Question

A circle is growing so that the radius is increasing at the rate of 2cm/min. How fast is the area of the circle changing at the instant the radius is 10cm? Include units in your answer.

Answer 1

$\bf \textit{area of a circle}\\\\ A=\pi r^2\qquad \qquad \implies \cfrac{dA}{dt}=\pi \cdot 2r\cdot \cfrac{dr}{dt}\impliedby chain-rule \\\\\\ \begin{cases} r=10\\ \frac{dr}{dt}=2 \end{cases}\qquad \cfrac{dA}{dt}=\pi \cdot 2(10)\cdot 2\implies \cfrac{dA}{dt}=40\pi~\frac{cm}{min}$

Answer 2

The answer to this question is:

A circle is growing so that the radius is increasing at the rate of 2cm/min. How fast is the area of the circle changing at the instant the radius is 10cm? Include units in your answer.?
✔️I assume here the linear scale is changing at the rato of 5cm/min
✔️dR/dt=5(cm/min) (R - is the radius.... yrs, of the circle (not the side)
✔️The rate of area change would be d(pi*R^2)/dt=2pi*R*dR/dt. 
✔️At the instant when R=20cm,this rate would be, 
✔️2pi*20*5(cm^2/min)=200pi (cm^2/min)  or, almost, 628 (cm^2/min) 

Hoped This Helped, Cello10
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