Question

Ar está sendo bombeado dentro de um cilindro cujo volume V cresce a uma taxa de 60\pi \textrm{cm}^3/\textrm{seg}. Analise e responda nas duas situações a seguir:

I) Se a altura h=60 \ \textrm{cm} permanecer fixa, quão rápido o raio r do cilindro está crescendo quando o diâmetro da base for 120 \ \textrm{cm}?

II) Se o raio r=60 \ \textrm{cm} permanecer fixo, quão rápido a altura h do cilindro está crescendo quando o diâmetro da base for 120 \ \textrm{cm}?

(Dica: o volume V de um cilindro de raio r e altura h é V=\pi r^2h).

Answer 1

$\large\boxed{\begin{array}{l}\sf Ar\,est\acute a\,sendo\,bombeado\,dentro\,de\,um\,cilindro\\\sf cujo\,volume\,V\,cresce\,a\,uma\,taxa\,de\,60\pi\,cm^3\!/s.\\\sf Analise\,e\,responda\,nas\,duas\,situac_{\!\!,}\tilde oes\,a\,seguir:\\\sf I)~Se\,a\,altura\,h=60\,cm\,permanecer\,fixa,qu\tilde ao\,r\acute apido\\\sf o\,raio\,r\,do\,cilindro\,est\acute a\,crescendo\,quando\,o\,di\hat ametro\\\sf da\,base\,for\,120\,cm?\end{array}}$

$\large\boxed{\begin{array}{l}\sf 2)~Se\,o\,raio\,r=60\,cm\,permanecer\,fixo,qu\tilde ao\,r\acute apido\\\sf a\,altura\,h\,do\,cilindro\,est\acute a\,crescendo\,quando\,o\,di\hat ametro\\\sf da\,base\,for\,120\,cm?\end{array}}$

$\large\boxed{\begin{array}{l}\underline{\sf soluc_{\!\!,}\tilde ao\!:}\\\rm 1)~\dfrac{dV}{dt}=60\pi\,cm^3\!/s~~d=120\,cm\longrightarrow r=60\,cm\\\\\rm V=\pi\cdot r^2\cdot h\\\rm \dfrac{dV}{dt}=2\pi rh\dfrac{dr}{dt}+\pi\cdot r^2\cdot\dfrac{dh}{dt}\\\\\rm 60\pi=2\pi\cdot 60\cdot 60\cdot\dfrac{dr}{dt}+\pi\cdot 60^2\cdot\dfrac{d}{dt}60\end{array}}$

$\large\boxed{\begin{array}{l}\rm\dfrac{dr}{dt}=\dfrac{\diagdown\!\!\!\!\!\!60\diagup\!\!\!\!\pi}{2\diagup\!\!\!\!\pi\cdot\diagdown\!\!\!\!\!\!60\cdot60}\\\\\rm\dfrac{dr}{dt}=\dfrac{1}{120}~cm/s\end{array}}$

$\large\boxed{\begin{array}{l}\rm 2)~\dfrac{dV}{dt}=60\pi\,cm^3/s~~\dfrac{dh}{dt}=?~~r=60~cm\\\\\rm V=\pi\cdot r^2\cdot h\\\rm V=2\pi r\dfrac{dr}{dt}\cdot h+\pi\cdot r^2\cdot\dfrac{dh}{dt}\\\\\rm 60\pi=120\pi\cdot\dfrac{d}{dt}60\cdot h+\pi\cdot 60^2\cdot\dfrac{dh}{dt}\\\\\rm 3600\pi\cdot\dfrac{dh}{dt}=60\pi\\\\\rm\dfrac{dh}{dt}=\dfrac{60\pi\div60}{3600\pi\div60}\\\\\rm\dfrac{dh}{dt}=\dfrac{1}{60}\,cm/s \end{array}}$