Calculating the height of a liquid in a sphere using secant method
์กฐํ ์: 8 (์ต๊ทผ 30์ผ)
์ด์ ๋๊ธ ํ์
Justin Michaellmore
2021๋
1์ 13์ผ
ํธ์ง: James Tursa
2021๋
1์ 14์ผ
The problem is to determine how much the spherical tank must be filled to contain 30m3 if
R = 3m. The volume is calculated according to equation: V = (๐โ^2*[3๐
โ โ])/3
Use the secant method. Print the approximations, the relative error. Ensure that the calculation ends when the error is less than a certain value.
Find the correct solutions using the appropriate matlab function. Print approximations and errors.
Bonus: show graphically the dependance of the Volume V on the Height h. f(x).
๋๊ธ ์: 2
James Tursa
2021๋
1์ 14์ผ
What have you done so far? What specific problems are you having with your code?
์ฑํ๋ ๋ต๋ณ
James Tursa
2021๋
1์ 14์ผ
ํธ์ง: James Tursa
2021๋
1์ 14์ผ
You need to use a function that is 0 at the h you are looking for. So you need that 30 in your function handle. E.g.,
func = @(h)(pi*h^2*(9-h))/3 - 30
If you want to compare to a MATLAB function answer, realize that you have a polynomial in h. So just make a vector of the coefficients and feed it to the roots( ) function.
๋๊ธ ์: 4
James Tursa
2021๋
1์ 14์ผ
ํธ์ง: James Tursa
2021๋
1์ 14์ผ
"Didn't work for me"
So, that doesn't tell me much. Running your exact code with the -30 addition produced the answer for me. What is it that didn't work for you? You should be getting the exact same answer I did.
clear all
clc
er=100;
es=0.001;
func = @(h)(pi*h^2*(9-h))/3 - 30;
h1=1;
h2=2;
f1=func(h1);
counter=0;
while abs(er) > es
counter=counter+1;
f2=func(h2);
er=((h2-h1)*f2)/(f2-f1);
fprintf('error:%f\n',er);
h1=h2;
f1=f2;
h2=h2-er;
fprintf('height (aprox):%f\n',h2);
end
As for the plot, just realize that the height can range from 0 to 2*R. So create that range and plug it into your vectorized V formula:
R = 3;
h = 0:0.01:2*R;
V = (pi*h.^2.*(9-h))/3;
Then plot(h,V)
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