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</o:shapelayout></xml><![endif]--></head><body lang=EN-US link=blue vlink=purple><div class=WordSection1><p class=MsoPlainText><o:p> </o:p></p><p class=MsoPlainText>>… On Behalf Of Ben Zaiboc<br><br></p><p class=MsoPlainText>Spike claimed: "... so she turns right pi/2 ..."<o:p></o:p></p><p class=MsoPlainText><o:p> </o:p></p><p class=MsoPlainText>>…Spike, NOBODY turns right pi/2 :D<o:p></o:p></p><p class=MsoPlainText><span style='color:black'><o:p> </o:p></span></p><p class=MsoPlainText><span style='color:black'>I do, nearly every day.<o:p></o:p></span></p><p class=MsoPlainText><o:p> </o:p></p><p class=MsoPlainText>>>…"If Mrs. Claus travels 4 km north, then 3 km east, what is the maximum distance she can be from the igloo?"<o:p></o:p></p><p class=MsoPlainText><o:p> </o:p></p><p class=MsoPlainText>>…5km, because east is 90 deg. from north, and the ground is flat over such short distances, and if she started out less than 4km from the north pole, she'd get stuck there, according to those directions, because once you get to it, you can't go any farther north.<o:p></o:p></p><p class=MsoPlainText><span style='color:black'><o:p> </o:p></span></p><p class=MsoPlainText><span style='color:black'>On the contrary, sir. Read on please.<o:p></o:p></span></p><p class=MsoPlainText><o:p> </o:p></p><p class=MsoPlainText>>…Ha! Take that, maths-boy! Ben Zaiboc<o:p></o:p></p><p class=MsoPlainText><span style='color:black'><o:p> </o:p></span></p><p class=MsoPlainText><span style='color:black'>If the term “that” represents “the derivative of the distance function” I did so. I took the derivative, set it equal to zero, solved for theta. Rather attempted to. Note the figure below, with Mrs. Claus’ path in red:<o:p></o:p></span></p><p class=MsoPlainText><o:p> </o:p></p><p class=MsoPlainText>_______________________________________________<o:p></o:p></p><p class=MsoPlainText><span style='color:black'><o:p> </o:p></span></p><p class=MsoPlainText><span style='color:black'><o:p> </o:p></span></p><p class=MsoPlainText><img width=404 height=643 id="Picture_x0020_1" src="cid:image003.jpg@01CF218A.B44CF6E0"><span style='color:black'><o:p></o:p></span></p><p class=MsoPlainText><span style='color:black'><o:p> </o:p></span></p><p class=MsoPlainText><span style='color:black'>The challenge is to find theta such that the distance from start is maximized given a path length of 7 (or the more general N+E). I can get up to about 6.274 km, with a theta of about 2.01695. (This is about 115.56 degrees, for the degree fans among us. Note that degrees are in short supply in the north pole region, even in our current era of global warming, and even if you can find some, most of them are negative.) <o:p></o:p></span></p><p class=MsoPlainText><span style='color:black'><o:p> </o:p></span></p><p class=MsoPlainText><span style='color:black'>spike<o:p></o:p></span></p></div></body></html>