Depth/tank/rate of change problem

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SUMMARY

The discussion focuses on solving a related rates problem involving a conical tank with a diameter of 10 feet and a height of 12 feet. Water flows into the tank at a rate of 10 cubic feet per minute. To find the rate of change of the water depth when it reaches 8 feet, participants emphasize the importance of using the volume formula for a cone and establishing a relationship between the radius and height through similar triangles. The solution involves differentiating the volume with respect to time to express the change in volume in terms of the change in height.

PREREQUISITES
  • Understanding of related rates in calculus
  • Knowledge of the volume formula for a cone: V = (1/3)πR²h
  • Familiarity with similar triangles and their properties
  • Basic differentiation techniques
NEXT STEPS
  • Study the application of related rates in calculus problems
  • Practice deriving the volume of a cone and its implications in real-world scenarios
  • Explore the concept of similar triangles in depth
  • Learn how to differentiate functions involving multiple variables
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Students studying calculus, particularly those focusing on related rates problems, as well as educators seeking to enhance their teaching methods in mathematical concepts involving geometry and calculus.

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Homework Statement


A conical tank, (with vertex down) is 10 feet across the top and twelve feet deep. If water is flowing into the tank at a rate of 10 cubic feet per minute, find the rate of change of the depth of the water when the water is 8 feet deep.


Homework Equations

? :frown:



The Attempt at a Solution

? :bugeye:
 
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Have you made any attempt to solve this problem at all? Surely, if someone expects you to do this problem, they expect you to know some basic formulas and some concepts of "related rates".

A very relevant equation would be the volume of a cone of radius R and height h. Also, as the water level goes up, both R and h change. Here's my recomendation: draw a picture, a triangle with base (at the top) of length 10ft and altitude (downward) 12 feet. Now draw a horizontal line across the triangle representing the surface of the water. You can get a relation between h and R using "similar triangles". Replace R by that function of h so you have the volume as a function of h only and differentiate to get dV/dt depending on dh/dt.
 

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