How fast is the distance changing between boy in point X from girl in point D?

Question

There's a square-shaped area, every corner is marked with letter from A to D. The boy is running from A to B with steady speed 8,7 m/s. The girl is in the corner D. The question is - how fast changes the boy's distance from girl (DX) at the moment, when boy is located at point X and X is from A 14,8 metres far away.. I also added a picture of the situation in order I didn't express myself enough well. Please help, I'm a bit confused how to solve that!

figure

Answer 1

There are two ways to get the answer. From a physics point of view, you can decompose the velocity of the boy along DX and perpendicular to DX If you call $\theta$ the angle ADX, you can easily show that the velocity along DX is the velocity of the boy multiplied by $\sin\theta$. From the ADX triangle $\sin\theta=\frac{14.8}{\sqrt{14.8^2+32^2}}$. The component perpendicular to DX does not change the length of DX, just the orientation.

You can get the same result in a mathematical way. Assume that after a small interval $\Delta t$ the boy is at $X_1=X+8.7\Delta t=14.8+8.7\Delta t$. The rate of change that you want is $\frac{DX_1-DX}{\Delta t}$. Applying Pythagoras' theorem we get $$v=\frac{\sqrt{32^2+(14.8+8.7\Delta t)^2}-\sqrt{32^2+14.8^2}}{\Delta t}$$ Ignoring second order terms in $\Delta t$ and using $\sqrt{1-x}\approx1-x/2$ we get $$v=\frac{\sqrt{32^2+14.8^2+2\cdot14.8\cdot 8.7\Delta t}-\sqrt{32^2+14.8^2}}{\Delta t}\\=\frac{\sqrt{32^2+14.8^2}}{\Delta t}\left(\sqrt{1+\frac{2\cdot14.8\cdot 8.7\Delta t}{32^2+14.8^2}}-1\right)\\=\frac{\sqrt{32^2+14.8^2}}{\Delta t}\left(1+\frac{14.8\cdot 8.7\Delta t}{32^2+14.8^2}-1\right)\\=8.7\sin\theta$$