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is there somebody good at math? no,2
- #1
-
- frank
- 2004/05/24 16:41
problem 2 A
find an equation tangent to the curve y=e^x that is parralel to the line x-4y=1
problem 2 B
find an equation of the tangent to the curve y=e^x that passes through the origin.
- #2
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- まぼろし探偵
- 2004/05/24 (Mon) 17:49
- Report
4:33pmから4:41pmまでのわずか10分足らずの間の連続5つものトピ連立。よほどのバカか、嫌がらせですか???
時間がわずかにズレているのは、登録コテハン使用の証拠。なのにダブったトピを消去する様子もなく、理解に苦しみます。マスだけに?
自己中というか、常識知らずというか、フランクくん、数学の前に、もう一度幼稚園へ行こう。
- #4
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- とる太
- 2004/05/24 (Mon) 18:34
- Report
もしかしたら宿題できなくてあせってるのかもしれませんよ。
- #5
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- エコエコアザラク
- 2004/05/24 (Mon) 18:42
- Report
おいおい!びびなび!何でこんな5個のアホトピ消去しないんだ?一つだけ残して消すとかしろよ。ねずみ講や下ネタなら速攻消すくせに。
- #6
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- SM男
- 2004/05/24 (Mon) 20:55
- Report
まあまあ、そういわずに・・・
微分ぐらい文句言わずにしてやれよ。
- #7
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- Recess
- 2004/05/24 (Mon) 21:12
- Report
そうですね、誰かに答えて欲しいかも。そろそろ期末テストだと思いますが、数学も最後の難しい段階にきているはず。教えて欲しいなら、そう書けばイタズラと思われないのに。出て来てよ。
- #8
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- まぼろし探偵
- 2004/05/24 (Mon) 21:15
- Report
4時間以上も過ぎているのに、本人も、管理人さんも、なぜ消さないのでしょうね。
もし本人が宿題で焦っているなら、何度もここを見返すはず。なのに消さない。
5つも同じスレを建てれば目立つ。目立てば誰かが答えてくれる?微分なんってもう何十年も前に忘れてしまったからボクはお助けできないけれど、目立つほどたくさんスレ建てて、自分が良ければそれでいいのですか?
- #9
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- SM男
- 2004/05/24 (Mon) 21:23
- Report
そういえば、y=e^xは、微分をしても、dy/dx=e^xだったような。表に載ってると思うから調べてくれ・・・
y=e^xをグラフに描いてみてみるとけっこう簡単なんじゃないかな?x-4y=1は結局はy=0.25x-0.25だからね。
頭の中でできるほどSM男は賢くない。
- #10
-
- なめたっけ
- 2004/05/25 (Tue) 01:27
- Report
frankさん勉強頑張ってますね!
あたしには全然力になれないのですが、
世界中の誰が見るかわからないこの掲示板に問題を載せて一生懸命になってるになってるなんてすごいなぁ。って思います。誰でもみんな誰かのパワーをもらわず見は生きてけないのだからね!
努力は絶対に報われますよ☆わたしもそう願ってます。
残り少ない今学期がんばってください。
- #15
-
- sed
- 2004/05/25 (Tue) 02:49
- Report
英語の言い回しが私にはよくわからないけれど、第一問目はy=e^xの接線で、傾きがx-4y=1と同じものを探せってことだろう。接点座標における傾きはd/dx e^x=e^x=0.25。両辺でlogをとればx=log0.25
それで、y=0.25。目的の直線をy=0.25x+bと置いて(x,y)を代入すると、0.25=0.25log0.25+b。結局b=0.25-0.25log0.25。答えy=0.25x+0.25-0.25log0.25。きれいに書けばy=0.25(x+1+log4)
第二問目はy=e^xに接する線で原点を通るもの。接点の座標をaという変数で表せば(a,e^a)。このとき傾きは(e^a)/a。一方原点を通る直線の傾きはd/dx e^x=e^xであり、x=aに置いてはe^a。よって(e^a)/a=e^a。aについて解けばa=1で結局傾きはe。答えy=e*x
- #16
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- sed
- 2004/05/25 (Tue) 02:50
- Report
#この掲示板、なんだか深夜は調子悪いね。バッチ処理でもしているのだろうか...
- #17
-
thank you. sed san.
however, there is one more question i do not get.
problem is
show that the length of the portion of any tangent line to the astroid x^2/3 + y^2/3 = a^2/3 cut off by the cordinate axes is constant.
- #19
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- sed
- 2004/05/25 (Tue) 10:52
- Report
#18日本でも区別しています。底はeと一言書けばよかったですね。
lnはエンジニアリング系の人しか使わないかと思っていました。
- #21
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- sed
- 2004/05/25 (Tue) 14:30
- Report
良く見たら#20で載せた式はxとyの展開式をコピペで作っているうちに間違っていましたので修正(^^;
x=a*cos^3(t)
y=a*sin^3(t)
0<=t<=2pi
x^2/3 + y^2/3 = (a^2/3)*cos^2(t) + (a^2/3)*sin^2(t)=a^2/3
として媒介変数表示(parameterize)すれば変数がt一つにまとまるので微分が楽。
それぞれtについて微分すると
dx/dt= -3a*cos^2(t)*sin(t)
dy/dt= 3a*sin^2(t)*cos(t)
dy/dx= -1/tan(t)
あるtにおける接線の方程式は
dx/dt*(x-x(t))-dy/dt*(y-y(t))=0
展開すると
dx/dt*(x-a*cos^3(t))-dy/dt*(y-a*sin^3(t))=0
(x-a*cos^3(t))+(1/tan(t))*(y-a*sin^3(t))=0
y軸と交わる点はx=0を代入して
-a*cos^3(t)+(1/tan(t))*(y-a*sin^3(t))=0
yについて整理すると
y=a*sin(t)
同様にx軸と交わる点はy=0を代入して
(x-a*cos^3(t))+(1/tan(t))*(-a*sin^3(t))=0
xについて整理すると
x=a*cos(t)
よって線分の長さlは
l=sqrt(x^2+y^2)=sqrt(a^2*cos^2(t)+a^2sin^2(t))
l=a
変数tは消去され、定数項aだけが残った。よって線分の長さは
常に一定である。
- #22
-
- とる太
- 2004/05/25 (Tue) 17:26
- Report
どうでもいいけど、なんかSedさんカッコいいです!
- #23
-
- frank
- 2004/05/26 (Wed) 00:55
- Report
sedサン、とても助かりました。ありがとうございました。
- #24
-
の式で表してから微分した方が楽では?(もっとエレガントなやり方あるかもしれんけど)
Astroidをparametric equation で表せば:
x=a cos^3(t)
y=a sin^3(t)
where a is constant
これにtangentな直線の傾きは(by chain rule):
dy/dx=dy/dt times dt/dx = (3asin^2(t)cos(t))/(-3acos^2(t)sin(t))=-tan(t)
よってこの直線の式は
y=-tan(t)x+b
a sin^3(t)=-tan(t)acos^3(t)+b
--> b=a sin(t)[sin^2(t)+cos^2(t)]=asin(t)
Therefore, y=-tan(t)x+asint(t)
この直線がcordinate axes (x and y)をcut-off する点は
@ x=0, y=a sin(t)
@ y=0, x=a cos(t)
よってこの直線の座標軸を通る(なおかつastroidにtangentearu)部分の長さは
L=sqrt(x^2+y^2)=sqrt(a^2cos^2(t)+a^2sin^2(t))=sqrt(a^2)=a
とconstantになる。
なんか日本語と英語混ぜると変な感じ。数学の問題解く時は英語で考える方が好きです(個人的には)。
- #25
-
- sed
- 2004/05/26 (Wed) 08:58
- Report
修正した#21も間違っていて済みませんfrankさんもう宿題提出しちゃったかな?(^^;
#24にあるようにdy/dx=-tan(t)が正しいですね。
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