Can someone please fix my matlab code? Don't know how to get rid of the warning: Unable to solve symbolically. Returning a numerical solution using vpasolve.

% Values of constants
syms r V L h
r = 0.3;
V = 0.35;
L = 3;
eqn = (0.5*pi*r^2-r^2*asin(h/r)-h*sqrt(r^2-h^2))*L == V;
answer = solve(eqn, h)
depth = r - answer

[Star Strider]

It is not possible to solve symbolically for ‘h’. A numerical solutionj is the best you can do.

Plotting it:

eqn = @(h) (0.5*pi*r^2-r^2*asin(h/r)-h*sqrt(r^2-h^2))*L - V;

figure
fplot(eqn,[-5, 5])

there only appears to be one root.

[Hamza Aboumaray]

I copied your code and still have the warning.

[Star Strider]

The warning is not the problem.

It is not possible to solve symbolically for ‘h’. A numerical solution is the best you can do.

[Hamza Aboumaray]

Thank you for your help!

Answer 1

[Walter Roberson]

r = 0.3;

You used one significant digit in your coefficients. In science that means that r is some undetermined value in the range 25/100 inclusive to 35/100 exclusive. And since your input coefficients are indeterminate, it is a category mistake to request an infinitely precise closed form solution, which is what solve() is designed to do. You should be using vpasolve or a numeric technique such as fzero()

[Hamza Aboumaray]

I tried using the vpasolve technique but i dont know how to use it. Can you please show me how to do it in a code?

[Walter Roberson]

% Values of constants
syms h
r = 0.3;
V = 0.35;
L = 3;
eqn = (0.5*pi*r^2-r^2*asin(h/r)-h*sqrt(r^2-h^2))*L == V;
answer = vpasolve(eqn, h)
depth = r - answer

[Hamza Aboumaray]

Thank you very much!