A viscious liquid is poured onto a flat surface. It forms a circular patch which grows at a steady rate of 8cm^3/s. Find in terms of π,
a)radius of the patch 25s after pouring has commenced;
b)rate of increase of radius at this instant
a)radius of the patch 25s after pouring has commenced;
b)rate of increase of radius at this instant
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Area of a Circle, A = πr²
Rate of change of A = dA/dt = 8
Then ∫dA/dt dt = A = 8t + k.....assume that when t = 0 then A = 0 thus k = 0 then A = 8t.
Equating the two expression for A gives : πr² = 8t : r = √(8t/π)
a) when t = 25 then r = √[(8x25)/π] = 7.98 cm (2dp)
b) By the chain rule dA/dt = dA/dr * dr/dt and thus dr/dt = (dA/dt ) ÷ (dA/dr)
A = πr² : dA/dr = 2πr
dr/dt = 8 ÷ 2πr = 4/πr
When r ≅ 7.98 then dr/dt = 4/7.98π = 0/16 cm/sec (2dp)
Rate of change of A = dA/dt = 8
Then ∫dA/dt dt = A = 8t + k.....assume that when t = 0 then A = 0 thus k = 0 then A = 8t.
Equating the two expression for A gives : πr² = 8t : r = √(8t/π)
a) when t = 25 then r = √[(8x25)/π] = 7.98 cm (2dp)
b) By the chain rule dA/dt = dA/dr * dr/dt and thus dr/dt = (dA/dt ) ÷ (dA/dr)
A = πr² : dA/dr = 2πr
dr/dt = 8 ÷ 2πr = 4/πr
When r ≅ 7.98 then dr/dt = 4/7.98π = 0/16 cm/sec (2dp)