How to find Centroid of the following curve?
curve =
Plot[Cos[x]/(x + Log[x]), {x, 1, 8}, Ticks -> None, PlotStyle -> {Blue, Thick}]
I tried:
Graphics[{curve, Red, Point[RegionCentroid[curve]]}]
but it doesn't work
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For those, who wonder what we would do without RegionCentroid, there is the simple way of calculating the centre by normal integration. When we consider your function $f$ as parametric curve the situation is clear.
$$x(t)=t\\y(t)=\frac{\cos(t)}{t+\log(t)}$$
Furthermore, we need the arc length $L(f)$ of the curve which can be calculated with
$$ds=\sqrt{x'(t)^2+y'(t)^2}\;dt$$
Then your centre $(x_m, y_m)$ is given by
$$x_m\cdot L(f)=\int x\;ds\qquad y_m\cdot L(f)=\int y\;ds$$
For your particular case this means the x-coordinate of your centroid is given by
$$x_m = \left(\int_1^8 t\; ds\right)/(\int_1^8ds)$$
In Mathematica code
x[t_] := t;
y[t_] := Cos[t]/(t + Log[t]);
ds = Sqrt[x'[t]^2 + y'[t]^2];
NIntegrate[{x[t], y[t]} ds, {t, 1, 8}]/NIntegrate[ds, {t, 1, 8}]
(* {4.38467, -0.00815125} *)
answered Dec 19 '16 at 12:08
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2$\begingroup$ It should be:
NIntegrate[{x*Sqrt[1 + dydx^2], y[x]*Sqrt[1 + dydx^2]}, {x, 1, 8}]/ NIntegrate[Sqrt[1 + dydx^2], {x, 1, 8}],{4.38467, -0.00815125}$\endgroup$ – Mariusz Iwaniuk Dec 19 '16 at 12:38 -
$\begingroup$ @MariuszIwaniuk I'm sorry. I didn't look it up. I'm fixing the post. Thanks for being attentive. $\endgroup$ – halirutan♦ Dec 19 '16 at 12:59
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To expand on J.M.'s comment, you will need to use the correct predicates to build the region (I used ImplicitRegion because I find it most flexible):
region = ImplicitRegion[y == (Cos[x]/(x + Log[x])) && x > 1 && x < 8, {x, y}];
Then you can use RegionCentroid on it:
centroid = RegionCentroid[DiscretizeRegion@region];
Show[RegionPlot[region], ListPlot[centroid], PlotLabel -> centroid]
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Simplest is the numerical approach for Graphics objects, which is to use DiscretizeGraphics:
curve = Plot[Cos[x]/(x + Log[x]), {x, 1, 8}, Ticks -> None, PlotStyle -> {Blue, Thick}];
RegionCentroid[DiscretizeGraphics[curve]]
{4.3846982827317245, -0.008143914969936251}
Note, while you can get lucky with analytical results too sometimes:
RegionCentroid[ParametricRegion[{x, x^2}, {{x, 1, 8}}]]
$$\left\{\frac{257 \sqrt{257}-5 \sqrt{5}}{3 \left(-2 \sqrt{5}+16 \sqrt{257}-\sinh ^{-1}(2)+\sinh ^{-1}(16)\right)},\frac{-18 \sqrt{5}+8208 \sqrt{257}+\sinh ^{-1}(2)-\sinh ^{-1}(16)}{16 \left(-2 \sqrt{5}+16 \sqrt{257}-\sinh ^{-1}(2)+\sinh ^{-1}(16)\right)}\right\}$$
Generally it is not the case:
answered Dec 19 '16 at 11:43
ParametricRegion[]?RegionCentroid[]works on regions (as the function name implies) and not onGraphics[]objects. $\endgroup$ – J. M.'s ennui♦ Dec 19 '16 at 11:28