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Chapter 3: Problem 21

Water is leaking out the bottom of a hemispherical tank of radius 8 feet at a rate of 2 cubic feet per hour. The tank was full at a certain time. How fast is the water level changing when its height \(h\) is 3 feet? Note: The volume of a segment of height \(h\) in a hemisphere of radius \(r\) is \(\pi h^{2}[r-(h / 3)] .\)

Short Answer

Expert verified
The water level decreases at \(-\frac{2}{39\pi}\) feet/hour when the height is 3 feet.

Step by step solution

01

Understanding the Problem

We have a hemispherical tank and water is leaking from it. We need to find the rate of change of the water level, which means we need to find \( \frac{dh}{dt} \) when \( h = 3 \) feet. We know \( \frac{dV}{dt} = -2 \) cubic feet per hour and the formula for the volume of the segment is given as \( V = \pi h^{2} [r - \frac{h}{3}] \).
02

Differentiate the Volume Formula

Given the volume formula \( V = \pi h^{2} \left(r - \frac{h}{3}\right) \), differentiate both sides with respect to time \( t \):\[ \frac{dV}{dt} = \frac{d}{dt}(\pi h^{2} [r - \frac{h}{3}]) \]Apply the product rule and chain rule to differentiate the equation.
03

Calculate the Derivative

Differentiate the equation:\[ \frac{dV}{dt} = \pi \left(2h (r - \frac{h}{3}) \frac{dh}{dt} + h^{2} \left(\frac{-1}{3}\right) \frac{dh}{dt}\right) \]Factor out \( \frac{dh}{dt} \) to simplify:\[ \frac{dV}{dt} = \pi \frac{dh}{dt} \left( 2h (r - \frac{h}{3}) - \frac{h^2}{3}\right) \]
04

Substitute Known Values

Substitute \( \frac{dV}{dt} = -2 \), \( r = 8 \), and \( h = 3 \) into the equation:\[ -2 = \pi \frac{dh}{dt} \left( 2(3)(8 - \frac{3}{3}) - \frac{3^2}{3}\right) \]Simplify the equation:
05

Solve for \( \frac{dh}{dt} \)

After simplification, solve for \( \frac{dh}{dt} \):\[ -2 = \pi \frac{dh}{dt} \left( 2(3)(7) - 3 \right) \]This becomes:\[ -2 = \pi \frac{dh}{dt} (42 - 3) \]\[ -2 = \pi \frac{dh}{dt} (39) \]Finally, solve for \( \frac{dh}{dt} \):\[ \frac{dh}{dt} = \frac{-2}{39\pi} \]
06

Interpret the Result

The negative sign indicates the water level is decreasing. Thus, the rate at which the water level is falling when \( h = 3 \) feet is \(-\frac{2}{39\pi}\) feet per hour.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

hemispherical tank
A hemispherical tank is a three-dimensional geometric container that is half of a sphere. Naturally, it has a circular base and a curved surface, just like half of a round ball. The radius of the hemisphere is fundamental in determining various calculations such as its volume or surface area. This type of tank is frequently utilized due to its efficient use of space and balance of pressure within.
  • The tank's radius is central to all calculations. In this context, the radius is 8 feet.
  • The hemispherical shape implies calculations for curvature and volume are different than for cylindrical or rectangular tanks.
  • Understanding the geometry will help in applying related mathematical formulas.
Knowing how water flows and functions within a hemispherical tank can be crucial in applying calculus, especially related rates.
volume of a segment
The volume of a segment in a hemisphere is an interesting concept. When we talk about "a segment of height \(h\)", it refers to the slice of the tank (part of the hemisphere) containing water up to a height \(h\). Calculating this volume correctly is key to solving the problem.
  • The formula provided is: \( V = \pi h^{2} \left( r - \frac{h}{3} \right) \), where \(r\) is the radius.
  • This formula helps calculate the occupied space by water within a specified height.
  • It鈥檚 essential to recognize that the volume considered here changes as the water level changes.
This understanding is crucial because the rate at which this volume changes is what ties into the related rates problem, focusing on how fast the height \(h\) is changing.
chain rule
The chain rule in calculus is a fundamental method used to find the derivative of composite functions. In the context of related rates, it assists in differentiating expressions with respect to time when variables are interconnected through functions.
  • For example, when differentiating the volume formula \( V = \pi h^{2} \left( r - \frac{h}{3} \right) \), we apply the chain rule to each part affected by \(h\).
  • Using the chain rule, we consider each component of the formula, treating it as a function of \(h\) and then differentiate in terms of \(t\).
  • This rule is invaluable when dealing with problems where rates of change are indirectly involved.
Mastery of the chain rule allows for the accurate linking of different rates of change, especially in problems involving volumes and heights as in this case.
differentiation
Differentiation is the process through which we find the rate at which one quantity changes with respect to another. It's an essential operation in calculus used to determine how variables behave over time, such as the water level inside the tank in this problem.
  • Here, the differentiation of the volume formula gives us the relationship between the rate of change of volume \(\frac{dV}{dt}\) and the rate of change of height \(\frac{dh}{dt}\).
  • The differentiated form \( \frac{dV}{dt} = \pi \frac{dh}{dt} \left( 2h \left(r - \frac{h}{3}\right) - \frac{h^2}{3} \right) \) incorporates the chain rule, enabling us to solve for \(\frac{dh}{dt}\).
  • It's the differentiation process that allows us to find specific rates like how fast the water level is dropping when the height is exactly 3 feet.
Differentiation not only aids in solving for unknown rate changes but also offers insight into the dynamic behavior of systems and their elements.

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