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Specimen 2017 H2 Mathematics Paper 1 Question 11
Differential Equations (DEs)
Answers
(a)
d
A
d
t
=
k
A
(
10
−
A
)
.
{\frac{\mathrm{d}A}{\mathrm{d}t} = k A (10 - A).}
d
t
d
A
=
k
A
(
10
−
A
)
.
t
=
14.13.
{t = 14.13.}
t
=
14.13.
(b)
9.65
m
2
.
{9.65 \textrm{ m}^2.}
9.65
m
2
.
(c)
A
=
10
4
e
−
1
5
ln
(
8
3
)
t
+
1
{A = \frac{10}{4 \mathrm{e}^{-{ \frac{1}{5} \ln (\frac{8}{3}) } t} + 1}}
A
=
4
e
−
5
1
l
n
(
3
8
)
t
+
1
10
Full solutions
(a)
d
A
d
t
=
k
A
(
10
−
A
)
■
1
A
(
10
−
A
)
d
A
d
t
=
k
∫
1
A
(
10
−
A
)
d
A
=
∫
k
d
t
∫
1
10
A
+
1
100
−
10
A
d
A
=
k
t
+
c
1
10
ln
∣
A
∣
−
1
10
ln
∣
10
−
A
∣
=
k
t
+
c
\begin{gather*} \frac{\mathrm{d}A}{\mathrm{d}t} = k A (10 - A) \; \blacksquare \\ \frac{1}{A(10 - A)} \frac{\mathrm{d}A}{\mathrm{d}t} = k \\ \int \frac{1}{A(10 - A)} \; \mathrm{d}A = \int k \; \mathrm{d}t \\ \int \frac{ 1 }{ 10 A } + \frac{ 1 }{ 100 - 10 A } \; \mathrm{d}A = kt + c \\ \frac{1}{10} \ln \left| A \right| - \frac{1}{10} \ln \left| 10 - A \right| = kt + c \\ \end{gather*}
d
t
d
A
=
k
A
(
10
−
A
)
■
A
(
10
−
A
)
1
d
t
d
A
=
k
∫
A
(
10
−
A
)
1
d
A
=
∫
k
d
t
∫
10
A
1
+
100
−
10
A
1
d
A
=
k
t
+
c
10
1
ln
∣
A
∣
−
10
1
ln
∣
10
−
A
∣
=
k
t
+
c
ln
∣
A
10
−
A
∣
=
10
k
t
+
10
c
∣
A
10
−
A
∣
=
e
10
k
t
+
10
c
∣
A
10
−
A
∣
=
e
10
c
e
10
k
t
A
10
−
A
=
b
e
10
k
t
\begin{align*} \ln \left| \frac{A}{10-A} \right | &= 10kt + 10c \\ \left| \frac{A}{10-A} \right | &= \mathrm{e}^{10kt + 10c} \\ \left| \frac{A}{10-A} \right | &= \mathrm{e}^{10c}\mathrm{e}^{10kt} \\ \frac{A}{10-A} &= b\mathrm{e}^{10kt} \\ \end{align*}
ln
10
−
A
A
10
−
A
A
10
−
A
A
10
−
A
A
=
10
k
t
+
10
c
=
e
10
k
t
+
10
c
=
e
10
c
e
10
k
t
=
b
e
10
k
t
When
t
=
0
,
A
=
2
,
{t=0, A=2,}
t
=
0
,
A
=
2
,
2
10
−
2
=
b
e
0
b
=
1
4
\begin{align*} \frac{2}{10-2} &= b \mathrm{e}^0 \\ b &= \frac{1}{4} \end{align*}
10
−
2
2
b
=
b
e
0
=
4
1
When
t
=
5
,
A
=
4
,
{t=5, A=4,}
t
=
5
,
A
=
4
,
4
10
−
4
=
1
4
e
10
k
(
5
)
k
=
ln
8
3
50
\begin{align*} \frac{4}{10-4} &= \frac{1}{4} \mathrm{e}^{10k(5)} \\ k &= \frac{\ln \frac{8}{3}}{50} \end{align*}
10
−
4
4
k
=
4
1
e
10
k
(
5
)
=
50
ln
3
8
When
A
=
8
,
{A=8,}
A
=
8
,
8
10
−
8
=
1
4
e
t
5
ln
8
3
e
t
5
ln
8
3
=
16
t
=
5
ln
16
ln
8
3
=
14.13
(2 d.p.)
■
\begin{align*} \frac{8}{10-8} &= \frac{1}{4} \mathrm{e}^{{ \frac{t}{5} \ln \frac{8}{3} } } \\ \mathrm{e}^{{ \frac{t}{5} \ln \frac{8}{3} } } &= 16 \\ t &= \frac{5 \ln 16}{\ln \frac{8}{3}} \\ &= 14.13 \textrm{ (2 d.p.)} \; \blacksquare \end{align*}
10
−
8
8
e
5
t
l
n
3
8
t
=
4
1
e
5
t
l
n
3
8
=
16
=
ln
3
8
5
ln
16
=
14.13
(2 d.p.)
■
(b)
When
t
=
24
,
{t=24,}
t
=
24
,
A
10
−
A
=
1
4
e
1
5
ln
(
8
3
)
(
24
)
A
=
(
10
−
A
)
27.707
A
=
27.71
m
2
(2 d.p.)
■
\begin{gather*} \frac{A}{10-A} = \frac{1}{4} \mathrm{e}^{{ \frac{1}{5} \ln (\frac{8}{3}) } (24)} \\ A = (10-A) 27.707 \\ A = 27.71 \textrm{ m}^2 \textrm{ (2 d.p.)} \; \blacksquare \end{gather*}
10
−
A
A
=
4
1
e
5
1
l
n
(
3
8
)
(
24
)
A
=
(
10
−
A
)
27.707
A
=
27.71
m
2
(2 d.p.)
■
(c)
A
10
−
A
=
1
4
e
t
5
ln
8
3
4
A
=
10
e
t
5
ln
8
3
−
A
e
t
5
ln
8
3
A
(
4
+
e
t
5
ln
8
3
)
=
10
e
t
5
ln
8
3
\begin{gather*} \frac{A}{10-A} = \frac{1}{4}\mathrm{e}^{{ \frac{t}{5} \ln \frac{8}{3} } } \\ 4A = 10\mathrm{e}^{{ \frac{t}{5} \ln \frac{8}{3} } }- A\mathrm{e}^{{ \frac{t}{5} \ln \frac{8}{3} } } \\ A(4+\mathrm{e}^{{ \frac{t}{5} \ln \frac{8}{3} } }) = 10\mathrm{e}^{{ \frac{t}{5} \ln \frac{8}{3} } } \end{gather*}
10
−
A
A
=
4
1
e
5
t
l
n
3
8
4
A
=
10
e
5
t
l
n
3
8
−
A
e
5
t
l
n
3
8
A
(
4
+
e
5
t
l
n
3
8
)
=
10
e
5
t
l
n
3
8
A
=
10
e
t
5
ln
8
3
4
+
e
t
5
ln
8
3
=
10
4
e
−
t
5
ln
8
3
+
1
=
10
4
e
ln
(
8
3
)
−
t
+
1
=
10
4
(
3
8
)
t
+
1
■
\begin{align*} A &= \frac{10 \mathrm{e}^{{ \frac{t}{5} \ln \frac{8}{3} } }}{4 + \mathrm{e}^{{ \frac{t}{5} \ln \frac{8}{3} } }} \\ &= \frac{10}{4 \mathrm{e}^{-{ \frac{t}{5} \ln \frac{8}{3} } } + 1} \\ & = \frac{10}{4 \mathrm{e}^{\ln (\frac{8}{3})^{-t}} + 1} \\ & = \frac{10}{4 (\frac{3}{8})^{t} + 1} \; \blacksquare \end{align*}
A
=
4
+
e
5
t
l
n
3
8
10
e
5
t
l
n
3
8
=
4
e
−
5
t
l
n
3
8
+
1
10
=
4
e
l
n
(
3
8
)
−
t
+
1
10
=
4
(
8
3
)
t
+
1
10
■
0
{0}
0
t
{t}
t
A
{A}
A
2
{2}
2
A
=
10
{A=10}
A
=
10
A
=
f
(
t
)
{A=f(t)}
A
=
f
(
t
)
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