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Exercise 2.1 · Q14

Q.A spherical soap bubble is expanding so that its radius is increasing at the rate of 0.020.02 cm/sec. At what rate is the surface area increasing, when its radius is 55 cm?

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Let rr be the radius and S=4πr2S=4\pi r^2 the surface area of the bubble. Differentiating w.r.t. time tt:

dSdt=8πrdrdt\dfrac{dS}{dt} = 8\pi r\dfrac{dr}{dt}

Given drdt=0.02\dfrac{dr}{dt}=0.02 cm/sec. At r=5r=5 cm: …

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