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Oil from a ruptured tanker spreads in a circular pattern. If the radius of the circle increases at the constant rate of 1.5 fee...Asked by Karol
Oil from a ruptured tanker spreads in a circular pattern. If the radius of the circle increases at the constant rate of 1.5 feet per second, how fast is the enclosed area increasing at the end of 2 hours?
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Answered by
MathMate
A=πr²
dr/dt=1.5 ft/s
r=1.5t
dA/dt = dA/dr * dr/dt
=2πr * 1.5 sq.ft/s
=3πr
=3π(1.5*2*3600)
=28800π sq.ft/s
dr/dt=1.5 ft/s
r=1.5t
dA/dt = dA/dr * dr/dt
=2πr * 1.5 sq.ft/s
=3πr
=3π(1.5*2*3600)
=28800π sq.ft/s
Answered by
Mona
Oil spills out of a tanker at rate of 50ft3 per minute. The oil spreads in a 3
circle with thickness of 1 ft. Determine the rate at which the radius of the oil is increasing 48
when the radius reaches 200 ft.
circle with thickness of 1 ft. Determine the rate at which the radius of the oil is increasing 48
when the radius reaches 200 ft.
Answered by
Me 2
But it's 10 mins instead of two hr
Answered by
Dam what ur were y'all in
They reuse the questions from ur generation??? Bruhhh my generation is messed up . Hs class of 293456677
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