NCEA Calculus 2019 ONE(e - Solution & Explanation

Question

The volume of a sphere is increasing.

At the instant when the sphere’s radius is

0.5

m, the surface area of the sphere is increasing at
a rate of

0.4

m

^2

s

^{-1}

.

Find the rate at which the volume of the sphere is increasing at this instant.

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

Official Answer

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

S=4\pi r^2\Rightarrow\dfrac{dS}{dr}=8\pi r

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

\dfrac{dS}{dt}=0.4

when

r=0.5

\dfrac{dV}{dt}=0.4\times\dfrac{1}{8\pi r}\times 4\pi r^2

=0.2r

When

r=0.5

,

\dfrac{dV}{dt}=0.1\,\mathrm{m}^3/\mathrm{s}

Grading Criteria

Achievement (u)

Correct expressions for $\dfrac{dS}{dr}$ and $\dfrac{dV}{dr}$ .

Merit (r)

Correct expression for $\dfrac{dV}{dt}$ .
Anything equivalent.
Line 5 is ok.

Excellence T1

Correct solution with correct derivatives.
Units not required.

Excellence T2

-

Video Explanation

NCEA Level 3 Calculus Differentiation 2019 NZQA Exam - Worked Answers by infinityplusone(starts at 14:30)

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2019 ONE(e)

Achievement (u)

Merit (r)

Excellence T1

Excellence T2