For simplicity let us take b 1 and then compute the

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Unformatted text preview: t the isotherms for the following: 0º –1 Insulation 1 D 20º 42 Answer Recall that the isotherms are the equipotentials. Call the above region D. We saw 20 -1 two sections ago that D maps to a nicer representation of H using F(z) = s in (z) + π π -1 10. The inverse of this is just F (z) = sin( [z-10]). Here again is H: 20 0º 20º –π/2 Insulation π/2 H On H, the isotherms are just vertical lines, which can be parameterized as x = K, y = t, where K is the temperature, and t ≥ 0. In other words, z(t) = K + it t≥0 Therefore, the corresponding isotherms in D are given by π -1 F [z(t)] = sin( [K-10 + it]) t≥0 20 To plot this, resolve into real and imaginary parts using the identity sin(x + iy) = sinx coshy + i cosx sinhy Therefore π π π π π sin( [K-10 + it]) = sin( [K-10]) cosh t + i cos( [K-10]) sinh t 20 20 20 20 20 Therefore, we have the following parameterization of the isotherms: π π x = sin( [K-10]) cosh t 20 20 π π y = cos( [K-10]) sinh t 20 20 K = temperature; 0 ≤ K ≤ 20, t ≥ 0. Here is what the lines look like in our little Excel plotter: (Compare with Exercise 2 in the preceding homework assignment.] The 7 lines are K = 2º, 4.7º, 7.3º, 10...
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This document was uploaded on 03/20/2014 for the course MATH 144 at Hofstra University.

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