NCEA Calculus 2016 TWO(d - Solution & Explanation

Question

A large spherical helium balloon is being inflated at a constant rate of

4800

cm

^3

s

^{-1}

.

At what rate is the radius of the balloon increasing when the volume of the balloon is

288\,000\pi

cm

^3

?

*You must use calculus and show any derivatives that you need to find when solving this
problem.*

Official Answer

\dfrac{dV}{dt}=4800\ \text{cm}^3\ \text{s}^{-1}

V=\dfrac{4}{3}\pi r^3

\dfrac{dV}{dr}=4\pi r^2

\dfrac{dr}{dt}=\dfrac{dr}{dV}\times\dfrac{dV}{dt}

=\dfrac{4800}{4\pi r^2}=\dfrac{1200}{\pi r^2}

V=288000\pi=\dfrac{4}{3}\pi r^3

288000=\dfrac{4}{3}r^3

r^3=216000

r=60\ \text{cm}

\therefore\ \dfrac{dr}{dt}=\dfrac{1200}{\pi\times 60^2}=0.106\ \text{cm}\ \text{s}^{-1}

Grading Criteria

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2016 TWO(d)