作者GSXSP (Gloria)
看板trans_math
标题Re: [考古] 交大88第八题
时间Mon Jul 2 22:09:00 2007
※ 引述《johnnyzsefb (AJ)》之铭言：
: ８．A heat-seeking particle is located at the point (2,-3) on a metal
: plate whose temperature at (x,y) is T(x,y)=20-4x^2-y^2.
: Find the path of the particle as it continuously moves in the direction
: of maximum temperature increase.
▽T = (-8x , -2y)
Let path be (t , f(t))
direction: (1 , f'(t)) = ▽T = (-8t , -2f(t)) = -8t(1 , f(t)/4t)
= (1 , f(t)/4t)
f'(t) = f(t)/4t ...解微方
4tf'(t) = f(t)
Let f(t) = K*t^r 代回原式
可以解到 r 然後再利用 (2,-3) 解 K
应该就解完了 ##
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※ 发信站: 批踢踢实业坊(ptt.cc)
◆ From: 140.113.140.51
1F：→ GSXSP:补一下 t要选范围 140.113.140.51 07/02 22:13
2F：推 johnnyzsefb:水喔!~方法真赞XD 219.70.25.20 07/02 22:40
3F：→ johnnyzsefb:但是选范围啥意思? 要说t为实数吗? 219.70.25.20 07/02 22:42
4F：→ johnnyzsefb:path (t,t^(1/4)-3-2^(1/4)) t为大於0 219.70.25.20 07/02 22:45
5F：→ johnnyzsefb:之实数? 219.70.25.20 07/02 22:49
6F：→ GSXSP:(2,3)起点 一个方向增加 一个方向减少 140.113.140.51 07/02 22:49
7F：→ GSXSP:(2,-3) 140.113.140.51 07/02 22:56
8F：→ GSXSP:还有我上面有一个等号稍欠考虑 140.113.140.51 07/02 23:05
9F：→ GSXSP:-8t提出来..所以t<0才可以 然而我们要从t=0 140.113.140.51 07/02 23:06
10F：→ GSXSP:开始 所以是从t=2 跑到t=0 140.113.140.51 07/02 23:07
11F：→ GSXSP:抱歉 上面手误 是从t=2开始 140.113.140.51 07/02 23:09
12F：→ GSXSP:然後可以看看(0,0)的确是最大值 140.113.140.51 07/02 23:10
13F：→ GSXSP:讲的有点混乱@@ 自行体会一下吧XD 140.113.140.51 07/02 23:12