Math Problem Statement

At a sand and gravel plant, sand is falling off a conveyor and onto a conical pile at a rate of 16 cubic feet per minute. The diameter of the base of the cone is approximately three times the altitude. At what rate (in ft/min) is the height of the pile changing when the pile is 22 feet high?

Solution

The question you're working on involves a related rates problem where sand is falling onto a conical pile.

Key Information from the Problem:

  • Sand is falling at a rate of 16cubic feet per minute16 \, \text{cubic feet per minute}, i.e., dVdt=16ft3/min\frac{dV}{dt} = 16 \, \text{ft}^3/\text{min}.
  • The diameter of the base of the cone is three times the height: d=3hd = 3h, which implies r=d2=3h2r = \frac{d}{2} = \frac{3h}{2}.
  • The height of the pile is 22feet22 \, \text{feet}.
  • The formula for the volume of a cone is: V=13πr2hV = \frac{1}{3} \pi r^2 h
  • We need to find dhdt\frac{dh}{dt} when h=22feeth = 22 \, \text{feet}.

Solution Outline:

  1. Volume Equation Substitution: From the relation r=3h2r = \frac{3h}{2}, substitute into the volume formula: V=13π(3h2)2h=13π(9h24)h=3π4h3V = \frac{1}{3} \pi \left( \frac{3h}{2} \right)^2 h = \frac{1}{3} \pi \left( \frac{9h^2}{4} \right) h = \frac{3\pi}{4} h^3 So, the volume in terms of hh is: V=3π4h3V = \frac{3\pi}{4} h^3

  2. Differentiate with respect to time tt: Use implicit differentiation to find the rate of change of volume with respect to time: dVdt=3π43h2dhdt=9π4h2dhdt\frac{dV}{dt} = \frac{3\pi}{4} \cdot 3h^2 \frac{dh}{dt} = \frac{9\pi}{4} h^2 \frac{dh}{dt}

  3. Solve for dhdt\frac{dh}{dt}: Since dVdt=16ft3/min\frac{dV}{dt} = 16 \, \text{ft}^3/\text{min}, plug that into the equation: 16=9π4h2dhdt16 = \frac{9\pi}{4} h^2 \frac{dh}{dt} When h=22feeth = 22 \, \text{feet}, substitute this value: 16=9π4(22)2dhdt16 = \frac{9\pi}{4} (22)^2 \frac{dh}{dt} Simplify: 16=9π4×484×dhdt16 = \frac{9\pi}{4} \times 484 \times \frac{dh}{dt} 16=1089πdhdt16 = 1089\pi \frac{dh}{dt} Solve for dhdt\frac{dh}{dt}: dhdt=161089π\frac{dh}{dt} = \frac{16}{1089\pi}

  4. Numerical Solution: Approximating the value of π\pi: dhdt161089×3.1416163421.970.00467ft/min\frac{dh}{dt} \approx \frac{16}{1089 \times 3.1416} \approx \frac{16}{3421.97} \approx 0.00467 \, \text{ft/min}

Thus, the rate at which the height of the pile is changing is approximately 0.00467ft/min0.00467 \, \text{ft/min}.

Would you like more details on any step or have any other questions?

Here are five related questions:

  1. How would the solution change if the diameter was twice the height instead of three times?
  2. What if the rate of sand falling was not constant but a function of time?
  3. Can you calculate the rate of change of the radius when the height is 22 feet?
  4. How would this problem change if it involved a pyramid instead of a cone?
  5. What if the base of the cone was not circular but elliptical?

Tip: Always double-check units and make sure they are consistent throughout your calculations to avoid errors.

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Math Problem Analysis

Mathematical Concepts

Related Rates
Differentiation
Cone Geometry

Formulas

Volume of a cone: V = (1/3)πr^2h
Relationship between diameter and height: d = 3h, which implies r = (3h)/2

Theorems

Implicit Differentiation

Suitable Grade Level

Undergraduate Calculus