J's answer to universe's Junior College 1 H1 Maths Singapore question.
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Let area of sheet be A. Let x be the length of one side of the sheet.
Initial length of square = 8 cm = 80mm
So x = 80 + 0.2t, where t is time in seconds.
dx/dt = 0.2
(Actually it was already given that dx/dt = 0.2 mm/s)
A = x²
dA/dx = 2x
dA/dt = dA/dx × dx/dt
dA/dt = 2x (0.2) = 0.4x
dA/dt = 0.4(80 + 0.2t)
dA/dt = 32 + 0.08t
When t = 5,
dA/dt = 32 + 0.08(5) = 32 + 0.4 = 32.4
Rate of change of area after 5 seconds = 32.4 mm/s
Initial length of square = 8 cm = 80mm
So x = 80 + 0.2t, where t is time in seconds.
dx/dt = 0.2
(Actually it was already given that dx/dt = 0.2 mm/s)
A = x²
dA/dx = 2x
dA/dt = dA/dx × dx/dt
dA/dt = 2x (0.2) = 0.4x
dA/dt = 0.4(80 + 0.2t)
dA/dt = 32 + 0.08t
When t = 5,
dA/dt = 32 + 0.08(5) = 32 + 0.4 = 32.4
Rate of change of area after 5 seconds = 32.4 mm/s
Date Posted:
3 years ago
Alternatively,
A = x²
A = (80 + 0.2t)²
A = 6400 + 32t + 0.04t²
dA/dt = 32 + 0.08t
When t = 5,
dA/dt = 32 + 0.08(5) = 32 + 0.4 = 32.4
Rate of change of area after 5 seconds = 32.4 mm/s
A = x²
A = (80 + 0.2t)²
A = 6400 + 32t + 0.04t²
dA/dt = 32 + 0.08t
When t = 5,
dA/dt = 32 + 0.08(5) = 32 + 0.4 = 32.4
Rate of change of area after 5 seconds = 32.4 mm/s
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