Volume of water in a swimming pool
A rectangular swimmin...

Question

Volume of water in a swimming pool

A rectangular swimming pool is 30 ft wide and 50 ft long. The accompanying table shows the depth h(x) of the water at 5-ft intervals from one end of the pool to the other. Estimate the volume of water in the pool using the Trapezoidal Rule with n = 10 applied to the integral

V = ∫ from 0 to 50 of 30 · h(x) dx.

depth1

Accepted Answer

Welcome back everyone. In this problem, a rectangular irrigation channel runs 60 ft in length and has a constant width of 20 ft. A surveyor measures the water depth day of x at 6 ft intervals along the channel from x equals 0 ft at the upstream end to x equals 60 ft at the downstream end. The measurements are shown in the table. Use the trapezoidal rule with n equals 10 to estimate the volume of water in the channel given by V equals the definite integral between 0 and 60 of 20 d of x with respect to x. Here we have our table, and for our answer choices A says it's 2,841 cubic feet, B 3,782 cubic feet, C, 5,682 cubic feet, and D, 7,564 cubic feet. No. How can we use the trapezoidal rule to help us estimate the volume of water in the channel? What do we know? Well, recall by the trapezoidal rule approximation. By this approximation. The integral between a and b of fx with respect to x is approximately equal to the width of each subinterval delta x divided by 2. Multiplied by f x0, where x0 x1 and so on all the way to xn are the points at which the function f x is evaluated where f x represents the values of the function at these given points. So it's going to be F X plus 2 multiplied by F of X1. Plus 2 multiplied by f of x 2. All the way up to 2 multiplied by F of XN minus 1 plus fXN, OK. Where we know that delta x represents the horizontal length of each trapezoid's base, OK, we also know that x0 x1 all the way up to XN are the end points of the subintervals, and FX0 sorry, all the way up to F xn represents the heights of the trapezoids at each point and FX0. And f xn are each counted once because they form one side of the first and last trapezoid respectively, while all the intermediate points between F of X1 f x2 all the way to F xn minus 1 are counted twice because each height. Because each is a height shared by two adjacent trapezites, that is why we have 2 multiplied by these terms, OK. No Let's see if we can identify a few of the variables to help us make this approximation. Now, if we think about it here, yeah. In our problem. We know, OK, that n, the number of subintervals equals 10. OK, we know that each subinterval has a width delta x of 6, OK, and we can also have our measured depths here, which would be our points x0 X1, and so on, or in this case D, D1, and so on of 1st 2.5, that would be d0 or first depth. Uh, D1 would be 3.1 our second depth, D2 3.8, D3, 4.2, and so on and so on all the way up to D10. OK. Or 10th depth, which would be equal to 6 as well. So, no, at this point, OK, that means if we apply the trapezoidal rule. 2 or a definite integral taken out a constant of 20, that means the integral between 0 and 60 of d of x with respect to x is going to be approximately equal. To delta x, which we know is 6 divided by 2, multiplied by d0 plus 2 multiplied by the sum between i equals 1 to 9 of dii plus d10. So let's put the values in that we know. Now if we, since we already know delta x, d0 D10, what about our sum here? Well, our sum from i equals 1 to 9 of dii is going to be equal to the sum of all of those points D1, D2, D3, and so on. So that will be 3.1 plus 3.8 plus 4.2 plus 4.6. Plus 5 + 5.3 + 5.5 plus 5.7 plus 5.9, OK, which would all be equal to 43.1. So since we have this sum and our other variables, that means then that our integral or approximate integral for d of x with respect to x. is going to be approximately 6 divided by 2 multiplied by 2.5 plus 2 multiplied by 43.1 plus 6. And when we go ahead and add here. That is going to be equal to 3 multiplied by 94.7, which is 284.1 square feet. Now remember we're trying to estimate the volume, and if we go back to our formula that told us how to find V using our integration for our channel width of 20 ft, then that means, OK, let me put this in black here. Therefore this means that v is going to be equal to 20 multiplied by or integral between 0 and 60 of the effects with respect to x, so that's approximately 20. Multiplied by 284.1 square feet, so rather I should say 20 ft, multiplied by approximately 284.1 square feet, and that will be equal to 5,682 cubic feet. Thus, this is going to be the approximate volume of the water in the channel. If we look back at our answer choices, that means C is the correct answer. Thanks a lot for watching everyone. I hope this video helped.

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