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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?

JTanNeedHelp
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    Perhaps I'm missing something, but I don't think this question has anything to do with square roots (referring to your chosen title and tag). Square centimeters is just the unit of area here. – amWhy Sep 16 '14 at 12:14
  • You're told how much corrosion there is now. Can you see that it tells you how much there will be in 5 days from now? Then in 10 days? Then.... – Gerry Myerson Sep 16 '14 at 12:16
  • Does the area double in 5 days or the linear dimension double in 5 days? In the edit you assume the linear dimension doubles. – Ross Millikan Sep 16 '14 at 13:39

2 Answers2

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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

Alice Ryhl
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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 $t=80$:

$$x=2e^{\ln2^4}=32\;cm$$