Problem 70 - Calculus for Biology and Medicine

Question

Suppose that we pump water into an inverted right circular conical tank at the rate of 5 cubic feet per minute (i.e., the tank stands with its point facing downward). The tank has a height of 6 ft and the radius on top is \(3 \mathrm{ft}\). What is the rate at which the water level is rising when the water is \(2 \mathrm{ft}\) deep? (Note that the volume of a right circular cone of radius \(r\) and height \(h\) is \(V=\frac{1}{3} \pi r^{2} h .\) )

Step-by-Step Solution

Answer

The water level is rising at a rate of \( \frac{5}{\pi} \) ft/min when the water is 2 ft deep.

Step 1: Understanding the Problem

We need to find how fast the water level (height of water, h) in the cone-shaped tank is rising when the current water depth is 2 ft. The tank is being filled at a rate of 5 cubic feet per minute.

Step 2: Relate Radius and Height in the Tank

Since the tank is a cone, the relationship between the radius and height of the water is proportional. For the full tank (height = 6 ft, radius = 3 ft), the ratio is \( \frac{r}{h} = \frac{3}{6} = \frac{1}{2} \). Thus, for any water level, \( r = \frac{1}{2}h \). This relation helps in expressing the radius in terms of height.

Step 3: Express Volume as a Function of Height

Using the relation obtained, substitute for \( r \) in the volume formula: \[ V = \frac{1}{3} \pi (\frac{1}{2}h)^2 h = \frac{1}{3} \pi \cdot \frac{h^3}{4} = \frac{\pi}{12} h^3 \].

Step 4: Differentiate Volume with Respect to Time

Take the derivative of the volume function with respect to time: \( \frac{dV}{dt} = \frac{d}{dt} \left( \frac{\pi}{12} h^3 \right) = \frac{\pi}{12} (3h^2 \frac{dh}{dt}) = \frac{\pi}{4} h^2 \frac{dh}{dt} \).

Step 5: Substitute Known Values to Solve for \( \frac{dh}{dt} \)

We know \( \frac{dV}{dt} = 5 \) cubic feet/min and want to find the rate of change of the height \( \frac{dh}{dt} \) when \( h = 2 \) ft. Substitute: \[ 5 = \frac{\pi}{4} (2)^2 \frac{dh}{dt} \].

Step 6: Solve for \( \frac{dh}{dt} \)

Simplify and solve for \( \frac{dh}{dt} \): \[ 5 = \frac{\pi}{4} \times 4 \times \frac{dh}{dt} = \pi \times \frac{dh}{dt} \]. Thus, \( \frac{dh}{dt} = \frac{5}{\pi} \).

Step 7: Final Answer

At the moment when the water in the tank is 2 feet deep, the rate at which the water height is rising is \( \frac{5}{\pi} \) ft/min.

Key Concepts

Calculus

Calculus is critical in working with problems involving changes, such as how the water height in a tank changes over time as it fills. One of calculus's primary roles is to provide tools for analyzing and modeling dynamic systems. Functions, derivatives, and integrals are the building blocks of calculus.

• Functions represent relationships, like how volume relates to cone height and radius in this scenario.
• Derivatives help you understand how quantities change; in our example, they show how water height increases as volume changes.
• Integrals, while not directly used in this problem, allow you to accumulate or combine small quantities, useful when calculating volumes or areas in certain situations.

By translating word problems into mathematical equations, calculus gives us a toolkit to systematically analyze these changes, making sense of how they interact and evolve over time.

In the example of our conical tank, calculus helped to derive how quickly water height increases by relating it first to the more directly observed volume change. By understanding the rate of volume change, calculus enabled the determination of height changes through differentiation.

Differentiation

Differentiation is a fundamental concept of calculus used to find the rate at which a quantity changes. This concept is particularly evident in "related rates" problems where you relate different changing quantities and find one rate based on another.

In our exercise with the conical tank, we needed to determine how fast the height of the water changes concerning time. This required setting up a differentiation equation based on volume and height. We expressed water volume as a function of height, giving:\[ V = \frac{\pi}{12} h^3\]Taking the derivative with respect to time, you get:\[\frac{dV}{dt} = \frac{\pi}{4} h^2 \frac{dh}{dt}\]
Taking derivatives like this allows you to determine the rate of height change (\frac{dh}{dt}) once you have the rate of volume change (\frac{dV}{dt}) and the current water height ( h ).

Differentiation provides a clear framework for engaging with how rates of change relate to functions at both a specific point and over intervals, ensuring you can accurately determine and predict dynamical behavior as in our tank problem.

Problem Solving

Problem solving in mathematics involves breaking down complex issues into simpler parts, as demonstrated in the exercise above. Here are steps taken that typify effective problem-solving strategies:

• Understanding the problem: Clearly define what is given and what needs to be found. In our case, the rate at which water level rises is sought.
• Establishing relationships: Use the known dimensions and relationships (like the similarity of triangles) in the cone to relate variables, such as height and radius.
• Applying differentiation: Form key equations (e.g., variable relationships) and differentiate them to relate rates.
• Substituting values: Use given numerical values and known rates to solve the equations.
• Verification: Review solutions to ensure they make sense within the problem's context.

Applying these steps systematically enhances accuracy and efficiency in mathematical problem-solving. With the cone problem, each step clarified a part of the challenge, eventually leading to a coherent answer about how fast the water level rises.