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The volume of a spherical balloon is decreasing at a rate of $20 cm^3/min$. How fast is the radius of the balloon decreasing when the volume is $1m^3$?

Basically I did this:

$$V=\frac 43\pi r^3$$

When the volume is $1m^3$, $r=\sqrt[3] {3000000 cm^3 \over 4\pi}$

Then I implicitly derived the volume equation and got:

$${dV \over dt} = 4\pi r^2 {dr\over dt}$$

Then I replaced

$$ 20 {cm^3 \over min} = 4\pi \sqrt[2/3] {3000000 cm^3 \over 4\pi} {dr \over dt}$$

And solving for ${dr \over dt}$ I finally got approximately $0.54 cm/min$

I'm not sure if it's right, or not. And I'm not sure if the final answer should have a minus sign since the radius -and the volume obviously- is decreasing.

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your only problem I see is squaring $r$, as for the negative sign that depends on your preference you can have it positive and say it is decreasing at $x$ centimeters a minute. – Deven Ware Nov 13 '12 at 17:34
up vote 1 down vote accepted

Your approach is fine, but when you insert $r$ it should be $\frac 23$ power or $\frac 32$ root. You should indicate the radius is decreasing, either with a minus sign or in the text.

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Ups! My bad, so then the answer should be $0.000413 cm/min$ approximately, and not 0.54 as I stated before. – ChairOTP Nov 13 '12 at 17:39
But you multiplied by $\pi$ instead of dividing. Look – ChairOTP Nov 13 '12 at 18:09
@ChairOTP: correct. I agree with your figure (though it seems to round to 0.000414) – Ross Millikan Nov 13 '12 at 18:12

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