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Here is the question:Two people start from the same point at the same time. One walks north at 2 mi/h and the other walks west at 4 mi/h. How fast is the distance between them changing after 30 minutes?.

I understand here I should use $z^2_t=x^2_t+y^2_t$ and use the chain rule to find $dz/dt$; but I suddenly thought if I use $t$=time, then $z^2=4t^2+16t^2=20t^2$, why can't I use this equation to find the change rate? Because $x,y$ are also changing spontaneously and I wrongly treated one of them as a constant or other reasons?

Sorry this may seem really stupid

elainehxw
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    You can. Do it both ways and see that you get the same answer. – Ethan Bolker Oct 12 '23 at 13:52
  • If you use the correct equation, $z=\sqrt{20}t$ then you will get the same answer – acat3 Oct 12 '23 at 13:55
  • I think this is a good idea. You can solve the problem more than one way. All solutions should give the same results, demonstrating that it is the fundamental problem statement itself that determines the answer and the methods are not arbitrary. – David K Oct 12 '23 at 13:56
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    I find the reason why I thought it was impossible---when I used $z^2=20t^2$, I forgot when I differentiated the left side it should be $2z.z'$. Thanks! – elainehxw Oct 12 '23 at 13:58
  • @RezhaAdrianTanuharja actually, for this one $z=\sqrt{20}t$, I wondered by this way the answer should be $\sqrt {20}$ instead of the correct answer $10/\sqrt{5}$ – elainehxw Oct 12 '23 at 14:02
  • Welcome to MSE! <> 1. You're welcome to post an answer yourself, especially the pedagogical insight about forgetting to use the chain rule, and/or the variety of methods (I count three) you found or that people suggested in the comments. 2. $\sqrt{20}$ is $10/\sqrt{5}$, and both are $2\sqrt{5}$. :) – Andrew D. Hwang Oct 12 '23 at 14:25
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    @AndrewD.Hwang Thanks for the kind words! I was too stupid to notice that $\sqrt20 $ is actually $10/\sqrt5$. Thanks again. – elainehxw Oct 12 '23 at 14:35
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    Just to note/emphasize, we're all human, and we all sometimes overlook or misunderstand things. Sometimes those things later seem "basic" or "obvious," but that doesn't mean overlooking or misunderstanding something deserves insult, scorn, or abuse. In other words, in a kind but firmly serious sense, you're not stupid, and saying otherwise -- no matter how jokingly -- is not without harm. <> At risk of calling out entire professions and professional cultures, math has work to do here. – Andrew D. Hwang Oct 12 '23 at 14:47
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    @AndrewD.Hwang The sentiment expressed in your words is both empathetic and insightful! Thank you so much for the encouragement and I'll remember it! Hope I can keep enjoying the fun of math while keeping exploring ( I do like math!) – elainehxw Oct 12 '23 at 15:13
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    What I like about this question is that you are actually doing math rather than just following someone's set procedure. – David K Oct 12 '23 at 15:43
  • @DavidK thanks! – elainehxw Oct 12 '23 at 16:11

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