# 울프람알파의 활용

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## 그래프 그리기

10.4.44.

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 $48\pi-96\alpha+128\cos \alpha+16\sin 2\alpha-(32\cos \alpha+16 \sin 2\alpha)=48\pi-96\alpha+96\cos \alpha\approx 204.16$

## 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