作者a039333780 (阿国~)
站内trans_math
标题[微分] 求切线斜率@@
时间Thu Apr 15 22:49:52 2010
求同时切於y=x^2+1和y=-x^2 两曲线之直线@@
答案是y=x^1/2+1/2 y=-x^1/2+1/2
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※ 发信站: 批踢踢实业坊(ptt.cc)
◆ From: 122.117.249.29
1F：推 dollarok99:两边的dy/dx相等 得解? 219.69.69.43 04/15 23:02
2F：→ dollarok99:好像不是= = 219.69.69.43 04/15 23:03
3F：推 n19860423:为何求切线斜率 答案会有根号x？！ 114.45.137.128 04/15 23:27
4F：推 midarmyman:两曲线无焦点 所以分别对两曲线假设 140.117.198.78 04/15 23:31
5F：→ midarmyman:参数式 然後利用两点斜率 和微分斜率 140.117.198.78 04/15 23:32
6F：→ midarmyman:求出参数 即求出点 140.117.198.78 04/15 23:32
7F：→ midarmyman:答案没打错吗? 切线怎麽会有x^1/2? 140.117.198.78 04/15 23:34
8F：推 n19860423:我是假设切点分别为(t,t^2+1)和(s,-s^2) 114.45.137.128 04/15 23:34
9F：→ n19860423:如果以微分求斜率，得2t=-2s 知t=-s 114.45.137.128 04/15 23:35
10F：→ n19860423:於是两切点分别为(s,s^2+1)和(s,-s^2) 114.45.137.128 04/15 23:36
11F：→ n19860423: (-s,s^2+1) 114.45.137.128 04/15 23:37
12F：→ n19860423:则两点连线斜率：(-2s^2-1)/2s 114.45.137.128 04/15 23:37
13F：推 midarmyman:原PO应该是要打x*根号2吧 140.117.198.78 04/15 23:39
14F：→ n19860423:斜率=(-2s^2-1)/2s=2s求得s=正负根号2 114.45.137.128 04/15 23:39
15F：→ n19860423: 1/(根号2) 114.45.137.128 04/15 23:39
16F：→ midarmyman:n大是对的 140.117.198.78 04/15 23:40
17F：→ n19860423:谢谢~XD 114.45.137.128 04/15 23:40
18F：→ a039333780:等等等等 打错答案@@@ 122.117.249.29 04/15 23:41
19F：→ a039333780:y=+-根号2*x+(1/2) 122.117.249.29 04/15 23:41
20F：→ a039333780:抱歉 算到眼残= = 122.117.249.29 04/15 23:41
21F：→ a039333780:嗯嗯~了解 谢啦:D 122.117.249.29 04/15 23:45
22F：→ a039333780:抱歉 眼残可能有害到一些人= =+ 122.117.249.29 04/15 23:45
23F：推 n19860423:哈~不会啦~这答案错得很明显~没被害到XD 114.45.137.128 04/15 23:46