J's answer to Sally Tan's Junior College 1 H2 Maths Singapore question.
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dz/dt ∝ 1/ z³/²
So the differential equation you would derive is:
dz/dt = k / z³/² , k is a constant.
From here, z³/² dz/dt = k
Integrate both sides with respect to t,
1/(3/2 + 1) · z³/² +¹ = kt
2/5 · z⁵/² = kt + c , c is a constant.
When z = 25, t = 0 ,
2/5 · (25)⁵/² = k(0) + c
2/5 · (3125) = c
c = 1250
When z = 16, t = 30,
2/5 · (16)⁵/² = k(30) + 1250
2/5 · (1024) = 30k + 1250
2048/5 = 30k + 1250
30k = -4202/5
k = -2101/75
So when z = 0,
2/5 · 0⁵/² = -2101/75 t + 1250
0 = -2101/75 t + 1250
2101/75 t = 1250
t = 44 1306/2101 ≈ 44.6 (3s.f)
So the differential equation you would derive is:
dz/dt = k / z³/² , k is a constant.
From here, z³/² dz/dt = k
Integrate both sides with respect to t,
1/(3/2 + 1) · z³/² +¹ = kt
2/5 · z⁵/² = kt + c , c is a constant.
When z = 25, t = 0 ,
2/5 · (25)⁵/² = k(0) + c
2/5 · (3125) = c
c = 1250
When z = 16, t = 30,
2/5 · (16)⁵/² = k(30) + 1250
2/5 · (1024) = 30k + 1250
2048/5 = 30k + 1250
30k = -4202/5
k = -2101/75
So when z = 0,
2/5 · 0⁵/² = -2101/75 t + 1250
0 = -2101/75 t + 1250
2101/75 t = 1250
t = 44 1306/2101 ≈ 44.6 (3s.f)
Date Posted:
3 years ago
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