Derivatives, rates of change (cone)

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Homework Help Overview

The problem involves rates of change related to the volume of a conical pile of gravel being formed as it is dumped from a conveyor belt. The relationship between the height and radius of the cone is specified, with the base diameter equal to the height.

Discussion Character

  • Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the relationship between the radius and height of the cone, with one participant attempting to clarify their approach to differentiation and volume calculation.

Discussion Status

There is an ongoing examination of the original poster's calculations and assumptions, with participants providing feedback on potential errors in differentiation and clarifying the geometric relationships involved.

Contextual Notes

Participants are addressing a discrepancy between the original poster's calculated rate of change and the textbook answer, indicating a need for further exploration of the assumptions made in the problem setup.

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1. Gravel is being dumped from a conveyor belt at a rate of 30 ft3/min, and its coarseness is such that it forms a pile in the shape of a cone whose base diameter and height are always equal. How fast is the height of the pile increasing when the pile is 10 ft high?

Homework Equations


$$V=\frac{\pi}{3}r^2h$$

The Attempt at a Solution



Diameter = height, so $$\frac{h}{2}=r$$
$$V=\frac{\pi}{3}\frac{h^2}{4}h = \frac{\pi}{12}h^3$$
$$\frac{dV}{dt}=\frac{\pi}{12}(3×h)\frac{dh}{dt}$$
$$30=\frac{\pi}{12}(3×10)\frac{dh}{dt}$$
$$\frac{dh}{dt}=\frac{12}{\pi}$$

The textbook's answer is $$\frac{6}{5\pi}$$ What did I do wrong?
 
Last edited:
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You incorrectly assumed that the diameter of the pile = half of the radius.
 
Sorry that was just a typo. The work should still follow r=h/2.
 
Check your differentiation.
 

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