NCEA Calculus 2015 TWO(e - Solution & Explanation

Question

A water container is constructed in the shape of a square-based pyramid. The height of the
pyramid is the same as the length of each side of its base.

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A vertical height of 20 cm is then cut off the top of the pyramid, and a new flat top added.
The pyramid is then inverted and water is poured in at a rate of

3000\text{ cm}^3

per minute.

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Find the rate at which the surface area of the water is increasing when the depth of the water
is

15\text{ cm}

.

Volume of pyramid

= \dfrac{1}{3} \times

base area

\times

height

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

Official Answer

Depth of water

= x

h=x+20

V=\frac{1}{3}h^3-\frac{1}{3}20^3

=\frac{1}{3}(x+20)^3-\frac{1}{3}20^3

\frac{dV}{dx}=(x+20)^2

A=(x+20)^2

\frac{dA}{dx}=2(x+20)

\frac{dV}{dt}=3000

\frac{dA}{dt}=\frac{dA}{dx}\times\frac{dx}{dV}\times\frac{dV}{dt}

=2(x+20)\times\frac{1}{(x+20)^2}\times3000

When

x=15

\frac{dA}{dt}=2\times35\times\frac{1}{35^2}\times3000=171.4\ \mathrm{cm}^2\ \mathrm{min}^{-1}

Grading Criteria

Achievement (u)

Correct $\dfrac{dV}{dx}$
OR
$\dfrac{dA}{dx}$

Merit (r)

Correct $\dfrac{dV}{dx}$
AND
$\dfrac{dA}{dx}$

Excellence T1

Correct solution.

Excellence T2

-

2015 TWO(e)

Achievement (u)

Merit (r)

Excellence T1

Excellence T2