NCEA Calculus 2024 THREE(d - Solution & Explanation

Question

Jamie is doing some baking and pouring
the flour to form a conical pile.

The height of the pile is always the same
as the diameter of the base of the cone.

If the flour is being added at a constant
rate of

3\ \mathrm{cm}^3

per second, at what rate is
the height increasing when the pile is

4\ \mathrm{cm}

in height?

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

Note that volume of a cone

= \dfrac{1}{3}\pi r^2 h

.

Loading diagram...

Official Answer

Method 1:

V=\frac{1}{3}\pi r^2h=\frac{2}{3}\pi r^3

\frac{dV}{dr}=2\pi r^2

We require

\frac{dh}{dt}=\frac{dh}{dr}\times\frac{dr}{dV}\times\frac{dV}{dt}

\frac{dh}{dt}=2\times\frac{1}{2\pi r^2}\times 3

=\frac{3}{\pi r^2}

When

h=4\ \mathrm{cm}

,

r=2\ \mathrm{cm}

, so

\frac{dh}{dt}=\frac{3}{4\pi}=0.2387\text{ cm sec}^{-1}

OR
Method 2:

V=\frac{1}{3}\pi r^2h=\frac{1}{12}\pi h^3

\frac{dV}{dh}=\frac{1}{4}\pi h^2

We require

\frac{dh}{dt}=\frac{dh}{dV}\times\frac{dV}{dt}

\frac{dh}{dt}=\frac{4}{\pi h^2}\times 3

=\frac{12}{\pi h^2}

When

h=4\ \mathrm{cm}

,

\frac{dh}{dt}=\frac{3}{4\pi}=0.2387\text{ cm sec}^{-1}

Units not required.

Grading Criteria

Achievement (u)

Correct expression for $\frac{dh}{dt}$ .

Merit (r)

Correct value for $\frac{dh}{dt}$ , with evidence of a calculus method.

Excellence T1

-

Excellence T2

-

Video Explanation

2024 NCEA L3 Calculus Exam Walkthrough by infinityplusone(starts at 49:28)

Subscribe

2024 THREE(d)

Achievement (u)

Merit (r)

Excellence T1

Excellence T2