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wel..again i face anther project for math..
which is calculus...any application in this topic..
i knw u guys will suggest to find volume and area using anti derivative..but i dont want it..others application??but i dun have idea..someone help to figure out what project should i do..and hope it is with the answers because easy for me to refer back....thanx

2007-09-24 18:48:58 · 4 answers · asked by Anonymous in Science & Mathematics Mathematics

4 answers

how about a project about draining odd shaped containers through a hole in the side. The water leaving the tank is proportional to the height from the hole to the top of water level. The water level is constantly changing. Start simple, with a rectangular tank, then a cylinder, etc...

Good Luck

2007-10-01 13:36:51 · answer #1 · answered by Mugwump 7 · 0 0

How about a demo of the related rates problem of someone walking away from a streetlight at constant speed and the rate of increase of the length of the shadow as he goes. You could make a small "lightpole", and a small model person that could move at a constant speed.

2007-09-24 18:55:13 · answer #2 · answered by cattbarf 7 · 0 0

Calculate the escape velocity of a space ship with a mass of 1000 kg from the moon.

2007-10-02 15:50:10 · answer #3 · answered by Will 4 · 0 0

Joe can run 4 cases as quickly as he can swim regularly if Joe angles around the circulate, 4/6*sqrt(2 hundred^2+ x^2) = time spent swimming (x = downstream ingredient to the swim) 2/12*(3 hundred-x) = time working t = 4/6sqrt(40,000+ x^2) +2/12*(3 hundred-x) d/tdx = 4/6*a million/2*2x/sqrt(40,000+x^2)-2/12 dtdx = 2/3*x/sqrt(40,000+x^2) -a million/6 set dt/dx to 0 a million/6 = 2/3*x/sqrt(40,000+x^2) a million = 4x/sqrt (40,000 +x^2) sqrt(40,000+x^2) =4x 40,000 +x^2 = 16x^2 40,000 = 15 x^2 2666.7 = x^2 fifty one.sixty 4 = x joe runs (3 hundred-fifty one.sixty 4)m at 12/2 m/sec = 1499sec joe swims sqrt (40,000 + fifty one.sixty 4^2)m at 6/4 m/sec =206.56m at 6/4m/sec = 309.eighty 4 sec entire = 1808.eighty 4 sec entire

2016-10-05 07:53:46 · answer #4 · answered by ? 4 · 0 0

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