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2015
P2 Q1
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DE
15 P2 Q1
2015 H2 Mathematics Paper 2 Question 1
Differential Equations (DEs)
Answers
(i)
Maximum
h
=
32
{h=32}
h
=
32
(ii)
t
=
160
−
40
16
−
1
2
h
{t = 160 - 40\sqrt{16 - \frac{1}{2} h}}
t
=
160
−
40
16
−
2
1
h
160
−
80
2
{160 - 80\sqrt{2}}
160
−
80
2
≈
46.9
years
{\approx 46.9 \textrm{ years}}
≈
46.9
years
Full solutions
(i)
At maximum height,
d
h
d
t
=
0
,
{\frac{\mathrm{d}h}{\mathrm{d}t} = 0,}
d
t
d
h
=
0
,
0
=
1
10
16
−
1
2
h
0
=
16
−
1
2
h
h
=
32
■
\begin{gather*} 0 = \frac{1}{10} \sqrt{16 - \frac{1}{2}h} \\ 0 = 16 - \frac{1}{2}h \\ h = 32 \; \blacksquare \end{gather*}
0
=
10
1
16
−
2
1
h
0
=
16
−
2
1
h
h
=
32
■
(ii)
d
h
d
t
=
1
10
16
−
1
2
h
1
16
−
1
2
h
d
h
d
t
=
1
10
∫
(
16
−
1
2
h
)
−
1
2
d
h
=
∫
1
10
d
t
(
16
−
1
2
h
)
1
2
1
2
(
−
1
2
)
=
1
10
t
+
c
−
4
(
16
−
1
2
h
)
1
2
=
1
10
t
+
c
\begin{gather*} \frac{\mathrm{d}h}{\mathrm{d}t} = \frac{1}{10} \sqrt{16 - \frac{1}{2} h} \\ \frac{1}{\sqrt{16 - \frac{1}{2} h}} \frac{\mathrm{d}h}{\mathrm{d}t} = \frac{1}{10} \\ \int \left(16 - \frac{1}{2} h\right)^{-\frac{1}{2}} \; \mathrm{d}h = \int \frac{1}{10} \; \mathrm{d}t \\ \frac{\left(16 - \frac{1}{2} h\right)^{\frac{1}{2}}}{\frac{1}{2}\left(-\frac{1}{2}\right)} = \frac{1}{10}t +c \\ - 4 \left(16 - \frac{1}{2} h\right)^{\frac{1}{2}} = \frac{1}{10}t + c \\ \end{gather*}
d
t
d
h
=
10
1
16
−
2
1
h
16
−
2
1
h
1
d
t
d
h
=
10
1
∫
(
16
−
2
1
h
)
−
2
1
d
h
=
∫
10
1
d
t
2
1
(
−
2
1
)
(
16
−
2
1
h
)
2
1
=
10
1
t
+
c
−
4
(
16
−
2
1
h
)
2
1
=
10
1
t
+
c
When
t
=
0
,
h
=
0
{t=0, h=0}
t
=
0
,
h
=
0
−
4
16
−
0
=
1
10
(
0
)
+
c
c
=
−
16
\begin{align*} -4 \sqrt{16-0} &= \frac{1}{10}(0) + c \\ c &= -16 \end{align*}
−
4
16
−
0
c
=
10
1
(
0
)
+
c
=
−
16
−
4
(
16
−
1
2
h
)
1
2
=
1
10
t
−
16
1
10
t
=
16
−
4
16
−
1
2
h
t
=
160
−
40
16
−
1
2
h
■
\begin{gather*} - 4 \left(16 - \frac{1}{2} h\right)^{\frac{1}{2}} = \frac{1}{10}t -16 \\ \frac{1}{10}t = 16 - 4\sqrt{16 - \frac{1}{2} h} \\ t = 160 - 40\sqrt{16 - \frac{1}{2} h} \; \blacksquare \end{gather*}
−
4
(
16
−
2
1
h
)
2
1
=
10
1
t
−
16
10
1
t
=
16
−
4
16
−
2
1
h
t
=
160
−
40
16
−
2
1
h
■
When the tree reaches half the maximum height,
h
=
16
,
{h=16,}
h
=
16
,
t
=
160
−
40
16
−
1
2
(
16
)
=
160
−
40
8
=
160
−
80
2
years
■
\begin{align*} t &= 160 - 40\sqrt{16-\frac{1}{2}(16)} \\ &= 160 - 40\sqrt{8} \\ &= 160 - 80\sqrt{2} \textrm{ years} \; \blacksquare \end{align*}
t
=
160
−
40
16
−
2
1
(
16
)
=
160
−
40
8
=
160
−
80
2
years
■
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