NCEA Calculus 2021 THREE(e - Solution & Explanation

Question

A lamp is suspended above the centre of a round table of radius

r

.

The height,

h

, of the lamp above the table is adjustable.

Loading diagram...

Point P is on the edge of the table.

At point P the illumination

I

is directly proportional to the cosine of angle

\theta

in the above diagram,
and inversely proportional to the square of the distance,

S

, to the lamp.

i.e.

I = \dfrac{k\cos\theta}{S^2}

, where

k

is a constant.

Prove that the edge of the table will have maximum illumination when

h = \dfrac{r}{\sqrt{2}}

.

*You do not need to prove that your solution gives the maximum value.*

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

Official Answer

\cos\theta=\dfrac{h}{S}

S^2=h^2+r^2

S=\sqrt{h^2+r^2}

k

and

r

are constant

I=\dfrac{k\cos\theta}{S^2}

I=\dfrac{k\dfrac{h}{S}}{S^2}

=\dfrac{kh}{S^3}

I=\dfrac{kh}{\left(h^2+r^2\right)^{\frac{3}{2}}}

\dfrac{dI}{dh}=\dfrac{\left(h^2+r^2\right)^{\frac{3}{2}}k-kh\left(\dfrac{3}{2}\right)\left(h^2+r^2\right)^{\frac{1}{2}}(2h)}{\left(h^2+r^2\right)^3}

\dfrac{dI}{dh}=\dfrac{k\left(h^2+r^2\right)^{\frac{3}{2}}-3kh^2\left(h^2+r^2\right)^{\frac{1}{2}}}{\left(h^2+r^2\right)^3}

\dfrac{dI}{dh}=\dfrac{k\left(h^2+r^2\right)^{\frac{1}{2}}\left(h^2+r^2-3h^2\right)}{\left(h^2+r^2\right)^3}

\dfrac{dI}{dh}=\dfrac{k\left(r^2-2h^2\right)}{\left(h^2+r^2\right)^{\frac{5}{2}}}

\dfrac{dI}{dh}=0 \Rightarrow k\left(r^2-2h^2\right)=0

2h^2=r^2

h^2=\dfrac{r^2}{2}

h=\dfrac{r}{\sqrt{2}}

Grading Criteria

Achievement (u)

-

Merit (r)

Correct expression for $\dfrac{dI}{dh}$

Excellence T1

-

Excellence T2

Correct proof with correct derivative

Video Explanation

NCEA Level 3 Calculus Differentiation 2021 NZQA Exam - Worked Answers by infinityplusone(starts at 75:34)

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2021 THREE(e)

Achievement (u)

Merit (r)

Excellence T1

Excellence T2