Project Euler 363
Project Euler 363
题目
Bézier Curves
A cubic Bézier curve is defined by four points: \(P_0, P_1, P_2,\) and \(P_3\).
The curve is constructed as follows:
On the segments \(P_0 P_1\), \(P_1 P_2\), and \(P_2 P_3\) the points \(Q_0, Q_1,\) and \(Q_2\) are drawn such that \(\dfrac{P_0 Q_0}{P_0 P_1} = \dfrac{P_1 Q_1}{P_1 P_2} = \dfrac{P_2 Q_2}{P_2 P_3} = t\), with \(t\) in \([0, 1]\).
On the segments \(Q_0 Q_1\) and \(Q_1 Q_2\) the points \(R_0\) and \(R_1\) are drawn such that
\(\dfrac{Q_0 R_0}{Q_0 Q_1} = \dfrac{Q_1 R_1}{Q_1 Q_2} = t\) for the same value of \(t\).
On the segment \(R_0 R_1\) the point \(B\) is drawn such that \(\dfrac{R_0 B}{R_0 R_1} = t\) for the same value of \(t\).
The Bézier curve defined by the points \(P_0, P_1, P_2, P_3\) is the locus of \(B\) as \(Q_0\) takes all possible positions on the segment \(P_0 P_1\). (Please note that for all points the value of \(t\) is the same.)
From the construction it is clear that the Bézier curve will be tangent to the segments \(P_0 P_1\) in \(P_0\) and \(P_2 P_3\) in \(P_3\).
A cubic Bézier curve with \(P_0 = (1, 0), P_1 = (1, v), P_2 = (v, 1),\) and \(P_3 = (0, 1)\) is used to approximate a quarter circle.
The value \(v \gt 0\) is chosen such that the area enclosed by the lines \(O P_0, OP_3\) and the curve is equal to \(\dfrac{\pi}{4}\) (the area of the quarter circle).
By how many percent does the length of the curve differ from the length of the quarter circle?
That is, if \(L\) is the length of the curve, calculate \(100 \times \dfrac{L - \frac{\pi}{2}}{\frac{\pi}{2}}\)
Give your answer rounded to \(10\) digits behind the decimal point.
解决方案
这个页面给出了这条三次贝塞尔曲线的参数方程：
\[{\displaystyle \mathbf {B} (t)=\mathbf {P} _{0}(1-t)^{3}+3\mathbf {P} _{1}t(1-t)^{2}+3\mathbf {P} _{2}t^{2}(1-t)+\mathbf {P} _{3}t^{3}{,}t\in [0,1]}\]
那么化简后，得到关于\(x,y\)的参数方程：
\[\begin{aligned} x(t)&=(1-t)^3+3t(1-t)^2+3t^2(1-t)v\\ y(t)&=3t(1-t)^2v+3t^2(1-t)+t^3 \end{aligned}\]
第一个问题：如何求贝塞尔曲线\(B\)和线段\(C:P_3O,D:OP_0)\)共同围绕而成的封闭曲线的面积？
使用补线法进行计算，可以列出公式为：
\[\dfrac{\pi}{4}=S=\dfrac{1}{2}\int_{B+C+D}xdy-ydx\]
注意线段\(B\)和线段\(C\)的积分值都为\(0\)，那么只需要计算\(\dfrac{1}{2}\int_Bxdy-ydx\)即可。
\[\dfrac{1}{2}(1+\dfrac{6v}{5}-\dfrac{3v^2}{10})=\dfrac{\pi}{4}\]
解这个一元二次方程，得到\(v=2\pm\sqrt{\dfrac{1}{3}(22-5\pi)}\)。
由于按照题目的需要，它是一个\(\dfrac{1}{4}\)圆的近似，因此取\(v=2-\sqrt{\dfrac{1}{3}(22-5\pi)}\)
第二个问题：如何求曲线的长度？
\[d_{P_0P_3}=\int_0^1\sqrt{x'^2(t)+y'^2(t)}dt\]
通过Mathematica直接计算出具体值即可。
代码
1 | x[t_] = (1 - t)^3 + 3 t (1 - t)^2 + 3 t^2 (1 - t) v |