每当在微积分中学到极值时,学生们常常会提出如下的问题(参看图1)图1 一只划艇与岸 SH的距离是b 某人在距岸边的最近点S是b公里的划艇中。并且点S距离他在岸边上的家(点H)是a公里,如果此人的划速是每小时7公里(常数),步行速度是ω公里(ω也是常数,且ω>γ),他为了尽早地回到家,问应该在S和H之间的那一点上岸? 本文将说明这个问题可以不用微积分来解。事实上,下面的解可以表明某些基本三角恒等式有着奇妙的应用。解:设a表示此人划经的路径与划艇到S所连直线组成的角,并且这里φ=arctg(a/b)。如果用T(a)表示他在旅程中所花费的时数,则 Whenever extremes are learned in calculus, students often ask the following questions (see Figure 1). Figure 1 The distance between a rowing boat and the bank SH is b The nearest point to someone on the bank is the distance of b kilometers. Rowing boat. And point S is a kilometer from his home on the shore (point H), if the person’s speed is 7 kilometers per hour (constant), and the walking speed is ω kilometers (ω is also a constant, and ω>γ), In order to return home as soon as possible, he asked that he should go ashore between S and H. This article will show that this problem can be solved without calculus. In fact, the following solution can show that some basic triangular identities have wonderful applications. Solution: Let a be the angle formed by the person’s chanting path and the straight line from the rowing boat to S, and here φ=arctg(a/b). If T(a) is used to represent the hours he spent on the journey, then