1. **Problem Statement:** We want to express the velocity vector $v$ of a swimmer in a canal flowing upward with speed $k$, where the swimmer swims at speed $c$ in still water, always facing point $q=(0,0)$ from position $(x,y)$. 2. **Velocity Components:** The swimmer's velocity relative to the water is $\mathbf{V_s} = c(\cos\theta, \sin\theta)$, where $\theta$ is the angle between the swimmer's direction and the positive $x$-axis. 3. **Direction of Swimming:** Since the swimmer always faces $q=(0,0)$ from $(x,y)$, the direction vector points from $(x,y)$ to $(0,0)$, which is $(-x,-y)$. 4. **Unit Direction Vector:** Normalize this direction: $$\hat{u} = \frac{(-x,-y)}{\sqrt{x^2 + y^2}}$$ 5. **Swimmer Velocity in Still Water:** Thus, $$\mathbf{V_s} = c \hat{u} = c \left(-\frac{x}{\sqrt{x^2 + y^2}}, -\frac{y}{\sqrt{x^2 + y^2}}\right)$$ 6. **Water Velocity:** Water flows upward with velocity $$\mathbf{V_w} = (0,k)$$ 7. **Total Velocity:** The swimmer's total velocity is $$\mathbf{v} = \mathbf{V_s} + \mathbf{V_w} = \left(-c \frac{x}{\sqrt{x^2 + y^2}}, -c \frac{y}{\sqrt{x^2 + y^2}} + k \right)$$ 8. **Differential Equations:** This gives $$\frac{dx}{dt} = -c \frac{x}{\sqrt{x^2 + y^2}}, \quad \frac{dy}{dt} = -c \frac{y}{\sqrt{x^2 + y^2}} + k$$ 9. **Classification of the Differential Equation:** The system is nonlinear due to the $\sqrt{x^2 + y^2}$ in the denominator and is not separable because $x$ and $y$ are coupled inside the square root. **Answer:** Option 4. Non-linear and non-separable. --- 10. **Integral Substitution:** For $$\int \frac{1}{\sqrt{1 + x^2}} dx$$ use the substitution $$u = \sinh^{-1}(x) \quad \text{or equivalently} \quad x = \sinh(u)$$ This works because $$\frac{d}{du} \sinh(u) = \cosh(u) = \sqrt{1 + \sinh^2(u)} = \sqrt{1 + x^2}$$ --- 11. **Swimmer Position Midway Across Canal:** Given $w=200$, $c=10$, $k=10$, the swimmer starts at $p=(200,0)$ and swims toward $q=(0,0)$. At midway $x=100$, the swimmer's velocity in $y$ is $$\frac{dy}{dt} = -10 \frac{y}{\sqrt{100^2 + y^2}} + 10$$ To find $y$ at $x=100$, note that the swimmer moves along the direction to $q$, so $$\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{-c \frac{y}{\sqrt{x^2 + y^2}} + k}{-c \frac{x}{\sqrt{x^2 + y^2}}} = \frac{-c y + k \sqrt{x^2 + y^2}}{-c x}$$ At $x=100$, solving numerically or by approximation, Assuming $y$ is small compared to $x$, approximate $\sqrt{x^2 + y^2} \approx x = 100$: $$\frac{dy}{dx} \approx \frac{-10 y + 10 \times 100}{-10 \times 100} = \frac{-10 y + 1000}{-1000} = -\frac{-10 y + 1000}{1000} = \frac{10 y - 1000}{1000}$$ Setting $\frac{dy}{dx} = 0$ for steady $y$ gives $$10 y - 1000 = 0 \implies y = 100$$ So the swimmer's position midway is approximately $$(100, 100)$$ 12. **Can the Swimmer Reach $(0,0)$?** Because the water flows upward at speed equal to the swimmer's speed, the swimmer is pushed upward as they swim left. The swimmer's velocity in $y$ is always increased by $k=10$, so they cannot reach $y=0$ at $x=0$. **Answer:** No, the swimmer will not be able to reach $(0,0)$ exactly.