Solving for a variable in equation

조회 수: 3 (최근 30일)
Maddie Sanders
Maddie Sanders 2020년 1월 23일
댓글: Star Strider 2020년 1월 23일
I want to solve for t in the equation: Tf=Ts+(T0-Ts)*exp(-k*t), but I can't figure out how to solve for t. I can only solve for Tf. Here is my script. What do I need to add/change to solve for t?
T0=120;
Ts=38;
k=0.45;
Tf=65;
Tf=Ts+(T0-Ts)*exp(-k*t)

채택된 답변

Star Strider
Star Strider 2020년 1월 23일
Use a root-finding function (I chose fzero here), then take the known variable values, create an anonymous function from the expression for ‘Tf’ and solve:
T0=120;
Ts=38;
k=0.45;
Tf=65;
% Tf=Ts+(T0-Ts)*exp(-k*t);
[tval,fval] = fzero(@(t) Ts+(T0-Ts)*exp(-k*t)-Tf, 1)
producing:
tval =
2.4686
  댓글 수: 3
Star Strider
Star Strider 2020년 1월 23일
My pleasure!
The ‘tval’ output is the value of ‘t’ where the expression equals ‘Tf’, and the ‘fval’ output is the value of the function at that point, indicating that it found a ‘t’ value that resulted in the anonymous function being very close to zero (within the tolerance fzero uses). Every nonlinear parameter estimation function requires an initial estimate for the parameter (or parameters) it is estimating, and I chose 1 here. The initial parameter estimates can be important in some problems, however not in this one, where widely differing iinitial estimates all produce the same result for ‘tval’.
Star Strider
Star Strider 2020년 1월 23일
If my Answer helped you solve your problem, please Accept it!

댓글을 달려면 로그인하십시오.

추가 답변 (0개)

카테고리

Help CenterFile Exchange에서 Solver Outputs and Iterative Display에 대해 자세히 알아보기

태그

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by