The-volume-of-a-sphere-is-increasing-at-the-constant-rate-of-10cm-3-sec-Calculate-the-rate-of-increase-of-the-surface-area-at-the-instant-when-the-radius-is-5cm-What-is-the-radius-of-the-sphere-whe

Question Number 141776 by 7770 last updated on 23/May/21

$$\boldsymbol{\mathrm{The}}\:\boldsymbol{\mathrm{volume}}\:\boldsymbol{\mathrm{of}}\:\boldsymbol{\mathrm{a}}\:\boldsymbol{\mathrm{sphere}}\:\boldsymbol{\mathrm{is}}\:\boldsymbol{\mathrm{increasing}} \\ $$$$\boldsymbol{\mathrm{at}}\:\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{constant}}\:\boldsymbol{\mathrm{rate}}\:\boldsymbol{\mathrm{of}}\:\mathrm{10}\boldsymbol{\mathrm{cm}}^{\mathrm{3}} /\boldsymbol{\mathrm{sec}}. \\ $$$$\boldsymbol{\mathrm{Calculate}}\:\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{rate}}\:\boldsymbol{\mathrm{of}}\:\boldsymbol{\mathrm{increase}}\:\boldsymbol{\mathrm{of}}\:\boldsymbol{\mathrm{the}}\: \\ $$$$\boldsymbol{\mathrm{surface}}\:\boldsymbol{\mathrm{area}}\:\boldsymbol{\mathrm{at}}\:\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{instant}}\:\boldsymbol{\mathrm{when}}\:\boldsymbol{\mathrm{the}} \\ $$$$\boldsymbol{\mathrm{radius}}\:\boldsymbol{\mathrm{is}}\:\mathrm{5}\boldsymbol{\mathrm{cm}}.\boldsymbol{\mathrm{What}}\:\boldsymbol{\mathrm{is}}\:\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{radius}}\:\boldsymbol{\mathrm{of}}\:\boldsymbol{\mathrm{the}} \\ $$$$\boldsymbol{\mathrm{sphere}}\:\boldsymbol{\mathrm{when}}\:\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{surface}}\:\boldsymbol{\mathrm{area}}\:\boldsymbol{\mathrm{is}}\:\boldsymbol{\mathrm{increasing}} \\ $$$$\boldsymbol{\mathrm{at}}\:\mathrm{2}\boldsymbol{\mathrm{cm}}^{\mathrm{2}} /\boldsymbol{\mathrm{sec}}. \\ $$$$\boldsymbol{\mathrm{Any}}\:\boldsymbol{\mathrm{step}}\:\boldsymbol{\mathrm{by}}\:\boldsymbol{\mathrm{step}}\:\boldsymbol{\mathrm{solution}}\:\boldsymbol{\mathrm{pls}}. \\ $$

Answered by physicstutes last updated on 23/May/21

V = (4/3)πr^3 , (dV/dt) = 4πr^2 (dr/dt) 10 = 4π(5)^2 (dr/dt) ⇒ (dr/dt) = ((10)/(25(4π))) A = 4πr^2 ⇒ (dA/dt) = 8π(dr/dt)

$$\:{V}\:=\:\frac{\mathrm{4}}{\mathrm{3}}\pi{r}^{\mathrm{3}} \:,\:\:\:\frac{{dV}}{{dt}}\:=\:\mathrm{4}\pi{r}^{\mathrm{2}} \frac{{dr}}{{dt}} \\ $$$$\mathrm{10}\:=\:\mathrm{4}\pi\left(\mathrm{5}\right)^{\mathrm{2}} \frac{{dr}}{{dt}}\:\Rightarrow\:\frac{{dr}}{{dt}}\:=\:\frac{\mathrm{10}}{\mathrm{25}\left(\mathrm{4}\pi\right)} \\ $$$${A}\:=\:\mathrm{4}\pi{r}^{\mathrm{2}} \:\Rightarrow\:\frac{{dA}}{{dt}}\:=\:\mathrm{8}\pi\frac{{dr}}{{dt}} \\ $$

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