Distance of closest aproach

Question

A particle is kept at rest at origin. Another particle starts from $(5,0)$ with a velocity of $-4i+3j$. Find the closest distance of approach.

Answer 1

The position over time is $(5,0)+(-4,3)t=(5-4t,3t)$, and the squared distance to the origin $(5-4t)^2+(3t)^2$.

Find the minimum value of this polynomial.

Answer 2

The particle is at $$\vec{p}_0=\begin{pmatrix}5 \\ 0\end{pmatrix}$$ And has a directional velocity vector of $$-4\vec{i}+3\vec{j} = \binom{-4}{3}$$ Therefore, the line this particle is traveling on is simply $$\vec{p}(t) = \binom{5}{0}+t\binom{-4}{3}$$ Or, as a line equation $y=ax+b$: We know through the velocity vector that the slope is $$a = \frac{\Delta y}{\Delta x} = \frac{3}{-4} = -\frac{3}{4}.$$

We know that $(5,0)$ is on the line, so: $$0 =-\frac{3}{4}\cdot 5+b$$ $$0 =-\frac{15}{4}+b$$ $$\frac{15}{4} =b$$ So we must have $$y=-\frac{3}{4}x+\frac{15}{4}$$

We setup a function that measures how far away the a point on the line $p(t)$ is away from the origin $(0,0)$. That is, we simply compute the vector-distance between $\vec{p}(t)$ and $\vec{0} =\binom{0}{0}$ as: $$d(t) = \left\vert\vec{p}(t) - \vec{0} \right\vert = |\vec{p}(t)| = |(5-4t)\vec{i} + (0+3t)\vec{j} |= \sqrt{(5-4t)^2+(3t)^2} $$

The distance has to be minimal. So, this is an optimization problem on finding the value of $t$ for which $d(t)$ will be minimal. But, we can also minimize the squared-distance, that is $$d(t)^2= (5-4t)^2+(3t)^2$$ We expand. $$d(t)^2= 25t^2-40t+25$$ We compute the derivative. $$(d^2(t))'= 50t-40$$ We find the zeroes. $$50t-40=0$$ $$50t=40$$ $$t=\frac{40}{50} = \frac{4}{5} = 0.8$$

We plug this value into $p(t)$ to find the point on the line:

$$\vec{p}(0.8) = \binom{1.8}{2.4}$$ And the distance minimal distance to $(0,0)$ is $$\left\vert \binom{1.8}{2.4}\right\vert = \sqrt{1.8^2 + 2.4^4} = \sqrt{9} = 3$$ Which is the final answer, the minimal distance of approach is equal to $3$.

Answer 3

The path meets the $y$-axis at some point $B$. The ratio $OB:OA$ is given as 3:4, so $OB=\frac{15}{4}$. By Pythagoras the ratio $OB:OA:AB$ is 3:4:5, so $AB=\frac{25}{4}$.

$D$ is the point of closest approach, so $\angle ODB=90^o$. Hence triangles $OBD,ABO$ are similar. Hence $\frac{OD}{OB}=\frac{OA}{AB}$ giving $OD=3$.