In Session 41 (Advanced Example), dealing with the Lagrange Multipliers. There was a triangular pyramid. To find the slant height of one of the sides, Dr. Auroux made it into the hypotenuse of the right triangle formed by the pyramid height (PQ) and a line on the base triangle to the side (which he called "u_
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I think you misunderstood me. I meant the u's as in the heights ON THE BASE TRIANGLE. Not the slant heights.
can you give me lecture no?
He does the example at 38:00. Thank you! I really appreciate it. http://www.youtube.com/watch?v=15HVevXRsBA&feature=BFa#t=38m
i'll try to rely as fast as i can ... if i can.
I just don't know how he makes the assumption that the segments from Q to the sides are perpendicular at 45:10.
|dw:1344017875015:dw| How are those segments on the base perpendicular?
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