一道数学应用题,用到初级微积分Aaboutleavesadockat2:00p.mandtravelsduesouthataspeedof20km/h.Anotherboathasbeenheadingdueeastat15km/handreachesthesamedockat3:00p.m.Atwahttimewerethetwoboatsclosesttog

Question

一道数学应用题,用到初级微积分
A about leaves a dock at 2:00 p.m and travels due south at a speed of 20 km/h.Another boat has been heading due east at 15km/h and reaches the same dock at 3:00p.m.At waht time were the two boats closest together?

Answer 1

时间T
船一到码头距离：20（T-2）
船二到码头距离：15（3-T）
两船距离平方：
L^2=［20(T－2)］^2+［15(3-T)^2
=625T^2-2950T+1600+3625
L最小,则L^2最小,则(625T^2-2950T+1600+3625)‘=0
2*625T-2950=0
T=2950/1250=2.36时
0.36*60=21.6
0.6*60=36
即2：21’36‘’时两船距离最近
此题也可用配方法解