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Um cone circular reto está crescendo. Enquanto cresce, sua altura permanece
igual ao seu diâmetro. O diâmetro está crescendo a uma taxa constante de \(2 \frac {\textrm {unidades}}{\textrm {s}}\).
A que taxa seu volume está crescendo no instante em que o diâmetro é
\(6 \textrm {unidades}\)?
SOLUÇÃO:
Seja \(V\) o volume, \(r\) o raio, \(D\) o diâmetro e \(h\) a altura do cone.
A taxa dada é
\[ \dd [\class{xm-answer}{\cssId{xm-answer-1}{\phantom{\rule{7em}{1.5em}}}}]{t}= 2 \frac {\textrm {units}}{\textrm {s}}. \]
A taxa a ser determinada é
\[ \left [ \dd [\class{xm-answer}{\cssId{xm-answer-2}{\phantom{\rule{7em}{1.5em}}}}]{t}\right ]_{D=6} \]
xmhintSabemos que \begin{align*} D &= 2r\\ V &= \frac {1}{3} \pi r^2 h\\ h &= D.\\ \end{align*}
xmhintEmbora pudéssemos derivar sem pensar, e tudo daria certo, vale a
pena fazer um pouco de álgebra antes, já que só nos interessa a relação
entre o volume e o diâmetro. Um pouco de manipulação fornece \[ V = \frac {1}{3} \pi \left (\frac {D}{2}\right )^2D. \] Assim,
\[ V = \frac {1}{12} \pi D^3. \]
xmhintDerivando em relação ao tempo, obtemos \[ \dd [V]{t} = \frac {1}{4} \pi D^2 \dd [D]{t}. \]
xmhintNo instante de interesse, temos \[ \Bigl [ \dd [V]{t} \Bigr ]_{D=6}= \frac {1}{4} \pi (6)^2 (2) = 18\pi \frac {\textrm {unidades}^3}{\textrm {s}}. \]
\[ \Bigl [ \dd [V]{t} \Bigr ]_{D=6} = \class{xm-answer}{\cssId{xm-answer-3}{\phantom{\rule{7em}{1.5em}}}}\frac {\textrm {units}^3}{\textrm {s}} \]
![[Picture]](/rendered/moreThanOneRate/exercises/relatedRatesSimple5-6847a81f21e430a35014a4ffd9f6bb9c.svg)