1. ## 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

2. Hello, i_zz_y_ill!

The radius of a circular ink blot is increasing at a rate inversely poportional to its area $A.$
Find an expression for $\frac{dA}{dt}$ . . . . In terms of what? . ${\color{blue}r? \hdots A? \hdots \text{both?}}$

Answer: . $\frac{k'}{\sqrt{A}}$
A strange way to write the answer . . .
Since $k$ is a constant, $k'$ is zero.
So I must assume $k'$ means "some other constant" . . . a sloppy answer!

We are told: . $\frac{dr}{dt} \:=\:\frac{k}{A}$ .[1]

Since $A \:=\:\pi r^2$, we have: . $r \:=\:\frac{A^{\frac{1}{2}}}{\sqrt{\pi}} \quad\Rightarrow\quad \frac{dr}{dt} \:=\:\frac{1}{2\sqrt{\pi}A^{\frac{1}{2}}}\cdot\fra c{dA}{dt}$ .[2]

Substitute [2] into [1]: . $\frac{1}{2\sqrt{\pi}{A^{\frac{1}{2}}}}\cdot\frac{d A}{dt} \:=\:\frac{k}{A} \quad\Rightarrow\quad \frac{dA}{dt} \;=\;\frac{2\sqrt{\pi}A^{\frac{1}{2}}}{1}\cdot\fra c{k}{A}$

Therefore: . $\frac{dA}{dt} \;=\;\frac{2k\sqrt{\pi}}{\sqrt{A}}$