Minimum distance and time between moving objects

Question

Object 1 is at the origin,object 2 is at the point (30ft,30ft).At exactly the same moment object 1 move along positive x-axis at a speed of 2 feet per second, and object 2 star moving toward the origin at a speed of 3 feet per second.

a) At what time are object 1 and object 2 closest?

b) What is their distance at that time?

What I have done is that applying Pythagorean theorem I can find distance between object 2 and origin at first. $x^2$+$y^2$ = $z^2$
z = $\sqrt{30^2 + 30^2}=30\sqrt{2}$

Using implicit differentiation I get: $2*30\sqrt{2}*3$ = $2.30.y'$ + $2.2.30$

$y'\approx 2.2426$

I only be able to find the rate that the distance is decreasing but not the minimum distance nor the time. I have no idea what to do next so any help would be appreciated.

Answer 1

This is an optimization problem. We know that a maximum or minimum will occur only if the derivative is equal to zero. Be careful about what we are minimizing. We aren't minimizing $y(t)$ but instead we are minimizing the distance between the objects. In your question, you called the distance between the objects $z$ so we will continue using that. Using the Pythagorean Theorem to express the distance between the objects is correct. If we let $(x_1,y_1)$ denote the position of object one and $(x_2,y_2)$ denote the position of object two then the distance between the objects will be $$z^2 = {(x_2 - x_1)^2 + (y_2 - y_1)^2}.$$

This is what we have to minimize. Taking the derivative with respect to time (and noting that $x_1,x_2,y_1,y_2$ are all functions of time) we find that

$$2z\frac{dz}{dt} = 2(x_2 - x_1)\bigg(\frac{dx_2}{dt} - \frac{dx_1}{dt}\bigg) + 2(y_2 - y_1)\bigg(\frac{dy_2}{dt} - \frac{dy_1}{dt}\bigg) \tag{1}.$$

This begs the question what are $(x_1,y_1)$ and $(x_2, y_2)?$ From the problem definition we have that object one starts at the origin and moves in the positive $x$ direction at 2 feet per second. Thus we have that $x_1 = 0+2t$ and $y_1 = 0.$ Object two starts at the point $(30 \text{ ft},30\text{ ft})$ and moves toward the origin at 3 feet per second. The velocity of the object is thus split equally between the -$y$ and -$x$ directions and by the Pythagorean Theorem we have that each component should be $3\sqrt{2}/2.$ This means that the position of object two is described by $x_2 = 30 - 3t\frac{\sqrt{2}}{2}$ and $y_2 = 30 - 3t\frac{\sqrt{2}}{2}. $ Plugging these values into $\textbf{(1)},$ we have that

$$2z\frac{dz}{dt} = 2(30 - 3t\frac{\sqrt{2}}{2} - 2t)\bigg(\frac{d}{dt}(30 - 3t\frac{\sqrt{2}}{2}) - \frac{d}{dt}(2t)\bigg) + 2(30 - 3t\frac{\sqrt{2}}{2})\bigg(\frac{d}{dt}(30 - 3t\frac{\sqrt{2}}{2}) \bigg) .$$

Recall that the minimum will occur at a point where $\frac{dz}{dt} =0$ which also implies that $2z\frac{dz}{dt}=0.$ Now all we must do is solve

$$2(30 - 3t\frac{\sqrt{2}}{2} - 2t)\bigg(\frac{d}{dt}(30 - 3t\frac{\sqrt{2}}{2}) - \frac{d}{dt}(2t)\bigg) + 2(30 - 3t\frac{\sqrt{2}}{2})\bigg(\frac{d}{dt}(30 - 3t\frac{\sqrt{2}}{2}) \bigg)=0.$$ This might look a little messy but it is first order once you've taken the derivatives.

Solving for $t$ tells you the time at which the minimum distance $z$ occurs. Plugging this $t$ value into $$z(t) = \sqrt{ (x_2(t) - x_1(t))^2 + (y_2(t) - y_1(t))^2 }$$ will tell you the minimum distance between the objects.

Let me know if you need any more clarification.

Answer 2

Depending on if the question is asking for time when you achieve the shortest distance or the shortest distance can decide the faster approach. In this case, it is asking for both so we could either use differentiation or the perpendicular distance from a static point using the relative velocity.

A) Using differentiation -

To start with Object1 is at the origin (O) and Object2 is at $A (30,30)$ on line $x = y$.

Velocity in direction $OA = -3 - 2 \cos45^0 = -(3+\sqrt2)$
Velocity in perp direction to $OA = 2 \sin45^0 = \sqrt2)$

Square of distance between them at time $t, \, d^2 = (30 \sqrt2 - (3 + \sqrt2)t)^2 + (\sqrt2t)^2$

Differentiation the square of distance wrt. time and equating to zero to find the extrema,

$2(3+\sqrt2)^2t - 60(2+3\sqrt2) + 4t = 0$

$(13 + 6 \sqrt2)t = 30(2+3\sqrt2) \implies \displaystyle t = \frac{90\sqrt2+60}{13 + 6 \sqrt2} \approx 8.717$

That gives min distance $\approx 12.944$.

B) Using vector

Parametric directional vector of Object2 motion is $ \vec{p} (-1,-1)$.

Relative velocity directional vector wrt point $O$ is $\vec{r} ((-\frac{3}{\sqrt2} - 2), -\frac{3}{\sqrt2})$

The shortest distance will be when they are perpendicular to each other which is $d\sin \theta$ where $d$ is the initial distance $(30\sqrt2)$ and $\theta$ is the angle between the above two vectors.

$\sin \theta = \displaystyle \frac{\vec{p} \times \vec{r}} {|p||r|} = \frac{\sqrt 2} {\sqrt{13+6\sqrt2}}$

So min distance $= \displaystyle \frac{60} {\sqrt{13+6\sqrt2}} \approx 12.944$

Now you can find time which comes to $\approx 8.717$.