BenA

The displacement of the midpoint from the equilibrium at the time 1.1s is -0.121. (redo)

My heat flow problem where I found the steady state temperature distribution: $$u_t=3u_{xx}+1$$

My boundary conditions were: $$u(0,t)=-3$$ and $$u(1,t)=4$$

Since $u_t=0$ when the bar is in steady state I used this to solve for $u_{xx}$, which I got was $u_{xx}=\frac{-1}{3}$. Then integrating twice: $$u(x)=\frac{-1}{6}x^2+Cx+D$$

I found these constants by using my boundary conditions: $C=\frac{43}{6}$ and $D=-3$

Finally I got: $$u(x)=\frac{-1}{6}x^2+\frac{43}{6}x-3$$

(Redo) A metal bar of length 1 lies along the unit interval. Its temperature distribution is given by: $$g(x)=4x^2-2x$$

At time t=0, its left end is set to t=-3 and the right end t=-1. Sketch of temperature distribution at times t=0, t=.01, t=.1, and t=10 is below:

The temperature at the midpoint of the insulated side of the triangle at 1s was 0.096.

My team matrix: [

  [0,3,3,2,2],

  [4,0,1,3,1],

  [1,4,0,4,2],

  [2,1,3,0,2],

  [3,1,4,2,0]

 }

Team 3: rating = 0.4848807961654844
Team 5: rating = 0.47317429197368976
Team 1: rating = 0.4632070845269747
Team 2: rating = 0.4163867448282543
Team 4: rating = 0.39122624975897735

Team 3 had the highest rating at 0.48488.

With my unique conditions I formed a polygon where $\kappa=0$ and $f=0$, meaning there was no external heat source present and no signs of growth/decay. After reaching steady state the temperature in the lower right hand corner of the polygon was $u=0.87677$.