write a function about ' v(t)=10^9+​10^8*(1-ex​p(-t./100)​)-1.5*10^7​*t '

[EDIT: 20111115 14:07 CST - merge duplicate - WDR]

question about matlab : I want to write a function about ' v(t)=10^9+10^8*(1-exp(-t./100))-1.5*10^7*t ' . In order to get each 't' , how can I write a function with 't' at the left side of the equal sign?(for example:t=......) Thanks for help!!

[Merge from duplicate]

I want to write a function about

 v(t)=10^9+10^8*(1-exp(-t./100))-1.5*10^7*t

In order to get each 't' , how can I write a function with 't' at the left side of the equal sign?(for example : t = ......)

 v(t)=10^9*(x./100) 
 x=(100:-5:40)

If I don't write a function with 't' at the left side of the equal sign, but I still have to write a to acquire each 't'.

Answer 1

[Walter Roberson]

If you have a particular v value and wish to solve for the corresponding t value, then the solution is

t = 100*LambertW(-(1/15)*exp((1/1500000000)*v-11/15))-(1/15000000)*v+220/3

LambertW is part of the Symbolic Toolbox, or you can use one of the contributions such as http://www.mathworks.com/matlabcentral/fx_files/6909/1/content/doc/lambertw.html

[kris mere]

thank you!!i will try to understand

though it seems to be complicated

[Daniel Shub]

Thank you Walter, I have learn a little new math today.

[Walter Roberson]

The solution for your rephrased question remains the same. At best the additional information tells you what value of v to use in the LambertW expression.

If you have the Symbolic Toolkit, the way to get this solution is:

syms v t

tsol = solve(10^9+10^8*(1-exp(-t./100))-3/2*10^7*t,t);

After that if you want to put in particular v, you can do that using subs(), such as

x=100:-5:40;

tnumeric = double( subs(tsol, v, 10^9*(x./100)) );

Myself, I find it difficult to understand how you can define v(t) in terms that do not involve t but do involve a previously unstated variable x.

Answer 2

[Daniel Shub]

I don't think you can. If you work through the math, I think you will eventually get something t+a*log(b*t) is equal to c. If you are willing to accept the approximation that t+a*log(b*t) is approximately t (for large enough t), then you can isolate t on the LHS.

[kris mere]

yeah

i think i can't

but i don't know how to get 't' without this method

Answer 3

[Friedrich]

Hi,

risky (violates some basic rules of numerical mathematics) but should work in the most cases

func = @(t) 10^9+10^8*(1-exp(-t./100))-1.5*10^7*t;

x0 = 0;
v = 3000;

t = fzero(@(x) func(x) - v,x0)

func(t)

[kris mere]

thank you!!i will try to understand