"울프람알파의 활용"의 두 판 사이의 차이

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<h5>그래프 그리기</h5>
 
<h5>그래프 그리기</h5>
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<math>r=\cos \frac{\theta}{3}</math>
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[/pages/4176465/attachments/2110527 MSP74719784071iad69b2900005cid5ibigi61dg94.gif]
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 [http://www.wolframalpha.com/input/?i=r%3Dcos+%28theta/3%29 http://www.wolframalpha.com/input/?i=r%3Dcos+(theta/3)]
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<math>r=1+2\cos\theta</math>
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  [/pages/4176465/attachments/2110537 MSP11271978443hf7aa7h2f00002bd5bh2db919dc1f.gif]
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 [http://www.wolframalpha.com/input/?i=r%3D1%2B2+cos+%28theta%29 http://www.wolframalpha.com/input/?i=r%3D1%2B2+cos+(theta)]
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<math>r=2+\sin 3\theta</math>
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[http://www.wolframalpha.com/input/?i=r%3D2%2Bsin%283thetA%29 http://www.wolframalpha.com/input/?i=r%3D2%2Bsin(3thetA)]
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 [/pages/4176465/attachments/2110539 MSP47019784919haafi5hc00002h164hgd33f4i46i.gif]
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<math>r=1+2\sin 3\theta</math>
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 [http://www.wolframalpha.com/input/?i=r%3D1%2B2sin%283theta%29 http://www.wolframalpha.com/input/?i=r%3D1%2B2sin(3theta)]
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[/pages/4176465/attachments/2110535 MSP6981978475e4574h83h00003a7ga0a9g5d06c13.gif]
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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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[/pages/4176465/attachments/2110589 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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'''10.4.55.'''
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(b)
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<math>r^2=\cos 2\theta</math>
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[/pages/4176465/attachments/2110541 MSP71919784049c09bb98700001095e8fadhf5gc4d.gif]
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[http://www.wolframalpha.com/input/?i=r%5E2%3Dcos+%282theta%29 http://www.wolframalpha.com/input/?i=r^2%3Dcos+(2theta)]
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2011년 1월 29일 (토) 18:17 판

이 항목의 스프링노트 원문주소

 

 

개요

 

문제풀이와 울프람알파

 

 

방정식의 풀이

 

 

숙제와 울프람알파

 

 

그래프 그리기

\(r=\cos \frac{\theta}{3}\)

[/pages/4176465/attachments/2110527 MSP74719784071iad69b2900005cid5ibigi61dg94.gif]

 http://www.wolframalpha.com/input/?i=r%3Dcos+(theta/3)

 

\(r=1+2\cos\theta\)

  [/pages/4176465/attachments/2110537 MSP11271978443hf7aa7h2f00002bd5bh2db919dc1f.gif]

 http://www.wolframalpha.com/input/?i=r%3D1%2B2+cos+(theta)

 

\(r=2+\sin 3\theta\)

http://www.wolframalpha.com/input/?i=r%3D2%2Bsin(3thetA)

 [/pages/4176465/attachments/2110539 MSP47019784919haafi5hc00002h164hgd33f4i46i.gif]

 

\(r=1+2\sin 3\theta\)

 http://www.wolframalpha.com/input/?i=r%3D1%2B2sin(3theta)

[/pages/4176465/attachments/2110535 MSP6981978475e4574h83h00003a7ga0a9g5d06c13.gif]

 

 

 

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]

 

 

10.4.55.

(b)

\(r^2=\cos 2\theta\)

[/pages/4176465/attachments/2110541 MSP71919784049c09bb98700001095e8fadhf5gc4d.gif]

http://www.wolframalpha.com/input/?i=r^2%3Dcos+(2theta)

 

 

 

 

사용예

 

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