Math144Notes

# For simplicity let us take b 1 and then compute the

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

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...
View Full Document

## This document was uploaded on 03/20/2014 for the course MATH 144 at Hofstra University.

Ask a homework question - tutors are online