"울프람알파의 활용"의 두 판 사이의 차이

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==개요==
  
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* 울프람 알파
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* 연산능력을 갖춘 지식엔진
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* http://www.wolframalpha.com
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* http://mathdl.maa.org/images/upload_library/23/karaali/alpha/index.html 
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==문제풀이와 울프람알파==
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* [http://blog.wolframalpha.com/2009/12/01/step-by-step-math/ Step-by-Step Math]
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* [http://castingoutnines.wordpress.com/2010/01/04/wolframalpha-as-a-self-verification-tool/ Wolfram|Alpha as a self-verification tool]
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==방정식의 풀이==
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* [http://blog.wolframalpha.com/2010/01/14/solving-equations-with-wolframalpha/ Solving Equations with Wolfram|Alpha]
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==숙제와 울프람알파==
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* http://blog.wolframalpha.com/2010/01/22/is-it-cheating-to-use-wolframalpha-for-math-homework/
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==그래프 그리기==
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'''10.4.44.'''
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[http://www.wolframalpha.com/input/?i=%28r-%288%2B8sin+theta%29%29%28r+sin+theta+-4%29%3D0 http://www.wolframalpha.com/input/?i=(r-(8%2B8sin+theta))(r+sin+theta+-4)%3D0]
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[[파일:4176465-1MSP761197847h1hg25e7he00004g3ehh9ag590ea81.gif]]
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Note that <math>2\sin^2 \alpha +2\sin \alpha-1=0</math>.
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Since <math>0< \alpha <\frac{\pi}{2}</math>, <math>\sin \alpha =\frac{\sqrt 3 -1}{2}</math> and <math>\cos \alpha =\sqrt{\frac{\sqrt 3}{2}}</math>
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<math>\frac{1}{2}\int_{\alpha}^{\pi-\alpha}r^2 d\theta=\int_{\alpha}^{\pi/2}r^2 d\theta=\int_{\alpha}^{\pi/2}64(1+\sin\theta)^2d\theta</math>
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<math>=64\int_{\alpha}^{\pi/2}(1+\sin\theta)^2d\theta=64\int_{\alpha}^{\pi/2}1+2\sin\theta+\sin^2\theta d\theta</math>
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<math>=64[\frac{3}{2}\theta - 2 \cos \theta- \frac{1}{4}\sin 2\theta]_{\alpha}^{\pi/2}=[96\theta - 128 \cos \theta- 16\sin 2\theta]_{\alpha}^{\pi/2}</math>
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<math>=[96\theta - 128 \cos \theta- 16\sin 2\theta]_{\alpha}^{\pi/2}=48\pi-96\alpha+128\cos \alpha+16\sin 2\alpha</math>
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We need to subtract from this the area of the triangle which is given by :
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<math>2\times \frac{1}{2}\cdot 4 (8+8\sin\alpha)\sin (\pi/2-\alpha)=32(1+\sin\alpha)\cos \alpha=32\cos \alpha+16 \sin 2\alpha</math>
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So the area will be given by :
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<math>48\pi-96\alpha+128\cos \alpha+16\sin 2\alpha-(32\cos \alpha+16 \sin 2\alpha)=48\pi-96\alpha+96\cos \alpha\approx 204.16</math>
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[http://www.wolframalpha.com/input/?i=N%5B48pi+-96%28arcsin%7B%28sqrt+3+-+1%29/2%7D%29%2B96%28sqrt%28%28sqrt+3%29/2%29%29,10%5D http://www.wolframalpha.com/input/?i=N[48pi+-96(arcsin{(sqrt+3+-+1)/2})%2B96(sqrt((sqrt+3)/2)),10]]
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==사용예==
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* [http://www.wolframalpha.com/input/?i=r%5E2 http://www.wolframalpha.com/input/?i=r^2]
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==parametric curves==
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*  parabola
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** plot x=t^2-2,y=5-2t where -3<t<4
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** [http://www.wolframalpha.com/input/?i=plot+x%3Dt%5E2-2,y%3D5-2t+where+-3%3Ct%3C4 output]
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*  ellipse
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** plot x=4cos t,y=5sin t where -pi/2<t<pi/2
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** [http://www.wolframalpha.com/input/?i=plot+x%3D4cos+t,y%3D5sin+t+where+-pi/2%3Ct%3Cpi/2 output]
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*  hyperbola
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** parametricplot x=sinh t,y=cosh t, -1<t<1
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** [http://www.wolframalpha.com/input/?i=parametricplot+x%3Dsinh+t,y%3Dcosh+t,+-1%3Ct%3C1 output]
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==polar coorinates==
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* [http://www.wolframalpha.com/input/?i=polar+plot+r+%3D+1+-+2sin+t polar plot r = 1 - 2sin t]
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==관련된 항목들==
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* [[프랙탈]]
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[[분류:계산 리소스]]

2020년 12월 28일 (월) 03:47 기준 최신판

개요


문제풀이와 울프람알파



방정식의 풀이



숙제와 울프람알파



그래프 그리기

10.4.44.

http://www.wolframalpha.com/input/?i=(r-(8%2B8sin+theta))(r+sin+theta+-4)%3D0

4176465-1MSP761197847h1hg25e7he00004g3ehh9ag590ea81.gif

Note that \(2\sin^2 \alpha +2\sin \alpha-1=0\).

Since \(0< \alpha <\frac{\pi}{2}\), \(\sin \alpha =\frac{\sqrt 3 -1}{2}\) and \(\cos \alpha =\sqrt{\frac{\sqrt 3}{2}}\)


\(\frac{1}{2}\int_{\alpha}^{\pi-\alpha}r^2 d\theta=\int_{\alpha}^{\pi/2}r^2 d\theta=\int_{\alpha}^{\pi/2}64(1+\sin\theta)^2d\theta\)

\(=64\int_{\alpha}^{\pi/2}(1+\sin\theta)^2d\theta=64\int_{\alpha}^{\pi/2}1+2\sin\theta+\sin^2\theta d\theta\)

\(=64[\frac{3}{2}\theta - 2 \cos \theta- \frac{1}{4}\sin 2\theta]_{\alpha}^{\pi/2}=[96\theta - 128 \cos \theta- 16\sin 2\theta]_{\alpha}^{\pi/2}\)

\(=[96\theta - 128 \cos \theta- 16\sin 2\theta]_{\alpha}^{\pi/2}=48\pi-96\alpha+128\cos \alpha+16\sin 2\alpha\)

We need to subtract from this the area of the triangle which is given by \[2\times \frac{1}{2}\cdot 4 (8+8\sin\alpha)\sin (\pi/2-\alpha)=32(1+\sin\alpha)\cos \alpha=32\cos \alpha+16 \sin 2\alpha\]

So the area will be given by \[48\pi-96\alpha+128\cos \alpha+16\sin 2\alpha-(32\cos \alpha+16 \sin 2\alpha)=48\pi-96\alpha+96\cos \alpha\approx 204.16\]

http://www.wolframalpha.com/input/?i=N[48pi+-96(arcsin{(sqrt+3+-+1)/2})%2B96(sqrt((sqrt+3)/2)),10]






사용예


parametric curves

  • parabola
    • plot x=t^2-2,y=5-2t where -3<t<4
    • output
  • ellipse
    • plot x=4cos t,y=5sin t where -pi/2<t<pi/2
    • output
  • hyperbola
    • parametricplot x=sinh t,y=cosh t, -1<t<1
    • output



polar coorinates


관련된 항목들

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