Math Problem Statement

The table (supplied by San Diego Gas and Electric) gives the power consumption in megawatts in San Diego County from midnight to 6:00 a.m. on December 8, 1999. Use Simpson's Rule to estimate the energy used during that time period, giving your answer correct to two decimal places. Use the fact that power is the derivative of energy and that the units of energy are megawatt-hours.

Time P (mw) 0:00 1814 0:30 1735 1:00 1686 1:30 1646 2:00 1637 2:30 1609 3:00 1604 3:30 1611 4:00 1621 4:30 1666 5:00 1745 5:30 1886 6:00 2052 Energy used ≈

megawatt-hours

Solution

We will apply Simpson’s Rule to estimate the energy used during the time period. Simpson's Rule is defined as:

abf(x)dxΔx3[f(x0)+4f(x1)+2f(x2)+4f(x3)++f(xn)],\int_{a}^{b} f(x) dx \approx \frac{\Delta x}{3} \left[f(x_0) + 4f(x_1) + 2f(x_2) + 4f(x_3) + \cdots + f(x_n)\right],

where:

  • nn (the number of subintervals) must be even,
  • Δx=ban\Delta x = \frac{b - a}{n},
  • and the coefficients alternate between 44 and 22, with 11 at both ends.

Step 1: Key Information

  • Time is from 0:000:00 to 6:006:00, so a=0a = 0 and b=6b = 6 hours.
  • There are 13 data points (at 30-minute intervals), so n=12n = 12 subintervals (even as required).
  • Δx=6012=0.5\Delta x = \frac{6 - 0}{12} = 0.5 hours.

Step 2: Assign Power Values to P(x)P(x)

The power values correspond to f(x)f(x), as follows: P(x)=[1814,1735,1686,1646,1637,1609,1604,1611,1621,1666,1745,1886,2052].P(x) = [1814, 1735, 1686, 1646, 1637, 1609, 1604, 1611, 1621, 1666, 1745, 1886, 2052].


Step 3: Apply Simpson’s Rule

Using the formula: 06P(t)dtΔx3[P0+4P1+2P2+4P3++2Pn2+4Pn1+Pn],\int_{0}^{6} P(t) \, dt \approx \frac{\Delta x}{3} \left[P_0 + 4P_1 + 2P_2 + 4P_3 + \cdots + 2P_{n-2} + 4P_{n-1} + P_n\right], we substitute the power values: 06P(t)dt0.53[1814+4(1735)+2(1686)+4(1646)+2(1637)+4(1609)+2(1604)+4(1611)+2(1621)+4(1666)+2(1745)+4(1886)+2052].\int_{0}^{6} P(t) \, dt \approx \frac{0.5}{3} \Big[1814 + 4(1735) + 2(1686) + 4(1646) + 2(1637) + 4(1609) + 2(1604) + 4(1611) + 2(1621) + 4(1666) + 2(1745) + 4(1886) + 2052\Big].


Step 4: Simplify the Expression

Evaluate each term: [ \begin{aligned} \int_{0}^{6} P(t) , dt &\approx \frac{0.5}{3} \Big[1814 + 6940 + 3372 + 6584 + 3274 + 6436 + 3208 + 6444 + 3242 + 6664 + 3490 + 7544 + 2052\Big] \ &\approx \frac{0.5}{3} \Big[59764\Big]. ]

Now compute: 06P(t)dt0.53×59764=2988239960.67megawatt-hours.\int_{0}^{6} P(t) \, dt \approx \frac{0.5}{3} \times 59764 = \frac{29882}{3} \approx 9960.67 \, \text{megawatt-hours}.


Final Answer:

The energy used during this time period is approximately: 9960.67megawatt-hours.\boxed{9960.67 \, \text{megawatt-hours}}.

Would you like more explanation or a breakdown of any step? 😊


Questions:

  1. What is Simpson’s Rule, and why is it accurate for numerical integration?
  2. How does the step size (Δx\Delta x) affect the accuracy of Simpson’s Rule?
  3. What is the physical relationship between power and energy in this problem?
  4. How would the result change if nn were not even? Could we still apply Simpson’s Rule?
  5. How can we use trapezoidal rule to solve this problem, and how does it compare to Simpson’s Rule?

Tip:

Simpson’s Rule requires the function to be smooth and well-behaved; it’s particularly accurate for functions resembling parabolas over small intervals.

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Math Problem Analysis

Mathematical Concepts

Numerical Integration
Simpson's Rule
Definite Integrals
Power-Energy Relationship

Formulas

Simpson's Rule formula: ∫_a^b f(x) dx ≈ (Δx/3) [f(x0) + 4f(x1) + 2f(x2) + ... + f(xn)]
Energy = ∫ Power(t) dt

Theorems

Simpson's Rule for Numerical Integration
Relationship between Power and Energy

Suitable Grade Level

Grades 11-12