울프람알파의 활용

이 항목의 스프링노트 원문주소

• 울프람알파의 활용

 

 

==개요

• 울프람 알파
• 연산능력을 갖춘 지식엔진
• http://www.wolframalpha.com
• http://mathdl.maa.org/images/upload_library/23/karaali/alpha/index.html
   
  

 

==문제풀이와 울프람알파

• Step-by-Step Math
• Wolfram|Alpha as a self-verification tool

 

 

==방정식의 풀이

• Solving Equations with Wolfram|Alpha

 

 

==숙제와 울프람알파

• http://blog.wolframalpha.com/2010/01/22/is-it-cheating-to-use-wolframalpha-for-math-homework/

 

 

==그래프 그리기

 

 

 

 

10.4.44.

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

[/pages/4176465/attachments/2110589 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

• http://www.wolframalpha.com/input/?i=r^2
  
• http://www.wolframalpha.com/input/?i=
  
• http://www.wolframalpha.com/input/?i=
  
• http://www.wolframalpha.com/input/?i=
  

 

==관련된 항목들

• 프랙탈