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2018
P1 Q10
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Maxima
18 P1 Q10
2018 H2 Mathematics Paper 1 Question 10
Differentiation II: Maxima, Minima, Rates of Change
Answers
(i)
Under the conditions that
V
{V}
V
is a constant
(ii)
C
=
4
L
R
2
{C = \frac{4L}{R^2}}
C
=
R
2
4
L
(iii)
Maximum value of
I
=
3
A
2
e
{I = \frac{3A}{2\mathrm{e}}}
I
=
2
e
3
A
(iv)
Full solutions
(i)
L
d
I
d
t
+
R
I
+
q
C
=
V
L\frac{\mathrm{d}I}{\mathrm{d}t} + RI + \frac{q}{C} = V
L
d
t
d
I
+
R
I
+
C
q
=
V
Differentiating w.r.t.
t
,
{t,}
t
,
L
d
2
I
d
t
2
+
R
d
I
d
t
+
1
C
d
q
d
t
=
d
V
d
t
L\frac{\mathrm{d}^{2}I}{\mathrm{d}t^{2}} + R \frac{\mathrm{d}I}{\mathrm{d}t} + \frac{1}{C} \frac{\mathrm{d}q}{\mathrm{d}t} = \frac{\mathrm{d}V}{\mathrm{d}t}
L
d
t
2
d
2
I
+
R
d
t
d
I
+
C
1
d
t
d
q
=
d
t
d
V
Under the conditions that
V
{V}
V
is a constant
■
{\; \blacksquare}
■
d
V
d
t
=
0
L
d
2
I
d
t
2
+
R
d
I
d
t
+
I
C
=
0
■
\begin{gather*} \frac{\mathrm{d}V}{\mathrm{d}t}=0 \\ L\frac{\mathrm{d}^{2}I}{\mathrm{d}t^{2}} + R \frac{\mathrm{d}I}{\mathrm{d}t} + \frac{I}{C} = 0 \; \blacksquare \end{gather*}
d
t
d
V
=
0
L
d
t
2
d
2
I
+
R
d
t
d
I
+
C
I
=
0
■
(ii)
I
=
A
t
e
−
R
t
2
L
d
I
d
t
=
A
e
−
R
t
2
L
−
A
t
R
2
L
e
−
R
t
2
L
=
A
e
−
R
t
2
L
−
R
2
L
I
d
2
I
d
t
2
=
−
A
R
2
L
e
−
R
t
2
L
−
R
2
L
d
I
d
t
\begin{align*} I &= At \mathrm{e}^{-\frac{Rt}{2L}} \\ \frac{\mathrm{d}I}{\mathrm{d}t} &= A \mathrm{e}^{-\frac{Rt}{2L}} - At \frac{R}{2L} \mathrm{e}^{-\frac{Rt}{2L}} \\ &= A \mathrm{e}^{-\frac{Rt}{2L}} - \frac{R}{2L} I \\ \frac{\mathrm{d}^{2}I}{\mathrm{d}t^{2}} &= -\frac{AR}{2L} \mathrm{e}^{-\frac{Rt}{2L}} - \frac{R}{2L}\frac{\mathrm{d}I}{\mathrm{d}t} \\ \end{align*}
I
d
t
d
I
d
t
2
d
2
I
=
A
t
e
−
2
L
Rt
=
A
e
−
2
L
Rt
−
A
t
2
L
R
e
−
2
L
Rt
=
A
e
−
2
L
Rt
−
2
L
R
I
=
−
2
L
A
R
e
−
2
L
Rt
−
2
L
R
d
t
d
I
Since
I
=
A
t
e
−
R
t
2
L
{I=At\mathrm{e}^{-\frac{Rt}{2L}}}
I
=
A
t
e
−
2
L
Rt
is a solution, we can substitute the above into the differential equation
L
(
−
A
R
2
L
e
−
R
t
2
L
−
R
2
L
d
I
d
t
)
+
R
d
I
d
t
+
I
C
=
0
−
A
R
2
e
−
R
t
2
L
+
R
2
d
I
d
t
+
I
C
=
0
−
A
R
2
e
−
R
t
2
L
+
R
2
(
A
e
−
R
t
2
L
−
R
2
L
I
)
+
I
C
=
0
\begin{gather*} L \left( -\frac{AR}{2L} \mathrm{e}^{-\frac{Rt}{2L}} - \frac{R}{2L}\frac{\mathrm{d}I}{\mathrm{d}t} \right) + R \frac{\mathrm{d}I}{\mathrm{d}t} + \frac{I}{C} = 0 \\ -\frac{AR}{2}\mathrm{e}^{-\frac{Rt}{2L}} + \frac{R}{2} \frac{\mathrm{d}I}{\mathrm{d}t} + \frac{I}{C} = 0 \\ -\frac{AR}{2}\mathrm{e}^{-\frac{Rt}{2L}} + \frac{R}{2} \left( A \mathrm{e}^{-\frac{Rt}{2L}} - \frac{R}{2L} I \right) + \frac{I}{C} = 0 \\ \end{gather*}
L
(
−
2
L
A
R
e
−
2
L
Rt
−
2
L
R
d
t
d
I
)
+
R
d
t
d
I
+
C
I
=
0
−
2
A
R
e
−
2
L
Rt
+
2
R
d
t
d
I
+
C
I
=
0
−
2
A
R
e
−
2
L
Rt
+
2
R
(
A
e
−
2
L
Rt
−
2
L
R
I
)
+
C
I
=
0
−
R
2
4
L
I
+
I
C
=
0
I
C
=
R
2
4
L
C
=
4
L
R
2
■
\begin{gather*} -\frac{R^2}{4L}I + \frac{I}{C} = 0 \\ \frac{I}{C} = \frac{R^2}{4L} \\ C = \frac{4L}{R^2} \; \blacksquare \end{gather*}
−
4
L
R
2
I
+
C
I
=
0
C
I
=
4
L
R
2
C
=
R
2
4
L
■
(iii)
I
=
A
t
e
−
4
t
2
(
3
)
=
A
t
e
−
2
3
t
d
I
d
t
=
A
e
−
2
3
t
−
2
3
A
t
e
−
2
3
t
=
(
1
−
2
3
t
)
A
e
−
2
3
t
d
2
I
d
t
2
=
−
2
3
A
e
−
2
3
t
−
2
3
(
1
−
2
3
t
)
A
e
−
2
3
t
\begin{align*} I &= At \mathrm{e}^{-\frac{4t}{2(3)}} \\ &= At \mathrm{e}^{-\frac{2}{3}t} \\ \frac{\mathrm{d}I}{\mathrm{d}t} &= A \mathrm{e}^{-\frac{2}{3}t} - \frac{2}{3} At \mathrm{e}^{-\frac{2}{3}t} \\ &= \left(1-\frac{2}{3}t\right)A \mathrm{e}^{-\frac{2}{3}t} \\ \frac{\mathrm{d}^{2}I}{\mathrm{d}t^{2}} &= -\frac{2}{3}A \mathrm{e}^{-\frac{2}{3}t} - \frac{2}{3}\left(1-\frac{2}{3}t\right)A \mathrm{e}^{-\frac{2}{3}t} \end{align*}
I
d
t
d
I
d
t
2
d
2
I
=
A
t
e
−
2
(
3
)
4
t
=
A
t
e
−
3
2
t
=
A
e
−
3
2
t
−
3
2
A
t
e
−
3
2
t
=
(
1
−
3
2
t
)
A
e
−
3
2
t
=
−
3
2
A
e
−
3
2
t
−
3
2
(
1
−
3
2
t
)
A
e
−
3
2
t
At maximum
I
,
d
I
d
t
=
0
{\displaystyle I, \frac{\mathrm{d}I}{\mathrm{d}t}=0}
I
,
d
t
d
I
=
0
(
1
−
2
3
t
)
A
e
−
2
3
t
=
0
1
−
2
3
t
=
0
t
=
3
2
\begin{gather*} \left(1-\frac{2}{3}t\right)A \mathrm{e}^{-\frac{2}{3}t} = 0\\ 1-\frac{2}{3}t = 0\\ t = \frac{3}{2} \end{gather*}
(
1
−
3
2
t
)
A
e
−
3
2
t
=
0
1
−
3
2
t
=
0
t
=
2
3
Maximum value of
I
=
A
(
3
2
)
e
−
2
3
(
3
2
)
=
3
2
A
e
−
1
=
3
A
2
e
■
\begin{align*} & \textrm{Maximum value of } I \\ & = A \left(\frac{3}{2}\right) \mathrm{e}^{-\frac{2}{3}\left(\frac{3}{2}\right)} \\ &= \frac{3}{2}A\mathrm{e}^{-1} \\ &= \frac{3A}{2\mathrm{e}} \; \blacksquare \end{align*}
Maximum value of
I
=
A
(
2
3
)
e
−
3
2
(
2
3
)
=
2
3
A
e
−
1
=
2
e
3
A
■
When
t
=
3
2
,
{t=\frac{3}{2},}
t
=
2
3
,
d
2
I
d
t
2
=
−
2
3
A
e
−
2
3
(
3
2
)
−
0
=
−
2
3
A
e
−
1
<
0
\begin{align*} \frac{\mathrm{d}^{2}I}{\mathrm{d}t^{2}} &= -\frac{2}{3} A \mathrm{e}^{-\frac{2}{3}\left(\frac{3}{2}\right)} - 0 \\ &= -\frac{2}{3} A \mathrm{e}^{-1} \\ &< 0 \end{align*}
d
t
2
d
2
I
=
−
3
2
A
e
−
3
2
(
2
3
)
−
0
=
−
3
2
A
e
−
1
<
0
Hence
I
=
3
A
2
e
{I=\frac{3A}{2\mathrm{e}}}
I
=
2
e
3
A
is a maximum
■
{\blacksquare}
■
(iv)
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