- #1
tangents
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- 0
Hey guys, I'm having some trouble figuring out this problem and was wondering if someone would be kind enough to look over my work so I know that I am doing it correctly =).
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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 =/
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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 =/