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
방정식의 풀이
숙제와 울프람알파
그래프 그리기
10.4.44.
http://www.wolframalpha.com/input/?i=(r-(8%2B8sin+theta))(r+sin+theta+-4)%3D0
Note that <math>2\sin^2 \alpha +2\sin \alpha-1=0</math>.
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>
<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>
<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>
<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>
<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>
We need to subtract from this the area of the triangle which is given by :
<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>
So the area will be given by :
<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>
http://www.wolframalpha.com/input/?i=N[48pi+-96(arcsin{(sqrt+3+-+1)/2})%2B96(sqrt((sqrt+3)/2)),10]
사용예
parametric curves
- parabola
- plot x=t^2-2,y=5-2t where -3<t<4
- output
- ellipse
- plot x=4cos t,y=5sin t where -pi/2<t<pi/2
- output
- hyperbola
- parametricplot x=sinh t,y=cosh t, -1<t<1
- output
polar coorinates