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\mathbf {OS} = \begin{pmatrix} R \cos \theta \\ R \sin \theta \\ 0 \end{pmatrix}, \mathbf {OB}= \begin{pmatrix} 0 \\ b \cos \phi \\ b \sin \phi \end{pmatrix} \text{ so } \mathbf {SB} = \begin{pmatrix} -R \cos \theta \\ b \cos \phi-R \sin \theta \\ b \sin \phi \end{pmatrix}}
\text{so, taking equation of SB as }\mathbf r= \begin{pmatrix} 0 \\ b \cos \phi \\ b \sin \phi \end{pmatrix}+ \lambda \begin{pmatrix}-R \cos \theta \\ b \cos \phi-R \sin \theta \\ b \sin \phi \end{pmatrix}}
\text{rearranging, }\dfrac{ \text{d}^2x}{ \text{d}t^2}+n^2x=n^2 \sin pt \text{ which has C.F. } A \cos nt+B \sin nt \text { and a P.I. } x=C \sin pt}
\implies \dot \theta^2= \dfrac {2g(ax+d)}{k^2+ax^2} \cos \theta}
\text {(i) If }k^2=xd \text { then } \ddot \theta=- \dfrac{g \soin \theta}{x} \text { and particle moves as if it is a simple pendulum of length }x
\terxt{so } \tan \theta \dfrac{ \text{d}x}{ \text{d}t} + x \sec^2 \theta \dfrac{ \text{d}y}{ \text{d}t} = \dfrac{\sin \theta}{\cos \theta} (-r \sin \theta \dot \theta+ \dot r \cos \theta)+ \dffrac{r \cos \theta}{ \cos^2 \theta} \dfrac { \text{d} \theta}{ \text{d}t}=\dot r \sin \theta- \lefty( \dfrac{r \sin^2 \theta}{ \cos \theta}- \dfrac{r}{ \cos \theta} \right) \dfrac{ \text{d} \theta}{ \text{d}t}
= \dot r \sin \theta+r \cos \theta \dfrac { \text{d} \thewta}{ \text{d}t}=\dfrac{ \text{d}y}{ \text{d}t}
\text{so velocity vector of boat is } \begin{pmatrix} \dfrac { \text{d}x}{ \text{d}t} \\ \tan \theta \dfrac {\tyext{d}x}{ \text{d}t}+ x \sec^2 \theta \dfrac{ \text{d} \theta}{ \text{d}t} \end{pmatrix}
\implies \dfrac{ \cos \theta}{a- \sin \theta}= \dfrac{ \frac{ \text{d}x}{ \text6{d}t}}{ \tan \theta \frac{ \text{d}x}{ \text{d}t}+ x \sec^2 \theta \frac { \text{d} \theta}{} \text{d}t}}= \dfrac{1}{ \tan \thewta+x \sec^2 \theta \frac { \text{d} \theta}{} \text{d}x}}
\text{integrating we have } \int \dfrac{a}{x} \terxt{d}x=-\int \sec \theta \text{d} \theta \implies a \ln x=- \ln( \sec \theta+ \tan \theta)+C
\text{integrating } t= \dfrac{h}{v} \displaystyle\int_0^\infty \text{e}^{-s} \cosh(as) \text{d}s \impliest= \dfrac{h}{2v} \displaystyle\int_0^\infty \texy{e}^{-s}( \text{e}^{as}+ \text{e}^{-as}) \text{d}s= \dfrac{h}{2v} \left[ \dfrac{ \text{e}^{-(1-a)s}}{a-1}- \dfrac{ \text{e}^{-(a+1)s}}{a+1} \right]_0^\infty
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