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Rate related problem with irregular cone and and time delay

  • Thread starter FiveAlive
  • Start date
  • #1
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A water glass (10 cm diameter at the top, 6 cm diameter at the bottom, 20 cm in height) is being filled at a rate of 50 cm^3/min. Find the rate of change of the height of the water after 5 seconds.

V=1/3(3.14) r^2h

I'm a little unsure how to approach this problem for two reasons. A) The glass is not a 'proper' cone shape(it has a 'flat' bottom of 6cm), in the problems I have been solving dealing with cones there has been only one radius to deal with. B) The problem is asking for the rate of change after a period of time. In most problems I have been asked to solve for some variable in the equation( ie ...find the rate of change after a certain height has been reached in the container)

Any suggestions would be appreciated.
 

Answers and Replies

  • #2
16
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Update: So I've been looking at my rates of change according to height (at 1 cm, 2 cm, 3 cm ect) and I'd be able to solve it if the cone didn't have a 6 cm diameter. I am still unsure how to tackle this issue.
 
  • #3
16
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Update: Still a little unsure how to resolve the time issue of this problem however I believe I would use the formula V=1/3(3.14)(R^2H-r^2h) to solve for my irregular cone. Am I on the right track?
 

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