# The area of a circle increases at a rate of $2\pi~\text{cm}^2\text{s}^{-1}$. Calculate the rate of increase of the radius when the radius is $6$ cm. [closed]

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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• $A = \pi r^2,\, dA/dt= 2 \pi r dr/dt \,$ can you compare and proceed? May 7 '18 at 17:47

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.
• We are given that the rate of increase of the area is $2\pi~\frac{\text{cm}^2}{\text{s}}$. May 7 '18 at 14:52
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.