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Application of differentiation
2013-10-15 12:24 am
回答 (1)
2013-10-15 7:51 am
✔ 最佳答案
(a) P = 5/r + 10/(10 - r)(b) dP/dr = -5/r^2 + 10/(10 - r)^2d^2P/dr^2 = 10/r^3 + 20/(10 - r)^3(c) Let dP/dr = 0-5/r^2 + 10/(10 - r)^25/r^2 = 10/(10 - r)^210r^2 = 5(10 - r)^22r^2 = r^2 - 20r + 100r^2 + 20r - 100 = 0r = 4.1421(d) Let the gain factor is G-5/6^2 + G/(10 - 6)^2 = 0-5/36 + G/16 = 0G = 80/36 G = 20/9
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