Using related rates, why can we ignore dimensions and consider rates of change, when in seemingly identical situations, we must consider both?

Question

So I was preparing a lesson on related rates for the calc 1 class I am a TA for and I realized that the two problems below in the photo are basically identical: Given a right triangle, x, x', y, y' are known, Find z' (or s').

Problem #1 and $4 are solved identically, but in problem #4, we can use a "cheat" and just consider a right triangle with legs x'=25 and y'=60 and hypotenuse=s'.

Solving for $s'... \\s'=\sqrt{x'^2+y'^2}=\sqrt{25^2+60^2}=\sqrt{4225}=65 $

This implies the distance between the cars is changing at constant rate, independent of the location of the cars. But this method does not work for the seemingly identical problem #1. I am conflicted... why is this "cheat" only viable for some instances of these problems and not all?

I checked back in my own notes from calc 1 and this "cheat" could be used on other problems too, so it's not something unique with the numbers in #4.

Answer 1

It is because position is proportional to a constant velocity ($x=tx',y=ty'$) so in the second example

$$\frac{xx'+yy'}{\sqrt{x^2+y^2}}=\frac{(tx')x'+(ty')y'}{\sqrt{(tx')^2+(ty')^2}}=\sqrt{x'^2+y'^2.}$$

This proportionality is not obeyed in the first example.

Answer 2

If $t$ denotes time, then $x$ and $y$ cannot generally be assumed to denote position: in the first example, the given couple $(x,y)=(5,12)$ indeed corresponds not to a common/single instant but to $(t_x,t_y)=(2.5,4).$

In the second example, the two objects start from the same point, so we can define $x$ and $y$ as their positions.

Furthermore, since all the rates of change in the second example are constant, the second and first triangle must be similar and thus the 'cheat' work as desired.