# Finding a point within the profile

Continuing my studies on solids of revolution ... I work daily with CAD software and created a random profile where I can define it only by coordinates $(x, y)$.

The image below illustrates the profile I created ... p1 = {0, 15}; p2 = {20, 30}; p3 = {x3, 25}; p4 = {80, 20};
points = {p1, p2, p3, p4}


I tried to find an $f(x)$, but I did not succeed:

eq = Fit[points, {1, x, x^2, x^3, x^4}, x]


My CAD software was able to find the value $x3 = 39.02120136$ because the profile is tangent at point $p2$, that is, there is a derivative $f'(x)$ there. Other information that can help is the volume that I can also get with my software. The value obtained was $V(x)=137761.98$ With this I could set $V(x)$:

 V[x_] := Integrate[f[x]^2*Pi, {x, 0, 80}]


I tried something very strange, but it sure would not work ...

 DSolve[V[x_] == 137761.98, f[x], x]


With this information someone got the idea of how to find $f(x)$?

It may be that the question has been confused, but according to the comments I will improve the question...

If I understand correctly, you have: a random profile, a few points along this, and the volume formed by it's revolution. And you want to somehow extract an analytical expression for the function of your profile?

I cannot see how this is generally possible! Your profile could be described by any function or indeed no function, why should it be possible to describe it by an analytical function?

You can indeed approximate it with a polynomial:

f[x_] := a x^3 + b x^2 + c x + d

sol = Solve[
f == 15 &&
f == 30 &&
f == 20 &&
f' == 0,
{a, b, c, d}
]

Plot[f[x] /. sol, {x, 0, 80}, PlotRange -> {All, {0, All}}, Epilog -> Point[{{0, 15}, {20, 30}, {80, 20}}]] But this will only be an approximation: • It's a beginning. It's more or less this idea. – Luciano Jul 14 '17 at 16:36
• Really enjoyed it. May even be a solution to my other issue attached – LCarvalho Jul 14 '17 at 16:40
• Perhaps having a higher degree could improve – LCarvalho Jul 14 '17 at 16:41
• @LCarvalho: A higher degree would likely help but only if the number of sample points is greatly increased. – JimB Jul 14 '17 at 17:38