MAT1613 SEMESTER 1 ASSIGNMENT 1 2021
Question 1
𝑟 = 10
𝑑𝑉
= 15
𝑑𝑡
4
𝑉 = 𝜋𝑟 3
3
𝑑𝑉 4 𝑑𝑟
= 3 × 𝜋𝑟 3−1 ×
𝑑𝑡 3 𝑑𝑡
𝑑𝑉 𝑑𝑟
= 4𝜋𝑟 2
𝑑𝑡 𝑑𝑡
𝑑𝑟
15 = 4𝜋(10)2
𝑑𝑡
𝑑𝑟
15 = 400𝜋
𝑑𝑡
15 𝑑𝑟
=
400𝜋 𝑑𝑡
𝑑𝑟 15
=
𝑑𝑡 400𝜋
𝑆𝐴 = 4𝜋𝑟 2
𝑑 𝑑𝑟
(𝑆𝐴) = 2 × 4𝜋𝑟 2−1 ×
𝑑𝑡 𝑑𝑡
𝑑 𝑑𝑟
(𝑆𝐴) = 8𝜋𝑟
𝑑𝑡 𝑑𝑡
𝑑 15
(𝑆𝐴) = 8𝜋 × 10 ×
𝑑𝑡 400𝜋
𝑑 15
(𝑆𝐴) = 8𝜋 × 1 ×
𝑑𝑡 40𝜋
𝑑 15 ÷ 5
(𝑆𝐴) = 8 ×
𝑑𝑡 40 ÷ 5
𝑑 3
(𝑆𝐴) = 8 ×
𝑑𝑡 8
𝑑
(𝑆𝐴) = 3
𝑑𝑡
𝑇ℎ𝑒 𝑠𝑢𝑟𝑓𝑎𝑐𝑒 𝑎𝑟𝑒𝑎 𝑜𝑓 𝑡ℎ𝑒 𝑏𝑎𝑙𝑙𝑜𝑛 𝑖𝑠 𝑖𝑛𝑐𝑟𝑒𝑎𝑠𝑖𝑛𝑔 𝑎𝑡 𝑎 𝑟𝑎𝑡𝑒 𝑜𝑓 3 𝑐𝑚2 ⁄𝑠𝑒𝑐
,Question 2
𝑓(𝑥) = 𝑥𝑒 −𝑥
(a) 𝑇ℎ𝑒 𝑦 − 𝑖𝑛𝑡𝑒𝑟𝑐𝑒𝑝𝑡.
𝑦 = 𝑥𝑒 −𝑥 (𝑡𝑜 𝑠𝑢𝑏𝑠𝑡𝑖𝑡𝑢𝑡𝑒 𝑥 = 0)
𝑦 = 0 × 𝑒 −0
𝑦 = 0 × 𝑒0
𝑦 =0×1
𝑦=0
𝑇ℎ𝑒 𝑐𝑜𝑜𝑟𝑑𝑖𝑛𝑎𝑡𝑒𝑠 𝑜𝑓 𝑡ℎ𝑒 𝑦 − 𝑖𝑛𝑡𝑒𝑟𝑐𝑒𝑝𝑡 𝑎𝑟𝑒 (0 , 0)
(b) 𝑇ℎ𝑒 ℎ𝑜𝑟𝑖𝑧𝑜𝑛𝑡𝑎𝑙 𝑎𝑠𝑦𝑚𝑝𝑡𝑜𝑡𝑒(𝑠).
𝑦 = lim 𝑓(𝑥)
𝑥→∞
𝑦 = lim 𝑥𝑒 −𝑥
𝑥→∞
𝑥
𝑦 = lim
𝑥→∞ 𝑒 𝑥
∞
𝐴𝑓𝑡𝑒𝑟 𝑖𝑚𝑚𝑒𝑑𝑖𝑎𝑡𝑒 𝑠𝑢𝑏𝑠𝑡𝑖𝑡𝑢𝑡𝑖𝑜𝑛, 𝑤𝑒 𝑔𝑒𝑡 𝑎𝑛 𝑖𝑛𝑑𝑒𝑟𝑡𝑒𝑚𝑖𝑛𝑎𝑡𝑒 𝑓𝑜𝑟𝑚 𝑜𝑓 . 𝑊𝑒 𝑚𝑎𝑦 𝑎𝑝𝑝𝑙𝑦 𝐿′ 𝐻𝑜𝑝𝑖𝑡𝑎𝑙 ′ 𝑠 𝑟𝑢𝑙𝑒.
∞
𝑑
(𝑥) = 1.
𝑑𝑥
𝑑 𝑥
(𝑒 ) = 𝑒 𝑥
𝑑𝑥
1
𝑦 = lim
𝑥→∞ 𝑒 𝑥
𝑦=0
𝑓 ℎ𝑎𝑠 𝑎 ℎ𝑜𝑟𝑖𝑧𝑜𝑛𝑡𝑎𝑙 𝑎𝑠𝑦𝑚𝑝𝑜𝑡𝑒 , 𝑦 = 0.
, 𝑇ℎ𝑒 𝑣𝑒𝑟𝑡𝑖𝑐𝑎𝑙 𝑎𝑠𝑦𝑚𝑝𝑡𝑜𝑡𝑒(𝑠).
𝑦 = 𝑥𝑒 −𝑥
𝑥
𝑦=
𝑒𝑥
𝑇ℎ𝑒 𝑑𝑒𝑛𝑜𝑚𝑖𝑛𝑎𝑡𝑜𝑟, 𝑒 𝑥 𝑐𝑎𝑛 𝑛𝑒𝑣𝑒𝑟 𝑏𝑒 𝑧𝑒𝑟𝑜, 𝑖𝑛 𝑓𝑎𝑐𝑡 𝑒 𝑥 > 0 𝑓𝑜𝑟 𝑎𝑛𝑦 𝑟𝑒𝑎𝑙 𝑛𝑢𝑚𝑏𝑒𝑟 𝑥.
𝑥
𝑆𝑜, 𝑖𝑠 𝑑𝑒𝑓𝑖𝑛𝑒𝑑 𝑜𝑛 𝑥 ∈ ℝ
𝑒𝑥
𝑓 𝑖𝑠 𝑑𝑒𝑓𝑖𝑛𝑒𝑑 𝑒𝑣𝑒𝑟𝑦𝑤ℎ𝑒𝑟𝑒 𝑜𝑛 𝑡ℎ𝑒 𝑟𝑒𝑎𝑙 𝑙𝑖𝑛𝑒.
𝑓 𝑑𝑜𝑒𝑠 𝑛𝑜𝑡 ℎ𝑎𝑣𝑒 𝑣𝑒𝑟𝑡𝑖𝑐𝑎𝑙 𝑎𝑠𝑦𝑚𝑝𝑡𝑜𝑡𝑒𝑠.
(c) 𝑓(𝑥) = 𝑥𝑒 −𝑥
𝑥
𝑓(𝑥) =
𝑒𝑥
𝑑 𝑑 𝑥
(𝑥) × 𝑒 𝑥 − (𝑒 ) × 𝑥
𝑓 ′ (𝑥) = 𝑑𝑥 𝑑𝑥
(𝑒 𝑥 )2
1 × 𝑒𝑥 − 𝑒𝑥 × 𝑥
𝑓 ′ (𝑥) =
(𝑒 𝑥 )2
𝑒 𝑥 − 𝑥𝑒 𝑥
𝑓 ′ (𝑥) =
(𝑒 𝑥 )2
𝑒 𝑥 (1 − 𝑥)
𝑓 ′ (𝑥) =
𝑒𝑥𝑒𝑥
(1 − 𝑥)
𝑓 ′ (𝑥) =
𝑒𝑥
1−𝑥
𝑓 ′ (𝑥) =
𝑒𝑥
(i)
𝐶𝑟𝑖𝑡𝑖𝑐𝑎𝑙 𝑝𝑜𝑖𝑛𝑡𝑠 𝑜𝑓 𝑓 ′ ∶
1 − 𝑥 = 0 𝑎𝑛𝑑 𝑒 𝑥 = 0
𝑥=1 𝑎𝑛𝑑 𝑛𝑜 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛
𝑇ℎ𝑒𝑟𝑒 𝑖𝑠 𝑜𝑛𝑙𝑦 𝑜𝑛𝑒 𝑐𝑟𝑖𝑡𝑖𝑐𝑎𝑙 𝑝𝑜𝑖𝑛𝑡, 𝑥=1
Question 1
𝑟 = 10
𝑑𝑉
= 15
𝑑𝑡
4
𝑉 = 𝜋𝑟 3
3
𝑑𝑉 4 𝑑𝑟
= 3 × 𝜋𝑟 3−1 ×
𝑑𝑡 3 𝑑𝑡
𝑑𝑉 𝑑𝑟
= 4𝜋𝑟 2
𝑑𝑡 𝑑𝑡
𝑑𝑟
15 = 4𝜋(10)2
𝑑𝑡
𝑑𝑟
15 = 400𝜋
𝑑𝑡
15 𝑑𝑟
=
400𝜋 𝑑𝑡
𝑑𝑟 15
=
𝑑𝑡 400𝜋
𝑆𝐴 = 4𝜋𝑟 2
𝑑 𝑑𝑟
(𝑆𝐴) = 2 × 4𝜋𝑟 2−1 ×
𝑑𝑡 𝑑𝑡
𝑑 𝑑𝑟
(𝑆𝐴) = 8𝜋𝑟
𝑑𝑡 𝑑𝑡
𝑑 15
(𝑆𝐴) = 8𝜋 × 10 ×
𝑑𝑡 400𝜋
𝑑 15
(𝑆𝐴) = 8𝜋 × 1 ×
𝑑𝑡 40𝜋
𝑑 15 ÷ 5
(𝑆𝐴) = 8 ×
𝑑𝑡 40 ÷ 5
𝑑 3
(𝑆𝐴) = 8 ×
𝑑𝑡 8
𝑑
(𝑆𝐴) = 3
𝑑𝑡
𝑇ℎ𝑒 𝑠𝑢𝑟𝑓𝑎𝑐𝑒 𝑎𝑟𝑒𝑎 𝑜𝑓 𝑡ℎ𝑒 𝑏𝑎𝑙𝑙𝑜𝑛 𝑖𝑠 𝑖𝑛𝑐𝑟𝑒𝑎𝑠𝑖𝑛𝑔 𝑎𝑡 𝑎 𝑟𝑎𝑡𝑒 𝑜𝑓 3 𝑐𝑚2 ⁄𝑠𝑒𝑐
,Question 2
𝑓(𝑥) = 𝑥𝑒 −𝑥
(a) 𝑇ℎ𝑒 𝑦 − 𝑖𝑛𝑡𝑒𝑟𝑐𝑒𝑝𝑡.
𝑦 = 𝑥𝑒 −𝑥 (𝑡𝑜 𝑠𝑢𝑏𝑠𝑡𝑖𝑡𝑢𝑡𝑒 𝑥 = 0)
𝑦 = 0 × 𝑒 −0
𝑦 = 0 × 𝑒0
𝑦 =0×1
𝑦=0
𝑇ℎ𝑒 𝑐𝑜𝑜𝑟𝑑𝑖𝑛𝑎𝑡𝑒𝑠 𝑜𝑓 𝑡ℎ𝑒 𝑦 − 𝑖𝑛𝑡𝑒𝑟𝑐𝑒𝑝𝑡 𝑎𝑟𝑒 (0 , 0)
(b) 𝑇ℎ𝑒 ℎ𝑜𝑟𝑖𝑧𝑜𝑛𝑡𝑎𝑙 𝑎𝑠𝑦𝑚𝑝𝑡𝑜𝑡𝑒(𝑠).
𝑦 = lim 𝑓(𝑥)
𝑥→∞
𝑦 = lim 𝑥𝑒 −𝑥
𝑥→∞
𝑥
𝑦 = lim
𝑥→∞ 𝑒 𝑥
∞
𝐴𝑓𝑡𝑒𝑟 𝑖𝑚𝑚𝑒𝑑𝑖𝑎𝑡𝑒 𝑠𝑢𝑏𝑠𝑡𝑖𝑡𝑢𝑡𝑖𝑜𝑛, 𝑤𝑒 𝑔𝑒𝑡 𝑎𝑛 𝑖𝑛𝑑𝑒𝑟𝑡𝑒𝑚𝑖𝑛𝑎𝑡𝑒 𝑓𝑜𝑟𝑚 𝑜𝑓 . 𝑊𝑒 𝑚𝑎𝑦 𝑎𝑝𝑝𝑙𝑦 𝐿′ 𝐻𝑜𝑝𝑖𝑡𝑎𝑙 ′ 𝑠 𝑟𝑢𝑙𝑒.
∞
𝑑
(𝑥) = 1.
𝑑𝑥
𝑑 𝑥
(𝑒 ) = 𝑒 𝑥
𝑑𝑥
1
𝑦 = lim
𝑥→∞ 𝑒 𝑥
𝑦=0
𝑓 ℎ𝑎𝑠 𝑎 ℎ𝑜𝑟𝑖𝑧𝑜𝑛𝑡𝑎𝑙 𝑎𝑠𝑦𝑚𝑝𝑜𝑡𝑒 , 𝑦 = 0.
, 𝑇ℎ𝑒 𝑣𝑒𝑟𝑡𝑖𝑐𝑎𝑙 𝑎𝑠𝑦𝑚𝑝𝑡𝑜𝑡𝑒(𝑠).
𝑦 = 𝑥𝑒 −𝑥
𝑥
𝑦=
𝑒𝑥
𝑇ℎ𝑒 𝑑𝑒𝑛𝑜𝑚𝑖𝑛𝑎𝑡𝑜𝑟, 𝑒 𝑥 𝑐𝑎𝑛 𝑛𝑒𝑣𝑒𝑟 𝑏𝑒 𝑧𝑒𝑟𝑜, 𝑖𝑛 𝑓𝑎𝑐𝑡 𝑒 𝑥 > 0 𝑓𝑜𝑟 𝑎𝑛𝑦 𝑟𝑒𝑎𝑙 𝑛𝑢𝑚𝑏𝑒𝑟 𝑥.
𝑥
𝑆𝑜, 𝑖𝑠 𝑑𝑒𝑓𝑖𝑛𝑒𝑑 𝑜𝑛 𝑥 ∈ ℝ
𝑒𝑥
𝑓 𝑖𝑠 𝑑𝑒𝑓𝑖𝑛𝑒𝑑 𝑒𝑣𝑒𝑟𝑦𝑤ℎ𝑒𝑟𝑒 𝑜𝑛 𝑡ℎ𝑒 𝑟𝑒𝑎𝑙 𝑙𝑖𝑛𝑒.
𝑓 𝑑𝑜𝑒𝑠 𝑛𝑜𝑡 ℎ𝑎𝑣𝑒 𝑣𝑒𝑟𝑡𝑖𝑐𝑎𝑙 𝑎𝑠𝑦𝑚𝑝𝑡𝑜𝑡𝑒𝑠.
(c) 𝑓(𝑥) = 𝑥𝑒 −𝑥
𝑥
𝑓(𝑥) =
𝑒𝑥
𝑑 𝑑 𝑥
(𝑥) × 𝑒 𝑥 − (𝑒 ) × 𝑥
𝑓 ′ (𝑥) = 𝑑𝑥 𝑑𝑥
(𝑒 𝑥 )2
1 × 𝑒𝑥 − 𝑒𝑥 × 𝑥
𝑓 ′ (𝑥) =
(𝑒 𝑥 )2
𝑒 𝑥 − 𝑥𝑒 𝑥
𝑓 ′ (𝑥) =
(𝑒 𝑥 )2
𝑒 𝑥 (1 − 𝑥)
𝑓 ′ (𝑥) =
𝑒𝑥𝑒𝑥
(1 − 𝑥)
𝑓 ′ (𝑥) =
𝑒𝑥
1−𝑥
𝑓 ′ (𝑥) =
𝑒𝑥
(i)
𝐶𝑟𝑖𝑡𝑖𝑐𝑎𝑙 𝑝𝑜𝑖𝑛𝑡𝑠 𝑜𝑓 𝑓 ′ ∶
1 − 𝑥 = 0 𝑎𝑛𝑑 𝑒 𝑥 = 0
𝑥=1 𝑎𝑛𝑑 𝑛𝑜 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛
𝑇ℎ𝑒𝑟𝑒 𝑖𝑠 𝑜𝑛𝑙𝑦 𝑜𝑛𝑒 𝑐𝑟𝑖𝑡𝑖𝑐𝑎𝑙 𝑝𝑜𝑖𝑛𝑡, 𝑥=1
