Question

1. Sabendo que o raio de uma bola esférica está crescendo a uma taxa de 2cm/min, determine:
a) A que taxa a área da superfície da bola está crescendo quando o diâmetro for 16cm?
b) Quão rápido o volume está aumentando quando o diâmetro for 12 cm?

Answer 1

A = 4πr^2

dA/dt = 8πr dr/dt. Mas: dr/dt = 2cm/min

dA/dt = 16πr

dA/dt (para r = 8cm) = 16π × 8 = 128π cm^s/min

V = 4/3 × (πr^3)

dV/dt = 4πr^2 dr/dt

dV/dt = 8πr^2 . Para r = 6cm. dV/dt = 8π×6^2

dV/dt = 288π cm^3/min

Answer 2

$\large\boxed{\begin{array}{l}\rm \dfrac{dr}{dt}=2~cm/min\\\\\sf a)~\rm A=4\pi r^2\\\rm \dfrac{dA}{dt}= 4\cdot2\pi r\dfrac{dr}{dt}\\\rm mas~2\pi r=D\\\rm substituindo\,temos:\\\rm\dfrac{dA}{dt}=4\cdot16\cdot2\\\\\rm \dfrac{dA}{dt}=128~cm^2/min\end{array}}$

$\large\boxed{\begin{array}{l}\sf b)~\rm V=\dfrac{4}{3}\cdot\pi\cdot r^3\\\\\rm\dfrac{dV}{dt}=\dfrac{4}{\backslash\!\!\!3}\cdot\pi\cdot \backslash\!\!\!3r^2\cdot\dfrac{dr}{dt}\\\\\rm\dfrac{dV}{dt}= 4\pi r^2\dfrac{dr}{dt}\\\\\rm D=12\,cm\longrightarrow r=\dfrac{12}{2}=6\,cm\\\\\rm \dfrac{dV}{dt}=4\pi\cdot 6^2\cdot2\\\\\rm\dfrac{dV}{dt}=288\pi cm^3/min\end{array}}$