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2017
P1 Q11
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DE
17 P1 Q11
2017 H2 Mathematics Paper 1 Question 11
Differential Equations (DEs)
Answers
(ia)
d
v
d
t
=
c
{\displaystyle \frac{\mathrm{d}v}{\mathrm{d}t}=c}
d
t
d
v
=
c
(ib)
c
=
10
{c = 10}
c
=
10
v
=
10
t
+
4
{v=10t+4}
v
=
10
t
+
4
(ii)
v
=
10
−
10
e
−
k
t
k
{v = \frac{10-10\mathrm{e}^{-kt}}{k}}
v
=
k
10
−
10
e
−
k
t
(iii)
4
ln
10
≈
9.21
s
{4 \ln 10 \approx 9.21 \textrm{ s}}
4
ln
10
≈
9.21
s
Full solutions
(ia)
d
v
d
t
=
c
\frac{\mathrm{d}v}{\mathrm{d}t}=c
d
t
d
v
=
c
(ib)
d
v
d
t
=
c
v
=
c
t
+
d
\begin{gather*} \frac{\mathrm{d}v}{\mathrm{d}t} = c \\ v = ct + d \end{gather*}
d
t
d
v
=
c
v
=
c
t
+
d
When
t
=
0
,
v
=
4
,
{t=0, v=4,}
t
=
0
,
v
=
4
,
4
=
c
(
0
)
+
d
d
=
4
\begin{gather*} 4 = c(0) + d \\ d = 4 \end{gather*}
4
=
c
(
0
)
+
d
d
=
4
When
t
=
2.5
,
v
=
29
,
{t=2.5, v=29,}
t
=
2.5
,
v
=
29
,
29
=
c
(
2.5
)
+
4
c
=
10
■
\begin{gather*} 29 = c(2.5) + 4 \\ c = 10 \; \blacksquare \end{gather*}
29
=
c
(
2.5
)
+
4
c
=
10
■
v
=
10
t
+
4
■
v = 10t + 4 \; \blacksquare
v
=
10
t
+
4
■
(ii)
d
v
d
t
=
10
−
k
v
1
10
−
k
v
d
v
d
t
=
1
∫
1
10
−
k
v
d
v
=
∫
1
d
t
−
1
k
ln
∣
10
−
k
v
∣
=
t
+
C
ln
∣
10
−
k
v
∣
=
−
k
t
−
k
C
∣
10
−
k
v
∣
=
e
−
k
t
−
k
C
∣
10
−
k
v
∣
=
e
−
k
C
e
−
k
t
10
−
k
v
=
A
e
−
k
t
\begin{gather*} \frac{\mathrm{d}v}{\mathrm{d}t} = 10 - kv \\ \frac{1}{10-kv} \frac{\mathrm{d}v}{\mathrm{d}t} = 1 \\ \int \frac{1}{10-kv} \;\mathrm{d}v = \int 1 \;\mathrm{d}t \\ -\frac{1}{k} \ln | 10 - kv | = t + C \\ \ln | 10 - kv | = -kt - kC \\ | 10 - kv | = \mathrm{e}^{-kt - kC} \\ | 10 - kv | = \mathrm{e}^{-kC}\mathrm{e}^{-kt} \\ 10 - kv = A \mathrm{e}^{-kt} \end{gather*}
d
t
d
v
=
10
−
k
v
10
−
k
v
1
d
t
d
v
=
1
∫
10
−
k
v
1
d
v
=
∫
1
d
t
−
k
1
ln
∣10
−
k
v
∣
=
t
+
C
ln
∣10
−
k
v
∣
=
−
k
t
−
k
C
∣10
−
k
v
∣
=
e
−
k
t
−
k
C
∣10
−
k
v
∣
=
e
−
k
C
e
−
k
t
10
−
k
v
=
A
e
−
k
t
When
t
=
0
,
{t=0, }
t
=
0
,
v
=
0
,
{v=0,}
v
=
0
,
10
−
k
(
0
)
=
A
e
−
k
(
0
)
A
=
10
\begin{gather*} 10 - k(0) = A \mathrm{e}^{-k(0)} \\ A = 10 \end{gather*}
10
−
k
(
0
)
=
A
e
−
k
(
0
)
A
=
10
10
−
k
v
=
10
e
−
k
t
v
=
10
−
10
e
−
k
t
k
■
\begin{gather*} 10 - kv = 10 \mathrm{e}^{-kt} \\ v = \frac{10-10\mathrm{e}^{-kt}}{k} \; \blacksquare \end{gather*}
10
−
k
v
=
10
e
−
k
t
v
=
k
10
−
10
e
−
k
t
■
(iii)
As
t
→
∞
,
e
−
k
t
→
0
{t \to \infty, \mathrm{e}^{-kt} \to 0}
t
→
∞
,
e
−
k
t
→
0
v
=
10
−
10
e
−
k
t
k
→
10
k
\begin{align*} v &= \frac{10-10\mathrm{e}^{-kt}}{k} \\ &\to \frac{10}{k} \end{align*}
v
=
k
10
−
10
e
−
k
t
→
k
10
Since the terminal velocity is
40
,
{40,}
40
,
10
k
=
40
k
=
1
4
\begin{align*} \frac{10}{k} &= 40 \\ k &= \frac{1}{4} \end{align*}
k
10
k
=
40
=
4
1
At
90
%
{90\% }
90%
of the terminal velocity,
v
=
36
,
{v = 36,}
v
=
36
,
36
=
4
(
10
−
10
e
−
1
4
t
)
9
=
10
−
10
e
−
1
4
t
e
−
1
4
t
=
1
10
−
1
4
t
=
ln
1
10
t
=
−
4
ln
1
10
=
4
ln
10
s
■
\begin{align*} 36 &= 4 \left( 10-10\mathrm{e}^{-\frac{1}{4}t} \right) \\ 9 &= 10-10\mathrm{e}^{-\frac{1}{4}t} \\ \mathrm{e}^{-\frac{1}{4}t} &= \frac{1}{10} \\ -\frac{1}{4}t &= \ln \frac{1}{10} \\ t &= -4 \ln \frac{1}{10} \\ &= 4 \ln 10 \textrm{ s} \; \blacksquare \end{align*}
36
9
e
−
4
1
t
−
4
1
t
t
=
4
(
10
−
10
e
−
4
1
t
)
=
10
−
10
e
−
4
1
t
=
10
1
=
ln
10
1
=
−
4
ln
10
1
=
4
ln
10
s
■
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