From 3691af780fe86b90c4e52c4c6124a8b2dd58d8a3 Mon Sep 17 00:00:00 2001
From: joshuab <>
Date: Mon, 12 Jul 2010 22:40:03 +0000
Subject: typo fixed

---
 ClassJune26.page | 4 ++--
 1 file changed, 2 insertions(+), 2 deletions(-)

diff --git a/ClassJune26.page b/ClassJune26.page
index b5454a4..b5dff79 100644
--- a/ClassJune26.page
+++ b/ClassJune26.page
@@ -216,7 +216,7 @@ angle, and $df$ is linear, its image must bisect too, that is, the
 little square must become a little *square*. Numerically, this
 means that $\left(\begin{array}{c}
 \frac{\partial u}{\partial y}\\
-\frac{\partial u}{\partial y}\end{array}\right)$ is $\left(\begin{array}{c}
+\frac{\partial v}{\partial y}\end{array}\right)$ is $\left(\begin{array}{c}
 \frac{\partial u}{\partial x}\\
 \frac{\partial v}{\partial x}\end{array}\right)$ rotated by $\pi/2$. If we write 
 $\left(\begin{array}{c}
@@ -226,7 +226,7 @@ b\end{array}\right)=\left(\begin{array}{c}
 \frac{\partial v}{\partial x}\end{array}\right)$
 , then $\left(\begin{array}{c}
 \frac{\partial u}{\partial y}\\
-\frac{\partial u}{\partial y}\end{array}\right)=\left(\begin{array}{c}
+\frac{\partial v}{\partial y}\end{array}\right)=\left(\begin{array}{c}
 -b\\
 a\end{array}\right).$
 Put another way, the linear transformation $df(z)$ has the form 
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