P3.4 Parametric modelling

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

Most candidates were successful in this question, showing that parametric differentiation was overall well understood by candidates. It was noted by examiners that part (b) was not attempted by a significant number of candidates. When it was attempted, it was generally not done well, with very many candidates not scoring the mark. Many achieved the correct value only to lose the mark for not stating the units. A common wrong answer was to use x = 0 in the tangent equation to give 5m. Another was to use t = -3 and then substitute this into y = 1 - t3 to give 28m, ignoring the set of values for which t was valid. Others set d 0 d y x = thinking this would give a maximum and then substituted their t value into y. Some wrongly assumed that the largest value of t for the given domain would lead to the maximum height. There was little evidence that candidates had used the sketch to help them find the maximum height.

Common mistakes

• greatest y is at t = −2: 9 m (units needed)
• t = −3 is outside the domain (28 m wrong)
• not x = 0 in the tangent (5 m)
• dy/dx = 0 does not give the maximum

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

Many candidates left the entire question or parts of this question completely blank and clearly found it difficult to access the content being assessed here, perhaps due to a lack of familiarity with integration using parametric equations. Many tried to establish the Cartesian equation to answer various parts. That said, there were many candidates who provided clear and concise solutions to each part. In part (b) many candidates who had found part (a) difficult gave up and lost the opportunity of scoring some easier marks. Many candidates who scored no marks elsewhere were still able to access at least the first two marks in part (b) by setting sin2t = 4.2 5 and solving, frequently getting at least one correct angle; but too many of these found their second value of t by subtracting their first value from pi instead of calculating pi – 𝑠𝑠𝑠𝑠𝑠𝑠-1( 0.84) before dividing by 2. The following mark was for finding two values of x, but many simply substituted their first value in to find x, thinking that this gave them the width of the walkway. Too many marks were lost by approximating too soon and hence not having values with the required accuracy. A few lost the last mark because they omitted the units.

Common mistakes

• 5sin 2t = 4.2 gives two values of t: 2t = π − arcsin 0.84 for the second (not π − first t)
• find x at both and subtract
• width 2.40 m (units needed)
• avoid early rounding