Related Rates - Frustum of a Cone

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SUMMARY

The discussion focuses on solving a related rates problem involving a frustum of a cone, specifically a container with a top radius of 9 meters, a bottom radius of 2 meters, and a height of 7 meters. The water is being filled at a rate of 4.2 cubic meters per minute, and the goal is to determine the rate at which the water level rises when the water is 1 meter deep. Participants suggest using the volume formula for a frustum of a cone, V = (π * h / 3) (R² + Rr + r²), and emphasize the importance of establishing a relationship between the height and radius through a coordinate system.

PREREQUISITES
  • Understanding of related rates in calculus
  • Familiarity with the volume formula for a frustum of a cone
  • Basic knowledge of coordinate geometry
  • Ability to differentiate functions
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  • Study the derivation of the volume formula for a frustum of a cone
  • Learn how to set up and solve related rates problems in calculus
  • Practice drawing and analyzing geometric shapes in coordinate systems
  • Explore the application of implicit differentiation in related rates scenarios
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Students studying calculus, particularly those focusing on related rates problems, as well as educators looking for effective teaching strategies for geometric applications in calculus.

Temp0
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Homework Statement



A large container has the shape of a frustum of a cone with top radius 9 metres , bottom radius 2 metres , and height 7 metres. The container is being filled with water at the constant rate of 4.2 cubic meters per minute.
At what rate is the level of water rising at the instant the water is 1 metre deep?

Homework Equations


V = 1/3∏ (r^2) * h
Internet Says the volume of a frustum of a cone is V = (∏ * h / 3) (R^2 + Rr + r^2)

The Attempt at a Solution


I don't know where to start with this one, I think I have to find some relationship between the height of the cone and the radius of the cone, however, I don't know which radius to use. Once I isolate h with V, I can just derive it and substitute 4.2 for dV/dt, 1 for h, and find dh/dt. Any tips on how to relate radius to height in this situation? Thanks in advance.
 
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Have you drawn a picture?
 
Looked at from the side, a frustrum of a cone is a trapezoid. If you set up a coordinate system with origin at the center of the smaller face, then the two sides are two lines, one passing through (r, 0) and (R, h). you can write the equation for that line and so get a relation between x (the radius) and y (the height) at each point.
 
Hmm, I've tried drawing a picture, but it didn't help.

That trapezoid idea might work, i'll try it, thanks!
 
Temp0 said:
Hmm, I've tried drawing a picture, but it didn't help.
A picture by itself probably won't help, but if you draw it on a coordinate system and identify the points on the picture by their coordinates, as HallsOfIvy suggests, it will be useful.

For problems like these it's always a good idea to draw a picture, as described above.
Temp0 said:
That trapezoid idea might work, i'll try it, thanks!
 
Last edited:
Temp0 said:
That trapezoid idea might work, i'll try it, thanks!
This is the picture I was wondering if you drew.
 
No picture.
 
I meant I was wondering if he had drawn the trapezoid picture Halls had described. I didn't attach a picture.
 

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