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Um inseto se move ao longo da curva \(y=\sqrt {x}\) e sua altitude está aumentando à taxa de
\(4ft\)/s. Aqui, \(x\) e \(y\) são ambos medidos em pés. Com que rapidez a distância do
inseto até a origem está variando no instante em que o inseto está a \(\sqrt {12}\) pés da
origem?
ETAPA 1
Introduza uma variável \(D\) que denote a distância do inseto até a origem. Esboce e rotule uma figura.
ETAPA 2
Complete as afirmações a seguir.
A taxa dada é \[ \dd t \class{xm-answer}{\cssId{xm-answer-1}{\phantom{\rule{7em}{1.5em}}}}= 4 ft/s \]
A taxa a ser determinada é \[ \dd t \class{xm-answer}{\cssId{xm-answer-2}{\phantom{\rule{7em}{1.5em}}}} \text { , quando } D=\sqrt {12} ft \]
Escreva uma equação que relacione as variáveis \(x\), \(y\) e \(D\).
\[ D^2=\class{xm-answer}{\cssId{xm-answer-3}{\phantom{\rule{7em}{1.5em}}}} \]
ETAPA 3
Derive.
\[ 2 D\cdot \dd t{\class{xm-answer}{\cssId{xm-answer-4}{\phantom{\rule{7em}{1.5em}}}}}=\class{xm-answer}{\cssId{xm-answer-5}{\phantom{\rule{7em}{1.5em}}}}\cdot x\cdot \dd t{\class{xm-answer}{\cssId{xm-answer-6}{\phantom{\rule{7em}{1.5em}}}}}+2\cdot \class{xm-answer}{\cssId{xm-answer-7}{\phantom{\rule{7em}{1.5em}}}}\cdot \dd t{\class{xm-answer}{\cssId{xm-answer-8}{\phantom{\rule{7em}{1.5em}}}}} \]
Simplifique.
\[ D\cdot \dd t{\class{xm-answer}{\cssId{xm-answer-9}{\phantom{\rule{7em}{1.5em}}}}}=x\cdot \dd t{\class{xm-answer}{\cssId{xm-answer-10}{\phantom{\rule{7em}{1.5em}}}}}+\class{xm-answer}{\cssId{xm-answer-11}{\phantom{\rule{7em}{1.5em}}}}\cdot \dd t{\class{xm-answer}{\cssId{xm-answer-12}{\phantom{\rule{7em}{1.5em}}}}} \]
ETAPA 4
Calcule. Para calcular \(\left [\dd t{D}\right ]_{D=\sqrt {12}}\), precisamos calcular
xmhint\[ D^2=x^2+y^2 \]
Portanto, quando \(D=\sqrt {12}\)
\[ 12=x^2+x \]
Resolva a equação quadrática
\[ x^2+x-12=0 \]
\[ \left [x\right ]_{D=\sqrt {12}}=\class{xm-answer}{\cssId{xm-answer-13}{\phantom{\rule{7em}{1.5em}}}} ft \]
\[ \left [y\right ]_{D=\sqrt {12}}=\class{xm-answer}{\cssId{xm-answer-14}{\phantom{\rule{7em}{1.5em}}}} ft \]
\begin{align} \left [\dd t{x}\right ]_{D=\sqrt {12}}&=\left [\dd t{y^2}\right ]_{D=\sqrt {12}}\\ &=\left [2y\cdot \dd t{y}\right ]_{D=\sqrt {12}}\\ &=\class{xm-answer}{\cssId{xm-answer-15}{\phantom{\rule{7em}{1.5em}}}}\text { ft/s}\\ \end{align}
Portanto,
\[ \sqrt {12}\cdot \left [\dd t{D}\right ]_{D=\sqrt {12}}=3\cdot 8\sqrt {3}+\sqrt {3}\cdot {\class{xm-answer}{\cssId{xm-answer-16}{\phantom{\rule{7em}{1.5em}}}}} \]
e
\[ \left [\dd t{D}\right ]_{D=\sqrt {12}}= \class{xm-answer}{\cssId{xm-answer-17}{\phantom{\rule{7em}{1.5em}}}} \text { ft/s} \]
![[Picture]](/rendered/appliedRelatedRates/exercises/relatedRatesApplied77-7eefa0bae6a1c08b0e6ee1d4387d127b.svg)