The radius of a circular ink black is increasing at a rate inversely poportional to its area A. find an expression for dA/dt.

k so A=(pie)r^2,

A'=2(pie)r dA/dt=dA/dr x dr/dt dr/dt proportionl to 1/A

dr/dt = k/A

dr/dt = k/(pie)r^2

therefore: dA/dt = 2(pie)r x k/(pie)r^2

=2k/r (A=(pie)r^2 IMPLIES r=a/(pie) squarerooted

=2k x squarroot of pie/squareroot of a

NOW, sorry aboout bad writing, i have answers and i cant get from here to

,,,,,,,,,,,,,,,,,,,,,,,,ans: k'/squareroot of a

as in the differential of k,if anyone could spare a moment to engage their brains i bit better than myself that wold be gr8.thnx