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Seja \(s(t) = t^3-2t^2\). O gráfico de \(s\) é dado abaixo.
Os pontos \(A\) e \(B\) são dois pontos no gráfico de \(s\). \[ A=\left (1,\class{xm-answer}{\cssId{xm-answer-1}{\phantom{\rule{7em}{1.5em}}}}\right )\quad B=\left (2,\class{xm-answer}{\cssId{xm-answer-2}{\phantom{\rule{7em}{1.5em}}}}\right ) \]
A inclinação da reta secante que passa pelos pontos \(A\) e \(B\) é \[ m_{\text {sec}}=\class{xm-answer}{\cssId{xm-answer-3}{\phantom{\rule{7em}{1.5em}}}}. \]
A reta tangente ao gráfico de \(s\) no ponto \(A\) é mostrada acima. Observando essa reta
com atenção, podemos aproximar sua inclinação por \[ m_{\text {tan}}\approx \class{xm-answer}{\cssId{xm-answer-4}{\phantom{\rule{7em}{1.5em}}}}. \]
Em termos da função \(s\), o limite que teríamos de calcular para obter a inclinação da
reta tangente a \(s\) em \(t=1\) é \[ m_{\text {tan}} = \lim _{t\to 1}\frac {s(\class{xm-answer}{\cssId{xm-answer-5}{\phantom{\rule{7em}{1.5em}}}})-s(1)}{\class{xm-answer}{\cssId{xm-answer-6}{\phantom{\rule{7em}{1.5em}}}}}. \] Quando substituímos a fórmula de \(s\), obtemos o seguinte
limite, escrito em termos de \(t\): \[ m_{\text {tan}} = \lim _{t\to 1}\frac {\class{xm-answer}{\cssId{xm-answer-7}{\phantom{\rule{7em}{1.5em}}}}}{\class{xm-answer}{\cssId{xm-answer-8}{\phantom{\rule{7em}{1.5em}}}}}. \]
Suponha que a função \(s(t) = t^3-2t^2\) represente a posição, no instante \(t\), de um objeto que se move
ao longo de uma reta, em que \(s\) é medida em pés e \(t\) é medido em segundos. A
velocidade média do objeto no intervalo de tempo \([1,2]\) é \[ v_{\text {méd}} = \class{xm-answer}{\cssId{xm-answer-9}{\phantom{\rule{7em}{1.5em}}}} ft/s. \]
Complete a tabela a seguir com as velocidades médias apropriadas (em \(ft/s\)).
\[ \begin{array}{|c|c|} \hline \text {Time interval} & \text {Average velocity} \\\hline \hline [1,2] & \class{xm-answer}{\cssId{xm-answer-10}{\phantom{\rule{7em}{1.5em}}}} \\\hline [1,1.5] & \class{xm-answer}{\cssId{xm-answer-11}{\phantom{\rule{7em}{1.5em}}}} \\\hline [1,1.1] & \class{xm-answer}{\cssId{xm-answer-12}{\phantom{\rule{7em}{1.5em}}}} \\\hline [0.5,1] & \class{xm-answer}{\cssId{xm-answer-13}{\phantom{\rule{7em}{1.5em}}}} \\\hline [0.9,1] & \class{xm-answer}{\cssId{xm-answer-14}{\phantom{\rule{7em}{1.5em}}}} \\\hline \end{array} \]
Com base nos valores acima, uma conjectura razoável para a velocidade instantânea
do objeto em \(t=1\) é \[ v_{\text {inst}}\approx {\class{xm-answer}{\cssId{xm-answer-15}{\phantom{\rule{7em}{1.5em}}}}} ft/s. \]
Em termos da função \(s\), o limite que teríamos de calcular para obter a velocidade
instantânea do objeto em \(t=1\) é \[ v_{\text {inst}} = \lim _{t\to 1}\frac {s(\class{xm-answer}{\cssId{xm-answer-16}{\phantom{\rule{7em}{1.5em}}}})-s(1)}{\class{xm-answer}{\cssId{xm-answer-17}{\phantom{\rule{7em}{1.5em}}}}}. \] Observe que essa é exatamente a mesma expressão que
precisaríamos calcular para obter a inclinação da reta tangente ao gráfico de \(s(t)\) em
\(t=1\).
![[Picture]](/rendered/definitionOfTheDerivative/exercises/tangentLineGraphical1-3bb333f07a3a61b519c6d3d9d2c7135e.svg)
![[Picture]](/rendered/definitionOfTheDerivative/exercises/tangentLineGraphical1-1d4b1563ceeeb491435f88683dfe2bc2.svg)