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Volume of sphere, $V = \dfrac43 \pi r^3$
Surface area of sphere $S = 4 \pi r^2$
If we know, $\dfrac{dV}{dt} = R$

Let us consider both volume and area as composite functions, thus $$\dfrac{dV}{dt} = \dfrac{dV}{dr} \times \dfrac{dr}{dt} = 4 \pi r^2 \times \dfrac{dr}{dt} = R$$ whence $\dfrac{dr}{dt} = \dfrac {R}{4 \pi r^2}$ since $\dfrac{dS}{dt} = 6 \pi r \dfrac{dr}{dt}$, let the value of the $\dfrac{dr}{dt}$ into the second equation, to get the answer. Is this approach logically correct? Thank you.

Ankit Saha
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Versteher
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1 Answers1

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Everything is fine until you did

whence $\dfrac{dr}{dt} = \dfrac {R}{4 \pi r^2}$ since $\dfrac{dS}{dt} = 6 \pi r \dfrac{dr}{dt}$

$$\frac{dS}{dt}=8 \pi r \frac{dr}{dt}$$ therefore $$\frac{dS}{dt}=8 \pi r \frac {R}{4 \pi r^2}$$ and finally $$\frac{dS}{dt}=\frac{2R}{r}$$

Raffaele
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  • thank You so much. Thus, logically that approach (making the volume a composite functions, and finding thence dr/dt and putting it to the area formula) had a sound foundation, or? – Versteher Jan 26 '21 at 13:27
  • by the way, how can I draw those nice formulae, in order not to bother Moderators with converting my writs to a proper form? – Versteher Jan 26 '21 at 13:28
  • Press Edit for your post and start studying how they formatted your question – Raffaele Jan 26 '21 at 16:33