Thread: simple differentiation??????!a level

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

Hello, i_zz_y_ill!

The radius of a circular ink blot is increasing at a rate inversely poportional to its area
Find an expression for . . . . In terms of what? .

Answer: .

A strange way to write the answer . . .
Since is a constant, is zero.
So I must assume means "some other constant" . . . a sloppy answer!

We are told: . .[1]

Since , we have: . .[2]


Substitute [2] into [1]: .

Therefore: .