Problem 16 - The Calculus with Analytic Geometry

Question

A horizontal trough is \(16 \mathrm{ft}\) long, and its ends are isosceles trapezoids with an altitude of \(4 \mathrm{ft}\), a lower base of \(4 \mathrm{ft}\), and an upper base of \(6 \mathrm{ft}\). Water is being poured into the trough at the rate of \(10 \mathrm{ft}^{3} / \mathrm{min}\). How fast is the water level rising when the water is \(2 \mathrm{ft}\) deep?

Step-by-Step Solution

Answer

The water level is rising at \ \( \frac{5}{24} \) feet per minute.

Step 1: Understand the Trough Dimensions

The trough is a prism with a length of 16 feet and cross-sectional area that is an isosceles trapezoid. The trapezoid has an altitude (height) of 4 feet, a lower base of 4 feet, and an upper base of 6 feet.

Step 2: Volume of Water in the Trough

The volume of water in the trough at any time can be found by multiplying the cross-sectional area of the trapezoid by the length of the trough. We need to express the area of the trapezoid as a function of the water depth.

Step 3: Area of Cross-Section

The area of a trapezoid is given by the formula: \[ \text{Area} = \frac{1}{2} \times (\text{Base}_1 + \text{Base}_2) \times \text{Height} \]

Step 4: Express Bases as a Function of Height (Depth)

Let the water depth be h feet. The lower base remains 4 feet. The upper base changes linearly from 4 feet to 6 feet as the height increases from 0 to 4 feet. Thus, at any height h, the length of the top base is: \[ \text{Base}_2 = 4 + \frac{h}{2} \]

Step 5: Expression of Cross-Sectional Area

Substitute the values into the area formula: \[ A = \frac{1}{2} \times (4 + 4 + h/2) \times h \] \[ A = \frac{1}{2} \times \frac{8 + h}{2} \times h \] \[ A = \frac{8 + h}{4} \times h = \frac{8h + h^2}{4} \]

Step 6: Volume of Water in the Trough

The volume V of water is the area A of the cross-section times the length of the trough (16 feet): \[ V = A \times 16 = \frac{16(8h + h^2)}{4} = 4(8h + h^2) = 32h + 4h^2 \]

Step 7: Rate of Volume Change

Given that water is being poured into the trough at 10 cubic feet per minute, we have: \[ \frac{dV}{dt} = 10 \]

Step 8: Relation Between Volume and Water Depth

To find the rate at which the water level is rising, differentiate the volume V with respect to time t: \[ V = 32h + 4h^2 \] \[ \frac{dV}{dt} = 32 \frac{dh}{dt} + 8h \frac{dh}{dt} \] \[ 10 = (32 + 8h) \frac{dh}{dt} \]

Step 9: Solve for dh/dt

Substitute the given water depth (2 feet) into the equation: \[ 10 = (32 + 8 \times 2) \frac{dh}{dt} \] \[ 10 = (32 + 16) \frac{dh}{dt} \] \[ 10 = 48 \frac{dh}{dt} \] \[ \frac{dh}{dt} = \frac{10}{48} = \frac{5}{24} \]

Step 10: Final Answer

The rate at which the water level is rising when the water is 2 feet deep is: \[ \frac{dh}{dt} = \frac{5}{24} \text{ feet per minute} \]

Key Concepts

Calculus

In this related rates problem, we use calculus to understand how one quantity changes in relation to another. Here, we need to find how fast the water level rises in a trough when water is being poured in at a known rate. The key steps in solving this problem involve using differentiation, specifically by differentiating the equation that expresses volume in terms of water depth with respect to time.
Given the volume of the trough as a function of the water depth, we differentiate both sides of the equation, using the chain rule to incorporate the rate at which the water is being poured. For example:
\( V = 32h + 4h^2 \) \[ \frac{dV}{dt} = 32 \frac{dh}{dt} + 8h \frac{dh}{dt} \] The right side of the equation represents how the volume changes with height (depth), while the left side (dV/dt) represents how the volume changes with time.
The connection between the time rate of change of the volume and the water depth rate (dh/dt) is the core aspect of related rates problems. By solving the equation, we identify how quickly the water height is increasing in the trough, given the rate at which water is added.

Analytic Geometry

Analytic geometry allows us to describe the shapes within this problem using algebraic equations. In this trough, the cross-sectional shape is an isosceles trapezoid. We use properties of geometric shapes to compute the area of the trapezoid, which is crucial for determining the volume of water in the trough.
To find the area of the trapezoid as a function of water depth, we need to express the change in the length of the bases as the depth increases. For instance, if the height \( h \) varies, the upper base will also vary and can be described as: \[ \text{Base}_2 = 4 + \frac{h}{2} \] With this relationship, we calculate the area \( A\) of the trapezoid at any given depth h: \[ A = \frac{1}{2} \left( 4 + 4 + \frac{h}{2} \right) h = \frac{2}{8} (8h + h^2) \] By using these geometric principles, we lay down a foundation that links the area of the cross-section with its changing dimensions. This fundamental understanding of geometry supports the calculus used to find the rate at which the water level rises.

Volume

Volume is a measure of the amount of space that a substance or object occupies. In this problem, the volume of water in the trough changes as water is added. We need to determine how the volume increases based on the given dimensions of the trough and the rate at which water pours in.
To find the volume, we calculate the cross-sectional area of the water level at a specific depth, then multiply this by the length of the trough: \[ V = A \times 16 = \frac{16(8h + h^2)}{4} = 4(8h + h^2) \] This gives us the volume as a function of depth \( h \): \[ V = 32h + 4h^2 \] With this relationship, we can find how quickly the volume increases by differentiating with respect to time, given a constant rate of water flow. By analyzing the volume in connection to depth and using the differentiation process, students understand the concept of how changing one measurement (e.g., water height) impacts another (e.g., volume). This is essential in solving practical problems involving varying volumes in real-time scenarios.