A ship is 5 km east and 7 km North of a lighthouse. It is moving North at a rate of 12 $kmh^{−1}$ and East at a rate of 16 $kmh^{−1}$. At what rate is its distance from the lighthouse changing?
Looking at how the net resultant velocity vector is $v = (12^2+16^2)^{0.5} = 20$ $kmh^{-1}$, we know that the ship moves exactly 20 km away from the lighthouse each hour. This is what I thought the answer would be, but for some reason, the answer involved the initial position of the ship given ("5 km east and 7 km North"). The correct answer rounded is given as $19.1$ $kmh^{-1}$, which I worked out to be $20 - \frac{(5^2+7^2)^{0.5}}{10}$. Why is this so? Why is the initial position taken into account for the ship's rate of change of distance (aka velocity) from the lighthouse?
Let $r$ be the distance from the lighthouse to the ship, so we have $$r=(x^2+y^2)^{1/2}$$ Where $(x,\,y)$ is the position of the ship, taking $x$ as East direction and $y$ as North direction. Then, the rate of change of the distance is $$\frac{\mathrm d r}{\mathrm d t}=\frac{x\frac{\mathrm d x}{\mathrm d t}+y\frac{\mathrm d y}{\mathrm d t}}{(x^2+y^2)^{1/2}}$$ Thus, at the given instant, \begin{align*} \frac{\mathrm d r}{\mathrm d t}&=\frac{(5 \text{ km})(16\text{ km/h})+(7 \text{ km})(12\text{ km/h})}{((5 \text{ km})^2+(7 \text{ km})^2)^{1/2}}\\[3pt] &=\frac{164}{\sqrt{74}}\text{ km/h}\\[4pt] &\approx \boxed{\color{blue}{19.06\text{ km/h}}} \end{align*}