A = {2,1,2,4}
B = {4,1,6,2}
angle = arccos( (A.B)/(norm(A)*norm(B)) )
area = (1/2)A.B sin(angle)
Is this correct? if yes, it is easy to generalize for Rn.
Find the area of the triangle with sides
A = (a1 ... an)
B = (b1 ... bn)
and A-B = (a1-b1 ... an-bn)
I don't even know where to start. I know how to do it in 3D with the cross product, but that obviously won't work for higher dimensions.
So I need help generalizing for Rn.
1. Homework Statement
Why does the following ODE ALWAYS have two linearly independent solutions?
x''(t) + a(t) x'(t) + b(t) x(t) = f(t)
The characteristic polynomial argument is not sufficient?
Hello fellow PF members
I was wondering how one would go about finding the lagrangian of a problem like the following:
A particle is constrained to move along the a path defined by y = sin(x).
Would you simply do this:
x = x
y = sin(x)
x'^2 = x'^2
y'^2 = x'^2 (cos(x))^2...
6 replies and all of them useless.
Fixated on something frivolous.
I gave you guys the correct differential equation in the first line anyway, so there's absolutely no problem imo.
Why are you guys so fixated on that line??
That's not even the crux of the question I'm asking!
lol
I obtained this differential equation from applying the Euler-Lagrange equation to the following function:
f(y,y') = (y')^2/y
That's how I know my reverse quotient rule is...
1. Homework Statement
Solve the following differential equation:
2. Homework Equations
2 y (y'')^2 + 2 y y''' y' -2 (y')^2 y'' = - (y')^2
3. The Attempt at a Solution
I don't know if the following is useful, but if you divide both sides by y^2, the LHS of the above...