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As in Sec. \(24.2,\) compute \(F\) using the trapezoidal rule and Simpson's 1/3 and Simpson's 3/8 rules but use the following force. Divide the mast into 5 -ft intervals. $$F=\int_{0}^{30} \frac{250 z}{6+z} e^{-z / 10} d z$$

Short Answer

Expert verified
Using the trapezoidal rule, Simpson's 1/3 rule, and Simpson's 3/8 rule, we have calculated approximations of the force acting on the mast. After evaluating the function \(\dfrac{250 z}{6+z} e^{-z / 10}\) at the required points and using the formulas for each numerical integration method, we found the approximate values for the integral as \(F_{Trapezoidal}\), \(F_{Simpson's 1/3}\), and \(F_{Simpson's 3/8}\). Compare the results to assess the accuracy of each method.

Step by step solution

01

Calculate the interval width h

First, we need to calculate the interval width h for each of the three methods. Since we are dividing the mast into 5-ft intervals, we have n = 6 sub-intervals in total. The interval width h can be calculated as: $$h = \frac{b-a}{n} = \frac{30-0}{6} = 5$$
02

Prepare the function

We are given the function: $$f(z) = \frac{250 z}{6+z} e^{-z / 10}$$ We need to evaluate this function at the points z = 0, 5, 10, 15, 20, 25, and 30.
03

Trapezoidal rule

Using the Trapezoidal rule, we can compute the force F as: $$F_{Trapezoidal} = \frac{h}{2} \left[f(a) + 2 \sum_{i=1}^{n-1} f(a+ih) + f(b)\right]$$ We need to evaluate the function at the required points and plug the values into the formula above.
04

Simpson's 1/3 rule

Using Simpson's 1/3 rule, we can compute the force F as: $$F_{Simpson's 1/3} = \frac{h}{3} \left[f(a) + 4 \sum_{i=1}^{n/2} f(a+(2i-1)h) + 2 \sum_{i=1}^{n/2-1} f(a+2ih) + f(b)\right]$$ Again, we need to evaluate the function at the required points and plug the values into the formula above.
05

Simpson's 3/8 rule

Using Simpson's 3/8 rule, we can compute the force F as: $$F_{Simpson's 3/8} = \frac{3h}{8} \left[f(a) + 3 \sum_{i=1}^{(3n-1)/2} f(a+(3i-2)h) + 3 \sum_{i=1}^{(3n-2)/2} f(a+(3i-1)h) + 2 \sum_{i=1}^{n/3-1} f(a+3ih) + f(b)\right]$$ Finally, we need to evaluate the function at the required points and plug the values into the formula above.
06

Conclusion

After we have applied all three methods to the given force function, we will obtain three different approximations for the integral. These approximations should be compared against each other, as well as against any available analytical results, to evaluate the accuracy and suitability of each numerical integration method in this specific context.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Trapezoidal Rule
The trapezoidal rule is a numerical integration method often favored for its simplicity and applicability to various functions. Imagine you wish to calculate the area under a curve between two points; the trapezoidal rule approximates this area by dividing it into trapezoid shapes. Here's the core principle: each segment of the curve is replaced with a trapezoid whose area can be easily computed.

The formula for the trapezoidal rule is: \[F_{Trapezoidal} = \frac{h}{2} \left[f(a) + 2 \sum_{i=1}^{n-1} f(a+ih) + f(b)\right]\] where \( h \) is the interval width, \( a \) and \( b \) are the lower and upper bounds of the integration, respectively, and \( n \) is the number of intervals. The starting and ending points are used only once, while the midpoints are each used twice, symbolizing the two bases of a trapezoid.
Simpson's 1/3 Rule
Simpson's 1/3 rule is another method of numerical integration that can provide a more accurate approximation than the trapezoidal rule for certain types of functions. It works best with functions which are nicely approximated by parabolas, as the intervals are approximated by quadratic functions. The rule gets its name because it essentially averages two trapezoidal estimates, one with double the interval width.

The general formula for Simpson's 1/3 rule is: \[F_{Simpson's 1/3} = \frac{h}{3} \left[f(a) + 4 \sum_{odd\, i} f(a+ih) + 2 \sum_{even\, i} f(a+ih) + f(b)\right]\] This formula calculates the area under the curve by taking the end points, the midpoints, and points in between midpoints, with the interval width being \( h \). The midpoints have higher coefficients, indicating their greater influence on the final result.
Simpson's 3/8 Rule
Simpson's 3/8 rule is an extension of Simpson's rule for situations where the interval number is not divisible by two. It uses cubic polynomials to approximate the curve within each subinterval, thus potentially offering increased accuracy over both the trapezoidal rule and Simpson's 1/3 rule for certain types of functions. The distinguishing factor of this rule is how it allows the approximation to capture the behavior of more complex functions.

The formula for Simpson's 3/8 rule is: \[F_{Simpson's 3/8} = \frac{3h}{8} \left[f(a) + 3 \sum_{3i-2\, segments} f(a+ih) + 3 \sum_{3i-1\, segments} f(a+ih) + 2 \sum_{others\, i} f(a+ih) + f(b)\right]\] In this, \( h \) is the width of the intervals and is leveraged in a combination with the evaluated function values to approximate the integral.
Numerical Methods for Engineers
Numerical methods are essential tools for engineers, providing practical means to solve complex problems that are often impossible to tackle analytically due to complicated functions or geometries. These methods, including the trapezoidal and Simpson's rules, allow engineers to approximate solutions of integrals, differential equations, and other calculations with a known margin of error.

These techniques are crucial as they ensure the reliability and safety of engineering designs by providing close approximations to ideal models. Engineers choose different numerical methods based on the required accuracy, computational resources, and the nature of the problem at hand.
Interval Width Calculation
Interval width calculation is a preliminary step crucial to the accuracy of numerical integration. In simpler terms, it's how thinly you slice the region under the curve before approximating its area. It's given by the formula: \[h = \frac{b-a}{n}\] where \( b \) and \( a \) are the integration bounds and \( n \) is the chosen number of intervals. The selection of \( n \) directly influences the approximation's accuracy—the higher the \( n \), typically the more accurate the approximation but at the cost of increased computational effort.
Function Evaluation
Function evaluation is the process of calculating the values of a function at specific input values. This step is crucial in numerical integration, as it provides the data points needed to apply the trapezoidal, Simpson's 1/3, and the Simpson's 3/8 rules. In practice, engineers and scientists often use computational tools to automate these evaluations, particularly for complex or transcendental functions.

For our instance, the function \[f(z) = \frac{250 z}{6+z} e^{-z / 10}\] must be evaluated at specific intervals to approximate the integral of force over the length of a mast. The accuracy of numerical integration significantly depends on the reliability of these function evaluations.

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Most popular questions from this chapter

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The standard technique for determining cardiac output is the indicator dilution method developed by Hamilton. One end of a small catheter is inserted into the radial artery and the other end is connected to a densitometer, which can automatically record the concentration of the dye in the blood. A known amount of dye, \(5.6 \mathrm{mg},\) is injected rapidly, and the following data is obtained: $$\begin{array}{cccc} \hline \begin{array}{c} \text { Time, } \\ \text { s } \end{array} & \begin{array}{c} \text { Concentration, } \\ \text { mg/L } \end{array} & \begin{array}{c} \text { Time, } \\ \text { s } \end{array} & \begin{array}{c} \text { Concentration, } \\ \text { mg/L } \end{array} \\ \hline 5 & 0 & 21 & 2.3 \\ 7 & 0.1 & 23 & 1.1 \\ 9 & 0.11 & 25 & 0.9 \\ 11 & 0.4 & 27 & 1.75 \\ 13 & 4.1 & 29 & 2.06 \\ 15 & 9.1 & 31 & 2.25 \\ 17 & 8 & 33 & 2.32 \\ 19 & 4.2 & 35 & 2.43 \\ \hline \end{array}$$ Plotting the above data results in the dye dilution curve in Fig. P24.10a. The concentration reaches a maximum value at about 15 seconds and then falls off, followed by a rise due to the recirculation of dye. The curve is replotted on a semilog graph in Fig. P24.10b. Notice that a straight line approximates the descending limb of the dilution curve. In order to separate out the recirculation effect, analysts extend the straight-line portion. The cardiac output can then be calculated from $$C=\frac{M}{A} \times 60 \mathrm{s} / \mathrm{min}$$ where \(C=\) cardiac output \([\mathrm{L} / \mathrm{min}], M=\) amount of injected dye \((\mathrm{mg}),\) and \(A=\) area under the curve with the linear correction. Calculate the cardiac output of this patient using the trapezoidal rule with a step size of 0.5 s.

Perform the same computation as in Sec. \(24.3,\) but for the current as specified by $$i(t)=5 e^{-1.25 t} \sin 2 \pi t \quad \text { for } 0 \leq t \leq T / 2$$ $$i(t)=0\quad\quad\quad\quad\quad\quad \text { for } T / 2 < t \leq T$$ where \(T=1\) s. Use five-point Gauss quadrature to estimate the integral. .

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