BDR

The displacement of the midpoint from the equilibrium is approximately 0.358 when t=1.7

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 temperature 2 and its right end to 0. Sketch the temperature distribution at times

t=0, t=0.01, t=0.1, and t=10

Per the screenshot above, the temperature near the midpoint of the insulated side of the triangle is approximately 0.04795. See my random questions for more details.

My matrix:[

[0, 4, 2, 2, 1],

[1, 0, 4, 4, 1],

[1, 3, 0, 4, 4],

[1, 3, 2, 0, 1],

[2, 1, 4, 1, 0]

]

Team 3: rating = 0.5317387251068555
Team 2: rating = 0.4743664736534873
Team 1: rating = 0.43963279846587106
Team 5: rating = 0.41258201033241837
Team 4: rating = 0.3587888852214044

Clearly, team 3 has the advantage. However, since this is bling archery, you can never tell who will pull out on top.

For my conditions, $k=0$ and $f=0$, the steady-state of my polygon is shown below. The temperature in the lower right hand corner $(28.69,0.93)$, is approximately $u=0.92895$.

Heat flow with a constant internal heat source is governed by:

$u_t=2u_{xx}+7$, $u(0,t)=−2$, $u(1,t)=4$.

Because this is a steady state temperature distribution, $u_t=0$ since $u(x,t)$ does not change with time. So,

$0=2u_{xx}+7$

Solving for $u_{xx}$ gives $u_{xx}=\frac{-7}{2}$. By integrating this equation twice we get:

$u(x,t)=\frac{-7}{4}x^2+c_1x+c_2$,

where $c_1$ and $c_2$ are our two constants. Solving for $c_1$ and $c_2$ using the initial conditions of $u(0,t)=-2$, $u(1,t)=4$ produces the steady state distribution of

$u(x,t)=\frac{-7}{4}x^2+\frac{3}{2}x+\frac{19}{4}$