Symbolic variables: Isolate a variable in an equation?
조회 수: 21 (최근 30일)
이전 댓글 표시
I always have trouble isolating symbolic variables in equations. See this for example:
8*x1 + 2*x2 + 3*x3 - 51 = 0
want to isolate x1, manually know is this:
x1 = (51 - 2*x2 - 3*x3)/8
how can I do this in matlab? Thanks in advance.
댓글 수: 0
채택된 답변
추가 답변 (1개)
Arash Moharjeri
2017년 11월 10일
편집: Walter Roberson
2017년 11월 10일
Hi all,
I am trying to rearrange the below equation to solve for R0.
This is how I have written the equation in MATLAB.
syms R0 H hs K_h rw P
EQN= (H - power((power(hs,2))+((P/K_h)*((power(R0,2)*log(R0/rw))-(power(R0,2)-power(rw,2))*0.5)),0.5) == 0);
EQN2=isolate (EQN, R0)
When I run isolate, it doesn't give me an error but it doesn't give me a correct answer either. It gives me the original equation.
Can anyone assist me with this issue? Is it because the equation is too complicated to rearrange ?
Thanks
댓글 수: 7
andrea giovannini
2018년 1월 8일
Hi, I can't say the ranges for the values, because I'm working to built a completely general Bayesian Network to estimate the scour depth. I'll try with the numerical method and see what happens. Thank a lot for your time and disponibility.
Walter Roberson
2018년 1월 8일
Estimating the scour depth given the depth of the river and the other aspects would be fairly different mathematically than attempting to find the depth of the river with that formula.
Dc = ((3*d*u+8*v)*exp(-RootOf(-2*n*Z*(72*exp(Z)*u*y-3*d*u-8*v)-9*y^(5/3)*sqrt(s)*sqrt(2)*(ln(2)+ln(5))*exp(Z),Z))-36*y*u)/(36*u)
where RootOf(-2*n*Z*(72*exp(Z)*u*y-3*d*u-8*v)-9*y^(5/3)*sqrt(s)*sqrt(2)*(ln(2)+ln(5))*exp(Z),Z) means the set of Z such that -2*n*Z*(72*exp(Z)*u*y-3*d*u-8*v)-9*y^(5/3)*sqrt(s)*sqrt(2)*(ln(2)+ln(5))*exp(Z) is 0 -- the roots of that expression.
That expression does not have a closed form solution, but it would be pretty tractable to solve numerically I think, whereas I think solving for y would be more difficult numerically.
참고 항목
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!