Uma rolha flutua num lago, movendo-se para cima e para baixo.<\/p>\n
A dist\u00e2ncia $d(t)$ do fundo do lago ao centro da rolha no instante $t \\geqslant 0$ \u00e9 dada por $$d(t) = \\cos \\left( {\\pi t} \\right) + 12$$ com $d(t)$ expresso em metros e $t$ em segundos.<\/p><\/blockquote>\n
\u00ad<\/p>\n
\n- A dist\u00e2ncia da rolha ao fundo do lago igual a 11,5 m para: $$\\begin{array}{*{20}{l}}
\n{d(t) = 11,5}& \\Leftrightarrow &{\\cos \\left( {\\pi t} \\right) + 12 = 11,5 \\wedge t \\geqslant 0} \\\\
\n{}& \\Leftrightarrow &{\\cos \\left( {\\pi t} \\right) =\u00a0 – 0,5 \\wedge t \\geqslant 0} \\\\
\n{}& \\Leftrightarrow &{\\pi t =\u00a0 \\pm \\frac{{2\\pi }}{3} + 2k\\pi ,k \\in \\mathbb{Z} \\wedge t \\geqslant 0} \\\\
\n{}& \\Leftrightarrow &{t =\u00a0 \\pm \\frac{2}{3} + 2k,k \\in \\mathbb{Z} \\wedge t \\geqslant 0} \\\\
\n{}& \\Leftrightarrow &{\\left( {t = \\frac{2}{3} + 2k \\vee t = \\frac{4}{3} + 2k,k \\in \\mathbb{Z}} \\right) \\wedge t \\geqslant 0} \\\\
\n{}& \\Leftrightarrow &{t = \\frac{2}{3} + 2k \\vee t = \\frac{4}{3} + 2k,k \\in \\mathbb{Z}_0^ + }
\n\\end{array}$$
\n\u00ad<\/li>\n - A dist\u00e2ncia m\u00ednima da rolha ao fundo do lago \u00e9 ${d_{m\u00edn}} =\u00a0 – 1 + 12 = 11$ metros e a dist\u00e2ncia m\u00e1xima \u00e9 ${d_{m\u00e1x}} = 1 + 12 = 13$ metros. Portanto, $11 \\leqslant d \\leqslant 13$, em metros.
\n\u00ad<\/li>\n - Como $$\\begin{array}{*{20}{l}}
\n{d(t + T) = d(t)}& \\Leftrightarrow &{\\cos \\left( {\\pi \\left( {t + T} \\right)} \\right) + 12 = \\cos \\left( {\\pi t} \\right) + 12} \\\\
\n{}& \\Leftrightarrow &{\\cos \\left( {\\pi t + \\pi T} \\right) = \\cos \\left( {\\pi t} \\right)} \\\\
\n{}& \\Leftrightarrow &{\\pi T = 2k\\pi ,k \\in \\mathbb{Z}} \\\\
\n{}& \\Leftrightarrow &{T = 2k,k \\in \\mathbb{Z}}
\n\\end{array}$$ o per\u00edodo positivo m\u00ednimo \u00e9 $T = 2$ segundos.
\n\u00ad<\/li>\n - Como $$\\begin{array}{*{20}{l}}
\n{d’(t)}& = &{\\left( {\\cos \\left( {\\pi t} \\right) + 12} \\right)’} \\\\
\n{}& = &{ – \\pi \\operatorname{sen} \\left( {\\pi t} \\right),\\forall t \\geqslant 0}
\n\\end{array}$$
\na velocidade da rolha nos instantes considerados \u00e9, respetivamente, $$d’\\left( {\\frac{7}{6}} \\right) =\u00a0 – \\pi \\operatorname{sen} \\left( {\\frac{7}{6}\\pi } \\right) =\u00a0 – \\pi\u00a0 \\times \\left( { – \\frac{1}{2}} \\right) = \\frac{\\pi }{2}$$ e $$d’\\left( {\\frac{{17}}{3}} \\right) =\u00a0 – \\pi \\operatorname{sen} \\left( {\\frac{{17}}{3}\\pi } \\right) =\u00a0 – \\pi \\operatorname{sen} \\left( {\\frac{5}{3}\\pi } \\right) =\u00a0 – \\pi\u00a0 \\times \\left( { – \\frac{{\\sqrt 3 }}{2}} \\right) = \\frac{{\\sqrt 3 }}{2}\\pi $$ metros por segundo.
\n\u00ad<\/li>\n - A rolha sobe nos intervalos onde a fun\u00e7\u00e3o $d$ \u00e9 crescente, isto \u00e9, nos intervalos onde $d’$ \u00e9 positiva: $$\\begin{array}{*{20}{l}}
\n{d’(t) > 0}& \\Leftrightarrow &{ – \\pi \\operatorname{sen} \\left( {\\pi t} \\right) > 0 \\wedge t \\geqslant 0} \\\\
\n{}& \\Leftrightarrow &{\\operatorname{sen} \\left( {\\pi t} \\right) < 0 \\wedge t \\geqslant 0} \\\\
\n{}& \\Leftrightarrow &{\\pi\u00a0 + 2k\\pi\u00a0 < \\pi t < 2\\pi\u00a0 + 2k\\pi ,k \\in \\mathbb{Z} \\wedge t \\geqslant 0} \\\\
\n{}& \\Leftrightarrow &{1 + 2k < t < 2 + 2k,k \\in \\mathbb{Z} \\wedge t \\geqslant 0} \\\\
\n{}& \\Leftrightarrow &{t \\in \\left] {1 + 2k,2 + 2k} \\right[,k \\in \\mathbb{Z}_0^ + }
\n\\end{array}$$
\n\u00ad<\/li>\n<\/ol>\n