[微分] 求切線斜率@@

看板trans_math作者 (阿國~)時間14年前 (2010/04/15 22:49), 編輯推噓6(6推 0噓 17→)
留言23則, 4人參與, 最新討論串1/1
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 -- ※ 發信站: 批踢踢實業坊(ptt.cc) ◆ From: 122.117.249.29
推
04/15 23:02, , 1F
兩邊的dy/dx相等 得解?
04/15 23:02, 1F
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04/15 23:03, , 2F
好像不是= =
04/15 23:03, 2F
推
04/15 23:27, , 3F
為何求切線斜率 答案會有根號x？！
04/15 23:27, 3F
推
04/15 23:31, , 4F
兩曲線無焦點 所以分別對兩曲線假設
04/15 23:31, 4F
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04/15 23:32, , 5F
參數式 然後利用兩點斜率 和微分斜率
04/15 23:32, 5F
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04/15 23:32, , 6F
求出參數 即求出點
04/15 23:32, 6F
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04/15 23:34, , 7F
答案沒打錯嗎? 切線怎麼會有x^1/2?
04/15 23:34, 7F
推
04/15 23:34, , 8F
我是假設切點分別為(t,t^2+1)和(s,-s^2)
04/15 23:34, 8F
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04/15 23:35, , 9F
如果以微分求斜率，得2t=-2s 知t=-s
04/15 23:35, 9F
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04/15 23:36, , 10F
於是兩切點分別為(s,s^2+1)和(s,-s^2)
04/15 23:36, 10F
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04/15 23:37, , 11F
(-s,s^2+1)
04/15 23:37, 11F
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04/15 23:37, , 12F
則兩點連線斜率：(-2s^2-1)/2s
04/15 23:37, 12F
推
04/15 23:39, , 13F
原PO應該是要打x*根號2吧
04/15 23:39, 13F
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04/15 23:39, , 14F
斜率=(-2s^2-1)/2s=2s求得s=正負根號2
04/15 23:39, 14F
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04/15 23:39, , 15F
1/(根號2)
04/15 23:39, 15F
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04/15 23:40, , 16F
n大是對的
04/15 23:40, 16F
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04/15 23:40, , 17F
謝謝~XD
04/15 23:40, 17F
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04/15 23:41, , 18F
等等等等 打錯答案@@@
04/15 23:41, 18F
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04/15 23:41, , 19F
y=+-根號2*x+(1/2)
04/15 23:41, 19F
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04/15 23:41, , 20F
抱歉 算到眼殘= =
04/15 23:41, 20F
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04/15 23:45, , 21F
嗯嗯~了解 謝啦:D
04/15 23:45, 21F
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04/15 23:45, , 22F
抱歉 眼殘可能有害到一些人= =+
04/15 23:45, 22F
推
04/15 23:46, , 23F
哈~不會啦~這答案錯得很明顯~沒被害到XD
04/15 23:46, 23F
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文章代碼(AID): #1BnoUIPd (trans_math)