2022 H2 Mathematics Paper 1 Question 3 Differentiation I: Tangents and Normals, Parametric Curves
Answers Gradient of normal
= − 1 3 . {= - \frac{1}{3}.} = − 3 1 . x 2 − y 2 = 2. {x^2 - y^2 = 2.} x 2 − y 2 = 2. x ≥ 2 . {x \geq \sqrt{2}.} x ≥ 2 . Full solutions
(a) d x d t = 1 2 ( 3 e 3 t − 6 e − 3 t ) d y d t = 1 2 ( 3 e 3 t + 6 e − 3 t ) d y d x = d y d t ÷ d x d t = 1 2 ( 3 e 3 t + 6 e − 3 t ) 1 2 ( 3 e 3 t − 6 e − 3 t ) = e 3 t + 2 e − 3 t e 3 t − 2 e − 3 t \begin{align*}
\frac{\mathrm{d}x}{\mathrm{d}t} &= \frac{1}{2} \left( 3\mathrm{e}^{3t} - 6 \mathrm{e}^{-3t} \right)
\\ \frac{\mathrm{d}y}{\mathrm{d}t} &= \frac{1}{2} \left( 3\mathrm{e}^{3t} + 6 \mathrm{e}^{-3t} \right)
\\ \frac{\mathrm{d}y}{\mathrm{d}x} &= \frac{\mathrm{d}y}{\mathrm{d}t} \div \frac{\mathrm{d}x}{\mathrm{d}t}
\\ &= \frac{\frac{1}{2} \left( 3\mathrm{e}^{3t} + 6 \mathrm{e}^{-3t} \right)}{\frac{1}{2} \left( 3\mathrm{e}^{3t} - 6 \mathrm{e}^{-3t} \right)}
\\ &= \frac{\mathrm{e}^{3t} + 2 \mathrm{e}^{-3t}}{\mathrm{e}^{3t}-2\mathrm{e}^{-3t}}
\end{align*} d t d x d t d y d x d y = 2 1 ( 3 e 3 t − 6 e − 3 t ) = 2 1 ( 3 e 3 t + 6 e − 3 t ) = d t d y ÷ d t d x = 2 1 ( 3 e 3 t − 6 e − 3 t ) 2 1 ( 3 e 3 t + 6 e − 3 t ) = e 3 t − 2 e − 3 t e 3 t + 2 e − 3 t
When
x = 1 3 ln 2 , {x = \frac{1}{3} \ln 2,} x = 3 1 ln 2 , d y d x = e ln 2 + 2 e − ln 2 e ln 2 − 2 e − ln 2 = 3 \begin{align*}
\frac{\mathrm{d}y}{\mathrm{d}x} &= \frac{\mathrm{e}^{\ln 2} + 2 \mathrm{e}^{-\ln 2}}{\mathrm{e}^{\ln 2}-2\mathrm{e}^{-\ln 2}}
\\ &= 3
\end{align*} d x d y = e l n 2 − 2 e − l n 2 e l n 2 + 2 e − l n 2 = 3 Gradient of normal = − 1 ÷ d y d x = − 1 3 ■ \begin{align*}
& \textrm{Gradient of normal}
\\ & \quad = -1 \div \frac{\mathrm{d}y}{\mathrm{d}x}
\\ & \quad = -\frac{1}{3} \; \blacksquare
\end{align*} Gradient of normal = − 1 ÷ d x d y = − 3 1 ■
(b) x 2 = 1 4 ( e 3 t + 2 e − 3 t ) 2 4 x 2 = e 6 t + 4 + 4 e − 6 t 4 x 2 − 4 = e 6 t + 4 e − 6 t \begin{gather*}
x^2 = \frac{1}{4} \left( \mathrm{e}^{3t} + 2 \mathrm{e}^{-3t} \right)^2
\\ 4x^2 = \mathrm{e}^{6t} + 4 + 4 \mathrm{e}^{-6t}
\\ 4x^2 - 4 = \mathrm{e}^{6t} + 4 \mathrm{e}^{-6t} \tag{1}
\end{gather*} x 2 = 4 1 ( e 3 t + 2 e − 3 t ) 2 4 x 2 = e 6 t + 4 + 4 e − 6 t 4 x 2 − 4 = e 6 t + 4 e − 6 t ( 1 ) y 2 = 1 4 ( e 3 t − 2 e − 3 t ) 2 4 y 2 = e 6 t − 4 + 4 e − 6 t 4 y 2 + 4 = e 6 t + 4 e − 6 t \begin{gather*}
y^2 = \frac{1}{4} \left( \mathrm{e}^{3t} - 2 \mathrm{e}^{-3t} \right)^2
\\ 4y^2 = \mathrm{e}^{6t} - 4 + 4 \mathrm{e}^{-6t}
\\ 4y^2 + 4 = \mathrm{e}^{6t} + 4 \mathrm{e}^{-6t} \tag{2}
\end{gather*} y 2 = 4 1 ( e 3 t − 2 e − 3 t ) 2 4 y 2 = e 6 t − 4 + 4 e − 6 t 4 y 2 + 4 = e 6 t + 4 e − 6 t ( 2 )
Equation
( 1 ) {(1)} ( 1 ) and
( 2 ) , {(2),} ( 2 ) , the cartesian equation of the curve is given by
4 x 2 − 4 = 4 y 2 + 4 x 2 − y 2 = 2 ■ \begin{align*}
4x^2 - 4 &= 4y^2 + 4
\\ x^2 - y^2 &= 2 \; \blacksquare
\end{align*} 4 x 2 − 4 x 2 − y 2 = 4 y 2 + 4 = 2 ■
Since
x 2 = y 2 + 2 , {x^2 = y^2 +2,} x 2 = y 2 + 2 , x 2 ≥ 2. {x^2 \geq 2.} x 2 ≥ 2.
Moreover,
x = 1 2 ( e 3 t + 2 e − 3 t ) {x = \frac{1}{2} \left( \mathrm{e}^{3t} + 2 \mathrm{e}^{-3t} \right)} x = 2 1 ( e 3 t + 2 e − 3 t ) indicates that
x > 0. {x > 0.} x > 0.
Hence the restriction on the values of
x {x} x is
x ≥ 2 . ■ {x \geq \sqrt{2}. \; \blacksquare} x ≥ 2 . ■ Question Commentary
Part (b) of this question is tricky. While it has some similarities to
2010 Paper 1 Question 11,
our attempt that year was to consider adding and subtracting the equations.
Instead, the hint for this part was to consider x 2 {x^2} x 2 y 2 , {y^2,} y 2 ,
The part on the restrictions on the values of x {x} x