Rate of Change

Question

A viscous liquid is poured onto a flat surface. It forms a circular patch which grows at a steady rate of 4cm^2/s. Find, in terms of pi. 

 

(a) The radius of the patch 16 seconds after pouring has commenced.

(b) The rate of increase of the radius at this instant.

Answer 1

 A = pi•r²

dA/dt = 2(pi)r(dr/dt)

But the rate of increase of area is given as 4 cm²/sec, so

4 = 2(pi)r(dr/dt)

2 = pi(r) (dr/dt)

2 dt = pi(r)(dr)

 

Antidifferention gives us 2t = pi•r²/2 so t = ¼(pi)r²

 

When t=16, 16 = ¼(pi)r²; taking square root, we get 4 = ½(r)√pi

r = 8/√pi which is the answer for question (a).

 

Substitute that r value into one of the earlier equations to obtain the value of dr/dt.

 

ALSO, please check my work carefully, as the chief difficulty here is typing. Check every step!