![calculus bridge calculus bridge](https://i.pinimg.com/736x/fe/47/52/fe4752dd796aa08b05186eb7c377b9d1--dental-caries-calculus.jpg)
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. We can now use this formula to find the required width of our flat sheet of iron. Use the above formula to find the required width of the metal sheet in our example. 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). 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. 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.
![calculus bridge calculus bridge](https://i.pinimg.com/736x/cb/0c/5a/cb0c5ad4c3f2216f474d10f63fe9647c--what-is-calculus.jpg)
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