BMAD 65 Posted November 21, 2018 Report Share Posted November 21, 2018 Find a function where the arc lenth and area between any two randomly defined points is the same. There are two. Quote Link to post Share on other sites

0 rocdocmac 9 Posted April 3, 2019 Report Share Posted April 3, 2019 (edited) Does this question mean finding a function where the arc length (XZ in the figure below) is the same as the area of the segment (shaded), notwithstanding the fact that arc length is in units and segment area in squared units? Edited April 3, 2019 by rocdocmac Quote Link to post Share on other sites

0 EventHorizon 16 Posted April 3, 2019 Report Share Posted April 3, 2019 I'm sure I know what BMAD was saying. Yes, ignore the units / squared units difference. The problem doesn't involve a circle though. You need to find a function f(x) such that the area under the curve between two points is the same as the arc length between those two points along the function. Here the arc length is like the length of a string representing the function cut at the two points. How to find the arc length of a function: Spoiler 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. Restatement of problem using previous spoiler: Spoiler Find a function f(x) such that F(x) = integrate( sqrt( f'(x)^2 + 1) ), where F(x) is the indefinite integral of f(x) and f'(x) is the derivative of f(x). One step further: Spoiler Taking the derivative of both sides gives f(x) = sqrt( f'(x)^2 + 1) One answer is simple, the other is less so. Quote Link to post Share on other sites

0 grace123 0 Posted April 4, 2019 Report Share Posted April 4, 2019 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 Quote Link to post Share on other sites

0 BMAD 65 Posted April 4, 2019 Author Report Share Posted April 4, 2019 20 hours ago, EventHorizon said: I'm sure I know what BMAD was saying. Yes, ignore the units / squared units difference. The problem doesn't involve a circle though. You need to find a function f(x) such that the area under the curve between two points is the same as the arc length between those two points along the function. Here the arc length is like the length of a string representing the function cut at the two points. How to find the arc length of a function: Hide contents 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. Restatement of problem using previous spoiler: Hide contents Find a function f(x) such that F(x) = integrate( sqrt( f'(x)^2 + 1) ), where F(x) is the indefinite integral of f(x) and f'(x) is the derivative of f(x). One step further: Hide contents Taking the derivative of both sides gives f(x) = sqrt( f'(x)^2 + 1) One answer is simple, the other is less so. yes, this is what I mean. Quote Link to post Share on other sites

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## BMAD 65

Find a function where the arc lenth and area between any two randomly defined points is the same. There are two.

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