i hate
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"cell_type": "code",
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"id": "af61cdf6-cb10-43e3-81b5-703acbc893a0",
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"cell_type": "code",
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"cell_type": "code",
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"id": "9f577d98-27db-451c-a60e-bb03d24070af",
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{
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"cell_type": "code",
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"execution_count": 19,
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"execution_count": 4,
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"id": "cf574d6a-da52-4bc3-8691-df9314c52376",
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{
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"cell_type": "code",
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"execution_count": 21,
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"execution_count": 5,
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"id": "ab38a583-abac-40a5-934d-f09bd2db8e7f",
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{
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"cell_type": "code",
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"execution_count": 25,
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"execution_count": 6,
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"id": "3119b774-40ab-4598-bfe1-517ab3cf549a",
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"metadata": {},
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"name": "stdout",
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"output_type": "stream",
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"text": [
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"0.02256498080663735\n",
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"\n",
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"=========FINAL ANSWER=========\n",
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"2A:\n",
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@ -234,7 +233,7 @@
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},
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{
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"cell_type": "code",
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"execution_count": 26,
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"execution_count": 7,
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"id": "5a353c72-3a91-4eb2-a30c-65af14628be3",
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"metadata": {},
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"outputs": [
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@ -269,7 +268,7 @@
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},
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{
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"cell_type": "code",
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"execution_count": 27,
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"execution_count": 8,
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"id": "35215138-3a47-42bc-b702-adba0373eace",
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"metadata": {},
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"outputs": [
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@ -319,7 +318,7 @@
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},
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{
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"cell_type": "code",
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"execution_count": 37,
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"execution_count": 9,
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"id": "fce5c667-36be-46c9-8183-fafe4b3f28a3",
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"metadata": {},
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"outputs": [],
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@ -336,8 +335,9 @@
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]
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},
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{
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"attachments": {},
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"cell_type": "markdown",
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"id": "950dbab5-ece9-4f26-a8bf-8aca2a4cd947",
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"id": "478f0c67-cd5b-4323-940d-fdf3ced0e2f6",
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"metadata": {},
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"source": [
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"## Part A\n",
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@ -346,41 +346,55 @@
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"\n",
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"But because we're steady state $\\frac{\\partial T}{\\partial t} = 0$:\n",
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"\n",
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"$$ -q''' = k \\nabla^2 T $$\n",
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"$$ 0 = k \\nabla^2 T + q''' $$\n",
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"\n",
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"And because we're in one dimension...\n",
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"\n",
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"$$ k \\frac{d^2 T}{dx^2} = -q'''$$\n",
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"$$ k \\frac{d^2 T}{dx^2} + q''' = 0$$\n",
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"\n",
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"Now we integrate:\n",
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"\n",
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"$$ k\\frac{dT}{dx} = -q'''x + C_1 $$\n",
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"$$ k\\frac{dT}{dx} + q'''x + C_1 = 0 $$\n",
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"\n",
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"$$ \\frac{dT}{dx} = -\\frac{ q'''x + C_1}{k} $$\n",
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"\n",
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"and integrate again:\n",
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"\n",
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"$$ kT(x) = -\\frac{q'''x^2}{2} + C_1 x + C_2 $$\n"
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"$$ T(x) = -(\\frac{ q'''}{2k} x^2 + \\frac{C_1}{k} x + C_2) $$\n"
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]
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},
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{
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"cell_type": "markdown",
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"id": "d00a9406-b75b-4227-a9c4-fe4cdc194e98",
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"id": "8bc55c33-7c5c-4596-ad22-b3482eac3a4f",
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"metadata": {},
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"source": [
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"This gives us our governing equation. Now we need to solve for our boundary conditions. We do this first on the left side:\n",
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"$$ k\\frac{dT}{dx}_{x=0} = h_\\text{left} (T(0) - T_\\text{left})$$\n",
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"$$ -k\\frac{dT}{dx}_{x=0} = h_\\text{left} ( T_\\text{left}- T(0))$$\n",
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"\n",
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"$$ C_1 = h_\\text{left} (C_2/k - T_\\text{left})$$\n",
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"$$ -k(-q''' \\times 0 / k - C_1/k) = h_\\text{left} (T_\\text{left} - C_2)$$\n",
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"\n",
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"$$ C_1 = h_\\text{left} (T_\\text{left} - C_2)$$"
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]
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},
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{
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"cell_type": "markdown",
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"id": "40964535-3127-4082-bdc3-9226e7d5a780",
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"metadata": {},
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"source": [
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"And now for our right side:\n",
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"\n",
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"$$ k\\frac{dT}{dx}_{x=0.2/12 \\text{[ft]}} = h_\\text{right} (T(0.2/12)\\text{[ft]} - T_\\text{right})$$\n",
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"$$ -k\\frac{dT}{dx}_{x=0.2/12 \\text{[ft]}} = h_\\text{right} (T_\\text{right} - T(0.2/12)\\text{[ft]}) $$\n",
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"\n",
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"$$ -k \\left(-\\frac{q''' x_\\text{right}+ C_1}{k}\\right) = h_\\text{right} (T_\\text{right} + (\\frac{ q'''}{2k} x_\\text{right}^2 + \\frac{C_1}{k} x_\\text{right} + C_2)) $$\n",
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"\n",
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"$$ q''' x_\\text{right}+ C_1 = h_\\text{right} (T_\\text{right} + (\\frac{ q'''}{2k} x_\\text{right}^2 + \\frac{C_1}{k} x_\\text{right} + C_2)) $$\n",
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"\n",
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"Now at this point I'm going to introduce SymPy to do the algebra heavy lifting. We can get away with this because I've organized all the units to be compatible when writing the code cell above:"
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]
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},
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{
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"cell_type": "code",
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"execution_count": 41,
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"execution_count": 37,
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"id": "0766e4d9-1f49-4449-a35d-05de86a3580e",
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"metadata": {},
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"outputs": [
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@ -396,10 +410,10 @@
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{
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"data": {
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"text/latex": [
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"$\\displaystyle C_{1} = 40 C_{2} - 280000$"
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"$\\displaystyle - C_{1} = 280000 - 40 C_{2}$"
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],
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"text/plain": [
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"Eq(C_1, 40*C_2 - 280000)"
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"Eq(-C_1, 280000 - 40*C_2)"
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]
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},
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"metadata": {},
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@ -417,10 +431,10 @@
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{
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"data": {
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"text/latex": [
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"$\\displaystyle C_{1} - 833333.333333333 = 0.5 C_{1} + 30 C_{2} - 418333.333333333$"
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"$\\displaystyle C_{1} + 833333.333333333 = 0.5 C_{1} + 300 C_{2} + 418333.333333333$"
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],
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"text/plain": [
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"Eq(C_1 - 833333.333333333, 0.5*C_1 + 30*C_2 - 418333.333333333)"
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"Eq(C_1 + 833333.333333333, 0.5*C_1 + 300*C_2 + 418333.333333333)"
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]
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},
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"metadata": {},
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@ -430,10 +444,10 @@
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"source": [
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"C_1, C_2 = sm.symbols('C_1, C_2')\n",
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"\n",
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"left_BC = sm.Eq(C_1, h_left*(C_2/k - T_left))\n",
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"left_BC = sm.Eq(C_1, h_left*(T_left - C_2/k ))\n",
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"display('Left BC', left_BC)\n",
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"\n",
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"right_BC = sm.Eq(k*(-q_ppprime*x_right + C_1)/k, h_right*((-q_ppprime*x_right**2/2 + C_1*x_right + C_2)/k - T_right))\n",
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"right_BC = sm.Eq(q_ppprime*x_right + C_1, h_right*(T_right + (q_ppprime*x_right**2/2/k + C_1/k*x_right + C_2)))\n",
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"display('Right BC', right_BC)"
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]
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},
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@ -447,14 +461,14 @@
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},
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{
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"cell_type": "code",
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"execution_count": 51,
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"execution_count": 38,
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"id": "765d2f17-7d1a-43a8-8b46-88ffed3ddaa3",
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"metadata": {},
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"outputs": [
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{
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"data": {
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"text/plain": [
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"{C_1: -2500000.00000000, C_2: -55500.0000000000}"
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"{C_1: -240714.285714286, C_2: 982.142857142858}"
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]
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},
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"metadata": {},
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@ -476,12 +490,71 @@
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},
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{
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"cell_type": "code",
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"execution_count": null,
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"execution_count": 39,
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"id": "736b23d6-b2b3-43fe-bc16-5b1e5ae5c07b",
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"metadata": {},
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"outputs": [
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{
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"data": {
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"text/latex": [
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"$\\displaystyle -997.420634920635$"
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],
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"text/plain": [
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"-997.420634920635"
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]
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},
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"execution_count": 39,
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"metadata": {},
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"output_type": "execute_result"
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}
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],
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"source": [
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"T = lambda x: (-q_ppprime*x**2/2 + soln[C_1]*x + soln[C_2])/k\n",
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"T(0.2/12)"
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]
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},
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{
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"cell_type": "markdown",
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"id": "38a944a7-43bd-425d-a5d9-b502f5fc1246",
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"metadata": {},
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"source": [
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"Just trying some stuff"
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]
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},
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{
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"cell_type": "code",
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"execution_count": 35,
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"id": "7af1e713-fc89-4679-99aa-f2bbacfb0c8e",
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"metadata": {},
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"outputs": [],
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"source": [
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"T = lambda x: -q_ppprime*x**2/2 + soln[C_1]*x + "
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"x = sm.symbols('x')\n",
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"T = sm.Function('T')(x)\n",
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"gov_eq = sm.Eq(0,T.diff(x)+q_ppprime)"
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]
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},
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{
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"cell_type": "code",
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"execution_count": 36,
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"id": "5edad66f-d4c8-4b51-8c92-343f83f65944",
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"metadata": {},
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"outputs": [
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{
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"data": {
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"text/latex": [
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"$\\displaystyle T{\\left(x \\right)} = C_{1} - 50000000.0 x$"
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],
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"text/plain": [
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"Eq(T(x), C1 - 50000000.0*x)"
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]
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},
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"execution_count": 36,
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"metadata": {},
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"output_type": "execute_result"
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}
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],
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"source": [
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"sm.dsolve(gov_eq)"
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]
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},
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{
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"## Part C"
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]
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},
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{
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"cell_type": "code",
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"execution_count": 24,
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"id": "ca0c2b96-652e-4f67-81b0-f98aecfb1707",
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"metadata": {},
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"outputs": [
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{
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"name": "stdout",
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"output_type": "stream",
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"text": [
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"\n",
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"=========FINAL ANSWER=========\n",
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"3C:\n",
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"At the left face: -5550 F \n",
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"At the right face:-10411 F\n",
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"=========FINAL ANSWER=========\n",
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"\n"
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]
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}
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],
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"source": [
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"answer_print('3C',f'At the left face: {T(x_left):.0f} F \\nAt the right face:{T(x_right):.0f} F')"
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]
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},
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{
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"cell_type": "markdown",
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"id": "ea13e251-33e4-4578-9ba1-27c31caff0b7",
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