Num determinado ano (ano zero) havia, em certo parque natural, 318 \u00e1guias.<\/p>\n
Passado um ano, o n\u00famero de \u00e1guias era 417.<\/p>\n
Sabendo que o n\u00famero $P$ de \u00e1guias existentes nessa reserva, quando \u00e9 decorrido o tempo $t$, contado do in\u00edcio dos registos, \u00e9 dado por uma fun\u00e7\u00e3o do tipo $$P(t) = \\frac{a}{{1 + b{e^{ – t}}}}$$ com $t$ expresso em anos.<\/p><\/blockquote>\n
\u00ad<\/p>\n
\n- Ora, $$\\begin{array}{*{20}{l}}
\n{\\left\\{ {\\begin{array}{*{20}{l}}
\n{P(0) = 318} \\\\
\n{P(1) = 417}
\n\\end{array}} \\right.}& \\Leftrightarrow &{\\left\\{ {\\begin{array}{*{20}{l}}
\n{\\frac{a}{{1 + b}} = 318} \\\\
\n{\\frac{a}{{1 + \\frac{b}{e}}} = 417}
\n\\end{array}} \\right.}& \\Leftrightarrow &{\\left\\{ {\\begin{array}{*{20}{l}}
\n{318 + 318b = 417 + 417\\frac{b}{e}} \\\\
\n{\\frac{a}{{1 + b}} = 318}
\n\\end{array}} \\right.}& \\Leftrightarrow\u00a0 \\\\
\n{}& \\Leftrightarrow &{\\left\\{ {\\begin{array}{*{20}{l}}
\n{b = \\frac{{99}}{{318 – \\frac{{417}}{e}}}} \\\\
\n{\\frac{a}{{1 + b}} = 318}
\n\\end{array}} \\right.}& \\Leftrightarrow &{\\left\\{ {\\begin{array}{*{20}{l}}
\n{b = \\frac{{33e}}{{106e – 139}}} \\\\
\n{a = 318\\left( {1 + \\frac{{33e}}{{106e – 139}}} \\right)}
\n\\end{array}} \\right.}& \\Leftrightarrow\u00a0 \\\\
\n{}& \\Leftrightarrow &{\\left\\{ {\\begin{array}{*{20}{l}}
\n{b = \\frac{{33e}}{{106e – 139}}} \\\\
\n{a = \\frac{{44202\\left( {e – 1} \\right)}}{{106e – 139}}}
\n\\end{array}} \\right.}&{}&{}&{}
\n\\end{array}$$
\nLogo, $a = \\frac{{44202\\left( {e – 1} \\right)}}{{106e – 139}} \\approx 509$ e $b = \\frac{{33e}}{{106e – 139}} \\approx 0,6$.
\n\u00ad<\/li>\n - Considerando $a=509$ e $b=0,6$, vem:$$P(t) = \\frac{{509}}{{1 + 0,6{e^{ – t}}}}$$
\nDois anos antes da primeira contagem realizada existiam aproximadamente $P( – 2) = \\frac{{509}}{{1 + 0,6{e^2}}} \\approx 94$ \u00e1guias.
\n\u00ad<\/li>\n - Como\u00a0\\[\\begin{array}{*{20}{l}}
\n{P’\\left( t \\right)}& = &{{{\\left( {\\frac{{509}}{{1 + 0,6{e^{ – t}}}}} \\right)}^\\prime }} \\\\
\n{}& = &{\\frac{{0,6 \\times 509{e^{ – t}}}}{{{{\\left( {1 + 0,6{e^{ – t}}} \\right)}^2}}}} \\\\
\n{}& = &{\\frac{{305,4 \\times {e^{ – t}}}}{{{{\\left( {1 + 0,6{e^{ – t}}} \\right)}^2}}}}
\n\\end{array}\\] ent\u00e3o $$P'(2) = \\frac{{305,4 \\times {e^{ – 2}}}}{{{{\\left( {1 + 0,6{e^{ – 2}}} \\right)}^2}}} \\approx 35$$ e $$P'(4) = \\frac{{305,4 \\times {e^{ – 4}}}}{{{{\\left( {1 + 0,6{e^{ – 4}}} \\right)}^2}}} \\approx 5$$
\nA popula\u00e7\u00e3o das \u00e1guias est\u00e1 a crescer (a derivada \u00e9 positiva), mas o ritmo de crescimento \u00e9 cada vez menor.
\n\u00ad<\/li>\n - Nessas circunst\u00e2ncias, daqui a muitos, muitos anos<\/em> \u00e9 esperado que o n\u00famero de \u00e1guias seja 509, pois $$\\mathop {\\lim }\\limits_{t \\to\u00a0 + \\infty } P(t) = \\mathop {\\lim }\\limits_{t \\to\u00a0 + \\infty } \\frac{{509}}{{1 + 0,6{e^{ – t}}}} = \\frac{{509}}{{1 + 0,6 \\times \\mathop {\\lim }\\limits_{t \\to\u00a0 + \\infty } {e^{ – t}}}} = \\frac{{509}}{{1 + 0}} = 509$$<\/li>\n<\/ol>\n