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Uma escada de \(12\) pés está apoiada em uma parede. No instante em que o pé da
escada está a \(3\) pés da parede, o pé da escada está se afastando da parede a
uma taxa de \(5 \frac {\textrm {pés}}{\textrm {s}}\). A que taxa o topo da escada está descendo pela parede nesse
instante?
xmhintIntroduzimos as variáveis \(x\), a distância do pé da escada à parede, e \(y\), a
distância do topo da escada ao chão. A taxa dada é \(\frac {dx}{dt}=5\) pés/s, e a taxa desconhecida é
\(\left [\frac {dy}{dt}\right ]_{x=3}\).
xmhint
\(x\) e \(y\) devem satisfazer a relação
\[ x^2 + y^2 = 12^2 \]
pelo teorema de Pitágoras.
xmhint
Derivamos em relação ao tempo e obtemos
\[ 2x(t)x'(t)+2y(t)y'(t) = 0. \]
xmhintCalculamos todas as quantidades no instante em que \(x=3\). Primeiro, temos que
encontrar o valor de \(y\) nesse instante. \begin{align*} 3^2 +y^2 &= 12^2\\ y &= \sqrt {144-9}\\ y &= \sqrt {135}. \end{align*}
xmhint
\begin{align*} 2(3)(5)+2\sqrt {135}\left [\frac {dy}{dt}\right ]_{x=3} &=0\\ \left [\frac {dy}{dt}\right ]_{x=3} = -\frac {15}{\sqrt {135}}\frac {\textrm {ft}}{\textrm {s}}. \end{align*}
Assim, o topo da escada está descendo a uma taxa de \(\frac {15}{\sqrt {135}}\) pés por segundo.
O topo da escada está descendo a uma taxa de \(\class{xm-answer}{\cssId{xm-answer-1}{\phantom{\rule{7em}{1.5em}}}}\) pés por segundo.
![[Picture]](/rendered/appliedRelatedRates/exercises/relatedRatesApplied1-4fb33346e198092603630c0941d7fc65.svg)