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2019
P1 Q11
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DE
19 P1 Q11
2019 H2 Mathematics Paper 1 Question 11
Differential Equations (DEs)
Answers
(ia)
d
θ
d
t
=
−
k
(
θ
−
16
)
{\frac{\mathrm{d}\theta}{\mathrm{d}t} = -k(\theta - 16)}
d
t
d
θ
=
−
k
(
θ
−
16
)
θ
=
16
+
64
(
1
4
)
t
30
{\theta = 16 + 64 \left(\frac{1}{4}\right)^{\frac{t}{30} }}
θ
=
16
+
64
(
4
1
)
30
t
(ib)
2
4
∘
C
{24 ^\circ \textrm{C}}
2
4
∘
C
(ii)
t
=
540
min
{t = 540 \textrm{ min}}
t
=
540
min
Full solutions
(ia)
d
θ
d
t
=
−
k
(
θ
−
16
)
■
1
θ
−
16
d
θ
d
t
=
−
k
∫
1
θ
−
16
d
θ
=
∫
−
k
d
t
ln
∣
θ
−
16
∣
=
−
k
t
+
c
∣
θ
−
16
∣
=
e
−
k
t
+
c
∣
θ
−
16
∣
=
e
c
e
−
k
t
θ
−
16
=
A
e
−
k
t
θ
=
16
+
A
e
−
k
t
\begin{gather*} \frac{\mathrm{d}\theta}{\mathrm{d}t} = -k(\theta - 16)\; \blacksquare \\ \frac{1}{\theta - 16} \frac{\mathrm{d}\theta}{\mathrm{d}t} = - k \\ \int \frac{1}{\theta - 16} \; \mathrm{d}\theta = \int -k \; \mathrm{d}t \\ \ln \left| \theta - 16\right| = -kt + c \\ |\theta - 16| = \mathrm{e}^{-kt+c} \\ |\theta - 16| = \mathrm{e}^c\mathrm{e}^{-kt} \\ \theta - 16 = A\mathrm{e}^{-kt} \\ \theta = 16 + A\mathrm{e}^{-kt} \\ \end{gather*}
d
t
d
θ
=
−
k
(
θ
−
16
)
■
θ
−
16
1
d
t
d
θ
=
−
k
∫
θ
−
16
1
d
θ
=
∫
−
k
d
t
ln
∣
θ
−
16
∣
=
−
k
t
+
c
∣
θ
−
16∣
=
e
−
k
t
+
c
∣
θ
−
16∣
=
e
c
e
−
k
t
θ
−
16
=
A
e
−
k
t
θ
=
16
+
A
e
−
k
t
When
t
=
0
,
θ
=
80
{t=0, \theta = 80}
t
=
0
,
θ
=
80
80
=
16
+
A
e
−
k
(
0
)
A
=
64
\begin{gather*} 80 = 16 + A\mathrm{e}^{-k(0)} \\ A = 64 \end{gather*}
80
=
16
+
A
e
−
k
(
0
)
A
=
64
When
t
=
30
,
θ
=
32
{t=30, \theta = 32}
t
=
30
,
θ
=
32
32
=
16
+
64
e
−
k
(
30
)
e
−
30
k
=
16
64
−
30
k
=
ln
1
4
k
=
−
1
30
ln
1
4
\begin{gather*} 32 = 16 + 64\mathrm{e}^{-k(30)} \\ \mathrm{e}^{-30k} = \frac{16}{64} \\ -30k = \ln \frac{1}{4} \\ k = -\frac{1}{30}\ln \frac{1}{4} \\ \end{gather*}
32
=
16
+
64
e
−
k
(
30
)
e
−
30
k
=
64
16
−
30
k
=
ln
4
1
k
=
−
30
1
ln
4
1
θ
=
16
+
64
e
(
1
30
ln
1
4
)
t
=
16
+
64
e
ln
1
4
t
30
θ
=
16
+
64
(
1
4
)
t
30
■
\begin{align*} \theta &= 16 + 64\mathrm{e}^{\left(\frac{1}{30} \ln \frac{1}{4}\right) t} \\ &= 16 + 64\mathrm{e}^{\ln \frac{1}{4}^{\frac{t}{30}} } \\ \theta &= 16 + 64 \left(\frac{1}{4}\right)^{\frac{t}{30} } \; \blacksquare \end{align*}
θ
θ
=
16
+
64
e
(
30
1
l
n
4
1
)
t
=
16
+
64
e
l
n
4
1
30
t
=
16
+
64
(
4
1
)
30
t
■
(ib)
When
t
=
45
,
{t=45,}
t
=
45
,
θ
=
16
+
64
(
1
4
)
45
30
=
2
4
∘
C
■
\begin{align*} \theta &= 16 + 64 \left(\frac{1}{4}\right)^{\frac{45}{30} } \\ &= 24 ^\circ \textrm{C} \; \blacksquare \end{align*}
θ
=
16
+
64
(
4
1
)
30
45
=
2
4
∘
C
■
(ii)
d
T
d
t
=
k
2
T
T
d
T
d
t
=
k
2
∫
T
d
T
=
∫
k
2
d
t
1
2
T
2
=
k
2
t
+
C
\begin{align*} \frac{\mathrm{d}T}{\mathrm{d}t} &= \frac{k_2}{T} \\ T \frac{\mathrm{d}T}{\mathrm{d}t} &= k_2 \\ \int T \; \mathrm{d}T &= \int k_2 \; \mathrm{d}t \\ \frac{1}{2} T^2 &= k_2 t + C \end{align*}
d
t
d
T
T
d
t
d
T
∫
T
d
T
2
1
T
2
=
T
k
2
=
k
2
=
∫
k
2
d
t
=
k
2
t
+
C
When
t
=
0
,
T
=
0
{t=0, T=0}
t
=
0
,
T
=
0
0
=
k
2
(
0
)
+
C
C
=
0
\begin{gather*} 0 = k_2(0) + C \\ C = 0 \end{gather*}
0
=
k
2
(
0
)
+
C
C
=
0
When
t
=
60
,
T
=
1
{t=60, T=1}
t
=
60
,
T
=
1
1
2
(
1
)
2
=
k
2
(
60
)
k
=
1
120
\begin{gather*} \frac{1}{2}(1)^2 = k_2(60) \\ k = \frac{1}{120} \end{gather*}
2
1
(
1
)
2
=
k
2
(
60
)
k
=
120
1
1
2
T
2
=
1
120
t
\frac{1}{2}T^2 = \frac{1}{120}t
2
1
T
2
=
120
1
t
When it is first safe to skate on,
T
=
3
{T=3}
T
=
3
1
2
(
3
)
2
=
1
120
t
t
=
540
min
■
\begin{gather*} \frac{1}{2}(3)^2 = \frac{1}{120} t \\ t = 540 \textrm{ min} \; \blacksquare \end{gather*}
2
1
(
3
)
2
=
120
1
t
t
=
540
min
■
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