CAIE FP1 2015 June — Question 9 11 marks

Exam BoardCAIE
ModuleFP1 (Further Pure Mathematics 1)
Year2015
SessionJune
Marks11
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicParametric integration
TypeParametric arc length calculation
DifficultyChallenging +1.2 This is a standard Further Maths parametric arc length question requiring differentiation of power functions, substitution into the arc length formula, and integration. The expressions simplify nicely (the square root becomes a perfect square), making it more routine than typical arc length problems. The mean value part is straightforward application of the formula once the arc length is found. Slightly above average difficulty due to being Further Maths content with fractional powers, but the algebraic simplification is manageable.
Spec1.03g Parametric equations: of curves and conversion to cartesian4.08e Mean value of function: using integral8.06b Arc length and surface area: of revolution, cartesian or parametric
9 The curve \(C\) has parametric equations $$x = 4 t + 2 t ^ { \frac { 3 } { 2 } } , \quad y = 4 t - 2 t ^ { \frac { 3 } { 2 } } , \quad \text { for } 0 \leqslant t \leqslant 4$$ Find the arc length of \(C\), giving your answer correct to 3 significant figures. Find the mean value of \(y\) with respect to \(x\) over the interval \(0 \leqslant x \leqslant 32\).
Question 9:
AnswerMarks Guidance
WorkingMarks Notes
\(\dot{x} = 4 + 3t^{\frac{1}{2}}\), \(\dot{y} = 4 - 3t^{\frac{1}{2}}\)B1
\(\dot{x}^2 + \dot{y}^2 = 32 + 18t\)B1
\(s = \int_0^4 (32+18t)^{\frac{1}{2}} \, dt = \left[\frac{1}{27}(32+18t)^{\frac{3}{2}}\right]_0^4\)M1A1
\(= \frac{1}{27}\left(104^{\frac{3}{2}} - 32^{\frac{3}{2}}\right) = 32.6\)M1A1 (6) (CAO)
\(\int_0^{32} y \, dx = \int_0^4 y\frac{dx}{dt} \, dt = \int_0^4 t\left(4-2\sqrt{t}\right)\left(4+3\sqrt{t}\right)dt = \int_0^4 \left(16t + 4t^{\frac{3}{2}} - 6t^2\right)dt\)M1
\(= \left[8t^2 + \frac{8}{5}t^{\frac{5}{2}} - 2t^3\right]_0^4 \quad (= 51.2)\)M1A1
\(MV = \frac{\int_0^a y \, dx}{b-a} = \frac{51.2}{32} = 1.6\)M1A1 (5) (CAO)
Total: 11 
## Question 9:

| Working | Marks | Notes |
|---------|-------|-------|
| $\dot{x} = 4 + 3t^{\frac{1}{2}}$, $\dot{y} = 4 - 3t^{\frac{1}{2}}$ | B1 | |
| $\dot{x}^2 + \dot{y}^2 = 32 + 18t$ | B1 | |
| $s = \int_0^4 (32+18t)^{\frac{1}{2}} \, dt = \left[\frac{1}{27}(32+18t)^{\frac{3}{2}}\right]_0^4$ | M1A1 | |
| $= \frac{1}{27}\left(104^{\frac{3}{2}} - 32^{\frac{3}{2}}\right) = 32.6$ | M1A1 | (6) (CAO) |
| $\int_0^{32} y \, dx = \int_0^4 y\frac{dx}{dt} \, dt = \int_0^4 t\left(4-2\sqrt{t}\right)\left(4+3\sqrt{t}\right)dt = \int_0^4 \left(16t + 4t^{\frac{3}{2}} - 6t^2\right)dt$ | M1 | |
| $= \left[8t^2 + \frac{8}{5}t^{\frac{5}{2}} - 2t^3\right]_0^4 \quad (= 51.2)$ | M1A1 | |
| $MV = \frac{\int_0^a y \, dx}{b-a} = \frac{51.2}{32} = 1.6$ | M1A1 | (5) (CAO) |
| | **Total: 11** | |
9 The curve $C$ has parametric equations

$$x = 4 t + 2 t ^ { \frac { 3 } { 2 } } , \quad y = 4 t - 2 t ^ { \frac { 3 } { 2 } } , \quad \text { for } 0 \leqslant t \leqslant 4$$

Find the arc length of $C$, giving your answer correct to 3 significant figures.

Find the mean value of $y$ with respect to $x$ over the interval $0 \leqslant x \leqslant 32$.

\hfill \mbox{\textit{CAIE FP1 2015 Q9 [11]}}
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