# How do I solve an equation with a vector term?

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Efaz Ejaz on 11 Nov 2019
Commented: Efaz Ejaz on 12 Nov 2019
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);

Basil C. on 11 Nov 2019
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.
Efaz Ejaz on 11 Nov 2019
Edited: Efaz Ejaz on 11 Nov 2019
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.

Walter Roberson on 11 Nov 2019
%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.
Efaz Ejaz on 12 Nov 2019
Thanks a lot, mate. I really do appreciate the help.