MATLAB: Finding the Maximum/Minimum value of a function

findingMATLABmaximumminimumsolving

This seems to be quite a simple thing to do, but for some ungodly reason i am having unimaginable trouble doing so, on my Casio calculator its simple as Pie, you put in the function and say find maximum between this point and this point, however I cant for the life of me figure out how to do it on this, and i tried a whole bunch of things that kept giving me errors, i was hoping if someone could help me with this it will be amazing.

What i would like to know is…. How to find the maximum of a function, I am not posting the whole code, just the thing that i am having trouble with….. and this is it….

x(t)=0.03106*exp(-21.43*t)*sin(32.2*t)
v(t)=diff(x(t));
a(t)=diff(v(t));
jerk(t)=diff(a(t));
 G=solve(jerk(t)==0,t>0)
G=vpa(G,8)
 ezplot (a(t),[0,0.5]) ; title a(t)

Now this is where the problem begins… to solve for maximum acceleration I am differentiating acceleration and then making that equal to ZERO, which should give me the time at which acceleration is ZERO However It is outputting a value for G which is not coherent with the ezplot, why? and how do I fix this?

Any help would be very greatfully appreciated and thanks in advance!!!

Just a little bit more info, It seems that whatever is "outisde" the ezplot the equation COMPLETELY diregards, it is wierd but i am telling the truth, at one point i wanted to find the maximum of x(t), which say it was at t=0.5 but it showed me the maximum to be t=0.46 JUST because the top of the graph was cut and 0.46 was the "maximum" of the graph that ezplot saw what is going on???? if someone can please confirm I am not insane that would be great.

Best Answer

What value of G is it outputting?

When you apply solve() to something that is not a polynomial, it is not certain that it will be able to find all of the roots; it might return only one numeric root and that one root could be anywhere. There are an infinite number of solutions spaced pi/32.2 apart. The first positive one is at approximately -(5/161)*arctan(10976809340/56816825393)+5/161*pi -- approximately because it depends on whether 0.03106 means 3106/10000 exactly or if the 0.03106 is to be interpreted as 0.03106 +/- 0.000005 .

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MATLAB: Is it the answer from MATLAB is different from the other solving software? Where did I go wrong

Usually when I see a confused question, where someone says "help, i got a strange answer", the reason is because they were not careful in what they wrote. Here, for example, the code you claim to have used in MATLAB would not actually run. It would produce an error message, because while you have defined cond1 and cond2, then you call dsolve with the variable cond. As well, you never even defined k. What is k? How should MATLAB respond when you don't define variables? So if you did get an answer, then the answer you did get was based on some things we do not see at all.

When you see a problem, start from the beginning. Verify that when you think is in the variables you used, is what you actually have in those variables.

Next, what you seem to be wanting to claim, is that the problem with TWO boundary values, that at T ==0, AND at T == 1, will yield the solution you have. And that claim is incorrect.

syms y(T)
syms k real
eqn = diff(y(T),T)==(k*(y-80));
ysol=dsolve(eqn,y(0) == 20)
ysol =
80 - 60*exp(T*k)

As you should see, With only ONE condition, at T == 0, we get the same solution that you claim the other code provides.

However, with a second boundary value at T == 1, the problem would be unsolvable, unless the second condition happens to be consistent with the first.

subs(ysol,T,0)
ans =
20
subs(ysol,T,1)
ans =
80 - 60*exp(k)

Now, we could solve for the value of k, such that this second equation happens to result in y(1)==35.

ksol = solve(subs(ysol,T,1) == 35)
ksol =
log(3/4)

We would then need to substiture that BACK into the equation, eliminating k.

yfinalsol = subs(ysol,k,ksol)
yfinalsol =
80 - 60*exp(T*log(3/4))
subs(yfinalsol,T,0)
ans =
20
subs(yfinalsol,T,1)
ans =
35

But as you see, IF your goal is to satisfy BOTH conditions, at T==0, AND at T == 1, then you will no longer have the solution you think you got from the other software.

Finally, the solution you claim to have gotten, happens to be the solution of a related problem:

dsolve(eqn,y(0) == 5)
ysol =
80 - 75*exp(T*k)

It was simple enough to recover what the initial condition needed to be, to have gotten the solution you claim to have gotten, since if you got that solution, then at T == 0, y would need to be 5.

subs(80 - 75*exp(T*k),T,0)
ans =
5

But in no respect is it the problem you have posed.

Mathematics is both simple as well as unforgiving.

MATLAB: Second order differential equation

syms c positive
syms g negative
eqn = diff(y(t),t,2)== g+(c*(diff(y(t),t))*(diff(y(t),t)));
cond = [y(0) == 1000, diff(y(0),t) == 0];
ySol(t) = dsolve(eqn, cond)

MATLAB finds four solutions.

Best Answer

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MATLAB: Is it the answer from MATLAB is different from the other solving software? Where did I go wrong

MATLAB: Second order differential equation

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