The radius of a spherical balloon is increasing by 5 cm/sec. At what rate is air being blown into the balloon at the moment when the radius is 13 cm?

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The radius of a spherical balloon is increasing by 5 cm/sec. At what rate is air being blown into the balloon at the moment when the radius is 13 cm?

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Answer 1 · 2015-04-09T18:42:48

This is a Related Rates (of change) problem.

The rate at which air is being blown in will be measured in volume per unit of time. That is a rate of change of volume with respect to time. The rate at which air is being blown in is the same as the rate at which the volume of the balloon is increasing.

$V = \frac{4}{3} π r^{3}$ $V=4/3 pi r^3$

We know $\frac{d r}{d t} = 5 cm/sec$ $(dr)/(dt) = 5" cm/sec"$ . We want $\frac{d V}{d t}$ $(dV)/(dt)$ when $r = 13 cm$ $r=13" cm"$ .

Differentiate $V = \frac{4}{3} π r^{3}$ $V=4/3 pi r^3$ implicitly with respect to $t$ $t$

$\frac{d}{d t} (V) = \frac{d}{d t} (\frac{4}{3} π r^{3})$ $d/(dt)(V)=d/(dt)(4/3 pi r^3)$

$\frac{d V}{d t} = \frac{4}{3} π \cdot 3 r^{2} \frac{d r}{d t} = 4 π r^{2} \frac{d r}{d t}$ $(dV)/(dt)=4/3 pi *3r^2 (dr)/(dt)=4 pi r^2 (dr)/(dt)$

Plug in what you know and solve for what you don't know.

$\frac{d V}{d t} = 4 π {(13 cm)}^{2} (5 cm/sec) = 20 \cdot 169 \cdot π {cm}^{3} /sec$ $(dV)/(dt)=4 pi (13 " cm")^2 (5 " cm/sec") = 20*169* pi " cm"^3"/sec"$

The air is being blown in at a rate of $3380 π {cm}^{3} /sec$ $3380 pi " cm"^3"/sec"$ .

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The radius of a spherical balloon is increasing by 5 cm/sec. At what rate is air being blown into the balloon at the moment when the radius is 13 cm?

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