{ "cells": [ { "cell_type": "code", "execution_count": 1, "metadata": {}, "outputs": [], "source": [ "%display latex # Mostrar os resultados renderizados (latex)" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "## Solução dos exercícios" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "1. Gaussiana\n", "\n", " a. Verifique que $\\displaystyle\\int\\limits_{-\\infty}^\\infty e^{-\\frac{1}{2}x^2}dx = \\sqrt{2\\pi}$\n", " \n", " b. Defina a variável simbólica \"$a$\". Assuma $a>0$ e determine $\\displaystyle\\int\\limits_{-\\infty}^\\infty e^{-\\frac{1}{2}ax^2}dx$" ] }, { "cell_type": "code", "execution_count": 2, "metadata": {}, "outputs": [ { "data": { "text/html": [ " $\\newcommand{\\Bold}[1]{\\mathbf{#1}}\\sqrt{2} \\sqrt{\\pi}$ " ], "text/latex": [ "\\begin{math}\n", "\\newcommand{\\Bold}[1]{\\mathbf{#1}}\\sqrt{2} \\sqrt{\\pi}\n", "\\end{math}" ], "text/plain": [ "sqrt(2)*sqrt(pi)" ] }, "execution_count": 2, "metadata": {}, "output_type": "execute_result" } ], "source": [ "integrate(exp(-1/2*x^2),x,-oo,oo)" ] }, { "cell_type": "code", "execution_count": 3, "metadata": {}, "outputs": [], "source": [ "var('a')\n", "assume(a>0)" ] }, { "cell_type": "code", "execution_count": 4, "metadata": {}, "outputs": [ { "data": { "text/html": [ " $\\newcommand{\\Bold}[1]{\\mathbf{#1}}\\frac{2 \\, \\sqrt{\\frac{1}{2}} \\sqrt{\\pi}}{\\sqrt{a}}$ " ], "text/latex": [ "\\begin{math}\n", "\\newcommand{\\Bold}[1]{\\mathbf{#1}}\\frac{2 \\, \\sqrt{\\frac{1}{2}} \\sqrt{\\pi}}{\\sqrt{a}}\n", "\\end{math}" ], "text/plain": [ "2*sqrt(1/2)*sqrt(pi)/sqrt(a)" ] }, "execution_count": 4, "metadata": {}, "output_type": "execute_result" } ], "source": [ "integrate(exp(-1/2*a*x^2),x,-oo,oo)" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "3. Oscilador amortecido\n", "\n", " a. Resolva a equação do oscilador amortecido $$m\\ddot{y}(t) +b\\dot{y}(t)= -ky(t).$$\n", "\n", "Para isso você precisará:\n", " - Definir as variáveis simbólicas e funções involvidas (ver guia de referencia rápida);\n", " - Estabelecer a equação diferencial;\n", " - Assuma $4mk-b^2>0$;\n", " - Resolver a equação diferencial usando o `desolve`." ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "A sintáxe do comando é `desolve(equação, dvar, ivar=t)`, onde `dvar` é a variável dependente e `ivar=t` indica que a variável independente e $t$. Use a ajuda `desolve?` se necessário." ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "#### Solução" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "*Definir as variáveis simbólicas e funções involvidas (ver guia de referencia rápida);*" ] }, { "cell_type": "code", "execution_count": 6, "metadata": {}, "outputs": [], "source": [ "var('m,b,k,t')\n", "y = function('y')(t)" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "*Estabelecer a equação diferencial;*" ] }, { "cell_type": "code", "execution_count": 8, "metadata": {}, "outputs": [ { "data": { "text/html": [ " $\\newcommand{\\Bold}[1]{\\mathbf{#1}}b \\frac{\\partial}{\\partial t}y\\left(t\\right) + m \\frac{\\partial^{2}}{(\\partial t)^{2}}y\\left(t\\right) = -k y\\left(t\\right)$ " ], "text/latex": [ "\\begin{math}\n", "\\newcommand{\\Bold}[1]{\\mathbf{#1}}b \\frac{\\partial}{\\partial t}y\\left(t\\right) + m \\frac{\\partial^{2}}{(\\partial t)^{2}}y\\left(t\\right) = -k y\\left(t\\right)\n", "\\end{math}" ], "text/plain": [ "b*diff(y(t), t) + m*diff(y(t), t, t) == -k*y(t)" ] }, "execution_count": 8, "metadata": {}, "output_type": "execute_result" } ], "source": [ "eq = m*diff(y,t,2)+b*diff(y,t) == -k*y\n", "eq" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "*Assuma $4mk-b^2>0$;*" ] }, { "cell_type": "code", "execution_count": 9, "metadata": {}, "outputs": [], "source": [ "assume(4*m*k-b^2>0)" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "*Resolver a equação diferencial usando o `desolve`.*" ] }, { "cell_type": "code", "execution_count": 10, "metadata": {}, "outputs": [ { "data": { "text/html": [ " $\\newcommand{\\Bold}[1]{\\mathbf{#1}}{\\left(K_{2} \\cos\\left(\\frac{1}{2} \\, t \\sqrt{-\\frac{b^{2}}{m^{2}} + \\frac{4 \\, k}{m}}\\right) + K_{1} \\sin\\left(\\frac{1}{2} \\, t \\sqrt{-\\frac{b^{2}}{m^{2}} + \\frac{4 \\, k}{m}}\\right)\\right)} e^{\\left(-\\frac{b t}{2 \\, m}\\right)}$ " ], "text/latex": [ "\\begin{math}\n", "\\newcommand{\\Bold}[1]{\\mathbf{#1}}{\\left(K_{2} \\cos\\left(\\frac{1}{2} \\, t \\sqrt{-\\frac{b^{2}}{m^{2}} + \\frac{4 \\, k}{m}}\\right) + K_{1} \\sin\\left(\\frac{1}{2} \\, t \\sqrt{-\\frac{b^{2}}{m^{2}} + \\frac{4 \\, k}{m}}\\right)\\right)} e^{\\left(-\\frac{b t}{2 \\, m}\\right)}\n", "\\end{math}" ], "text/plain": [ "(_K2*cos(1/2*t*sqrt(-b^2/m^2 + 4*k/m)) + _K1*sin(1/2*t*sqrt(-b^2/m^2 + 4*k/m)))*e^(-1/2*b*t/m)" ] }, "execution_count": 10, "metadata": {}, "output_type": "execute_result" } ], "source": [ "desolve(eq,y,ivar=t)" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "b. Suponha que um sistema mecânico seja governado por pela equação diferencial $$\\ddot{y}(t) +\\frac{1}{5}\\dot{y}(t)= -\\frac{1}{4}y(t).$$\n", "Encontre a solução assumindo as condições iniciais $$y(0) = \\frac{1}{2}, \\dot{y}(0) = \\frac{7}{4}.$$" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "Para incluir as condições iniciais, use `desolve(equação, dvar, ics=[t_0, y(t_0), y'(t_0)])`, onde `t_0, y(t_0), y'(t_0)` são as tais condições." ] }, { "cell_type": "code", "execution_count": 11, "metadata": {}, "outputs": [ { "data": { "text/html": [ " $\\newcommand{\\Bold}[1]{\\mathbf{#1}}\\frac{1}{5} \\, \\frac{\\partial}{\\partial t}y\\left(t\\right) + \\frac{\\partial^{2}}{(\\partial t)^{2}}y\\left(t\\right) = -\\frac{1}{4} \\, y\\left(t\\right)$ " ], "text/latex": [ "\\begin{math}\n", "\\newcommand{\\Bold}[1]{\\mathbf{#1}}\\frac{1}{5} \\, \\frac{\\partial}{\\partial t}y\\left(t\\right) + \\frac{\\partial^{2}}{(\\partial t)^{2}}y\\left(t\\right) = -\\frac{1}{4} \\, y\\left(t\\right)\n", "\\end{math}" ], "text/plain": [ "1/5*diff(y(t), t) + diff(y(t), t, t) == -1/4*y(t)" ] }, "execution_count": 11, "metadata": {}, "output_type": "execute_result" } ], "source": [ "eq2 = diff(y,t,2)+1/5*diff(y,t) == -1/4*y; eq2" ] }, { "cell_type": "code", "execution_count": 15, "metadata": {}, "outputs": [ { "data": { "text/html": [ " $\\newcommand{\\Bold}[1]{\\mathbf{#1}}\\frac{1}{2} \\, {\\left(3 \\, \\sqrt{6} \\sin\\left(\\frac{1}{5} \\, \\sqrt{6} t\\right) + \\cos\\left(\\frac{1}{5} \\, \\sqrt{6} t\\right)\\right)} e^{\\left(-\\frac{1}{10} \\, t\\right)}$ " ], "text/latex": [ "\\begin{math}\n", "\\newcommand{\\Bold}[1]{\\mathbf{#1}}\\frac{1}{2} \\, {\\left(3 \\, \\sqrt{6} \\sin\\left(\\frac{1}{5} \\, \\sqrt{6} t\\right) + \\cos\\left(\\frac{1}{5} \\, \\sqrt{6} t\\right)\\right)} e^{\\left(-\\frac{1}{10} \\, t\\right)}\n", "\\end{math}" ], "text/plain": [ "1/2*(3*sqrt(6)*sin(1/5*sqrt(6)*t) + cos(1/5*sqrt(6)*t))*e^(-1/10*t)" ] }, "execution_count": 15, "metadata": {}, "output_type": "execute_result" } ], "source": [ "sol = desolve(eq2,y,ivar=t,ics=[0,1/2,7/4]); sol" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "c. Plote um gráfico da solução encontrada no intervalo $05*sqrt(6)*t))*e^(-1/10*t)\n", "t Expression t\n", "y Expression y(t)\n" ] } ], "source": [ "whos" ] }, { "cell_type": "code", "execution_count": null, "metadata": {}, "outputs": [], "source": [] } ], "metadata": { "kernelspec": { "display_name": "SageMath 9.2", "language": "sage", "name": "sagemath" }, "language_info": { "codemirror_mode": { "name": "ipython", "version": 3 }, "file_extension": ".py", "mimetype": "text/x-python", "name": "python", "nbconvert_exporter": "python", "pygments_lexer": "ipython3", "version": "3.8.5" } }, "nbformat": 4, "nbformat_minor": 4 }