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

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Prime Newtons

Prime Newtons

4 ай бұрын

In this video, I showed how to find the length of an arc if no bounds of integration are given for asum of functions

Пікірлер: 14
@MathFromAlphaToOmega
@MathFromAlphaToOmega 4 ай бұрын
I thought I'd seen most of the curves with nice arc lengths, but this is a new one!
@adamkucera9094
@adamkucera9094 4 ай бұрын
Me too.
@blackovich
@blackovich 4 ай бұрын
You are a good-looking guy. I'm sorry, I had to type that on a math channel.
@dirklutz2818
@dirklutz2818 4 ай бұрын
Terrific!
@vitotozzi1972
@vitotozzi1972 4 ай бұрын
Fantastic job!
@benchapple1583
@benchapple1583 4 ай бұрын
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 4 ай бұрын
Just takes less space on my small board. I personally prefer f(x)
@benchapple1583
@benchapple1583 4 ай бұрын
@@PrimeNewtons Thanks for taking the time to answer.
@surendrakverma555
@surendrakverma555 4 ай бұрын
Thanks Sir 👍
@florianbasier
@florianbasier 4 ай бұрын
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
@pauljackson3491
@pauljackson3491 4 ай бұрын
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 4 ай бұрын
1. I call it 'curly ell'. 2. Didn't he say that, about comparing squares? I thought he did.
@justabunga1
@justabunga1 4 ай бұрын
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.
@aminimam5118
@aminimam5118 4 ай бұрын
y=5-x/2 and the other side is 5-x (5-x/2)^2+(5-x)^2=25 it is a straightforward problem,
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