Problem 20 - Calculus of a Single Variable

Question

A trough is 12 feet long and 3 feet across the top (see figure). Its ends are isosceles triangles with altitudes of 3 feet. (a) Water is being pumped into the trough at 2 cubic feet per minute. How fast is the water level rising when the depth \(h\) is 1 foot? (b) The water is rising at a rate of \(\frac{3}{8}\) inch per minute when \(h=2 .\) Determine the rate at which water is being pumped into the trough.

Step-by-Step Solution

Answer

The water level is rising at \( \frac{1}{12} \) ft/min when the depth of the water is 1 foot. The rate at which the water is being pumped into the trough when the water level is rising at \( \frac{3}{8} \) inches per minute and the depth of the water is 2 feet is 1.5 cubic ft/min.

Step 1: Establishing the relationship

The trough has the shape of a triangular prism, so the volume of water in the trough can be calculated by multiplying the area of a cross-sectional triangle by the length of the trough, \( V = A l \). The area of the isosceles triangle cross section can be represented by \( A = \frac{1}{2} b h \), where \( b \) is the base and \( h \) is the height. Substituting this into the initial volume formula, we have \( V = \frac{1}{2} b h l \). Considering that the ratio of the base, \( b \), to the altitude of the isosceles triangle, \( h \), is given by the Pythagorean theorem as 2:1, we substitute \( b = 2h \), to get \( V = \frac{1}{2} (2h) h l = h^2 l \). Thus, the relationship between volume and height settled is \( V = l h^2 \).

Step 2: Derive both sides with respect to time

On differentiating both sides with respect to time (t), we have \( dV/dt = l * 2h (dh/dt) \). From this equation, we can solve for the rate of change of the water level, \( dh/dt \), remembering that \( dV/dt \) represents the rate at which water is being pumped into the trough, \( 2h \) is twice the current water level and \( l \) is the length of the trough.

Step 3: Substituting the given values

For part (a), we know that the rate at which water is being pumped into the trough is \( dV/dt = 2 \) and that we need to find \( dh/dt \) when the depth of the water \( h = 1 \). Thus, we plug these values into the equation to get \( dh/dt = \frac{dV/dt}{2lh} = \frac{2}{2*12*1} = \frac{1}{12} \) ft/min. For part (b), we're given that the water level is rising at a rate of \( dh/dt = \frac{3}{8} \) inches per minute when \( h = 2 \). We must convert this rate to feet per minute before substituting the given values into the equation. Therefore, \( dh/dt = \frac{3}{8} * \frac{1}{12} \) ft/min = \( \frac{1}{32} \) ft/min. Substituting \( h = 2 \) and that \( dh/dt = \frac{1}{32} \) into the equation, we find \( dV/dt = 2*12*2*\frac{1}{32} = 1.5 \) cubic ft/min.

Key Concepts

Understanding Triangular Prism Volume

A triangular prism is a three-dimensional geometric shape, and its volume can be found by using a specific formula. The volume of a prism is generally equal to the area of its base multiplied by its height (or length).
The cross-section of our problem's prism is an isosceles triangle.

• The volume formula for a triangular prism: \[ V = A \times l \]where:
  • \( V \) is the volume
  • \( A \) is the area of the triangular base
  • \( l \) is the length of the prism

This formula is significant in contexts involving problems like water flows or storage capacities. Properly applying it provides a clear understanding of how physical dimensions affect volume.

Characteristics of Isosceles Triangle

An isosceles triangle is unique because it has two sides that are equal in length.
It also has two angles that are the same.
In the context of the exercise, the isosceles shape forms the cross-section of the trough.

• Area formula of an isosceles triangle: \[ A = \frac{1}{2} \times b \times h \]Where:
  • \( b \) is the base
  • \( h \) is the height (altitude)
• The relationship between base and height as per the problem: \( b = 2h \)

Understanding the properties of an isosceles triangle aids in calculating dimensions consistently as the water level changes, specifically in dynamic environments like in a trough being filled with water.

Differentiation in Related Rates

Differentiation is a critical mathematical concept used in calculus to determine how a function changes as its inputs change.

• In this problem, we use differentiation to analyze how the volume of water in the trough changes over time.
• The derivative \( \frac{dV}{dt} \) represents the rate at which the volume changes over time.
• Similarly, \( \frac{dh}{dt} \) shows how the water level height changes over time.

Applying differentiation to problems involving related rates helps understand and predict how different variables are interdependent.
It's especially useful for real-time applications where rates are not constant, such as the filling of a trough.

Explaining Cubic Feet Per Minute

"Cubic feet per minute" (CFM) is a measure of flow rate that describes how much volume moves through a system in one minute.
In fluid dynamics, CFM indicates how quickly a fluid, such as water, is moving.

• In our exercise, \( \frac{dV}{dt} \) is given in cubic feet per minute.
  • This shows the volume of water added to the trough each minute.
  • It aids in visualizing the speed of water accumulation.

Understanding flow rate units like CFM is crucial for effective analysis and problem-solving in scenarios where time and volume are essential, such as engineering and environmental calculations.