由y??t?yt?y11,所以kAQ?21?,kBQ?22??2t2?1?1,
2x?1t?12t2?1?1t?1所以yt1??1,y322t2?2t?3t, 所以AB?|2t3?3t?t2?12t|?2t3?52t?12t(t?0). 令f(t)?2t3?52t?12t,t?0,则f?(t)?6t2?5112t4?5t2?12?2t2?2t2,
由f?(t)?0得t??5?7324,由f?(t)?0得0?t??5?7324, 所以f(t)在区间(0,?5?7324)单调递减,在(?5?7324,??)单调递增, 所以当t??5?7324时,f(t)取得极小值也是最小值,即AB取得最小值,
此时s?t2?1?19?7324.
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