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arturop Group Title

Hi, I'm looking at the ladder problem in the recitation session for lesson 31, related to rates of change. I can understand the solution being presented and I can't find any flaw with it. However I have what I believe is a simpler solution. The problem is the solution I come up with is different and I can't see the flaw of my solution either. My solution calls x to the base of the bigger triangle, y to its height. By Pithagorean Theorem and deriving we get to y' = - xx' / y (with ' notation meaning d/dt). At t=0, using triangle geometry you get x = 12 and y = 16. So it follows y'= -15/4.

  • 2 years ago
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  1. arturop Group Title
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    To follow up, what is wrong with my solution if anything :-)?

    • 2 years ago
  2. MattBenjamins Group Title
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    The rate of change dx/dt = 5 is the rate at which the distance between the base of the ladder and the wall changes. This is not the same as the rate at which the horizontal distance between the base and the top of the ladder (the base of the bigger triangle) changes. In your solution you assume it is and that's where your problem lies.

    • 2 years ago
  3. arturop Group Title
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    Matt, thanks very much for your answer. What you say makes sense and I was suspecting something like that. The problem I find is that it is very difficult to see it and I might do it wrong in a different context in the future. A bit of a rethorical question but any advice? :)

    • 2 years ago
  4. MattBenjamins Group Title
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    The best advice I can give is to carefully sketch everything out and label everything with the appropriate values, just like Joel Lewis does in the video. Then consider exactly what the rate of change you've been given means (in this case, moving the ladder away from the wall). Then try to figure out how the change you know about influences your picture and the various objects, points, distances , speeds, etc. in it. Other that that it's just a matter of practicing, making mistakes and learning from them. Finally, if you're unsure about an assumption you've made, plug some values into your equations and see if they really work.

    • 2 years ago
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