A stone is dropped into a still pond sends out a circular ripple who’s radius increases at a constant rate of 2 m/sec. How rapidly is the area enclosed by the ripple increasing at the end of 20 sec?

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A stone is dropped into a still pond sends out a circular ripple who’s radius increases at a constant rate of 2 m/sec. How rapidly is the area enclosed by the ripple increasing at the end of 20 sec?

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Answer 1 · 2018-03-06T20:09:33

$160 π$ $160pi$ mtrs^2 $/ sec$ $/sec$ .

Explanation:

Area of circle = $π r^{2}$ $pir^2$ ....... $A = π r^{2}$ $A=pir^2$ ...... $[1]$ $[1]$

Differentiating .... $[1]$ $[1]$ with respect to time, $\frac{d A}{d t} = 2 π r \frac{d r}{d t}$ $[dA]/[dt]=2pir[dr]/[dt]$ ............ $[2]$ $[2]$ . If the radius is increasing at a constant rate of $\frac{2 m}{sec}$ $[2m]/ [sec]$ then after 20 seconds radius = $40 m t r s$ $40 mtrs$ .

We know $\frac{d r}{d t}$ $[dr]/[dt$ = $\frac{2 m t r s}{sec}$ $[2mtrs]/[sec]$ and so $\frac{d A}{d t} = [2] π [40] [2]$ $[dA]/[dt]=[2]pi[40][2]$ = $160 π m^{2}$ $160pim^2$ sec

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A stone is dropped into a still pond sends out a circular ripple who’s radius increases at a constant rate of 2 m/sec. How rapidly is the area enclosed by the ripple increasing at the end of 20 sec?

1 Answer

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