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Exercise 7.10 · Q1

Q.The volume of a sphere is increasing in volume at the rate of 3pi,textcm3/textsec3\\pi\\,\\text{cm}^3/\\text{sec}. The rate of change of its radius when radius is dfrac12\\dfrac12 cm

(1) 3 cm/s
(2) 2 cm/s
(3) 1 cm/s
(4) 12\dfrac12 cm/s
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✓ Free question

Differentiate the volume formula and substitute the given rate and radius.

Step 1. Differentiate. V=43πr3⇒dVdt=4πr2drdtV=\dfrac43\pi r^3\Rightarrow\dfrac{dV}{dt}=4\pi r^2\dfrac{dr}{dt}.

Step 2. Substitute r=12r=\dfrac12, dVdt=3π\dfrac{dV}{dt}=3\pi.

3π=4π(14)drdt=πdrdt ⇒ drdt=3.3\pi=4\pi\left(\frac14\right)\frac{dr}{dt}=\pi\frac{dr}{dt}\ \Rightarrow\ \frac{dr}{dt}=3.

✓Final answer

(1) 3 cm/s

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