Solving for a variable in equation

Question

Maddie Sanders 2020-1-23

0
链接

此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/501553-solving-for-a-variable-in-equation

评论： Star Strider 2020-1-23

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)

0 个评论
显示 -2更早的评论隐藏 -2更早的评论

请先登录，再进行评论。

请先登录，再回答此问题。

Answer 1

Star Strider 2020-1-23

0
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/501553-solving-for-a-variable-in-equation#answer_411539

在 MATLAB Online 中打开

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

3 个评论
显示 1更早的评论隐藏 1更早的评论

Star Strider 2020-1-23

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’.