Pre-U Pre-U 9794/1 2015 June — Question 10 11 marks

Exam BoardPre-U
ModulePre-U 9794/1 (Pre-U Mathematics Paper 1)
Year2015
SessionJune
Marks11
TopicParametric differentiation
TypeShow dy/dx simplifies to given form
DifficultyStandard +0.3 Part (i) requires standard parametric differentiation with chain rule and simplification of surds—routine A-level technique. Part (ii) involves binomial expansion of (1-t²)^(-1/2) and substitution, which is straightforward application of series methods. The algebraic manipulation is slightly more involved than average but follows standard procedures without requiring novel insight.
Spec1.07s Parametric and implicit differentiation4.08b Standard Maclaurin series: e^x, sin, cos, ln(1+x), (1+x)^n
10 A curve has parametric equations given by $$x = - \sqrt { ( 1 - t ) ^ { 3 } } \quad y = \sqrt { ( 1 + t ) ^ { 3 } } \quad \text { for } - 1 < t < 1$$
  1. Show that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 1 + t } { \sqrt { 1 - t ^ { 2 } } }\).
  2. Write \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) as a series of ascending powers of \(t\) up to and including the term in \(t ^ { 4 }\), and hence estimate the gradient of the curve when \(t = 0.5\).
(i)
Use \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{\mathrm{d}y}{\mathrm{d}t} \times \dfrac{\mathrm{d}t}{\mathrm{d}x}\) [M1]
Obtain either \(\dfrac{\mathrm{d}y}{\mathrm{d}t} = \dfrac{3}{2}(1+t)^{\frac{1}{2}}\) or \(\dfrac{\mathrm{d}x}{\mathrm{d}t} = \dfrac{3}{2}(1-t)^{\frac{1}{2}}\) [B1]
Obtain correct \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{1.5(1+t)^{0.5}}{1.5(1-t)^{0.5}}\) [A1]
Show multiplication of denominator and numerator by \(\sqrt{1+t}\) or equivalent on correct derivative. [M1]
Clearly derive \(\dfrac{1+t}{\sqrt{1-t^2}}\) [A1]
Subtotal: [5]
(ii)
Show \((1-t^2)^{-0.5} = 1 + \left(-\dfrac{1}{2}\right)(-t^2) + \left(\dfrac{1}{2}\right)\left(-\dfrac{1}{2}\right)\left(-\dfrac{3}{2}\right)(-t^2)^2\) [M1]
Obtain \(1 + \dfrac{t^2}{2} + \dfrac{3t^4}{8}\) [A1]
Show multiplication of their \(1 + \dfrac{t^2}{2} + \dfrac{3t^4}{8}\) by \((1+t)\) [M1]
Obtain \(1 + t + \dfrac{t^2}{2}\) [A1]
Obtain \(\dfrac{t^3}{2} + \dfrac{3t^4}{8}\) [A1]
Substitute \(t = 0.5\) and obtain \(1.71\) or better (\(1.7109275\)) [A1]
Subtotal: [6]
Total: [11]
**(i)**
Use $\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{\mathrm{d}y}{\mathrm{d}t} \times \dfrac{\mathrm{d}t}{\mathrm{d}x}$ [M1]
Obtain either $\dfrac{\mathrm{d}y}{\mathrm{d}t} = \dfrac{3}{2}(1+t)^{\frac{1}{2}}$ or $\dfrac{\mathrm{d}x}{\mathrm{d}t} = \dfrac{3}{2}(1-t)^{\frac{1}{2}}$ [B1]
Obtain correct $\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{1.5(1+t)^{0.5}}{1.5(1-t)^{0.5}}$ [A1]
Show multiplication of denominator and numerator by $\sqrt{1+t}$ or equivalent on correct derivative. [M1]
Clearly derive $\dfrac{1+t}{\sqrt{1-t^2}}$ [A1]
**Subtotal: [5]**

**(ii)**
Show $(1-t^2)^{-0.5} = 1 + \left(-\dfrac{1}{2}\right)(-t^2) + \left(\dfrac{1}{2}\right)\left(-\dfrac{1}{2}\right)\left(-\dfrac{3}{2}\right)(-t^2)^2$ [M1]
Obtain $1 + \dfrac{t^2}{2} + \dfrac{3t^4}{8}$ [A1]
Show multiplication of their $1 + \dfrac{t^2}{2} + \dfrac{3t^4}{8}$ by $(1+t)$ [M1]
Obtain $1 + t + \dfrac{t^2}{2}$ [A1]
Obtain $\dfrac{t^3}{2} + \dfrac{3t^4}{8}$ [A1]
Substitute $t = 0.5$ and obtain $1.71$ or better ($1.7109275$) [A1]
**Subtotal: [6]**
**Total: [11]**
10 A curve has parametric equations given by

$$x = - \sqrt { ( 1 - t ) ^ { 3 } } \quad y = \sqrt { ( 1 + t ) ^ { 3 } } \quad \text { for } - 1 < t < 1$$

(i) Show that $\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 1 + t } { \sqrt { 1 - t ^ { 2 } } }$.\\
(ii) Write $\frac { \mathrm { d } y } { \mathrm {~d} x }$ as a series of ascending powers of $t$ up to and including the term in $t ^ { 4 }$, and hence estimate the gradient of the curve when $t = 0.5$.

\hfill \mbox{\textit{Pre-U Pre-U 9794/1 2015 Q10 [11]}}
This paper (11 questions)
View full paper
Q1 3 Q2 5 Q3 3 Q4 6 Q5 9 Q6 6 Q7 9 Q8 11 Q9 7 Q10 11 Q11 10