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There is an open right cylinder cone that has a heigh of 10cm and diameter of 10cm. the depth h is changing at -3/10cm/hr. V=1/3(pi)r^2(h).

a) find volume of water at h=5

V= 1/3(pi)(5^2)(5)= 125/3(pi)cm^3

b) find dv/dt when h=5cm

dv/dt= 1/3(pi)(5^2)(-3/10)=-5/2cm^3/hr

c) show that rate of change of volume of water is directly proportional to the surface area of water. what is the constant of proportionality

dv/dt=1/3(pi)r^2(dh/dt)

surface area=(pi)r^2

1/3(pi)r^2(dh/dt)=(pi)r^2

constant is (pi)r^2

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My main problem is with parts b and c as I'm not sure if I am getting the concept of related rates correctly =/