# The radius of a balloon is increasing at the rate of 10 cm/sec. At what rate is the surface area of the balloon increasing when the radius is 15 cm?

Last updated date: 18th Mar 2023

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Answer

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Hint: Here, we need to know the formula for finding the surface area of balloon = $4\pi {r^2}$ sq. units

Let $r$ be the radius and $S$ be the surface area of the balloon at any time t. Then,

$S = 4\pi {r^2}$ ... (1)

It is given that the radius of the balloon is increasing at a rate of 10 cm/sec

$ \Rightarrow \frac{{dr}}{{dt}} = 10cm/\sec $

As the radius of the balloon changes the surface area of the balloon also changes.

That means rate of change in surface area is (Differentiating equation (1) w.r.t time)

$ \Rightarrow \frac{{dS}}{{dt}} = 8\pi r\frac{{dr}}{{dt}}$

Substituting the value of $\frac{{dr}}{{dt}}$

$ \Rightarrow \frac{{dS}}{{dt}} = 80\pi r$

Finding the rate of change in surface area of when $r = 15cm$

${\left( {\frac{{dS}}{{dt}}} \right)_{r = 15}} = 80\pi (15) = 1200\pi $ $c{m^2}/\sec $

$\therefore $ Hence surface area of balloon is increasing at a rate of $1200\pi $ $c{m^2}/\sec $, when radius of the balloon is 15cm.

Note: The surface area of a balloon is $4\pi {r^2}$ as same as the surface area of the sphere, where $r$ is the radius of the balloon. The radius of the balloon is increasing with some rate means the amount of air inside the balloon is increasing with time which in turn increases the physical size of the balloon. This results in increasing surface area of the balloon with respect to the radius.

Let $r$ be the radius and $S$ be the surface area of the balloon at any time t. Then,

$S = 4\pi {r^2}$ ... (1)

It is given that the radius of the balloon is increasing at a rate of 10 cm/sec

$ \Rightarrow \frac{{dr}}{{dt}} = 10cm/\sec $

As the radius of the balloon changes the surface area of the balloon also changes.

That means rate of change in surface area is (Differentiating equation (1) w.r.t time)

$ \Rightarrow \frac{{dS}}{{dt}} = 8\pi r\frac{{dr}}{{dt}}$

Substituting the value of $\frac{{dr}}{{dt}}$

$ \Rightarrow \frac{{dS}}{{dt}} = 80\pi r$

Finding the rate of change in surface area of when $r = 15cm$

${\left( {\frac{{dS}}{{dt}}} \right)_{r = 15}} = 80\pi (15) = 1200\pi $ $c{m^2}/\sec $

$\therefore $ Hence surface area of balloon is increasing at a rate of $1200\pi $ $c{m^2}/\sec $, when radius of the balloon is 15cm.

Note: The surface area of a balloon is $4\pi {r^2}$ as same as the surface area of the sphere, where $r$ is the radius of the balloon. The radius of the balloon is increasing with some rate means the amount of air inside the balloon is increasing with time which in turn increases the physical size of the balloon. This results in increasing surface area of the balloon with respect to the radius.

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