作者a039333780 (阿國~)
站內trans_math
標題[微分] 求切線斜率@@
時間Thu Apr 15 22:49:52 2010
求同時切於y=x^2+1和y=-x^2 兩曲線之直線@@
答案是y=x^1/2+1/2 y=-x^1/2+1/2
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◆ From: 122.117.249.29
1F：推 dollarok99:兩邊的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:為何求切線斜率 答案會有根號x？！ 114.45.137.128 04/15 23:27
4F：推 midarmyman:兩曲線無焦點 所以分別對兩曲線假設 140.117.198.78 04/15 23:31
5F：→ midarmyman:參數式 然後利用兩點斜率 和微分斜率 140.117.198.78 04/15 23:32
6F：→ midarmyman:求出參數 即求出點 140.117.198.78 04/15 23:32
7F：→ midarmyman:答案沒打錯嗎? 切線怎麼會有x^1/2? 140.117.198.78 04/15 23:34
8F：推 n19860423:我是假設切點分別為(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:於是兩切點分別為(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:則兩點連線斜率：(-2s^2-1)/2s 114.45.137.128 04/15 23:37
13F：推 midarmyman:原PO應該是要打x*根號2吧 140.117.198.78 04/15 23:39
14F：→ n19860423:斜率=(-2s^2-1)/2s=2s求得s=正負根號2 114.45.137.128 04/15 23:39
15F：→ n19860423: 1/(根號2) 114.45.137.128 04/15 23:39
16F：→ midarmyman:n大是對的 140.117.198.78 04/15 23:40
17F：→ n19860423:謝謝~XD 114.45.137.128 04/15 23:40
18F：→ a039333780:等等等等 打錯答案@@@ 122.117.249.29 04/15 23:41
19F：→ a039333780:y=+-根號2*x+(1/2) 122.117.249.29 04/15 23:41
20F：→ a039333780:抱歉 算到眼殘= = 122.117.249.29 04/15 23:41
21F：→ a039333780:嗯嗯~了解 謝啦:D 122.117.249.29 04/15 23:45
22F：→ a039333780:抱歉 眼殘可能有害到一些人= =+ 122.117.249.29 04/15 23:45
23F：推 n19860423:哈~不會啦~這答案錯得很明顯~沒被害到XD 114.45.137.128 04/15 23:46