Derivatives, rates of change (cone)

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SUMMARY

The discussion focuses on a problem involving the rate of change of the height of a conical pile of gravel being dumped at a rate of 30 ft³/min. The relationship between the volume of the cone, its height, and radius is established using the formula \( V = \frac{\pi}{3} r^2 h \). The correct differentiation leads to the conclusion that the height of the pile is increasing at a rate of \( \frac{12}{\pi} \) ft/min when the height is 10 ft. A common error identified was the misinterpretation of the relationship between the radius and height of the cone.

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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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