A viscous liquid is poured onto a flat surface it forms a circular patch whose area grows at a steady rate of 5cm^2s^-1 find in term of pi (a) the radius of patch 20 sec after pouring has commenced (b) the rate of increase of the radius at this instant?

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Answer 1 · 2018-03-09T20:30:06

$[a]$ $[a]$ ...... $r = \frac{\sqrt{100}}{π}$ $r=sqrt100/pi$ cm, #[b] $..........$ [dr]/[dt] $=$ 1/[4sqrt[pi^3]]cmsec^-1#

We are given that $[dA]/[dt]=[5cm^2]/sec$ ............ $[1]$ [the rate of change of the area with respect to time t.

But area of circle A = $pir^2$ , so differentiating this with respect to time, implicitly.

$[dA]/[dt]=[2pir][dr]/[dt]$ so from what we know from.......... $[1]$

$5=[2pir][dr]/[dt]$ and thus, $[dr]/[dt]=[5]/[2pir]$ .............. $[2]$

After 20 sec the area will be $[20][5]=100cm^2$ [since the area is increasing at a constant rate].

When area A = $100=pir^2$ , $r=sqrt[100/pi]$ and substituting this vale for $r$ in .......... $[2]$

$dr/dt$ = $5/[2pisqrt[100/pi]$ = $5/[ 20pi/sqrtpi]$ = $1/[4pi^[3/2]$ = $1/[4sqrt[pi^3]$ .

Hope this was helpful.