![]() ![]() So we expect the curved distance OD to be around 12 cm. We zoom in near the center of the segment OA and we see the curve is almost straight.įor this portion, the curve EF is getting quite close to the straight line segment EF.Ĭurved length EF `= r ≈ int_a^bsqrt(1^2+0.57^2)=1.15` We'll use calculus to find the 'exact' value. If the horizontal distance is " dx" (or "a small change in x") and the vertical height of the triangle is " dy" (or "a small change in y") then the length of the curved arc " dr" is approximated as: Of course, the real curved length is slightly more than 1.15. Now, if we move point E very close to point F, we will have a very good approximation for the length of the curve in that local region. We need to add all those infinitesimally small lengths. We use integration, as it represents the sum of such infinitely small distances. The arc length of the curve y = f( x) from x = a to x = b is given by: We have for the distance between where `x = a` to `x = b`:īy performing simple surd manipulation, we can express this in a more familiar form as follows. Of course, we are assuming the function `y = f(x)` is continuous in the region `x = a` to `x = b` (otherwise, the formula won't work). Use the above formula to find the required width of the metal sheet in our example. We can now use this formula to find the required width of our flat sheet of iron. Substituting these into our formula gives us one wavelength of the curved material: Remember, we're finding the width needed for one wave, then we'll multiply by the number of waves.įor one period, the lower limit is x = 0 and the upper limit is x = 10.67. ![]()
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