How do I solve an equation with a vector term?

2 次查看（过去 30 天）
Efaz Ejaz
Efaz Ejaz 2019-11-11
评论： Efaz Ejaz 2019-11-12
How do I solve an equation that has one vector term with the rest being constant values? If you look at the part after "syms t", I am attempting to solve for "t" but there are 1001 values for m_0 in eqn1 and a 1001 values for every value of b, h(b) and v(b). So, I'm hoping to get a vector T1 containing 1001 values of t for 1001 values of m_0 and the same for T2 for every value of b, h(b) and v(b).
clear, clc
%Write the given values, u, m_e, b, q, and g.
u = 8000; m_e = 1500; g = 32.2;q = 15; t_0 = 0; b = 0:0.1:100;
%Compute m_0, h_b, and v_b.
m_0 = m_e + q.*b;
h_b = ((u.*m_e)./q)*log(m_e./(m_e+q.*b))+u.*b - 0.5.*g.*b.^2;
v_b = u*log(m_0/m_e) - g.*b;
%Part I'm not so sure about.
syms t
eqn1 = 50000 == u./q.*(m_0-q.*t).*log(m_0-q.*t)+u.*(log(m_0)+1).*t-0.5.*g.*t.^2-((m_0.*u)./q).*log(m_0);
T1 = solve(eqn1, t);
eqn2 = 50000 == h_b+v_b.*(t-b)-0.5.*32.2.*(t-b).^2;
T2 = solve(eqn2, t);
T2_desired = double(T2(2));
T = T2_desired - T1
The output I get is as follows:
Warning: Unable to find explicit solution. For options, see help.
> In solve (line 317)
In ROCKETMAN (line 15)
Index exceeds the number of array elements (0).
Error in sym/subsref (line 900)
R_tilde = builtin('subsref',L_tilde,Idx);

请先登录，再回答此问题。

回答（2 个）

Basil C.
Basil C. 2019-11-11
The line
eqn1 = 50000 == u./q.*(m_0-q.*t).*log(m_0-q.*t)+u.*(log(m_0)+1).*t-0.5.*g.*t.^2-((m_0.*u)./q).*log(m_0);
will define 1001*1001 equations, you can try indexing instead.
syms t
for i=1:numel(b)
eqn1 = -50000 + u/q*(m_0(i)-q*t)*log(m_0(i)-q*t)+u*(log(m_0(i))+1)*t-0.5*g*t^2-((m_0(i)*u)/q)*log(m_0(i));
T1(i)= solve(eqn1, t);
end
The solution for each will be stored in T, you can do the same for eqn2. Hope this is what you are looking for.
  6 个评论
显示 4更早的评论
Efaz Ejaz
Efaz Ejaz 2019-11-11
I am looking to get two separate arrays T1 and T2, both consisting of only positive real numbers. Thanks.
Efaz Ejaz
Efaz Ejaz 2019-11-11
编辑：Efaz Ejaz 2019-11-11
Sorry for the pain but I just wanted to add that I want to be able to find the difference between the values in T2 and the corresponding values in T1.
So, say when b is 100, I'll get one solution for t in eqn1 (let's call it t1). When b is 100, there'll be a particular value for h_b and v_b and I should get two solutions for t in eqn2(t21 and t22). I want to discard the first solution for t in eqn2 (t21), and do (t22-t1) and put all those differences into another array.
I was hoping to do this for all the values of b.

请先登录，再进行评论。

Walter Roberson
Walter Roberson 2019-11-11
%Write the given values, u, m_e, b, q, and g.
u = 8000; m_e = 1500; g = 32.2;q = 15; t_0 = 0; b = 0:0.1:100;
%Compute m_0, h_b, and v_b.
m_0 = m_e + q.*b;
h_b = ((u.*m_e)./q)*log(m_e./(m_e+q.*b))+u.*b - 0.5.*g.*b.^2;
v_b = u*log(m_0/m_e) - g.*b;
%Part I'm not so sure about.
syms t
nb = numel(b);
T1 = zeros(1,nb);
t0 = 1;
for i=1:nb
eqn1 = -50000 + u/q*(m_0(i)-q*t)*log(m_0(i)-q*t)+u*(log(m_0(i))+1)*t-0.5*g*t^2-((m_0(i)*u)/q)*log(m_0(i));
T1(i)= vpasolve(eqn1, t, t0);
t0 = T1(i);
end
%%
T2 = zeros(2,nb);
T0 = 1;
for i = 1:nb
eqn2 = -50000 + h_b(i)+v_b(i)*(t-b(i))-0.5*32.2*(t-b(i))^2;
k = vpasolve(eqn2, t, t0);
T2(1,i) = k(1);
T2(2,i) = k(2);
t0 = T2(1,i);
end
The above is not all that fast.
Note: b = 37.3 is the first b value for which the T2() are real-valued. With smaller b, both T2 are complex valued.
  3 个评论
显示 1更早的评论
Walter Roberson
Walter Roberson 2019-11-11
Even if you had deleted the comment about not so sure, then the line you indicate would be on line 22, not line 20. You must not have used the same code that I posted, so it is difficult to know what might have happened. Also, you did not mention which MATLAB release you are using.
Here is a modified version that takes into account the possibility of too few solutions:
%Write the given values, u, m_e, b, q, and g.
u = 8000; m_e = 1500; g = 32.2;q = 15; t_0 = 0; b = 0:0.1:100;
%Compute m_0, h_b, and v_b.
m_0 = m_e + q.*b;
h_b = ((u.*m_e)./q)*log(m_e./(m_e+q.*b))+u.*b - 0.5.*g.*b.^2;
v_b = u*log(m_0/m_e) - g.*b;
syms t
nb = numel(b);
T1 = zeros(1,nb);
t0 = 1;
wb = waitbar(0, 'T1');
for i=1:nb
waitbar(i./nb, wb);
eqn1 = -50000 + u/q*(m_0(i)-q*t)*log(m_0(i)-q*t)+u*(log(m_0(i))+1)*t-0.5*g*t^2-((m_0(i)*u)/q)*log(m_0(i));
T1(i)= vpasolve(eqn1, t, t0);
t0 = T1(i);
end
%%
T2 = zeros(2,nb);
T0 = 1;
waitbar(0, wb, 'T2');
for i = 1:nb
waitbar(i./nb, wb);
eqn2 = -50000 + h_b(i)+v_b(i)*(t-b(i))-0.5*32.2*(t-b(i))^2;
k = vpasolve(eqn2, t, t0);
if length(k) >= 1
T2(1,i) = k(1);
t0 = T2(1,i);
else
T2(1,i) = nan;
end
if length(k) >= 2
T2(2,i) = k(2);
else
T2(2,i) = nan;
end
end
delete(wb)
Efaz Ejaz
Efaz Ejaz 2019-11-12
Thanks a lot, mate. I really do appreciate the help.

请先登录，再进行评论。

类别

在 Help Center 和 File Exchange 中查找有关 Conversion Between Symbolic and Numeric 的更多信息

标签

Community Treasure Hunt

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

Start Hunting!

Translated by