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How could I calculate the point of a curve were the tangent has a slope of 0.5?

Y <- c(0.39, 0.40, 0.42, 0.43, 0.45, 0.46, 0.48, 0.50, 0.52, 0.54, 0.56, 0.58, 0.60, 0.62, 0.64, 0.67, 0.69, 0.71, 0.74, 0.77, 0.80, 0.82, 0.85, 0.89, 0.92, 0.95, 0.99, 1.02, 1.06, 1.10, 1.14, 1.18, 1.22, 1.27, 1.31, 1.36, 1.41, 1.46, 1.51, 1.57, 1.62, 1.68, 1.74, 1.81, 1.87, 1.94, 2.01, 2.08, 2.16, 2.23, 2.31, 2.40, 2.48, 2.57, 2.66, 2.76, 2.86, 2.96, 3.06, 3.17, 3.28, 3.40, 3.52, 3.65, 3.78, 3.91, 4.05, 4.19, 4.34, 4.49, 4.65, 4.81, 4.98, 5.15, 5.33, 5.51, 5.70, 5.90, 6.11, 6.32, 6.53, 6.76, 6.99, 7.23, 7.47, 7.72, 7.99, 8.26, 8.53, 8.82, 9.11, 9.42, 9.73, 10.05, 10.38, 10.72, 11.07, 11.43, 11.80, 12.18, 12.58, 12.98, 13.39, 13.81, 14.25, 14.70, 15.15, 15.62, 16.11, 16.60, 17.11, 17.62, 18.15, 18.70, 19.25, 19.82, 20.40, 20.99, 21.59, 22.21, 22.84, 23.48, 24.14, 24.80, 25.48, 26.17, 26.88, 27.59, 28.32, 29.06, 29.81, 30.57, 31.34, 32.12, 32.91, 33.71, 34.53, 35.35, 36.17, 37.01, 37.86, 38.71, 39.57)

X <- c(2168,  2178,  2188,  2198,  2208,  2218,  2228,  2238,  2248, 2258,  2268,  2278,  2288,  2298,  2308,  2318,  2328,  2338, 2348,  2358,  2368,  2378,  2388,  2398,  2408,  2418,  2428, 2438,  2448,  2458,  2468,  2478,  2488,  2498,  2508,  2518, 2528,  2538,  2548,  2558,  2568,  2578,  2588,  2598,  2608, 2618,  2628,  2638,  2648,  2658,  2668,  2678,  2688,  2698, 2708,  2718,  2728,  2738,  2748,  2758,  2768,  2778,  2788, 2798,  2808,  2818,  2828,  2838,  2848,  2858,  2868,  2878, 2888,  2898,  2908,  2918,  2928,  2938,  2948,  2958,  2968, 2978,  2988,  2998,  3008,  3018,  3028,  3038,  3048,  3058, 3068,  3078,  3088,  3098,  3108,  3118,  3128,  3138,  3148, 3158,  3168,  3178,  3188,  3198,  3208,  3218,  3228,  3238, 3248,  3258,  3268,  3278,  3288,  3298,  3308,  3318,  3328, 3338,  3348,  3358,  3368,  3378,  3388,  3398,  3408,  3418, 3428,  3438,  3448,  3458,  3468,  3478,  3488,  3498,  3508, 3518,  3528,  3538,  3548,  3558,  3568,  3578,  3588)
Ruben
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  • Run `diff(Y) > 0` first, and see what it is giving you. – Axeman Jul 20 '17 at 12:04
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    Not clear for me. I can't see any inflection point in the curve you posted. – nicola Jul 20 '17 at 12:09
  • @Axeman, there is no change with that modification of the code – Ruben Jul 20 '17 at 14:04
  • @nicola, I would have the same problem if Y would be Y^5, where the curve is more pronounced – Ruben Jul 20 '17 at 14:05
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    @Ruben I don't get what you want to calculate. Please notice that the value you specified **is not** an inflection point; it's just a nondescript point like any other. You should better describe the properties of the point you are seeking. – nicola Jul 20 '17 at 14:08
  • You are right, it is not the inflection point. What I would like to calculate is the point of the curve where the slope of the tangent is 0.5 – Ruben Jul 21 '17 at 08:05
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    @ruben then you're going to have to edit your post and its title, because that's something completely different. – Mike 'Pomax' Kamermans Jul 21 '17 at 23:10
  • I agree with @ Mike 'Pomax' Kamermans , I adapted the post to my specific question. – Ruben Jul 24 '17 at 10:20

1 Answers1

0

Try with inflection::ese(X, Y, 0). It returns 142 which corresponds to X = 3578.

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