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How can you find the length of arc? - Answers
If you want to find the lenght of a curve y = f(x) between two values of x, lets say x1 and x2, you must compute this integral : Intx1 to x2[sqrt(dx2 + dy2)] You can either express the original function in terms of y or in terms of x, but it is much simpler to express it in a way such that the integral will not be improper. For example, lets say we want to find the lenght of arc of the curve y = x2 between x = 0 and x = 1. We could express this function in terms of y but we will keep it this way because if we change it, we will have to compute an improper integral, which can sometimes be very tedious. The differential of y = x2 is dy = 2x dx. We now need to square the differential : (dy)2 = (2x dx)2 = 4x2 (dx)2 We now have to compute this integral: Int0 to 1[sqrt(dx2 + dy2)] = Int0 to 1[sqrt(dx2 + 4x2 dx2)] = Int0 to 1[sqrt(1 + 4x2) dx] This last integral is easy to compute using a trigonometric substitution.
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How can you find the length of arc? - Answers
If you want to find the lenght of a curve y = f(x) between two values of x, lets say x1 and x2, you must compute this integral : Intx1 to x2[sqrt(dx2 + dy2)] You can either express the original function in terms of y or in terms of x, but it is much simpler to express it in a way such that the integral will not be improper. For example, lets say we want to find the lenght of arc of the curve y = x2 between x = 0 and x = 1. We could express this function in terms of y but we will keep it this way because if we change it, we will have to compute an improper integral, which can sometimes be very tedious. The differential of y = x2 is dy = 2x dx. We now need to square the differential : (dy)2 = (2x dx)2 = 4x2 (dx)2 We now have to compute this integral: Int0 to 1[sqrt(dx2 + dy2)] = Int0 to 1[sqrt(dx2 + 4x2 dx2)] = Int0 to 1[sqrt(1 + 4x2) dx] This last integral is easy to compute using a trigonometric substitution.
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How can you find the length of arc? - Answers
If you want to find the lenght of a curve y = f(x) between two values of x, lets say x1 and x2, you must compute this integral : Intx1 to x2[sqrt(dx2 + dy2)] You can either express the original function in terms of y or in terms of x, but it is much simpler to express it in a way such that the integral will not be improper. For example, lets say we want to find the lenght of arc of the curve y = x2 between x = 0 and x = 1. We could express this function in terms of y but we will keep it this way because if we change it, we will have to compute an improper integral, which can sometimes be very tedious. The differential of y = x2 is dy = 2x dx. We now need to square the differential : (dy)2 = (2x dx)2 = 4x2 (dx)2 We now have to compute this integral: Int0 to 1[sqrt(dx2 + dy2)] = Int0 to 1[sqrt(dx2 + 4x2 dx2)] = Int0 to 1[sqrt(1 + 4x2) dx] This last integral is easy to compute using a trigonometric substitution.
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