"울프람알파의 활용"의 두 판 사이의 차이
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Pythagoras0 (토론 | 기여) |
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| (사용자 2명의 중간 판 32개는 보이지 않습니다) | |||
| 1번째 줄: | 1번째 줄: | ||
| − | + | ==개요== | |
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| − | + | * 울프람 알파 | |
| + | * 연산능력을 갖춘 지식엔진 | ||
| + | * http://www.wolframalpha.com | ||
| + | * http://mathdl.maa.org/images/upload_library/23/karaali/alpha/index.html | ||
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| − | + | ==문제풀이와 울프람알파== | |
| − | + | * [http://blog.wolframalpha.com/2009/12/01/step-by-step-math/ Step-by-Step Math] | |
| + | * [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== | |
| − | [http:// | + | * [http://blog.wolframalpha.com/2010/01/14/solving-equations-with-wolframalpha/ Solving Equations with Wolfram|Alpha] |
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| − | + | ==숙제와 울프람알파== | |
| − | + | * 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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| − | '''10. | ||
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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] | |
| − | [ | + | [[파일:4176465-1MSP761197847h1hg25e7he00004g3ehh9ag590ea81.gif]] |
| − | Note | + | Note that <math>2\sin^2 \alpha +2\sin \alpha-1=0</math>. |
| − | + | 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> | <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> | ||
| 97번째 줄: | 63번째 줄: | ||
<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> | <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> | ||
| − | We need to subtract from | + | We need to subtract from this the area of the triangle which is given by : |
<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> | <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> | ||
| 105번째 줄: | 71번째 줄: | ||
<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> | <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> | ||
| − | http://www.wolframalpha.com/input/?i=N[48pi+-96(arcsin{(sqrt+3+-+1)/2})%2B96(sqrt((sqrt+3)/2)),10] | + | [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== | |
| − | + | * [http://www.wolframalpha.com/input/?i=r%5E2 http://www.wolframalpha.com/input/?i=r^2] | |
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| − | + | ==parametric curves== | |
| − | + | * parabola | |
| + | ** plot x=t^2-2,y=5-2t where -3<t<4 | ||
| + | ** [http://www.wolframalpha.com/input/?i=plot+x%3Dt%5E2-2,y%3D5-2t+where+-3%3Ct%3C4 output] | ||
| − | + | * ellipse | |
| + | ** plot x=4cos t,y=5sin t where -pi/2<t<pi/2 | ||
| + | ** [http://www.wolframalpha.com/input/?i=plot+x%3D4cos+t,y%3D5sin+t+where+-pi/2%3Ct%3Cpi/2 output] | ||
| + | * hyperbola | ||
| + | ** parametricplot x=sinh t,y=cosh t, -1<t<1 | ||
| + | ** [http://www.wolframalpha.com/input/?i=parametricplot+x%3Dsinh+t,y%3Dcosh+t,+-1%3Ct%3C1 output] | ||
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| − | + | ==polar coorinates== | |
| − | + | * [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== | |
| − | * | + | * [[프랙탈]] |
| − | + | [[분류:계산 리소스]] | |
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2020년 12월 28일 (월) 03:47 기준 최신판
개요
- 울프람 알파
- 연산능력을 갖춘 지식엔진
- http://www.wolframalpha.com
- http://mathdl.maa.org/images/upload_library/23/karaali/alpha/index.html
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
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