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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').

enter image description here

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.

Golden_Ratio
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Haphazardly Proven
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  • It is because the initial position in #1 is not parallel to the velocity, whereas it is in #4. Notice that the rate of change of the distance from the origin is equal to the speed times the cosine of the angle between the position vector and the velocity vector. So at constant speed the only source of a time dependence is the time dependence of this angle. – Ian Mar 03 '22 at 04:36
  • @Ian, I don't quite follow your reasoning with cosine and theta, but I think you're saying that there's more going on in 1 than there is in 4. And particularly, the situation in 1 could not occur if our two points traveling along the x and y axis started from the origin at the same time. Since it would take 4 t units for the point on the y axis to arrive at 12, but at t=4, the other point would be at x=8, not x=5. – Haphazardly Proven Mar 03 '22 at 04:46
  • What I mean by "position" in the setting of #4 is relative position, which starts at the origin and then moves at a constant velocity. In #1 the position starts at $(5,12)$ and then moves at velocity $(2,3)$, and these aren't parallel, so the angle between the position and velocity changes over time. – Ian Mar 03 '22 at 10:24

2 Answers2

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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.

Golden_Ratio
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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.

ryang
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  • $(x,y)$ can still physically be a position at a single time in the first problem. You just can't have gotten there by going at constant velocity from the origin. Honestly I don't really understand what you're trying to say here. – Ian Mar 03 '22 at 15:17
  • Yes by position I mean with reference to the origin. In other words, it is not generally the case that $x$ is a position vector, i.e., a scalar whose sign indicates the object’s side of the reference point. – ryang Mar 03 '22 at 15:27
  • Not separately, no, but you can view $(x,y)$ as the position in the plane of an object in #1, and then the difference between this and the relative position of the cars in #4 is that the object in #1 can't have gone at constant velocity from the origin to get to its current state, while the relative position in #4 can. – Ian Mar 03 '22 at 15:31
  • When i say “cannot generally be blah blah”, I do not mean “is always not blah blah”. 2. Yes, the first example’s given information collectively (including the constant-velocity specifications) means that $x$ is not a position vector.
    • – ryang Mar 03 '22 at 15:38
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    Oh, I understand your point now, you're trying to map #1 into a pair of cars, one moving horizontally and the other moving vertically, and pointing out that for these numbers to happen then they can't have started at the same point. I didn't understand that you were trying to describe #1 in terms of horizontal and vertical motion of two objects, like in #4. I was implicitly thinking of $(x,y)$ in #1 as the 2D position of one object. – Ian Mar 03 '22 at 16:01
  • (Cont.) Anyway I think a better illustration of why they're different is to show a way that #1 can be described like #4, rather than a way it can't. One way to do that is to have an eastbound car start at $(-3,0)$ and a northbound car start at $(0,0)$. Then you would arrive at this scenario at $t=4$. – Ian Mar 03 '22 at 16:01