Answered: 0 25 50 20 Depth 20 19 19 18 16 14 12…

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

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Question

A pond is approximately circular, with a diameter of 400 feet. Starting at the center, the depth of the water is measured every 25 feet and recorded in the table (a) Use the regression capabilities of a graphing utility to find a quadratic model for the depths recorded in the table. Use the graphing utility to plot the depths and graph the model. (b) Use the integration capabilities of a graphing utility and the model in part (a) to approximate the volume of water in the pond. (c) Use the result of part (b) to approximate the number of gallons of water in the pond.

0 25 50
20
Depth 20
19
19
18
16
14
12
75
100
125
10
Depth| 17
15
14
2-
50 175 200
150
50
100
150
200
Distance from center
Depth
10
(in feet)
Depth (in feet)

Expert Solution

Step 1

The given data,

x	0	25	50	75	100	125	150	175	200
Depth	20	19	19	17	15	14	10	6	0

We have to find,

(a) The regression equation for a quadratic model for the depths recorded in the table.

(b) Using the integration capabilities of a graphing utility and the model in part (a) we have to approximate the volume of water in the pond.

Step 2

The given data,

x	0	25	50	75	100	125	150	175	200
Depth	20	19	19	17	15	14	10	6	0

Using the regression calculator,

The quadratic model for the depths is recorded in the table is $d = - 0.00056 x^{2} + 0.01887 x + 19.393939$ .

Graph of the depths and the quadratic model.

Step 3

The formula for the volume of water in the pond is,

$V = 2 π \int_{a}^{b} p (x) h (x) dx$ .

Here $p (x) = x, h (x) = d, a = 0$ and $b = 200$ .

Now,

$V = 2 π \int_{0}^{200} x (- 0.00056 x^{2} + 0.01887 x + 19.393939) dx$

Solve this integral we get

$V \approx 1345850.6$

The volume of water in the pond is $1345850.6 f t^{3}$ .

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