NCEA Calculus 2013 THREE(e - Solution & Explanation

Question

A spherical balloon is being inflated with helium.

The balloon is being inflated in such a way that its volume is increasing at a constant rate of

300\text{ cm}^3\text{ s}^{-1}

.

The material that the balloon is made of is of limited strength, and the balloon will burst when
its surface area reaches

7\,500\text{ cm}^2

.

Find the rate at which the surface area of the balloon is increasing when it reaches bursting
point.

*Show any derivatives that you need to find when solving this problem.*

Official Answer

\dfrac{dV}{dt}=300

A=4\pi r^2\ \ \Rightarrow\ \dfrac{dA}{dr}=8\pi r

V=\dfrac{4}{3}\pi r^3\ \ \Rightarrow\ \dfrac{dV}{dr}=4\pi r^2

\dfrac{dA}{dt}=\dfrac{dV}{dt}\cdot\dfrac{dA}{dr}\cdot\dfrac{dr}{dV}

=\dfrac{2400\pi r}{4\pi r^2}

=\dfrac{600}{r}

A=7500\ \Rightarrow\ 4\pi r^2=7500

r=\sqrt{\dfrac{7500}{4\pi}}=24.43\ \text{cm}

\therefore\ \dfrac{dA}{dt}=\dfrac{600}{24.43}=24.56\ \text{cm}^2\ \text{s}^{-1}

Grading Criteria

Achievement (u)

Correct expressions for $\dfrac{dV}{dr}$ and $\dfrac{dA}{dr}$

Merit (r)

Correct expressions for $\dfrac{dV}{dr}$ , $\dfrac{dA}{dr}$ and $\dfrac{dA}{dt}$

Excellence T1

Correct solution along with correct expressions for $\dfrac{dV}{dr}$ , $\dfrac{dA}{dr}$ and $\dfrac{dA}{dt}$

Excellence T2

-

2013 THREE(e)

Achievement (u)

Merit (r)

Excellence T1

Excellence T2