"울프람알파의 활용"의 두 판 사이의 차이

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*  울프람 알파<br>
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** 연산능력을 갖춘 지식엔진
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** http://www.wolframalpha.com
  
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<h5 style="line-height: 2em; margin-top: 0px; margin-right: 0px; margin-bottom: 0px; margin-left: 0px; color: rgb(34, 61, 103); font-family: 'malgun gothic', dotum, gulim, sans-serif; font-size: 1.166em; background-image: ; background-color: initial; background-position: 0px 100%;">samp</h5>
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'''10.3. 56'''
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(a) http://www.wolframalpha.com/input/?i=r%3Dsqrt(theta)
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[http://math53.springnote.com/pages/4144713/attachments/2109931 MSP119619783h6c32cf74c8000022832f47eg5f3a38.gif]
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(b) http://www.wolframalpha.com/input/?i=r%3Dsqrt(theta)^(4.01)
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[http://math53.springnote.com/pages/4144713/attachments/2109929 MSP18619784d1302aic6fe00005784bie94h0515cd.gif]
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(c) http://www.wolframalpha.com/input/?i=r%3Dcos+(theta/3)
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[http://math53.springnote.com/pages/4144713/attachments/2109915 MSP74719784071iad69b2900005cid5ibigi61dg94(1).gif]
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(d) http://www.wolframalpha.com/input/?i=r%3D1%2B2+cos+(theta)
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[http://math53.springnote.com/pages/4144713/attachments/2109919 MSP11271978443hf7aa7h2f00002bd5bh2db919dc1f.gif]
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(e) http://www.wolframalpha.com/input/?i=r%3D2%2Bsin(3thetA)
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[http://math53.springnote.com/pages/4144713/attachments/2109921 MSP47019784919haafi5hc00002h164hgd33f4i46i.gif]
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(f) http://www.wolframalpha.com/input/?i=r%3D1%2B2sin(3theta)
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[http://math53.springnote.com/pages/4144713/attachments/2109923 MSP6981978475e4574h83h00003a7ga0a9g5d06c13.gif]
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'''10.3. 69'''
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<math>r=a \sin \theta+ b\cos \theta</math>
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<math>r^2=a r\sin \theta+ br\cos \theta</math>
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<math>x^2+y^2=ay+bx</math>
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<math>(x-\frac{b}{2})^2+(y-\frac{a}{2})^2=\frac{a^2}{4}+\frac{b^2}{4}</math>
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center : <math>(\frac{b}{2},\frac{a}{2})</math>
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radius : <math>\frac{\sqrt{a^2+b^2}}{2}</math>
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'''10.3. 83.'''
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Use the following set of identities
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<math>x = r(\theta) \cos \theta</math>
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<math>y = r(\theta)\sin \theta</math>
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<math>\tan{\phi}=\frac{dy}{dx}={\frac{dy}{d\theta}}/{\frac{dx}{d\theta}}</math>
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Express <math>\tan \psi</math> in terms of <math>\theta</math> using the above identity.
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Now use the hint <math>\psi=\phi-\theta</math>. So <math>\tan \psi=\tan (\phi-\theta)</math>.
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Use
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<math>\tan \psi=\tan (\phi-\theta)=\frac{\tan \phi - \tan\theta}{1+\tan\phi \tan\theta}</math>
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Simplify this. 
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'''10.4.44.'''
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[http://math53.springnote.com/pages/4144713/attachments/2109769 MSP26619784bhf76f4e70900004e38231fbie4a798.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[48pi+-96(arcsin{(sqrt+3+-+1)/2})%2B96(sqrt((sqrt+3)/2)),10]
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'''10.4.47.'''
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<math>r=\theta^2</math>
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<math>L=\int_0^{2\pi}\sqrt{\theta^4+(2\theta)^2}\,d\theta=\int_0^{2\pi}\theta\sqrt{\theta^2+4}\,d\theta</math>
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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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[http://math53.springnote.com/pages/4144713/attachments/2109941 MSP71919784049c09bb98700001095e8fadhf5gc4d.gif]
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<h5 style="line-height: 3.428em; margin-top: 0px; margin-right: 0px; margin-bottom: 0px; margin-left: 0px; color: rgb(34, 61, 103); font-family: 'malgun gothic', dotum, gulim, sans-serif; font-size: 1.166em; background-image: ; background-color: initial; background-position: 0px 100%;">encyclopedia</h5>
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*  http://www.wolframalpha.com/input/?i=<br>
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*  http://www.wolframalpha.com/input/?i=<br>
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*  http://www.wolframalpha.com/input/?i=<br>
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*  http://www.wolframalpha.com/input/?i=<br>  <br>

2009년 9월 9일 (수) 13:51 판

 

samp

10.3. 56

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

MSP119619783h6c32cf74c8000022832f47eg5f3a38.gif

(b) http://www.wolframalpha.com/input/?i=r%3Dsqrt(theta)^(4.01)

MSP18619784d1302aic6fe00005784bie94h0515cd.gif

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

MSP74719784071iad69b2900005cid5ibigi61dg94(1).gif

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

MSP11271978443hf7aa7h2f00002bd5bh2db919dc1f.gif

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

MSP47019784919haafi5hc00002h164hgd33f4i46i.gif

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

MSP6981978475e4574h83h00003a7ga0a9g5d06c13.gif

 

10.3. 69

\(r=a \sin \theta+ b\cos \theta\)

\(r^2=a r\sin \theta+ br\cos \theta\)

\(x^2+y^2=ay+bx\)

\((x-\frac{b}{2})^2+(y-\frac{a}{2})^2=\frac{a^2}{4}+\frac{b^2}{4}\)

center : \((\frac{b}{2},\frac{a}{2})\)

radius : \(\frac{\sqrt{a^2+b^2}}{2}\)

 

10.3. 83.

Use the following set of identities

 

\(x = r(\theta) \cos \theta\)

\(y = r(\theta)\sin \theta\)

\(\tan{\phi}=\frac{dy}{dx}={\frac{dy}{d\theta}}/{\frac{dx}{d\theta}}\)

Express \(\tan \psi\) in terms of \(\theta\) using the above identity.

Now use the hint \(\psi=\phi-\theta\). So \(\tan \psi=\tan (\phi-\theta)\).

Use

\(\tan \psi=\tan (\phi-\theta)=\frac{\tan \phi - \tan\theta}{1+\tan\phi \tan\theta}\)

Simplify this. 

 

10.4.44.

MSP26619784bhf76f4e70900004e38231fbie4a798.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]

 

10.4.47.

\(r=\theta^2\)

\(L=\int_0^{2\pi}\sqrt{\theta^4+(2\theta)^2}\,d\theta=\int_0^{2\pi}\theta\sqrt{\theta^2+4}\,d\theta\)

 

 

10.4.55.

(b)

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

MSP71919784049c09bb98700001095e8fadhf5gc4d.gif

 

 

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