Container has shape of an open right circular cone. Height of container is 10 cm and diameter of top opening is 10 cm. water is evaporating so its depth, h, is changing at constant rate of -3/10 cm/hr.
a. find the volume V of the container when h=5 cm. ((Volume of cone = 1/3pie*r^2*h))
b. Find rate of change in the volume of the water with respect to time when h = 5 cm.
c. show that rate of change of volume of water due to evaporation is directly proportional to the exposed surface area of the water. what is the constnat of proportionality??
i did the first one, but i am not sure if it is right. i got (31.25*pie)/3
is this right?
also, part b, i took the derivative of...the volume equation?
and then i put in -3/10 cm for dh/dt, and 2.25 cm for r, but then i dont have dr/dt?
im confused. What is the equation i should be using for this part?
i am using dV/dt = 1/3*pie*2r*dr/dt*dh/dt
is this right...i dont think it is?
ANY help is greatly appreciated!! THANKS!
2006-10-24
10:54:37
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3 answers
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asked by
Anonymous
in
Mathematics