The steady state temperature u ( r , z ) such that ∂ 2 u ∂ r 2 + 1 r ∂ u ∂ r − u r 2 + ∂ 2 u ∂ z 2 = 0 , u ( 1 , z ) = 0 , u ( r , 0 ) = 0 and u ( r , 1 ) = r − r 3 .

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Chapter 13, Problem 20RE

To determine

The steady state temperature u(r,z) such that ∂2u∂r2+1r∂u∂r−ur2+∂2u∂z2=0, u(1,z)=0, u(r,0)=0 and u(r,1)=r−r3.

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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?

Answer 2

Question

Chapter 13, Problem 20RE

To determine

The steady state temperature u(r,z) such that ∂2u∂r2+1r∂u∂r−ur2+∂2u∂z2=0, u(1,z)=0, u(r,0)=0 and u(r,1)=r−r3.

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Answer 3

Question

Chapter 13, Problem 20RE

To determine

The steady state temperature u(r,z) such that ∂2u∂r2+1r∂u∂r−ur2+∂2u∂z2=0, u(1,z)=0, u(r,0)=0 and u(r,1)=r−r3.

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Answer 4

Question

Chapter 13, Problem 20RE

To determine

The steady state temperature u(r,z) such that ∂2u∂r2+1r∂u∂r−ur2+∂2u∂z2=0, u(1,z)=0, u(r,0)=0 and u(r,1)=r−r3.

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Answer 5

Chapter 13, Problem 20RE

To determine

The steady state temperature u(r,z) such that ∂2u∂r2+1r∂u∂r−ur2+∂2u∂z2=0, u(1,z)=0, u(r,0)=0 and u(r,1)=r−r3.

Answer 6

Chapter 13, Problem 20RE

To determine

The steady state temperature u(r,z) such that ∂2u∂r2+1r∂u∂r−ur2+∂2u∂z2=0, u(1,z)=0, u(r,0)=0 and u(r,1)=r−r3.

Answer 7

Chapter 13, Problem 20RE

To determine

The steady state temperature u(r,z) such that ∂2u∂r2+1r∂u∂r−ur2+∂2u∂z2=0, u(1,z)=0, u(r,0)=0 and u(r,1)=r−r3.

Answer 8

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Answer 11

The steady state temperature u ( r , z ) such that ∂ 2 u ∂ r 2 + 1 r ∂ u ∂ r − u r 2 + ∂ 2 u ∂ z 2 = 0 , u ( 1 , z ) = 0 , u ( r , 0 ) = 0 and u ( r , 1 ) = r − r 3 .

Solution Summary: The author explains the steady state temperature u(r,z) and the boundary conditions.

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The steady state temperature u(r,z) such that ∂2u∂r2+1r∂u∂r−ur2+∂2u∂z2=0, u(1,z)=0, u(r,0)=0 and u(r,1)=r−r3.

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