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]#......#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.