"울프람알파의 활용"의 두 판 사이의 차이
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2009년 9월 9일 (수) 14:11 판
- 울프람 알파
- 연산능력을 갖춘 지식엔진
- http://www.wolframalpha.com
샘플
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 \(\psi=\phi-\theta\).
\(\tan \psi=\tan (\phi-\theta)\).
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