Math144Notes

# 20041k25sint1 1k20081k2 20041k25kcost02sint and

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Unformatted text preview: (homework) -K iK z= ± 4π 2 2+1 16π If K= 0, (no rotation) we have stagnation points at z = ±1. As k increases, the stagnation points creep up the unit circle. When K reaches 4π, the quantity under the square root turns imaginary, and so z creeps up the z-axis. Note on finding streamlines: Inverting this function is not possible analytically. However, to draw the streamlines, all we really need to do is choose a small value of ∆t, start at some point, find the velocity vector there, take a small step in that direction, and then continue. (We are actually solving a system of two differential equations in two unknowns: x and y as functions of t, numerically using Euler's method.) Here is what you get using Excel with 30 points and ∆t = 0.12 (a very large value): 2.5 2 1.5 1 0.5 0 -3 -2 -1 -0.5 0 1 2 3 -1 -1.5 -2 -2.5 This uses v = F'(z) , where F(z) = z + 1/z. The Cartesian coordinates of F'(z) are 2 Real part: vx = 22 2 2 22 2 (x +y ) - x + y (x +y ) The increments are then given by (vx∆t, vy∆t). Imaginary part:...
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## This document was uploaded on 03/20/2014 for the course MATH 144 at Hofstra University.

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