Math Problem Statement

A swimming pool is constructed to have a uniform cross-section from an aerial view as shown below.The pool has a uniform depth of 1·6 metres. By using TWO applications of the trapezoidal rule, estimate the capacity of the pool to the nearest litre.

Solution

To estimate the capacity of the swimming pool using the trapezoidal rule, we'll first need to calculate the area of the cross-section shown in the image. The trapezoidal rule will help us approximate the area under the curve. The pool's cross-section consists of two segments that can be evaluated using the trapezoidal rule.

The given cross-sectional dimensions are:

  • Height segments: 5.4m and 3.4m
  • Widths of segments: 8m and 12m

Step 1: Apply the Trapezoidal Rule to Each Segment

We need to break down the cross-section into smaller segments where the trapezoidal rule can be applied effectively.

Segment 1: From 0 to 8m

The heights at the endpoints are:

  • h0=5.4h_0 = 5.4
  • h1=5.4h_1 = 5.4

The width of the segment (aa) is 8m.

Using the trapezoidal rule: A1=12×(5.4+5.4)×8=12×10.8×8=43.2 m2A_1 = \frac{1}{2} \times (5.4 + 5.4) \times 8 = \frac{1}{2} \times 10.8 \times 8 = 43.2 \text{ m}^2

Segment 2: From 8m to 20m

The heights at the endpoints are:

  • h2=5.4h_2 = 5.4
  • h3=3.4h_3 = 3.4

The width of the segment (bb) is 12m.

Using the trapezoidal rule: A2=12×(5.4+3.4)×12=12×8.8×12=52.8 m2A_2 = \frac{1}{2} \times (5.4 + 3.4) \times 12 = \frac{1}{2} \times 8.8 \times 12 = 52.8 \text{ m}^2

Step 2: Total Area of the Cross-Section

The total area of the cross-section is the sum of the areas of the two segments: A=A1+A2=43.2+52.8=96 m2A = A_1 + A_2 = 43.2 + 52.8 = 96 \text{ m}^2

Step 3: Calculate the Volume of the Pool

The pool has a uniform depth of 1.6 meters. Therefore, the volume VV of the pool is given by: V=Cross-sectional area×DepthV = \text{Cross-sectional area} \times \text{Depth} V=96 m2×1.6 m=153.6 m3V = 96 \text{ m}^2 \times 1.6 \text{ m} = 153.6 \text{ m}^3

Step 4: Convert Volume to Litres

Since 1 m3=1000 litres1 \text{ m}^3 = 1000 \text{ litres}: V=153.6 m3×1000=153600 litresV = 153.6 \text{ m}^3 \times 1000 = 153600 \text{ litres}

Thus, the estimated capacity of the pool is 153,600 litres.

Would you like more details or have any questions?

Here are 8 related questions to expand on this information:

  1. How would the estimated capacity change if the depth of the pool was 2 meters instead of 1.6 meters?
  2. Can you apply Simpson's rule to estimate the pool capacity and compare the results?
  3. What would be the effect of having an uneven depth in the pool?
  4. How does the trapezoidal rule differ from other numerical integration methods?
  5. What is the significance of accurately estimating the pool capacity?
  6. How can we calculate the surface area of the water in the pool?
  7. What other methods can be used to measure the volume of irregular shapes?
  8. Can the trapezoidal rule be used for 3D shapes, and if so, how?

Tip: When using the trapezoidal rule, increasing the number of intervals can improve the accuracy of the approximation.

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Math Problem Analysis

Mathematical Concepts

Numerical Integration
Trapezoidal Rule
Volume Calculations

Formulas

Trapezoidal Rule for Numerical Integration
Volume of a Pool

Theorems

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Suitable Grade Level

Grades 10-12