Q1
let 𝑢 = 𝑒 𝑥
𝑢′ = 𝑒 𝑥
Let 𝑣 = cos⁡ 𝑥
𝑣 ′ = −sin⁡ 𝑥
𝑓 ′ (𝑥)⁡= 𝑢𝑣 ′ + 𝑣𝑢′
⁡= 𝑒 𝑥 (−sin⁡ 𝑥) + 𝑒 𝑥 (cos⁡ 𝑥)
⁡= 𝑒 𝑥 (cos⁡ 𝑥 − sin⁡ 𝑥)
Q2
let 𝑢 = sin⁡(𝑒 3𝑥 )
𝑢′ ⁡= cos⁡(𝑒 3𝑥 ) × 3𝑒 3𝑥
⁡= 3𝑒 3𝑥 cos⁡(𝑒 3𝑥 )
let 𝑣 = ln⁡(cos⁡ 𝑥)
1
× −sin⁡ 𝑥
cos⁡ 𝑥
sin⁡ 𝑥
⁡= −
cos⁡ 𝑥
⁡= −tan⁡ 𝑥
𝑣 ′ ⁡=
𝑣𝑢′ − 𝑢𝑣 ′
𝑣2
(ln⁡(cos⁡ 𝑥) × 3𝑒 3𝑥 cos⁡(𝑒 3𝑥 )) − (sin⁡(𝑒 3𝑥 ) × −tan⁡(𝑥))
⁡=
ln2 ⁡(cos⁡ 𝑥)
3𝑥
3𝑥
3𝑒 cos⁡(𝑒 )ln⁡(cos⁡ 𝑥) + tan⁡(𝑥)sin⁡(𝑒 3𝑥 )
⁡=
ln2 ⁡(cos⁡ 𝑥)
𝑓 ′ (𝑥)⁡=
Q3
function is even if 𝑓(𝑥) = 𝑓(−𝑥)
function is odd if 𝑓(𝑥) = −𝑓(−𝑥)
𝑓(𝑥) = 𝑥 3 + 2𝑥 2
𝑓(−𝑥)⁡= (−𝑥)3 + 2(−𝑥)2
⁡= −𝑥 3 + 2𝑥 2
⁡≠ 𝑓(𝑥)
−𝑓(−𝑥)⁡= −(−𝑥 3 + 2𝑥 2 )
⁡= 𝑥 3 − 2𝑥 2
⁡≠ 𝑓(𝑥)
∴ 𝑓(𝑥) is neither even or odd
Q4
𝑦 = 𝑥 3 + 2𝑥 2 + 1, Domain [−4,1]
Boundary Points:
when 𝑥 = −4
𝑦⁡= (−4)3 + 2(−4)2 + 1
⁡= −31
When 𝑥 = 1
𝑦⁡= (1)3 + 2(1)2 + 1
⁡= 4
𝑑𝑦
Stationary Points (when 𝑑𝑥 = 0):
𝑑𝑦
= 3𝑥 2 + 4𝑥
𝑑𝑥
Let
𝑑𝑦
𝑑𝑥
=0
3𝑥 2 + 4𝑥 = 0
𝑥(3𝑥 + 4) = 0
𝑥 = 0⁡⁡⁡⁡and⁡⁡⁡⁡3𝑥 + 4⁡= 0
3𝑥⁡= −4
4
𝑥⁡= −
3
Test for Maximum, Minimum:
𝑑2 𝑦
= 6𝑥 + 4
𝑑𝑥 2
When 𝑥 = 0
𝑑2 𝑦
⁡= 6(0) + 4
𝑑𝑥 2
⁡= 4
4 > 0 ∴ minimum point
Find 𝑦 when 𝑥 = 4
𝑦⁡= (4)3 + 2(4)2 + 1
⁡= 97
∴ minimum point at (4,97)
4
When 𝑥 = − 3
𝑑2 𝑦
4
⁡=
6
(−
)+4
𝑑𝑥 2
3
⁡= −4
−4 < 0 ∴ maximum point
4
Find 𝑦 when 𝑥 = − 3
4 3
4 2
𝑦⁡= (− ) + 2 (− ) + 1
3
3
59
⁡=
27
4 59
∴ maximum point at (− 3 , 27)
Global Maximum:
59
The 𝑦 value of the boundary point (𝑦 = 4)is > the 𝑦 value of the maximum point (𝑦 = 27)
∴ global maximum at boundary point (1,4)
Q5
Q6
Q7
𝑑𝑉
= 80 mL/min
𝑑𝑡
Let 𝑉 = Volume of a sphere
4
⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡= 𝜋𝑟 3
3
𝑑𝑉
= 4𝜋𝑟 2
𝑑𝑟
𝑑𝑟 𝑑𝑟 𝑑𝑉
⁡=
×
𝑑𝑡 𝑑𝑉 𝑑𝑡
1
⁡=
× 80
4𝜋𝑟 2
80
⁡=
4𝜋𝑟 2
20
⁡= 2
𝜋𝑟
Substitute 𝑟 = 3cm
𝑑𝑟 20
⁡=
𝑑𝑡 𝜋𝑟 2
20
⁡=
𝜋 × 32
20
⁡=
9𝜋
20
∴ radius is increasing at a rate of 9𝜋cm/min when the radius is 3cm.
Q8
𝑥 2 = 122 − 4.52
𝑥 2 = 123.75
𝑥 = √123.75
𝑑𝑦
⁡= 5 cm/s
𝑑𝑡
⁡= 0.05 m/s
Displacement:
𝑥 2 + 𝑦 2 = 122
Velocity:
𝑥 2 + 𝑦 2 ⁡= 122
𝑑𝑥 2
𝑑𝑦 2
(𝑥 ) +
(𝑦 )⁡= 0
𝑑𝑡
𝑑𝑡
𝑑𝑥
𝑑𝑦
2𝑥 ( ) + 2𝑦 ( )⁡= 0
𝑑𝑡
𝑑𝑡
𝑑𝑥
𝑑𝑦
2𝑥 ( )⁡= 0 − 2𝑦 ( )
𝑑𝑡
𝑑𝑡
𝑑𝑥 −2𝑦 𝑑𝑦
⁡=
( )
𝑑𝑡
2𝑥 𝑑𝑡
𝑑𝑥
𝑦 𝑑𝑦
⁡= − ( )
𝑑𝑡
𝑥 𝑑𝑡
𝑑𝑦
Substitute = 0.05
𝑑𝑡
𝑑𝑥
𝑦 𝑑𝑦
⁡= − ( )
𝑑𝑡
𝑥 𝑑𝑡
−4.5
⁡=
× (0.05)
√123.75
−0.225
⁡=
√123.75
⁡= −0.02 (rounded to 2 dp)
∴ the other end will be moving at 0.02ms⁡−1 in the opposite direction to the vertical lift.