Question

A spherical soap bubble expanding so that its radius is increasing at rate 0.02 cm/sec. At what rate is the surface area increasing when its radius is 5 cm ?
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Answer 1

Answer:

Let r be the radius

S be the surface area of Soap bubbles at time,t

Then S = 4πr²

Differentiating w.r.t. t, we get

$\frac{dS}{dt} = 4\pi \times 2r \frac{dr}{dt} \\ \therefore\frac{dS}{dt} = 8\pi r \frac{dr}{dt} \: \: \: .......1$

$now \frac{dr}{dt} = \frac{0.02cm}{sec} \\ \: r = 5cm$

$\therefore(1) \: gives \: \frac{dS}{dt} = 8\pi(5)(0.02) \\ = 0.8\pi$

Hence,the surface area of the Soap bubble is increasing at the rate 0.8πcm²/sec

Answer 2

Step-by-step explanation:

Let r be the radius

S be the surface area of Soap bubbles at time,t

Then S = 4πr²

Differentiating w.r.t. t, we get

\begin{gathered}\frac{dS}{dt} = 4\pi \times 2r \frac{dr}{dt} \\ \therefore\frac{dS}{dt} = 8\pi r \frac{dr}{dt} \: \: \: .......1\end{gathered}dtdS=4π×2rdtdr∴dtdS=8πrdtdr.......1

\begin{gathered}now \frac{dr}{dt} = \frac{0.02cm}{sec} \\ \: r = 5cm\end{gathered}nowdtdr=sec0.02cmr=5cm

\begin{gathered}\therefore(1) \: gives \: \frac{dS}{dt} = 8\pi(5)(0.02) \\ = 0.8\pi\end{gathered}∴(1)givesdtdS=8π(5)(0.02)=0.8π

Hence,the surface area of the Soap bubble is increasing at the rate 0.8πcm²/sec