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2020
P2 Q3
Topical
Tangents
20 P2 Q3
2020 H2 Mathematics Paper 2 Question 3
Differentiation I: Tangents and Normals, Parametric Curves
Answers
(i)
2
x
+
y
=
39
{2x+y=39}
2
x
+
y
=
39
(ii)
2057
18
≈
114.28
units
2
{\frac{2057}{18} \approx 114.28 \textrm{ units}^2}
18
2057
≈
114.28
units
2
(iiia)
2057
3
≈
685.67
units
2
{\frac{2057}{3} \approx 685.67 \textrm{ units}^2}
3
2057
≈
685.67
units
2
(iiib)
x
=
(
y
+
3
)
2
54
+
4
{x = \frac{(y+3)^2}{54} + 4}
x
=
54
(
y
+
3
)
2
+
4
Full solutions
(i)
d
x
d
t
=
6
t
d
y
d
t
=
6
\begin{align*} \frac{\mathrm{d}x}{\mathrm{d}t} &= 6 t \\ \frac{\mathrm{d}y}{\mathrm{d}t} &= 6 \\ \end{align*}
d
t
d
x
d
t
d
y
=
6
t
=
6
d
y
d
x
=
d
y
d
t
÷
d
x
d
t
=
6
6
t
=
1
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{6}{6 t} \\ &= \frac{1}{t} \end{align*}
d
x
d
y
=
d
t
d
y
÷
d
t
d
x
=
6
t
6
=
t
1
At point
(
14
,
11
)
,
{(14,11),}
(
14
,
11
)
,
6
t
−
1
=
11
t
=
2
d
y
d
x
=
1
2
\begin{align*} 6t-1 &= 11 \\ t &= 2 \\ \frac{\mathrm{d}y}{\mathrm{d}x} &= \frac{1}{2} \end{align*}
6
t
−
1
t
d
x
d
y
=
11
=
2
=
2
1
Hence the gradient of
N
{N}
N
is
−
2
{-2}
−
2
Cartesian equation of
N
:
{N:}
N
:
y
−
11
=
−
2
(
x
−
14
)
y
−
11
=
−
2
x
+
28
2
x
+
y
=
39
■
\begin{gather*} y - 11 = -2 \left( x - 14 \right) \\ y-11 = -2x+28 \\ 2x+y=39 \; \blacksquare \end{gather*}
y
−
11
=
−
2
(
x
−
14
)
y
−
11
=
−
2
x
+
28
2
x
+
y
=
39
■
(ii)
Area under curve from
t
=
1
6
to
(
14
,
11
)
=
∫
1
6
2
(
6
t
−
1
)
(
6
t
)
d
t
=
∫
1
6
2
(
36
t
2
−
6
t
)
d
t
=
[
12
t
3
−
3
t
2
]
1
6
2
=
3025
36
\begin{align*} & \textrm{Area under curve from } {\textstyle t=\frac{1}{6}} \textrm{ to } (14,11) \\ & = \int_{\frac{1}{6}}^{2} (6t-1) (6t) \; \mathrm{d}t \\ & = \int_{\frac{1}{6}}^{2} (36t^2 - 6t) \; \mathrm{d}t \\ & = \left[ 12 t^3 - 3 t^2 \right]_{\frac{1}{6}}^2 \\ & = \frac{3025}{36} \end{align*}
Area under curve from
t
=
6
1
to
(
14
,
11
)
=
∫
6
1
2
(
6
t
−
1
)
(
6
t
)
d
t
=
∫
6
1
2
(
36
t
2
−
6
t
)
d
t
=
[
12
t
3
−
3
t
2
]
6
1
2
=
36
3025
When the normal cuts the
x
-
{x\textrm{-}}
x
-
axis,
x
=
39
2
x=\frac{39}{2}
x
=
2
39
Area under normal from
(
14
,
11
)
to
(
39
2
,
0
)
=
1
2
×
(
39
2
−
14
)
×
11
=
121
4
\begin{align*} & \textrm{Area under normal from } (14,11) \textrm{ to } {\textstyle \left( \frac{39}{2}, 0 \right)} \\ & = \frac{1}{2} \times \left( \frac{39}{2} - 14 \right) \times 11 \\ & = \frac{121}{4} \end{align*}
Area under normal from
(
14
,
11
)
to
(
2
39
,
0
)
=
2
1
×
(
2
39
−
14
)
×
11
=
4
121
Area enclosed by
C
,
N
and the
x
-axis
=
3025
36
+
121
4
=
2057
18
units
2
■
\begin{align*} & \textrm{Area enclosed by } C, N \textrm{ and the } x\textrm{-axis} \\ & = \frac{3025}{36} + \frac{121}{4} \\ & = \frac{2057}{18} \textrm{ units}^2 \; \blacksquare \end{align*}
Area enclosed by
C
,
N
and the
x
-axis
=
36
3025
+
4
121
=
18
2057
units
2
■
(iiia)
Area enclosed by
D
,
M
and the
x
-axis
=
2057
18
×
2
×
3
=
2057
3
units
2
■
\begin{align*} & \textrm{Area enclosed by } D, M \textrm{ and the } x\textrm{-axis} \\ & = \frac{2057}{18} \times 2 \times 3 \\ & = \frac{2057}{3} \textrm{ units}^2 \; \blacksquare \end{align*}
Area enclosed by
D
,
M
and the
x
-axis
=
18
2057
×
2
×
3
=
3
2057
units
2
■
(iiib)
Replacing
x
{x}
x
with
x
2
{\frac{x}{2}}
2
x
and
y
{y}
y
with
y
3
,
{\frac{y}{3},}
3
y
,
x
2
=
3
t
2
+
2
y
3
=
6
t
−
1
\begin{align} && \quad \frac{x}{2} = 3t^2 + 2 \\ && \quad \frac{y}{3} = 6t-1 \\ \end{align}
2
x
=
3
t
2
+
2
3
y
=
6
t
−
1
From
(
2
)
,
{(2),}
(
2
)
,
6
t
=
y
3
+
1
=
y
+
3
3
t
=
y
+
3
18
\begin{align*} 6t &= \frac{y}{3} + 1 \\ &= \frac{y+3}{3} \\ t &= \frac{y+3}{18} \\ \end{align*}
6
t
t
=
3
y
+
1
=
3
y
+
3
=
18
y
+
3
Substituting into
(
1
)
,
{(1),}
(
1
)
,
x
2
=
3
(
y
+
3
18
)
2
+
2
=
(
y
+
3
)
2
108
+
2
x
=
(
y
+
3
)
2
54
+
4
■
\begin{align*} \frac{x}{2} &= 3 \left( \frac{y+3}{18} \right)^2 + 2 \\ &= \frac{(y+3)^2}{108} + 2 \\ x &= \frac{(y+3)^2}{54} + 4 \; \blacksquare \end{align*}
2
x
x
=
3
(
18
y
+
3
)
2
+
2
=
108
(
y
+
3
)
2
+
2
=
54
(
y
+
3
)
2
+
4
■
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