NCEA Calculus 2018 TWO(e - Solution & Explanation

Question

A water tank is in the shape of an inverted right-circular cone.

The height of the cone is 200 cm and the radius of the cone is 80 cm.

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The tank is being filled with water at a rate of

150\ \mathrm{cm}^3

per second.

At what rate will the surface area of the water in the tank be increasing when the depth of
water in the tank is 125 cm?

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

Official Answer

\dfrac{dV}{dt}=150\,\text{cm}^3/\text{s}

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

h=2.5r

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

=\dfrac{5}{6}\pi r^3

\dfrac{dV}{dr}=2.5\pi r^2

SA=\pi r^2

\dfrac{dSA}{dr}=2\pi r

\dfrac{dSA}{dt}=150\times\dfrac{1}{2.5\pi r^2}\times 2\pi r

=\dfrac{120}{r}

When

h=125\,\text{cm}

,

r=50\,\text{cm}

\dfrac{dSA}{dt}=\dfrac{120}{50}=2.4\,\text{cm}^2/\text{s}

Grading Criteria

Achievement (u)

Correct expression for $\dfrac{dV}{dr}$ in terms of one variable.

Merit (r)

Correct expression for $\dfrac{dV}{dr}$ and $\dfrac{dSA}{dr}$ in terms of $r$ , and an attempt to relate two (or more) derivatives.

Excellence T1

Correct solution.

Excellence T2

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Video Explanation

NCEA Level 3 Calculus Differentiation 2018 NZQA Exam - Worked Answers by infinityplusone(starts at 30:47)

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2018 TWO(e)

Achievement (u)

Merit (r)

Excellence T1

Excellence T2