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Vivinavi Los Angelesis 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.
- #21
-
- 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。
Posting period for “ is there somebody good at math? no,2 ” has been closed.
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