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Specimen 2017 H2 Mathematics Paper 1 Question 8
Differentiation I: Tangents and Normals, Parametric Curves
Definite Integrals: Areas and Volumes
Answers
(a)
y
=
−
x
tan
p
+
a
sin
p
.
{y = - x \tan p + a \sin p.}
y
=
−
x
tan
p
+
a
sin
p
.
(b)
Length
A
B
=
a
.
{AB = a.}
A
B
=
a
.
(c)
x
2
3
+
y
2
3
=
1.
{x^{\frac{2}{3}} + y^{\frac{2}{3}} = 1.}
x
3
2
+
y
3
2
=
1.
(d)
16
105
π
units
2
.
{\frac{16}{105}\pi \textrm{ units}^2.}
105
16
π
units
2
.
Full solutions
(a)
d
y
d
x
=
d
y
d
t
÷
d
x
d
t
=
3
a
cos
t
sin
2
t
−
3
a
sin
t
cos
2
t
=
−
sin
t
cos
t
=
−
tan
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{3a \cos t \sin^2 t}{-3a \sin t \cos^2 t} \\ &= -\frac{\sin t}{\cos t} \\ &= -\tan t \end{align*}
d
x
d
y
=
d
t
d
y
÷
d
t
d
x
=
−
3
a
sin
t
cos
2
t
3
a
cos
t
sin
2
t
=
−
cos
t
sin
t
=
−
tan
t
At point
P
,
{P,}
P
,
equation of tangent:
y
−
a
sin
3
p
=
(
−
tan
p
)
(
x
−
a
cos
3
p
)
y
=
−
x
tan
p
+
a
sin
p
cos
2
p
+
a
sin
3
p
y
=
−
x
tan
p
+
a
sin
p
(
cos
2
p
+
sin
2
p
)
y
=
−
x
tan
p
+
a
sin
p
■
\begin{gather*} y - a \sin^3 p = (-\tan p) (x - a\cos^3 p) \\ y = - x \tan p + a \sin p \cos^2 p + a \sin^3 p \\ y = - x \tan p + a \sin p (\cos^2 p + \sin^2 p) \\ y = - x \tan p + a \sin p \; \blacksquare \end{gather*}
y
−
a
sin
3
p
=
(
−
tan
p
)
(
x
−
a
cos
3
p
)
y
=
−
x
tan
p
+
a
sin
p
cos
2
p
+
a
sin
3
p
y
=
−
x
tan
p
+
a
sin
p
(
cos
2
p
+
sin
2
p
)
y
=
−
x
tan
p
+
a
sin
p
■
(b)
At point
B
,
x
=
0
,
{B, x=0,}
B
,
x
=
0
,
y
=
a
sin
p
{y=a\sin p}
y
=
a
sin
p
At point
A
,
y
=
0
,
{A, y=0,}
A
,
y
=
0
,
0
=
−
x
tan
p
+
a
sin
p
x
=
a
sin
p
tan
p
x
=
a
cos
p
\begin{gather*} 0 = -x \tan p + a \sin p \\ x = \frac{a \sin p}{\tan p} \\ x = a \cos p \end{gather*}
0
=
−
x
tan
p
+
a
sin
p
x
=
tan
p
a
sin
p
x
=
a
cos
p
Length
A
B
=
(
a
cos
p
)
2
+
(
a
sin
p
)
2
=
a
2
(
cos
2
p
+
sin
2
p
)
=
a
which depends only on
a
■
\begin{align*} & \textrm{Length } AB \\ & = \sqrt{(a\cos p)^2 + (a \sin p)^2} \\ & = \sqrt{a^2 (\cos^2 p+ \sin^2 p)} \\ & = a \\ & \textrm{which depends only on } a \; \blacksquare \end{align*}
Length
A
B
=
(
a
cos
p
)
2
+
(
a
sin
p
)
2
=
a
2
(
cos
2
p
+
sin
2
p
)
=
a
which depends only on
a
■
(c)
cos
t
=
x
1
3
sin
t
=
y
1
3
\begin{align} && \quad \cos t &= x^{\frac{1}{3}} \\ && \quad \sin t &= y^{\frac{1}{3}} \\ \end{align}
cos
t
sin
t
=
x
3
1
=
y
3
1
Since
cos
2
t
+
sin
2
t
=
1
,
{\cos^2 t + \sin^2 t = 1,}
cos
2
t
+
sin
2
t
=
1
,
Cartesian equation of
C
:
{C:}
C
:
x
2
3
+
y
2
3
=
1
■
x^{\frac{2}{3}} + y^{\frac{2}{3}} = 1 \; \blacksquare
x
3
2
+
y
3
2
=
1
■
(d)
Volume
=
π
∫
0
1
x
2
d
y
=
π
∫
0
1
(
1
−
y
2
3
)
3
d
y
=
π
∫
0
1
1
−
3
y
2
3
+
3
y
4
3
−
y
2
d
y
=
π
[
y
−
9
5
y
5
3
+
9
7
y
7
3
−
1
3
y
3
]
0
2
=
16
105
π
units
2
■
\begin{align*} & \textrm{Volume} \\ & = \pi \int_0^1 x^2 \; \mathrm{d}y \\ & = \pi \int_0^1 \left( 1-y^{\frac{2}{3}} \right)^3 \; \mathrm{d}y \\ & = \pi \int_0^1 1 - 3 y^{\frac{2}{3}} + 3 y^{\frac{4}{3}} - y^2 \; \mathrm{d}y \\ & = \pi \left[ y - \frac{9}{5} y^{\frac{5}{3}} + \frac{9}{7} y^{\frac{7}{3}} - \frac{1}{3} y^3 \right]_0^2 \\ & = \frac{16}{105} \pi \textrm{ units}^2 \; \blacksquare \end{align*}
Volume
=
π
∫
0
1
x
2
d
y
=
π
∫
0
1
(
1
−
y
3
2
)
3
d
y
=
π
∫
0
1
1
−
3
y
3
2
+
3
y
3
4
−
y
2
d
y
=
π
[
y
−
5
9
y
3
5
+
7
9
y
3
7
−
3
1
y
3
]
0
2
=
105
16
π
units
2
■
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