A container has the shape of an open right circular cone, as shown. The height of the container is 10 cm and the diameter of the opening is 10 cm. water in the container is evaporating so that its depth h is changing at a constant rate of - 3/10cm/hr. (v=1/3pir^2h). A) find the volume when h=5cm and B) fine the rate of change of the volume of water when h=5

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Mathematics

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23 February, 01:07

A container has the shape of an open right circular cone, as shown. The height of the container is 10 cm and the diameter of the opening is 10 cm. water in the container is evaporating so that its depth h is changing at a constant rate of - 3/10cm/hr. (v=1/3pir^2h). A) find the volume when h=5cm and B) fine the rate of change of the volume of water when h=5

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Onyx · Answer 1

A) Volume = (1/12) pi*h^3, with height = 5cm.

b) You should be able to differentiate V = (1/12) pi*h^3 with respect to h, and you were given dh/dt = - 0.3 cm/hr.

does that make sense?