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A spherical balloon remains as spherical as it is slowly inflated. The radius (r) and the time (t) satisfies the equation;

$$\frac{dr}{dt} =\frac{K}{r^2}$$

When $r=3,$ the radius id increasing at $$\frac{5}{36}\pi$$, so $$K=\frac{5}{4}\pi$$

I have the general solution in explicit form;

$$r=\sqrt[3]{3Kt+C}$$
When r=1 and t=0 I have the particular solution;
$$r=\sqrt[3]{3Kt+1}$$
I now need to calculate the time at which r=6, to 3 significant figures,

$$6=\sqrt[3]{3Kt+1}$$
$$6^3 =3Kt+1$$
$$\frac{215}{3K} =t$$
However know this is wrong as haven't got it to 3 s.f.Should I have plugged in the original value of $K$ for when $r=3$? meaning $$t=516\pi$$

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    Is small $k$ the same as $K$? – 2017-02-22
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    Sorry, yes it is. – 2017-02-22
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    yes you should plug in the value for $K$ since this assumed to be constant for the duration of this problem. If it was a function of $r$ then you would have a different problem (a harder one than this ;) ) – 2017-02-24

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