Homework4
问题一
1
由题意可得，商船与海盗的位置与速度有关，有
\[\begin{aligned}
\frac{dx_a}{dt}&=v_a(t)\cdot cos(\alpha(t)) \\
\frac{dy_a}{dt}&=v_a(t)\cdot sin(\alpha(t)) \\
\frac{dy_b}{dt}&=v_b(t)\cdot cos(\beta(t)) \\
\frac{dy_b}{dt}&=v_b(t)\cdot sin(\beta(t))
\end{aligned}\]
有初值\(x_a(0)=y_a(0)=0,x_b(0)=x_0,y_b(0)=y_0\)
2
当海盗的方向指向商船时，\(r(t)\)减小最快。有
\[\begin{aligned}
\beta(t)=arctan\frac{y_a(t)-y_b(t)}{x_a(t)-x_b(t)}=\theta(t)
\end{aligned}\]
3
由于商船方向不变，那么将海盗的速度投影到商船的方向上，再减去商船的速度，可以得到\(r(t)\)的变化率，有
\[\begin{aligned}
\frac{dr}{dt}=v_a\cdot cos(\theta(t)-\alpha(t))-\lambda v_a=v_a(cos(\theta(t)-\alpha(t))-\lambda)
\end{aligned}\]
对于 \(\theta(t)\) 有
\[
\frac{d\theta(t)}{dt} = \frac{d}{dt} \arctan\frac{y_a(t) - y_b(t)}{x_a(t) - x_b(t)}
\]
代入 \(\beta(t) = \theta(t)\)，我们有
\[
\frac{d\theta(t)}{dt} = \frac{\nu_a(t) \sin(\alpha(t) - \theta(t))}{r(t)} = \frac{\nu_a \sin(\alpha(t) - \theta(t))}{r(t)}
\]
4
当\(\alpha(t) \equiv 0\)时，商船沿x轴正向行驶，因此\(\cos \alpha(t) = 1\)和\(\sin \alpha(t) = 0\)
此时有
\[
\frac{dr(t)}{dt} = \lambda \nu_a - \nu_a \cos \theta(t)
\]
\[
\frac{d\theta(t)}{dt} = \frac{\nu_a \sin \theta(t)}{r(t)}
\]
在\(\lambda = 1\)时，距离\( r(t) \)的微分方程简化为：
\[
\frac{dr(t)}{dt} = \nu_a (1 - \cos \theta(t))
\]
解微分方程
\[
\frac{d\theta(t)}{dt} = \frac{\nu_a \sin \theta(t)}{r(t)}
\]
\[
\frac{d\theta(t)}{\sin \theta(t)} = \frac{\nu_a}{r(t)} dt
\]
积分两边
\[
\int \frac{d\theta(t)}{\sin \theta(t)} = \int \frac{\nu_a}{r(t)} dt
\]
\[
\ln \left| \tan \frac{\theta(t)}{2} \right| = \nu_a \int \frac{1}{r(t)} dt + C_1
\]
对于距离\( r(t) \)的微分方程
\[
\frac{dr(t)}{dt} = \nu_a (1 - \cos \theta(t))
\]
\[
\frac{dr(t)}{1 - \cos \theta(t)} = \nu_a dt
\]
积分两边
\[
\int \frac{dr(t)}{1 - \cos \theta(t)} = \nu_a \int dt
\]
\[
\int \frac{dr(t)}{1 - \cos \theta(t)} = \nu_a t + C_2
\]
由于 \(\cos \theta(t) = 1 - 2 \sin^2 \frac{\theta(t)}{2}\)，我们有
\[
\frac{1}{1 - \cos \theta(t)} = \frac{1}{2 \sin^2 \frac{\theta(t)}{2}} = \frac{1}{2} \csc^2 \frac{\theta(t)}{2}
\]
\[
\int \frac{1}{2} \csc^2 \frac{\theta(t)}{2} dr(t) = \nu_a t + C_2
\]
\[
-\frac{1}{2} \cot \frac{\theta(t)}{2} = \nu_a t + C_2
\]
问题二
1
物体 \(B\) 的位置可以表示为
\[ x = a \cos \theta - \rho \cos \omega \]
\[ y = a \sin \theta - \rho \sin \omega \]
对于 \(\frac{dx}{d\theta}\) 和 \(\frac{dy}{d\theta}\)有
\[ \frac{dx}{d\theta} = \frac{d}{d\theta} (a \cos \theta - \rho \cos \omega) = -a \sin \theta - \frac{d\rho}{d\theta} \cos \omega + \rho \sin \omega \frac{d\omega}{d\theta} \]
\[ \frac{dy}{d\theta} = \frac{d}{d\theta} (a \sin \theta - \rho \sin \omega) = a \cos \theta - \frac{d\rho}{d\theta} \sin \omega - \rho \cos \omega \frac{d\omega}{d\theta} \]
2
物体 \(B\) 的运动轨迹在 \((x, y)\) 处的切线斜率为 \(\frac{dy}{dx}\)，有
\[ \frac{dy}{dx} = \frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}} = \frac{a \cos \theta - \frac{d\rho}{d\theta} \sin \omega - \rho \cos \omega \frac{d\omega}{d\theta}}{-a \sin \theta - \frac{d\rho}{d\theta} \cos \omega + \rho \sin \omega \frac{d\omega}{d\theta}} \]
切线方程为：
\[ y - y_0 = \frac{dy}{dx} (x - x_0) =\frac{a \cos \theta - \frac{d\rho}{d\theta} \sin \omega - \rho \cos \omega \frac{d\omega}{d\theta}}{-a \sin \theta - \frac{d\rho}{d\theta} \cos \omega + \rho \sin \omega \frac{d\omega}{d\theta}}(x-x_0)\]
其中 \((x_0, y_0) = (a \cos \theta - \rho \cos \omega, a \sin \theta - \rho \sin \omega)\)。
法线方程的斜率为 \(-\frac{1}{\frac{dy}{dx}}\)，因此法线方程为：
\[ y - y_0 = -\frac{1}{\frac{dy}{dx}} (x - x_0) =-\frac{-a \sin \theta - \frac{d\rho}{d\theta} \cos \omega + \rho \sin \omega \frac{d\omega}{d\theta}}{a \cos \theta - \frac{d\rho}{d\theta} \sin \omega - \rho \cos \omega \frac{d\omega}{d\theta}}(x-x_0)\]
3
设物体 \(B\) 的速度为 \(v\)，则在时间 \(t\) 内，物体 \(B\) 移动的距离为 \(vt\)，而物体 \(A\) 移动的距离为 \(\frac{vt}{n}\)。
根据几何关系，我们有
\[ \frac{d\rho}{dt} = -v \]
\[ \frac{d\theta}{dt} = \frac{v}{na} \]
因此，我们有
\[ \frac{d\rho}{d\theta} = \frac{\frac{d\rho}{dt}}{\frac{d\theta}{dt}} = -v \cdot \frac{na}{v} = -na \]
根据几何关系，我们有
\[ \frac{d\omega}{dt} = \frac{v \sin \omega}{\rho} \]
因此，我们有
\[ \frac{d\omega}{d\theta} = \frac{\frac{d\omega}{dt}}{\frac{d\theta}{dt}} = \frac{v \sin \omega}{\rho} \cdot \frac{na}{v} = \frac{na \sin \omega}{\rho} \]
所以，\(\omega\) 和 \(\rho\) 满足的微分方程组为
\[ \frac{d\rho}{d\theta} = -na \]
\[ \frac{d\omega}{d\theta} = \frac{na \sin \omega}{\rho} \]