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grace123

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Everything posted by grace123

  1. If the function was a straight line, you could use the Pythagorean theorem to find it. sqrt( (y2-y1)^2 + (x2-x1)^2 ) If you assume the arc length was a straight line, and slowly bring the points on the function closer together, eventually you get the change in y approaching the slope of the line times the difference in x's. The arc length for that infinitesimally small section is then sqrt( ( f'(x)dx )^2 + dx^2 ) = sqrt( dx^2 * (f'(x)^2 + 1 ) = sqrt( f'(x)^2 + 1 ) dx. Where f'(x) is the derivative of f(x). Integrate that between two points to get the arc length between those two points. Example: f(x)=2x f'(x) = 2 integrate( sqrt( 2^2 + 1 ) dx ) = integrate( sqrt(5) dx) = x*sqrt(5) + c If you evaluate the integral between 0 and 1 you end up with arc length = sqrt(5). Which is the same as if you just used the Pythagorean theorem. Yeah, I'm too lazy right now to find a function that ends up with a nice arc length function that isn't just a straight line
  2. if 12=6 , 6=3 10=what answer is 10=5 because 6 x 2 = 12 3 x 2 = 6 its just half of the number x 2 example half of 12 = 6 x 2 = 12
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