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Sorry if this question is easy because I'm quite new to this concept (learned today). Could someone please explain how do I calculate the rate of increase of the radius in this case?

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  • $\begingroup$ Welcome to MathSE. When you pose a question on this site, you should include your attempt at solving the problem and explain where you are stuck so that you receive responses that address the specific difficulties you are encountering. This tutorial explains how to typeset mathematics on this site. $\endgroup$ May 7 '18 at 14:50
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    $\begingroup$ $ A = \pi r^2,\, dA/dt= 2 \pi r dr/dt \, $ can you compare and proceed? $\endgroup$
    – Narasimham
    May 7 '18 at 17:47
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Start with area formula for a circle: $$A=\pi r^2$$ And you differentiate both sides with respect to time(now imagine area and radius are functions of time!), $$\frac{dA}{dt}=2\pi r\frac{dr}{dt}$$ In the problem you are given $\frac{dA}{dt}=2\pi, r=6$ with proper units, so plug them into the equation above to get $\frac{dr}{dt}$, namely, the rate of increase of radius with respect to time.

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  • $\begingroup$ We are given that the rate of increase of the area is $2\pi~\frac{\text{cm}^2}{\text{s}}$. $\endgroup$ May 7 '18 at 14:52
  • $\begingroup$ Oh sorry, typo! :) $\endgroup$
    – Macrophage
    May 7 '18 at 14:52
  • $\begingroup$ Thank you so much! $\endgroup$ May 8 '18 at 6:13
  • $\begingroup$ You are welcome! $\endgroup$
    – Macrophage
    May 8 '18 at 7:04
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The area $A$ of a disk of radius $r$ is given by $A(t)=\pi r^2(t)$, so the rate of increase of the area for a disk of radius $r$ is (by definition) $$A'(t)=\frac{dA}{dt}=2 \pi r(t)\frac{dr}{dt}.$$ Plugging in $A'(t)=2\pi$ cm$^2$s$^{-1}$ directly gives you $dr/dt=1/r(t)=1/6$ cm/s $\simeq0.16666$ cm/s.

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  • $\begingroup$ You are calculating the rate of increase of area with respect to radius, not what the OP wanted. $\endgroup$
    – Macrophage
    May 7 '18 at 14:52

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