$$A(t) = {t_0} + \\left( {{A_0} – {t_0}} \\right){e^{kt}},\\,\\,\\,t \\in \\left[ {0, + \\infty } \\right[$$ em que ${t_0}$ representa a temperatura ambiente, ${{A_0}}$ a temperatura de aquecimento (em \u00baC) e $k$ \u00e9 uma constante negativa<\/p><\/blockquote>\n
\u00ad<\/p>\n
\n- De acordo com os dados, ${t_0} = 20$ e ${A_0} = 230$.
\nLogo: $$\\begin{array}{*{20}{l}}
\n{A(t)}& = &{20 + \\left( {230 – 20} \\right){e^{kt}}} \\\\
\n{}& = &{20 + 210{e^{kt}}}
\n\\end{array}$$
\n\u00ad<\/li>\n - Dado que passados 5 minutos, a piza se encontra \u00e0 temperatura de 150\u00ba C, ser\u00e1 ${A(5) = 150}$.
\nLogo:\u00a0$$\\begin{array}{*{20}{l}}
\n{A(5) = 150}& \\Leftrightarrow &{20 + 210{e^{5k}} = 150} \\\\
\n{}& \\Leftrightarrow &{{e^{5k}} = \\frac{{130}}{{210}}} \\\\
\n{}& \\Leftrightarrow &{5k = \\ln \\frac{{13}}{{21}}} \\\\
\n{}& \\Leftrightarrow &{k = \\frac{1}{5}\\ln \\frac{{13}}{{21}}} \\\\
\n{}& \\Leftrightarrow &{k = \\ln \\sqrt[5]{{\\frac{{13}}{{21}}}}}
\n\\end{array}$$
\nPortanto, $k = \\ln \\sqrt[5]{{\\frac{{13}}{{21}}}} \\approx\u00a0 – 0,096$.
\n\u00ad<\/li>\n - Considerando $k =\u00a0 – 0,096$, vem $A(t) = 20 + 210{e^{ – 0,096t}}$.
\nOra, $$\\begin{array}{*{20}{l}}
\n{A(t) = 60}& \\Leftrightarrow &{20 + 210{e^{ – 0,096t}} = 60} \\\\
\n{}& \\Leftrightarrow &{{e^{^{ – 0,096t}}} = \\frac{{40}}{{210}}} \\\\
\n{}& \\Leftrightarrow &{ – 0,096t = \\ln \\frac{4}{{21}}} \\\\
\n{}& \\Leftrightarrow &{t =\u00a0 – \\frac{1}{{0,096}}\\ln \\frac{4}{{21}}}
\n\\end{array}$$
\nComo $t =\u00a0 – \\frac{1}{{0,096}}\\ln \\frac{4}{{21}} \\approx 17,273$ e $0,273 \\times 60 = 16,38$, decorre 17 minutos e 16 segundos, aproximadamente, entre o instante em que a piza \u00e9 retirada do forno e o instante em que a sua temperatura \u00e9 60\u00ba C.
\n\u00ad<\/li>\n - Como\u00a0\\[A’\\left( t \\right) = {\\left( {20 + 210{e^{ – 0,096t}}} \\right)^\\prime } =\u00a0 – 0,096 \\times 210{e^{ – 0,096t}} =\u00a0 – 20,16{e^{ – 0,096t}}\\]
\nent\u00e3o $A'(t) < 0,\\forall t \\in \\mathbb{R}_0^ + $, isto \u00e9, a fun\u00e7\u00e3o $A$ \u00e9 estritamente decrescente no seu dom\u00ednio. Consequentemente, a taxa m\u00e9dia de varia\u00e7\u00e3o da fun\u00e7\u00e3o $A$ \u00e9 negativa em qualquer intervalo do seu dom\u00ednio, pois $$tm{v_{\\left[ {{x_0},{x_0} + h} \\right]}} = \\frac{{A({x_0} + h) – A({x_0})}}{h} < 0,\\forall {x_0} \\in \\mathbb{R}_0^ + $$ com $h > 0$.
\n\u00ad<\/li>\n<\/ol>\n