Corrosion is attacking the inside of a water tank. Today a 2cm x 2cm size patch is measured. We know the corrosion will grow at rate of doubling size every 5 days. What will its size be in sq/cms be in 25 days?
(Edit) Knowing it is 64cm by 64cm there is another question which is If anti corrosion pain was use, we can slow the growth to be doubling every 20 days. In this case how long will it take the patch to cover an area of 64sq/cm
Will it be 80 days?
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I am trying a differential equation based approach for solution, though I am not sure about it. So it is better someone correct me if I am wrong.
Let's say grow rate is:
$$\frac{dx}{dt}=kx$$
Therefore:
$$x=Ae^{kt}$$
$t=0\:\Longrightarrow\:x=2$ thus we find $A=2$.
$t=5\:\Longrightarrow\:x=4$ thus we find $k=\frac{ln2}{5}$
If $t=25$:
$$x=2e^{\ln2^5}=64\:cm$$
If doubling rate is 20 days:
$$x=2e^{\frac{\ln2}{20}t}$$
For $t=80$:
$$x=2e^{\ln2^4}=32\;cm$$
Every five days it doubles
25 days go
This means it doubles $\frac{25}5=5$ times
It starts as $2$ on each side
Let's double the value $2$, and do that $5$ times
$$2\cdot2\cdot2\cdot2\cdot2\cdot2=64$$
The final size is then 64 cm by 64 cm