2018 H2 Maths Paper 2 Q9 — Worked Solution

Part (i)

2 marks

The points do not lie close to a straight line — they curve upwards more and more steeply as $R$ increases, bending away from any single straight line through them. This suggests the relationship between $P$ and $R$ is not well modelled by $P = aR + b$ ; a curved (non-linear) model fits the pattern of the data better.

Part (ii)

6 marks

Approach

to compare the two proposed models, $P = aR + b$ and $P = aR^{2} + b$ , we compute the product moment correlation coefficient (PMCC) for each — the first between $P$ and $R$ , the second between $P$ and $R^{2}$ (since $P = aR^{2} + b$ is linear in $R^{2}$ , not in $R$ ) — and take the model whose $r$ is closer to $1$ .

By GC,

r_{P,\, R} = 0.969\ \text{(3 s.f., by GC)}

r_{P,\, R^{2}} = 0.993\ \text{(3 s.f., by GC)}

Explanation: $r_{P,R^{2}} = 0.993$ is closer to $1$ than $r_{P,R} = 0.969$ , so $P = aR^{2} + b$ gives a stronger linear relationship (between $P$ and $R^{2}$ ) than $P = aR + b$ does (between $P$ and $R$ ), and is therefore the better model.

By GC, the regression line of $P$ on $R^{2}$ is

P = 2.85 \times 10^{- 8}R^{2} - 0.283\ \text{(3 s.f., by GC).}

$\therefore$ $P = aR^{2} + b$ is the better model, with $a = 2.85 \times 10^{- 8}$ and $b = - 0.283$ , both correct to 3 s.f.

Part (iii)

2 marks

Using the equation from (ii), we solve for $R$ when $P = 0.9$ :

0.9 = \left( 2.845744\ldots \times 10^{- 8} \right)R^{2} - 0.282608\ldots

\Rightarrow R^{2} = \frac{0.9 + 0.282608\ldots}{2.845744\ldots \times 10^{- 8}}

\Rightarrow R = 6450\ \text{(3 s.f., by GC),}

taking the positive root since speed $R > 0$ .

Explanation: $R = 6450$ lies within the range of the given data ( $3600 \leq R \leq 9900$ ), so this is an interpolation, and the estimate is reliable.

$\therefore$ the estimated speed is $R = 6450\ \text{rpm}$ (3 s.f.), and this estimate is reliable.

Part (iv)

2 marks

Using the equation from (ii), at $R = 3300$ ,

P = \left( 2.845744\ldots \times 10^{- 8} \right)(3300)^{2} - 0.282608\ldots

= 0.0273\ \text{(3 s.f., by GC).}

Explanation: $R = 3300$ lies outside the range of the given data ( $3600 \leq R \leq 9900$ ), so this is an extrapolation, and the estimate is unreliable — the equation is not known to hold below $R = 3600$ .

$\therefore$ the estimated power used is $P = 0.0273\ \text{watts}$ (3 s.f.), but this estimate is unreliable since it extrapolates beyond the data.

Part (v)

Approach

the equation in (ii) uses $R$ in revolutions per minute. If $r$ denotes the speed in revolutions per second, then $R = 60r$ (there are $60$ seconds in a minute), so we substitute $R = 60r$ into the equation from (ii).

Let $r$ be the fan speed in revolutions per second, so $R = 60r$ . Substituting into $P = aR^{2} + b$ from (ii),

P = a(60r)^{2} + b

= 3600ar^{2} + b

By GC, $3600a = 3600 \times 2.845744\ldots \times 10^{- 8} = 1.02 \times 10^{- 4}$ (3 s.f.).

$\therefore P = 1.02 \times 10^{- 4}r^{2} - 0.283$ , correct to 3 s.f., where $r$ is the fan speed in revolutions per second.

2018 H2 Maths Paper 2, Question 9

Part (i)

Part (ii)

Part (iii)

Part (iv)

Part (v)

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