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Find the rate of change of the area of a circle per second with respect to its radius when radius=5cm.

Source

$A= \pi r^2$$\frac{{\rm d}A}{{\rm d}r} =2rπ$

So when $r=5$cm, $\frac{{\rm d}A}{{\rm d}r}= 10 \pi$ cm

But the answer is $10\pi$ cm$^2$/sec. I don't understand how time comes into picture when we are only talking about radius and area? Why is "per second" even mentioned in the question?

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  • $\begingroup$ You are correct that this question does not make sense on itself, you need additional information. Probably whoever asked that question imagined a circle growing at a certain speed, i.e. $\frac{\partial r}{\partial t} = v$, and knowing $v$ you can solve the problem. $\endgroup$
    – Wouter
    Commented Mar 3, 2016 at 13:02

2 Answers 2

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Find the rate of change of the area of a circle per second when its radius of 5 cm is expanding at rate of 1 cm per second.

... such a question would be better

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You need to find the rate of change of the area which is $\frac{dA}{dt}$ and not $\frac{dA}{dr}$.

$$\frac{dA}{dt}=\frac{d}{dt}(\pi r(t)^2)=2\pi r(t) \cdot \frac{dr(t)}{dt} $$

The "dimension" of $\frac{d}{dt}$ is $s^{-1}$ and you will get the correct units, but note that you need to know how the radius changes in time, i.e. $r=r(t)$.

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  • $\begingroup$ The question says "rate of change of the area of a circle per second with respect to its radius": it seems like three variables are dependent on each other. How am I even supposed to interpret it? Why it is dA/dt and not dA/dr $\endgroup$
    – bluebellae
    Commented Mar 3, 2016 at 12:48
  • $\begingroup$ dA/dr is the rate of change of the area with respect to its radius, that's fine. Either there is something wrong with the question, or we are missing something trivial, but I cannot see how you can get the units cm/s unless you differentiate A with respect to time. $\endgroup$
    – eyedropper
    Commented Mar 3, 2016 at 12:58
  • $\begingroup$ Sorry i wrote the units wrong.; the answer is 10π cm^2/sec $\endgroup$
    – bluebellae
    Commented Mar 3, 2016 at 13:01
  • $\begingroup$ yes, yes, that is what I meant, since r has units of cm and dr/dt has units of cm/s, their product has units of cm^2/s. $\endgroup$
    – eyedropper
    Commented Mar 3, 2016 at 13:13

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