Homework 08

Question 01

本题没有明确定义“夹角”，这里直接视为函数看待

（1）\(x_a(t),y_a(t),x_b(t),y_b(t)\) 满足以下微分方程组：

\[ \begin{cases} \frac{dx_a(t)}{dt}=v_a(t)\cos\alpha(t)\\ \frac{dy_a(t)}{dt}=v_a(t)\sin\alpha(t)\\ \frac{dx_b(t)}{dt}=v_b(t)\cos\beta(t)\\ \frac{dy_b(t)}{dt}=v_b(t)\sin\beta(t) \end{cases} \]

（2）令 \(X(t)=x_b(t)-x_a(t), Y(t)=y_b(t)-y_a(t)\)，则有 \(r(t)=\sqrt{X^2(t)+Y^2(t)}\)

则 \(\frac{dr(t)}{dt}=\frac{X(t)(v_b(t)\cos\beta(t)-v_a(t)\cos\alpha(t))+Y(t)(v_b(t)\sin\beta(t)-v_a(t)\sin\alpha(t))}{r(t)}\)

我们又有 \(X(t)=r(t)\cos\theta(t),Y(t)=-r(t)\sin\theta(t)\)，代入可得：

\[ \begin{aligned} \frac{dr(t)}{dt}&=\cos\theta(t)(v_b(t)\cos\beta(t)-v_a(t)\cos\alpha(t))-\sin\theta(t)(v_b(t)\sin\beta(t)-v_a(t)\sin\alpha(t))\\ &=v_b(t)cos(\theta(t)+\beta(t))-v_a(t)cos(\theta(t)+\alpha(t)) \end{aligned} \]

因为 \(\alpha(t)\) 是未定的，想要让 \(r(t)\) 下降的最快即使前一项绝对值最大，即 \(\theta(t)=\pi-\beta(t)\)，即海盗应选择始终朝着商船的方向

（3）代入即可得 \(\frac{dr(t)}{dt}=-\lambda v_a-v_a\cos(\theta(t)+\alpha(t))\)

由 \(\cos\theta(t)=\frac{x_b(t)-x_a(t)}{r(t)}\)，有 \(-\sin\theta(t)·\frac{d\theta(t)}{dt}=\frac{\lambda v_a\cos\beta(t)-v_a\cos\alpha(t)-\cos\theta(t)\frac{dr(t)}{dt}}{r(t)}\)

同理由 \(\sin\theta(t)=\frac{y_b(t)-y_a(t)}{r(t)}\)，有 \(cos\theta(t)·\frac{d\theta(t)}{dt}=\frac{\lambda v_a\sin\beta(t)-v_asin\alpha(t)-\sin\theta(t)\frac{dr(t)}{dt}}{r(t)}\)

消去 \(\frac{dr}{dt}\) 即可得 \(\frac{d\theta(t)}{dt}=\frac{\lambda v_a\sin(\beta(t)-\theta(t))-v_a\sin(\alpha(t)-\theta(t))}{r(t)}=\frac{v_a\sin(\theta(t)-\alpha(t))}{r(t)}\)

（4）代入 \(\alpha(t)=0\)，有 \(\frac{dr(t)}{dt}=-\lambda v_a-v_a\cos\theta(t)\)，\(\frac{d\theta(t)}{dt}=\frac{v_a\sin\theta(t)}{r(t)}\)

两式相除有 \(\frac{dr(t)}{d\theta(t)}=\frac{r(t)(-\lambda-\cos\theta(t))}{\sin\theta(t)}\)，即 \(\frac{1}{r(t)}dr(t)=-\frac{\lambda+\cos\theta(t)}{\sin\theta(t)}d\theta(t)\)

两边积分，有 \(\ln r(t)-\ln r_0=-\lambda\ln|\frac{\tan\frac{\theta(t)}{2}}{\tan\frac{\theta_0}{2}}|-\ln\frac{\sin\theta(t)}{\sin\theta_0}\)

即 \(r(t)=\frac{r_0}{(|\frac{\tan\frac{\theta(t)}{2}}{\tan\frac{\theta_0}{2}}|)^{\lambda}\frac{\sin\theta(t)}{\sin\theta_0}}\)，其中 \(\tan\theta_0=\frac{y_0}{x_0},r_0=\sqrt{x_0^2+y_0^2}\)

当 \(\lambda=1\) 时，\(r(t)=\frac{r_0}{|\frac{\tan\frac{\theta(t)}{2}}{\tan\frac{\theta_0}{2}}|\frac{\sin\theta(t)}{\sin\theta_0}}=r_0\frac{\sin^2\frac{\theta_0}{2}}{\sin^2\frac{\theta(t)}{2}}=r_0\frac{1-\cos\theta_0}{1-\cos\theta(t)}=\frac{r_0-x_0}{1-\cos\theta(t)}\)

代入 \(\frac{d\theta(t)}{dt}=\frac{v_a\sin\theta(t)(1-\cos\theta(t))}{r_0-x_0}\)，积分解得 \(t=\frac{r_0-x_0}{2v_a}(\ln|\frac{\tan\frac{\theta(t)}{2}}{\tan\frac{\theta_0}{2}}|-\frac{1}{4\sin^2\frac{\theta(t)}{2}}+\frac{1}{4\sin^2\frac{\theta_0}{2}})\)，其中 \(\tan\theta_0=\frac{y_0}{x_0}\)

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Question 02

（1）\(\frac{dx}{d\theta}=\frac{dx}{dt}\times\frac{dt}{d\theta}=\frac{nv_A\cos\omega}{\frac{v_A}{a}}=na\cos\omega\)，同理 \(\frac{dy}{d\theta}=na\sin\omega\)

（2）切线方程 \(y=\tan\omega(x-x(t))+y(t)\)，法线方程 \(y=-\frac{1}{\tan\omega}(x-x(t))+y(t)\)

（3）令 \(X=a\cos\theta-x,Y=a\sin\theta-y\)，\(\rho=\sqrt{X^2+Y^2}\)，则：

\[ \begin{aligned} \frac{d\rho}{d\theta}&=\frac{(a\cos\theta-x)(-a\sin\theta-na\cos\omega)+(a\sin\theta-y)(a\cos\theta-na\sin\omega)}{\rho}\\ &=\frac{xa\sin\theta-ya\cos\theta-na^2cos(\theta+\omega)+xna\cos\omega+yna\sin\omega}{\rho} \end{aligned} \]

由 \(\cos\omega=\frac{X}{\rho}\)，有 \(-\sin\omega\frac{d\omega}{d\theta}=\frac{-a\sin\theta-na\cos\omega-\cos\omega\frac{d\rho}{d\theta}}{\rho}\)

同理 \(\sin\omega=\frac{Y}{\rho}\Rightarrow \cos\omega\frac{d\omega}{d\theta}=\frac{a\cos\theta-na\sin\omega-\sin\omega\frac{d\rho}{d\theta}}{\rho}\)

两式联立消去 \(\frac{d\rho}{d\theta}\) 可得 \(\frac{d\omega}{d\theta}=\frac{a\cos(\theta-\omega)}{\rho}\)

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