Arclentgh of y= sqrt(x-x^2) + arccos(sqrtx)

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  • Опубликовано: 12 сен 2024
  • In this video, I showed how to find the length of an arc if no bounds of integration are given for asum of functions

Комментарии • 14

  • @MathFromAlphaToOmega
    @MathFromAlphaToOmega 5 месяцев назад +15

    I thought I'd seen most of the curves with nice arc lengths, but this is a new one!

  • @blackovich
    @blackovich 5 месяцев назад +12

    You are a good-looking guy. I'm sorry, I had to type that on a math channel.

  • @florianbasier
    @florianbasier 5 месяцев назад +2

    Nothing critical but your explanations on the domains were a bit unprecise to say the least. the "a quadratic is less than 0 between its two roots" is true in this case, but only because the coefficient of x2 is positive. if you have a quadratic with a negative leading coefficient, it is negative outside of the roots. The explanation for the second domain is as well lacking discipline. You can't just "change the -1 to 0". Since you are working in R, you can say that 0

  • @benchapple1583
    @benchapple1583 5 месяцев назад +2

    This is not a criticism, rather it's a question.
    Why do you favour the form "y = whatever" as opposed to "f(x) = whatever"? I have always seen this written as f(x) or g(x) etc. and their derivatives as f'(x) g''(x) etc.

    • @PrimeNewtons
      @PrimeNewtons  5 месяцев назад +1

      Just takes less space on my small board. I personally prefer f(x)

    • @benchapple1583
      @benchapple1583 5 месяцев назад +1

      @@PrimeNewtons Thanks for taking the time to answer.

  • @dirklutz2818
    @dirklutz2818 5 месяцев назад

    Terrific!

  • @vitotozzi1972
    @vitotozzi1972 5 месяцев назад +1

    Fantastic job!

  • @pauljackson3491
    @pauljackson3491 5 месяцев назад

    What is the name of the fancy 'L' the integral is equal to?
    Also, you can't use squares in greater or lesser than.
    -3 < 2 but it isn't true that (-3)^2 < (2)^2 = 9 < 4

    • @rogerkearns8094
      @rogerkearns8094 5 месяцев назад

      1. I call it 'curly ell'.
      2. Didn't he say that, about comparing squares? I thought he did.

    • @justabunga1
      @justabunga1 5 месяцев назад

      Here is another example of y=(sin(x))^2. If you look at the function as y=sin(x), it ranges between -1 and 1. When you square the sine function, the range has to be in between 0 and 1 because (-1)^2=1, we cannot have some values that are less than 0 in that range. That's why we have to have values that are between 0 and 1.

  • @surendrakverma555
    @surendrakverma555 5 месяцев назад

    Thanks Sir 👍

  • @aminimam5118
    @aminimam5118 5 месяцев назад

    y=5-x/2 and the other side is 5-x
    (5-x/2)^2+(5-x)^2=25
    it is a straightforward problem,