울프람알파의 활용

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  • 울프람알파의 활용

 

 

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]

 

 

 

 

 

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