Explanation Newton’s approximation is given by, x n = x n − 1 − f ( x n ) f ′ ( x n ) Calculation: The given equation of altitude is, 0.00926 h 3 − 0.0833 h 2 + s = 0 Note that the value of s is 0.786, hence use s = 0.786 . Therefore, 0.00926 h 3 − 0.0833 h 2 + 0.786 = 0 f ( h ) = 0.00926 h 3 − 0.0833 h 2 + 0.786 … … ( 1 ) Differentiate the equation (1) with respect to h f ′ ( h ) = 0.02778 h 2 − 0.1666 h … … ( 2 ) Find the values of the equation 0.00926 h 3 − 0.0833 h 2 + 0.786 at the boundary points. Note that the solid sphere s sinks. Therefore, h is negative, Substitute h = − 2 in equation (1). f ( − 2 ) = 0.00926 ( − 2 ) 3 − 0.0833 ( − 2 ) 2 + 0.786 = − 0.07408 − 0.3332 + 0.786 = 0.37872 Substitute h = − 3 in equation (1). f ( − 3 ) = 0.00926 ( − 3 ) 3 − 0.0833 ( − 3 ) 2 + 0.786 = − 0.25002 − 0.7497 + 0.786 = − 0.21372 Note that the value of h lies between − 2 and − 3 hence use the value of h 1 = − 2.6 Step 1: Find the second approximation h 2 with h 1 = − 2.6 . Substitute h 1 = − 2

Basic Technical Mathematics with Calculus (11th Edition)

11th Edition

ISBN: 9780134437736

Author: Allyn J. Washington, Richard Evans

Publisher: PEARSON

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Chapter 24.2, Problem 28E

To determine

The depth h(in cm) of solid sphere of specific gravity s=0.786 sinks in water. The depth h is given by 0.00926h3−0.0833h2+s=0.

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Chapter 24.2, Problem 27E

Chapter 24.2, Problem 29E

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Problem 11 (a) A tank is discharging water through an orifice at a depth of T meter below the surface of the water whose area is A m². The following are the values of a for the corresponding values of A: A 1.257 1.390 x 1.50 1.65 1.520 1.650 1.809 1.962 2.123 2.295 2.462|2.650 1.80 1.95 2.10 2.25 2.40 2.55 2.70 2.85 Using the formula -3.0 (0.018)T = dx. calculate T, the time in seconds for the level of the water to drop from 3.0 m to 1.5 m above the orifice. (b) The velocity of a train which starts from rest is given by the fol- lowing table, the time being reckoned in minutes from the start and the speed in km/hour: | † (minutes) |2|4 6 8 10 12 14 16 18 20 v (km/hr) 16 28.8 40 46.4 51.2 32.0 17.6 8 3.2 0 Estimate approximately the total distance ran in 20 minutes.

- Let n = 7, let p = 23 and let S be the set of least positive residues mod p of the first (p − 1)/2 multiple of n, i.e. n mod p, 2n mod p, ..., p-1 2 -n mod p. Let T be the subset of S consisting of those residues which exceed p/2. Find the set T, and hence compute the Legendre symbol (7|23). 23 32 how come? The first 11 multiples of 7 reduced mod 23 are 7, 14, 21, 5, 12, 19, 3, 10, 17, 1, 8. The set T is the subset of these residues exceeding So T = {12, 14, 17, 19, 21}. By Gauss' lemma (Apostol Theorem 9.6), (7|23) = (−1)|T| = (−1)5 = −1.

Let n = 7, let p = 23 and let S be the set of least positive residues mod p of the first (p-1)/2 multiple of n, i.e. n mod p, 2n mod p, ..., 2 p-1 -n mod p. Let T be the subset of S consisting of those residues which exceed p/2. Find the set T, and hence compute the Legendre symbol (7|23). The first 11 multiples of 7 reduced mod 23 are 7, 14, 21, 5, 12, 19, 3, 10, 17, 1, 8. 23 The set T is the subset of these residues exceeding 2° So T = {12, 14, 17, 19, 21}. By Gauss' lemma (Apostol Theorem 9.6), (7|23) = (−1)|T| = (−1)5 = −1. how come?

Chapter 24 Solutions

Basic Technical Mathematics with Calculus (11th Edition)

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The depth h ( i n c m ) of solid sphere of specific gravity s = 0.786 sinks in water. The depth h is given by 0.00926 h 3 − 0.0833 h 2 + s = 0 .

The depth h(in cm) of solid sphere of specific gravity s=0.786 sinks in water. The depth h is given by 0.00926h3−0.0833h2+s=0.

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Chapter 24 Solutions