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Let's say I have this function
f(t) = 4*sin(a(t)) + x(t)*y(t) + h + cos(y(t))*sin(x(t))
How would I compute its derivative with respect to time?
You need to declare the variables and the functions inside it as being symbolic and then use diff:
clear
clc
syms a x y t h
a(t) = symfun(sym('a(t)'), t)
x(t) = symfun(sym('x(t)'), t)
y(t) = symfun(sym('y(t)'), t)
F = 4*sin(a(t)) + x(t)*y(t) + h + cos(y(t))*sin(x(t))
DerF_t = diff(F,t)
Giving the following (messy) output:
F = h + 4*sin(a(t)) + cos(y(t))*sin(x(t)) + x(t)*y(t)
DerF_t = x(t)*diff(y(t), t) + y(t)*diff(x(t), t) + 4*cos(a(t))*diff(a(t), t) + cos(x(t))*cos(y(t))*diff(x(t), t) - sin(x(t))*sin(y(t))*diff(y(t), t)
Note that since a(t),x(t) and y(t) are simply defined as functions of 't' we are stuck with their 'symbolic' derivative (I don't know the term for that sorry)...i.e. diff(a(t)) for instance.
Hope it's what you were after!
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I have this trigonometric equation;
cos(2*pi*50*t)+cos(2*pi*100*t)
I want to graphic of equation and I want to find field for a period. How can I do?
Graph:
>> f = #(t) cos(2*pi*50*t) + cos(2*pi*100*t);
>> x = linspace(0, 1/50, 100);
>> y = f(x);
>> plot(x,y)
Area over 1 period:
>> integral(f, 0, 1/50)
or just do it manually:
∫ ( cos(2π·50t) + cos(2π·100t) ) dt =
-1/2π·( 1/50·sin(2π·50t) + 1/100·sin(2π·100t) )
which, evaluated between 0 and 1/50, equals 0.
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I have a function (the hill function) when X and K are the same (x==k) the output should be 0.5, when I test the function it gives a result of 0.5, when I try to plot it, I do not get 0.5 for my Y. Can anyone explain what I am doing wrong?
n = 1;
k = 1;
x = [0:0.01:2].';
y = (x.^n)/((x.^n)+(k.^n));
plot(x,y);
n = 1;
k = 1;
x = [0:0.01:2].';
y = (x.^n)./((x.^n)+(k.^n));
plot(x,y);
You missed a dot before the division.
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function f = objfun(x)
f = exp(x(1)) * (4*x(1)^2 + 2*x(2)^2 + 4*x(1)*x(2) + 2*x(2) + 1);
x0=[-1,1];
options = optimoptions(#fminunc,'Algorithm','quasi-newton');
[x,fval,exitflag,output] = fminunc(#objfun,x0,options);
x,fval,exitflag,output
end
Can you please help me in running the code?
Convert f to a function handle as
fun = #(x) exp(x(1)) * (4*x(1)^2 + 2*x(2)^2 + 4*x(1)*x(2) + 2*x(2) + 1);
then call fminunc with
[x,fval,exitflag,output] = fminunc(fun,x0,options);
As a side note, don't ever call fminunc from within the objective function.
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I've two equations as:
x = c1 - y;
y = c2*c3*x / (1+c3*x);
where c1, c2 and c3 are constants. How to solve these equations in MATLAB? Please help.
Since I'm in a good mood this morning:
x = c1 - y;
y = c2*c3*x / (1+c3*x);
Now, pen and paper:
y = c1 - x
c1 - x = c2*c3*x / (1 + c3*x)
(c1 - x) * (1 + c3*x) = (c2 * c3 * x)
(c1 - x) * (1 + c3*x) - c2*c3*x = 0
You should be able to use fzero or roots to solve this by yourself.
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I have these two equations and I want to find the values of these two parameters:
9.393e(16) = ((N*K)/(K + 0.0045))*(1 - exp (-(K + 0.0045)*120))
1.376e (17) = ((N*K)/(K + 0.0045))*(1 - exp (-(K + 0.0045)*240))
How can I solve it in matlab or wolfram please
I guess a hand calculator is sufficient for that.
Call:
a = 9.393e(16)
b = 1.376e (17)
Q = (N*K)/(K + 0.0045)
f = exp (-(K + 0.0045)*120) => exp (-(K + 0.0045)*240) = f^2
You have:
a = Q (1 - f)
b = Q (1 - f^2)
so
a/b = (1 - f) / (1 - f^2) = 1 / (1 + f)
thus
f = b/a - 1
You can take the log at both sides and solve for K.
-(K + 0.0045)*120 = log(b/a - 1)
To find N the equation is again just linear.
You can solve simultaneous non-linear equations in MATLAB via FSOLVE or LSQNONLIN. However, this requires the Optimization Toolbox.
See this MathWorks knowledgebase article.
Given the magnitude of the LHS of your equations, I would not be surprised if you see some numerical instability. You might want to do this problem by hand as suggested by Acorbe.