\n| <\/td>\n | <\/td>\n | Note que o modelo considerado n\u00e3o \u00e9
\nestritamente crescente em\u00a0$\\mathbb{R}_0^ + $.<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/li>\n\\[tm{v_{\\left[ {0,1} \\right]}} = \\frac{{d\\left( 1 \\right) – d\\left( 0 \\right)}}{{1 – 0}} = \\frac{{\\left( {1,3 \\times {1^2} – 1 + 2} \\right) – 2}}{1} = 0,3\\]
\nNo 1.\u00ba segundo de movimento, a bola desceu o plano inclinado com velocidade m\u00e9dia\u00a0$0,3$ cm\/s.<\/p>\n<\/li>\nPara ${h > 0}$, vem:
\n\\[\\begin{array}{*{20}{l}}
\n{\\mathop {tm{v_{\\left[ {2,2 + h} \\right]}}}\\limits_{h > 0} }& = &{\\frac{{1,3{{\\left( {2 + h} \\right)}^2} – \\left( {2 + h} \\right) + 2 – \\left( {1,3 \\times {2^2} – 2 + 2} \\right)}}{{2 + h – 2}}} \\\\
\n{}& = &{\\frac{{1,3 \\times {2^2} + 1,3 \\times 4h + 1,3{h^2} – 2 – h + 2 – 1,3 \\times {2^2}}}{h}} \\\\
\n{}& = &{\\frac{{1,3{h^2} + 4,2h}}{h}} \\\\
\n{}& = &{1,3h + 4,2}
\n\\end{array}\\]<\/p>\nPara ${h < 0}$, vem:
\n\\[\\begin{array}{*{20}{l}}
\n{\\mathop {tm{v_{\\left[ {2 + h,2} \\right]}}}\\limits_{h < 0} }& = &{\\frac{{\\left( {1,3 \\times {2^2} – 2 + 2} \\right) – \\left[ {1,3{{\\left( {2 + h} \\right)}^2} – \\left( {2 + h} \\right) + 2} \\right]}}{{2 – h – 2}}} \\\\
\n{}& = &{\\frac{{ – 1,3{h^2} – 4,2h}}{{ – h}}} \\\\
\n{}& = &{1,3h + 4,2}
\n\\end{array}\\]<\/p>\n Ora,\u00a0se $h \\to {0^ + }$, ent\u00e3o $\\mathop {tm{v_{\\left[ {2,2 + h} \\right]}}}\\limits_{h > 0}\u00a0 = 1,3h + 4,2 \\to 4,2$ e se $h \\to {0^ – }$, ent\u00e3o $\\mathop {tm{v_{\\left[ {2 + h,2} \\right]}}}\\limits_{h < 0}\u00a0 = 1,3h + 4,2 \\to 4,2$.<\/p>\n ![\"Gr\u00e1fico\"](\"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2014\/02\/64-g3.jpg\") <\/a>Portanto, a taxa de varia\u00e7\u00e3o da fun\u00e7\u00e3o $d$, para $t = 2$ \u00e9 $4,2$.<\/p>\nAssim, no instante $t = 2$ s a velocidade da bola \u00e9 $4,2$ cm\/s.<\/p>\n Tamb\u00e9m se pode obter esse valor na calculadora gr\u00e1fica, ativando a op\u00e7\u00e3o\u00a0“Derivative”, conforme se ilustra na imagem ao lado.<\/p>\n<\/li>\n Seja ${t_0} \\in {D_d}$ e calculemos $d’\\left( {{t_0}} \\right)$:
\n\\[\\begin{array}{*{20}{l}}
\n{d’\\left( {{t_0}} \\right)}& = &{\\mathop {\\lim }\\limits_{t \\to 0} \\frac{{d\\left( {{t_0} + h} \\right) – d\\left( {{t_0}} \\right)}}{h}} \\\\
\n{}& = &{\\mathop {\\lim }\\limits_{t \\to 0} \\frac{{1,3{{\\left( {{t_0} + h} \\right)}^2} – \\left( {{t_0} + h} \\right) + 2 – \\left( {1,3 \\times {t_0}^2 – {t_0} + 2} \\right)}}{h}} \\\\
\n{}& = &{\\mathop {\\lim }\\limits_{t \\to 0} \\frac{{1,3{t_0}^2 + 2,6{t_0} \\times h + 1,3{h^2} – {t_0} – h + 2 – 1,3 \\times {t_0}^2 + {t_0} – 2}}{h}} \\\\
\n{}& = &{\\mathop {\\lim }\\limits_{t \\to 0} \\frac{{2,6{t_0} \\times h + 1,3{h^2} – h}}{h}} \\\\
\n{}& = &{\\mathop {\\lim }\\limits_{t \\to 0} \\left( {2,6{t_0} + 1,3h – 1} \\right)} \\\\
\n{}& = &{2,6{t_0} – 1}
\n\\end{array}\\]<\/p>\nComo
\n\\[\\begin{array}{*{20}{l}}
\n{d’\\left( {{t_0}} \\right) = 30}& \\Leftrightarrow &{2,6{t_0} – 1 = 30} \\\\
\n{}& \\Leftrightarrow &{{t_0} = \\frac{{31}}{{2,6}}} \\\\
\n{}& \\Leftrightarrow &{{t_0} = \\frac{{155}}{{13}}} \\\\
\n{}&{}&{{t_0} = 11,92}
\n\\end{array}\\]
\nent\u00e3o a bola tem uma velocidade de $30$ cm\/s\u00a0no instante $t = 11,9$ s, aproximadamente.<\/p>\n Tamb\u00e9m se pode obter um valor aproximado desse instante recorrendo \u00e0 calculadora gr\u00e1fica:<\/p>\n
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![\"Plano](\"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2014\/02\/11-2pag-64-5.gif\")
<\/a>Uma bola desce um plano inclinado, onde foi espalhado um gel que dificulta o movimento.<\/p>\n