Question

A soap bubbles has a radius of 2cm if its radius increases by 1cm find by how much does it volume increase
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Answer 1

$Radius \: of \: a \: soap \: bubble \\ is \: \: r$

$\frac{dr}{dt} = 2$

$\frac{dr}{dt} = \frac{4}{3} \pi \frac{d}{dt} ( {r)}^{3} \\ \\ = \frac{4}{3} \pi ( {3r)}^{3} \frac{dr}{dt} \\ \\ = 4\pi \: {r}^{2} \frac{dr}{dt}$

$When \: \: r = 2 \: cm$

$\frac{dv}{dt} = 4\pi \: {(2)}^{2} (1) \\ \\ = 16 \times 1\pi \\ \\ = 16\pi \: {cm}^{3}$

Answer 2

$\mathsf {Radius \: of \: a \: soap \: bubble\: is \: r}$

$\mathsf {\dfrac{dr}{dt} = 2 }$

$\mathsf { \dfrac{dr}{dt} = \dfrac{4}{3} \pi \frac{d}{dt} ( {r)}^{3}}$

$\mathsf { = \dfrac{4}{3} \pi ( {3r)}^{3} \dfrac{dr}{dt}}$

$\mathsf { = 4\pi \: {r}^{2} \dfrac{dr}{dt} }$

$\mathsf {When \: r = 2 \: cm}$

$\mathsf {\dfrac{dv}{dt} = 4\pi \: {(2)}^{2} (1)}$

$\mathsf {= 16 \times 1\pi}$

$\mathsf {= 16\pi \: {cm}^{3}}$