣𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑎 𝑠𝑝ℎ𝑒𝑟𝑒)
3
𝑑𝐴 𝑑𝐴 𝑑𝑟 𝑑𝑉
= × ×
𝑑𝑡 𝑑𝑟 𝑑𝑉 𝑑𝑡
4
𝑉 = 𝜋𝑟 3
3
𝑑𝑉
= 4𝜋𝑟 2
𝑑𝑟
𝑑𝑟 1
=
𝑑𝑉 4𝜋𝑟 2
𝐴 = 4𝜋𝑟 2
𝑑𝐴
= 8𝜋𝑟
𝑑𝑟
𝑑𝐴 1 𝑑𝑉
= (8𝜋𝑟) × ( 2
)×
𝑑𝑡 4𝜋𝑟 𝑑𝑡
𝑑𝐴 2 𝑑𝑉 𝑑𝑉
= × ∴ 𝑟 = 10, = 15 𝑐𝑚3 ⁄𝑠𝑒𝑐
𝑑𝑡 𝑟 𝑑𝑡 𝑑𝑡
𝑑𝐴 2
= × 15
𝑑𝑡 (10)
𝑑𝐴
= 3 𝑐𝑚2 ⁄𝑠𝑒𝑐
𝑑𝑡
𝑅𝑎𝑡𝑒 𝑜𝑓 𝑐ℎ𝑎𝑛𝑔𝑒 𝑜𝑓 𝑠𝑢𝑟𝑓𝑎𝑐𝑒 𝑎𝑟𝑒𝑎 = 3 𝑐𝑚2 ⁄𝑠𝑒𝑐
QUESTION 2
−𝑥
𝑓(𝑥) = 𝑥𝑒
a).
,𝑦 − 𝑖𝑛𝑡𝑒𝑟𝑐𝑒𝑝𝑡 = 𝑓(0)
= (0)𝑒 −(0)
=0
b).
𝐻𝑜𝑟𝑖𝑧𝑜𝑛𝑡𝑎𝑙 𝑎𝑠𝑦𝑚𝑝𝑡𝑜𝑡𝑒:
lim 𝑓(𝑥) = lim 𝑥𝑒 −𝑥
𝑥↔∞ 𝑥↔∞
𝑥
lim 𝑓(𝑥) = lim ( 𝑥 )
𝑥↔∞ 𝑥↔∞ 𝑒
𝑈𝑠𝑖𝑛𝑔 𝐿′ℎ𝑜𝑠𝑝𝑖𝑡𝑎𝑙
𝑑
[𝑥]
lim 𝑓(𝑥) = lim 𝑑𝑥
𝑥↔∞ 𝑥↔∞ 𝑑
[𝑒 𝑥 ]
𝑑𝑥
1
= lim
𝑥↔∞ 𝑒 𝑥
1
=
𝑒∞
1
=
∞
=0
𝐻𝑜𝑟𝑖𝑧𝑜𝑛𝑡𝑎𝑙 𝑎𝑠𝑦𝑚𝑝𝑡𝑜𝑡𝑒: 𝑦 = 0
𝑉𝑒𝑟𝑡𝑖𝑐𝑎𝑙 𝑎𝑠𝑦𝑚𝑝𝑡𝑜𝑡𝑒:
𝑥
𝑓(𝑥) = 𝑥𝑒 −𝑥 =
𝑒𝑥
∴ 𝑒 𝑥 > 0 𝑎𝑙𝑤𝑎𝑦𝑠 , 𝑚𝑒𝑎𝑛𝑖𝑛𝑔 𝑡ℎ𝑒𝑟𝑒 𝑖𝑠 𝑛𝑜 𝑣𝑒𝑟𝑡𝑖𝑐𝑎𝑙 𝑎𝑠𝑦𝑚𝑝𝑡𝑜𝑡𝑒
c).
i).
𝑓(𝑥) = 𝑥𝑒 −𝑥
𝐿𝑒𝑡: 𝑢 = 𝑥 𝑎𝑛𝑑 𝑣 = 𝑒 −𝑥
, 𝑑𝑣 𝑑𝑢
= −𝑒 −𝑥 𝑎𝑛𝑑 =1
𝑑𝑥 𝑑𝑥
𝑈𝑠𝑖𝑛𝑔 𝑡ℎ𝑒 𝑝𝑟𝑜𝑑𝑢𝑐𝑡 𝑟𝑢𝑙𝑒:
𝑑𝑣 𝑑𝑢
𝑓 ′ (𝑥) = 𝑢 +𝑣
𝑑𝑥 𝑑𝑥
= (𝑥)(−𝑒 −𝑥 ) + (𝑒 −𝑥 )(1)
= −𝑥𝑒 −𝑥 + 𝑒 −𝑥
= 𝑒 −𝑥 − 𝑥𝑒 −𝑥
= 𝑒 −𝑥 (1 − 𝑥)
(1 − 𝑥)
=
𝑒𝑥
∴ 𝑓(𝑥) 𝑟𝑖𝑠𝑒𝑠 𝑤ℎ𝑒𝑛 𝑓 ′ (𝑥) > 0
𝑓 ′ (𝑥) > 0
(1 − 𝑥)
>0 ∴ 𝑒 𝑥 𝑖𝑠 𝑎𝑙𝑤𝑎𝑦𝑠 𝑔𝑟𝑒𝑎𝑡𝑒𝑟 𝑡ℎ𝑎𝑛 𝑧𝑒𝑟𝑜
𝑒𝑥
(1 − 𝑥) > 0
𝑥<1
𝐼𝑛𝑡𝑒𝑟𝑣𝑎𝑙 𝑤ℎ𝑒𝑛 𝑓(𝑥) 𝑟𝑖𝑠𝑒𝑠 = (−∞, 1)
∴ 𝑓(𝑥) 𝑓𝑎𝑙𝑙𝑠 𝑤ℎ𝑒𝑛 𝑓 ′ (𝑥) < 0
𝑓 ′ (𝑥) < 0
(1 − 𝑥)
<0
𝑒𝑥
(1 − 𝑥) < 0
𝑥>1
𝐼𝑛𝑡𝑒𝑟𝑣𝑎𝑙 𝑤ℎ𝑒𝑛 𝑓(𝑥) 𝑓𝑎𝑙𝑙𝑠 = (1, ∞)
ii).
𝑙𝑜𝑐𝑎𝑙 𝑒𝑥𝑡𝑟𝑒𝑚𝑎:
𝑓 ′ (𝑥) = 0
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