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?

Question

4500 views around the world

You can reuse this answer
Creative Commons License

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.