Problem 32 - Calculus Volume 1

Question

Consider a pool shaped like the bottom half of a sphere, that is being filled at a rate of 25 \(\mathrm{ft}^{3} / \mathrm{min}\). The radius of the pool is \(10 \mathrm{ft}\). Find the rate at which the depth of the water is changing when the water has a depth of \(5 \mathrm{ft}\).

Step-by-Step Solution

Answer

The rate of change of the water depth is \( \frac{1}{\pi} \approx 0.318 \mathrm{ft/min} \).

Step 1: Understand the Problem

The pool is a hemisphere, and we are filling it with water. The rate of change of volume is given as \( \frac{dV}{dt} = 25 \, \mathrm{ft}^3/\mathrm{min} \). We need to find \( \frac{dh}{dt} \), the rate of change of the water depth, when the depth \( h = 5 \, \mathrm{ft} \) and the pool radius is \( R = 10 \, \mathrm{ft} \).

Step 2: Volume of Water

The volume of the water in the hemispherical pool as a function of its depth \( h \) is given by the formula: \[ V = \frac{1}{3}\pi h^2(3R - h) \]where \( R \) is the radius of the hemisphere. For a depth \( h \), this formula represents how the volume changes as the depth changes.

Step 3: Differentiate the Volume Formula

To find \( \frac{dh}{dt} \), we need to use the chain rule and differentiate the volume with respect to time: \[ \frac{dV}{dt} = \frac{d}{dt} \left( \frac{1}{3}\pi h^2(3R - h) \right) \]Using the chain rule, this becomes \[ \frac{dV}{dt} = \pi h (R - h)\frac{dh}{dt} \] Insert the given value, \( \frac{dV}{dt} = 25 \).

Step 4: Substitute Known Values

Given \( R = 10 \) and \( h = 5 \), substitute into the formula: \[ 25 = \pi \cdot 5 \cdot (10 - 5)\frac{dh}{dt} \]which simplifies to \[ 25 = 25\pi\frac{dh}{dt} \].

Step 5: Solve for \( \frac{dh}{dt} \)

Rearrange the equation to solve for \( \frac{dh}{dt} \):\[ \frac{dh}{dt} = \frac{25}{25\pi} \]which simplifies to \[ \frac{dh}{dt} = \frac{1}{\pi} \approx 0.318 \mathrm{ft/min} \].

Key Concepts

Hemispherical Volume

A hemisphere is essentially half of a full sphere, similar to how a bowl is half of a ball. Understanding its volume is crucial, especially when dealing with problems involving water filling such a shape, like a pool. The formula to calculate the volume of a liquid in a hemispherical container depends on its height.For our hemisphere with radius \( R \), the volume \( V \) as a function of depth \( h \) can be given by:\[ V = \frac{1}{3}\pi h^2(3R - h) \]This equation represents the relationship between the height of the water and the amount of space the water occupies in the hemisphere. If you picture pouring water into a rounded bowl, the volume increases with the depth of water, up to the midpoint of the sphere. It's important to understand this formula, as it serves as the foundational step for calculating how quickly the pool water level is rising.

Rate of Change

The concept of 'rate of change' revolves around how one quantity changes with respect to another. In this exercise, we're interested in how the depth of water, \( h \), in our hemispherical pool changes over time, which is expressed as \( \frac{dh}{dt} \).When you hear "rate of change," envision it as a speedometer for one changing factor related to another. Here, water being added to the pool is like pressing on a car's gas pedal — the pool's volume increases steadily at \( 25 \, \mathrm{ft}^3/\mathrm{min} \). Our job is to decipher how fast the depth of the water changes, given this constant fill rate.Knowing that the water volume changes at a constant rate helps us use differentiation to find how the height is affected. It’s about linking one rate (volume increase) to another (height increase) and seeing how they scale with each other.

Differentiation

Differentiation is a technique which is part of calculus and is used to understand how a function changes as its inputs change. It is a key tool for working out rates of change, like we've seen with our hemispherical pool. By differentiating, we take the derivative of a function, which gives us this rate information.In this context, we differentiate the volume formula with respect to time to find \( \frac{dh}{dt} \). The chain rule helps by linking the rates of multiple related functions:\[ \frac{dV}{dt} = \pi h (R - h)\frac{dh}{dt} \]This relationship lets us plug in known values, differentiating directly from the formula for volume to get a handle on \( \frac{dh}{dt} \). Solving these types of problems often involves manipulating such derivative relations to isolate the quantity of interest, here \( \frac{dh}{dt} \), reflecting how the pool's water depth changes over time. Differentiation simplifies this dynamic change into a solvable equation.