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2007
P1 Q11
Topical
Tangents
07 P1 Q11
2007 H2 Mathematics Paper 1 Question 11
Differentiation I: Tangents and Normals, Parametric Curves
Answers
(i)
(iii)
2
5
units
2
{\frac{2}{5} \textrm{ units}^2}
5
2
units
2
Full solutions
(i)
(ii)
d
x
d
t
=
−
2
cos
t
sin
t
d
y
d
t
=
3
sin
2
t
cos
t
\begin{align*} \frac{\mathrm{d}x}{\mathrm{d}t} &= -2\cos t \sin t \\ \frac{\mathrm{d}y}{\mathrm{d}t} &= 3 \sin^2 t \cos t \\ \end{align*}
d
t
d
x
d
t
d
y
=
−
2
cos
t
sin
t
=
3
sin
2
t
cos
t
d
y
d
x
=
d
y
d
t
÷
d
x
d
t
=
3
sin
2
t
cos
t
−
2
cos
t
sin
t
=
−
3
2
sin
t
\begin{align*} \frac{\mathrm{d}y}{\mathrm{d}x} &= \frac{\mathrm{d}y}{\mathrm{d}t} \div \frac{\mathrm{d}x}{\mathrm{d}t} \\ &=\frac{3 \sin^2 t \cos t}{-2\cos t \sin t} \\ &=-\frac{3}{2} \sin t \\ \end{align*}
d
x
d
y
=
d
t
d
y
÷
d
t
d
x
=
−
2
cos
t
sin
t
3
sin
2
t
cos
t
=
−
2
3
sin
t
At point
(
cos
2
θ
,
sin
3
θ
)
,
{\left( \cos^2 \theta, \sin^3 \theta \right), }
(
cos
2
θ
,
sin
3
θ
)
,
t
=
θ
{t=\theta}
t
=
θ
d
y
d
x
=
−
3
2
sin
θ
\frac{\mathrm{d}y}{\mathrm{d}x} = -\frac{3}{2} \sin \theta
d
x
d
y
=
−
2
3
sin
θ
Tangent at
(
cos
2
θ
,
sin
3
θ
)
:
{\left( \cos^2 \theta, \sin^3 \theta \right):}
(
cos
2
θ
,
sin
3
θ
)
:
y
−
sin
3
θ
=
−
3
2
sin
θ
(
x
−
cos
2
θ
)
y - \sin^3 \theta = -\frac{3}{2} \sin \theta \left( x - \cos^2 \theta \right)
y
−
sin
3
θ
=
−
2
3
sin
θ
(
x
−
cos
2
θ
)
At point
Q
,
{Q, }
Q
,
substituting
y
=
0
{y=0}
y
=
0
into equation of tangent
0
−
sin
3
θ
=
−
3
2
sin
θ
(
x
−
cos
2
θ
)
x
−
cos
2
θ
=
2
3
sin
2
θ
x
=
cos
2
θ
+
2
3
sin
2
θ
\begin{gather*} 0 - \sin^3 \theta = -\frac{3}{2} \sin \theta \left( x - \cos^2 \theta \right) \\ x - \cos^2 \theta = \frac{2}{3} \sin^2 \theta \\ x = \cos^2 \theta + \frac{2}{3} \sin^2 \theta \\ \end{gather*}
0
−
sin
3
θ
=
−
2
3
sin
θ
(
x
−
cos
2
θ
)
x
−
cos
2
θ
=
3
2
sin
2
θ
x
=
cos
2
θ
+
3
2
sin
2
θ
At point
R
,
{R, }
R
,
substituting
x
=
0
{x=0}
x
=
0
into equation of tangent
y
−
sin
3
θ
=
−
3
2
sin
θ
(
0
−
cos
2
θ
)
y
=
sin
3
θ
+
3
2
sin
θ
cos
2
θ
\begin{align*} y - \sin^3 \theta &= -\frac{3}{2} \sin \theta \left( 0 - \cos^2 \theta \right) \\ y &= \sin^3 \theta + \frac{3}{2}\sin \theta \cos^2 \theta \\ \end{align*}
y
−
sin
3
θ
y
=
−
2
3
sin
θ
(
0
−
cos
2
θ
)
=
sin
3
θ
+
2
3
sin
θ
cos
2
θ
Hence coordinates of
Q
{Q}
Q
and
R
{R}
R
are:
Q
(
cos
2
θ
+
2
3
sin
2
θ
,
0
)
R
(
0
,
sin
3
θ
+
3
2
sin
θ
cos
2
θ
)
\begin{align*} & Q \left( \cos^2 \theta + \frac{2}{3} \sin^2 \theta , 0 \right) \\ & R \left( 0 , \sin^3 \theta + \frac{3}{2}\sin \theta \cos^2 \theta \right) \end{align*}
Q
(
cos
2
θ
+
3
2
sin
2
θ
,
0
)
R
(
0
,
sin
3
θ
+
2
3
sin
θ
cos
2
θ
)
Area of
△
O
Q
R
=
1
2
(
cos
2
θ
+
2
3
sin
2
θ
)
(
sin
3
θ
+
3
2
sin
θ
cos
2
θ
)
=
1
2
(
1
3
)
(
1
2
)
sin
θ
(
3
cos
2
θ
+
2
sin
2
θ
)
(
2
sin
2
θ
+
3
cos
2
θ
)
=
1
12
sin
θ
(
3
cos
2
θ
+
2
sin
2
θ
)
2
■
\begin{align*} & \textrm{Area of } \triangle OQR \\ & = \frac{1}{2} \left( \cos^2 \theta + \frac{2}{3} \sin^2 \theta \right) \left( \sin^3 \theta + \frac{3}{2}\sin \theta \cos^2 \theta \right) \\ &= \frac{1}{2}\left( \frac{1}{3} \right)\left( \frac{1}{2} \right) \sin \theta \left( 3\cos^2 \theta + 2 \sin^2 \theta \right) \left( 2 \sin^2 \theta + 3\cos^2 \theta \right) \\ &= \frac{1}{12}\sin \theta \left( 3\cos^2 \theta + 2 \sin^2 \theta \right)^2 \; \blacksquare \end{align*}
Area of
△
OQR
=
2
1
(
cos
2
θ
+
3
2
sin
2
θ
)
(
sin
3
θ
+
2
3
sin
θ
cos
2
θ
)
=
2
1
(
3
1
)
(
2
1
)
sin
θ
(
3
cos
2
θ
+
2
sin
2
θ
)
(
2
sin
2
θ
+
3
cos
2
θ
)
=
12
1
sin
θ
(
3
cos
2
θ
+
2
sin
2
θ
)
2
■
(iii)
When
x
=
0
,
{x=0,}
x
=
0
,
cos
2
t
=
0
t
=
π
2
\begin{align*} \cos^2 t &= 0 \\ t &= \frac{\pi}{2} \end{align*}
cos
2
t
t
=
0
=
2
π
When
x
=
1
,
{x=1,}
x
=
1
,
cos
2
t
=
1
t
=
0
\begin{align*} \cos^2 t &= 1 \\ t &= 0 \end{align*}
cos
2
t
t
=
1
=
0
Area under curve
=
∫
0
1
y
d
x
=
∫
π
2
0
sin
3
t
(
−
2
cos
t
sin
t
)
d
t
=
−
2
∫
π
2
0
cos
t
sin
4
t
d
t
=
2
∫
0
π
2
cos
t
sin
4
t
d
t
■
\begin{align*} & \textrm{Area under curve} \\ & = \int_0^1 y \; \mathrm{d}x \\ & = \int_{\frac{\pi}{2}}^{0} \sin^3 t \left( -2 \cos t \sin t \right) \; \mathrm{d}t \\ &= -2 \int_{\frac{\pi}{2}}^{0} \cos t \sin^4 t \; \mathrm{d}t \\ &= 2 \int^{\frac{\pi}{2}}_{0} \cos t \sin^4 t \; \mathrm{d}t \; \blacksquare \end{align*}
Area under curve
=
∫
0
1
y
d
x
=
∫
2
π
0
sin
3
t
(
−
2
cos
t
sin
t
)
d
t
=
−
2
∫
2
π
0
cos
t
sin
4
t
d
t
=
2
∫
0
2
π
cos
t
sin
4
t
d
t
■
u
=
sin
t
d
u
d
t
=
cos
t
\begin{align*} u &= \sin t \\ \frac{\mathrm{d}u}{\mathrm{d}t} &= \cos t \end{align*}
u
d
t
d
u
=
sin
t
=
cos
t
When
t
=
0
,
{t=0,}
t
=
0
,
u
=
sin
0
=
0
\begin{align*} u &= \sin 0 \\ &= 0 \end{align*}
u
=
sin
0
=
0
When
t
=
π
2
,
{t=\frac{\pi}{2},}
t
=
2
π
,
u
=
sin
π
2
=
1
\begin{align*} u &= \sin \frac{\pi}{2} \\ &= 1 \end{align*}
u
=
sin
2
π
=
1
2
∫
0
π
2
cos
t
sin
4
t
d
t
=
2
∫
0
1
cos
t
u
4
1
cos
t
d
u
=
2
∫
0
1
u
4
d
u
=
2
[
u
5
5
]
0
1
=
2
5
units
2
■
\begin{align*} & 2 \int^{\frac{\pi}{2}}_{0} \cos t \sin^4 t \; \mathrm{d}t \\ & = 2 \int_0^1 \cos t \; u^4 \; \frac{1}{\cos t} \; \mathrm{d}u \\ & = 2 \int_0^1 u^4 \; \mathrm{d} u \\ & = 2 \left[ \frac{u^5}{5} \right]_0^1 \\ & = \frac{2}{5} \textrm{ units}^2 \; \blacksquare \end{align*}
2
∫
0
2
π
cos
t
sin
4
t
d
t
=
2
∫
0
1
cos
t
u
4
cos
t
1
d
u
=
2
∫
0
1
u
4
d
u
=
2
[
5
u
5
]
0
1
=
5
2
units
2
■
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