Arc length calculus12/28/2023 We use integration, as it represents the sum of such infinitely small distances. We need to add all those infinitesimally small lengths. 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. 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. 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. So we expect the curved distance OD to be around 12 cm. The distances AB, BC, CD are all equal, so we can say: We then use Pythagoras' Theorem to find the length OA: We plot the points O (0, 0), A (2.65, 1.35), B (5.33, 0), C (7.99, -1.35) and D (10.65, 0), which are key points on the curve (at the mid-points, maximum and minimum values), and join the line segments. (It's always good practice to estimate your answer first, and in this topic, it helps us understand the concept better). We'll find the width needed for one wave, then multiply by the number of waves.Īpproximate answer: Next, let's approximate the length of the curve so we've got a rough idea what our exact length should be. For background on this, see Period of a sine curve.) (Within the sine expression, we use 2π/10.67 = 0.589 for the coefficient of x. This has the required amplitude 1.35 and period 10.67. We model the corrugations using the curve
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