To approximate the flow f(t) of the Lochsa River in its flood stage, we can use a function of the formg(t)=c1+c2sin(t-90)π105-2π,where the coefficients c1 and c2 are found by evaluating the following two integrals:c1=1105∫90195f(t)dt, c2=19.95∫90195f(t)sin(t-90)π105-2πdt.(a) Use the&#...

Q. 49 - Calculus | StudyQuestionHub

Q. 49

Question

$To approximate the flow f (t) of the Lochsa River in its flood stage, we can use a function of the form g (t) = c_{1} + c_{2} (\sin (\frac{(t - 90) π}{105}) - \frac{2}{π}), where the coefficients c 1 and c 2 are found by evaluating the following two integrals : c_{1} = \frac{1}{105} \int_{90}^{195} f (t) dt, c_{2} = \frac{1}{9.95} \int_{90}^{195} f (t) (\sin (\frac{(t - 90) π}{105}) - \frac{2}{π}) dt . (a) Use the data points (t, f (t)) = (90, 2100), (120, 6300), (150, 11000), and (180, 4000), and left Riemann sums to approximate the values of the integrals for c 1 and c 2 . (b) Now that you have found c 1 and c 2, plot the resulting function g (t) against the data points from Exercise 48 .$

Step-by-Step Solution

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Answer

Part (a) $c_{1} = 5, 542.857, c_{2} = - 35, 900.917$

Part (b)

1Step 1. Given information

$To approximate the flow f (t) of the Lochsa River in its flood stage, we can use a function of the form g (t) = c_{1} + c_{2} (\sin (\frac{(t - 90) π}{105}) - \frac{2}{π}), where the coefficients c 1 and c 2 are found by evaluating the following two integrals : c_{1} = \frac{1}{105} \int_{90}^{195} f (t) dt, c_{2} = \frac{1}{9.95} \int_{90}^{195} f (t) (\sin (\frac{(t - 90) π}{105}) - \frac{2}{π}) dt . T he data points (t, f (t)) = (90, 2100), (120, 6300), (150, 11000), and (180, 4000)$

2Part (a) Step 2. Formula used and calculation .

$Formula used : If f is defined on a closed interval [a, b] and c_{k} is any point in [x_{k - 1}, x_{k}] then a Riemann sum is defined as \sum_{k = 1}^{n} f (c_{k}) ∆ x . Calculation : From the given data points, we can write as (t_{1} . f (t_{1})) = (90, 2100) (t_{2}, f (t_{2})) = (120, 6300) (t_{3}, f (t_{3})) = (150, 11000) (t_{4} . f (t_{4})) = (180, 4000) Here we use the left end points of 4 rectangles on the interval [90.210] . The width of each rectangle is differ by 30 and the last rectangle by 15 . The left Riemann sum is related to the definite integral as \int_{a}^{b} f (t) dt = \sum_{k = 1}^{n} f (t_{k}) ∆ t . where ∆ t = \frac{b - a}{n} . Now, \int_{190}^{195} f (t) dt = \sum_{k = 1}^{3} f (t_{k}) ∆ t \sum_{k = 1}^{3} f (t_{k}) ∆ t_{k} = [∆ t_{1} f (t_{1}) + ∆ t_{2} f (t_{2}) + ∆ t_{3} f (t_{3})] = 30 (2100) + 30 (6300) + 30 (11000) = 582000 c_{1} = \frac{1}{105} \int_{90}^{195} f (t) dt, = \frac{1}{105} (582000) = 5542.857 c_{2} = \frac{1}{9.95} \int_{90}^{195} f (t) (\sin (\frac{(t - 90) π}{105}) - \frac{2}{π}) dt . = \frac{1}{9.95} \int_{90}^{195} f (t) (\sin (\frac{(t - 90) π}{105})) dt - \frac{2}{π} \frac{1}{9.95} \int_{90}^{195} f (t) dt . let h (t) = f (t) \sin (\frac{(t - 90) π}{105}) . h (90) = f (90) \sin (\frac{(90 - 90) π}{105}) = 0 h (120) = f (120) \sin (\frac{(120 - 90) π}{105}) = 98.69 h (150) = f (150) \sin (\frac{(150 - 90) π}{105}) = 344.596$ $Now, \int_{90}^{195} f (t) (\sin (\frac{(t - 90) π}{105})) dt = \int_{90}^{195} h (t) dt = \sum_{k = 1}^{3} h (t_{k}) ∆ t \sum_{k = 1}^{3} h (t_{k}) ∆ t_{k} = [∆ t_{1} h (t_{1}) + ∆ t_{2} h (t_{2}) + ∆ t_{3} h (t_{3})] = 30 (0) + 30 (98.69) + 30 (344.596) = 13298.58 c_{2} = \frac{1}{9.95} \int_{90}^{195} f (t) (\sin (\frac{(t - 90) π}{105}) - \frac{2}{π}) dt . = \frac{1}{9.95} \int_{90}^{195} f (t) (\sin (\frac{(t - 90) π}{105})) dt - \frac{2}{π} \frac{1}{9.95} \int_{90}^{195} f (t) dt . = \frac{1}{9.95} (13298.58) - \frac{2}{π} \frac{1}{9.95} (582000) = 35900.917$

3Part (b) Step 1. Graph

The graph of the function $g (t) = 5, 542.857 - 35, 900.917 (\sin (\frac{(t - 90) π}{105}) - \frac{2}{π})$ is given by

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