2018 H2 Maths Paper 1 Q10 — Worked Solution

Part (i)

2 marks

Approach

Differentiating the given equation with respect to $t$ turns $\frac{q}{C}$ into $\frac{1}{C}\frac{dq}{dt} = \frac{I}{C}$ (using $I = \frac{dq}{dt}$ ) and turns $V$ into $\frac{dV}{dt}$ . The target equation has no term on the right, so this only works if $\frac{dV}{dt} = 0$ — the condition to state is that $V$ is constant.

Since $L\frac{dI}{dt} + RI + \frac{q}{C} = V$ , differentiating both sides with respect to $t$ ,

L\frac{d^{2}I}{dt^{2}} + R\frac{dI}{dt} + \frac{1}{C}\frac{dq}{dt} = \frac{dV}{dt}.

Using $I = \frac{dq}{dt}$ ,

L\frac{d^{2}I}{dt^{2}} + R\frac{dI}{dt} + \frac{I}{C} = \frac{dV}{dt}.

Provided $V$ is constant (does not vary with time), $\frac{dV}{dt} = 0$ , so

L\frac{d^{2}I}{dt^{2}} + R\frac{dI}{dt} + \frac{I}{C} = 0\ \ \left( \text{shown} \right)

Part (ii)

5 marks

Approach

Substitute $I = \text{Ate}^{- \frac{Rt}{2L}}$ into the differential equation of (i) and collect the result as a polynomial in $t$ multiplying $e^{- \frac{Rt}{2L}}$ . Since $e^{- \frac{Rt}{2L}} \neq 0$ , this polynomial must vanish for every $t$ , so each of its coefficients (the constant term and the coefficient of $t$ ) is separately zero; the coefficient of $t$ is the equation that gives $C$ .

Let $k = \frac{R}{2L}$ , so $I = \text{Ate}^{- kt}$ . Then

\frac{dI}{dt} = Ae^{- kt} - \text{Akte}^{- kt}

= Ae^{- kt}(1 - kt),

and

\frac{d^{2}I}{dt^{2}} = - \text{kAe}^{- kt}(1 - kt) + Ae^{- kt}( - k)

= Ae^{- kt}\left\lbrack - k(1 - kt) - k \right\rbrack

= Ae^{- kt}\left( k^{2}t - 2k \right).

Substituting into $L\frac{d^{2}I}{dt^{2}} + R\frac{dI}{dt} + \frac{I}{C} = 0$ ,

\text{LAe}^{- kt}\left( k^{2}t - 2k \right) + \text{RAe}^{- kt}(1 - kt) + \frac{1}{C}\text{Ate}^{- kt} = 0.

Since $A \neq 0$ and $e^{- kt} \neq 0$ , we can divide throughout by $Ae^{- kt}$ :

L\left( k^{2}t - 2k \right) + R(1 - kt) + \frac{t}{C} = 0

Collecting this as a polynomial in $t$ ,

\left( Lk^{2} - Rk + \frac{1}{C} \right)t + (R - 2Lk) = 0.

This must hold for every $t$ , so both the constant term and the coefficient of $t$ are zero. Since $k = \frac{R}{2L}$ , the constant term $R - 2Lk = R - 2L \cdot \frac{R}{2L} = 0$ automatically. The coefficient of $t$ then gives

Lk^{2} - Rk + \frac{1}{C} = 0.

Substituting $k = \frac{R}{2L}$ ,

L\left( \frac{R}{2L} \right)^{2} - R\left( \frac{R}{2L} \right) + \frac{1}{C} = 0

\frac{R^{2}}{4L} - \frac{R^{2}}{2L} + \frac{1}{C} = 0

- \frac{R^{2}}{4L} + \frac{1}{C} = 0

\frac{1}{C} = \frac{R^{2}}{4L}.

$\therefore C = \frac{4L}{R^{2}}\ \ \left( \text{shown} \right)$

Part (iii)

4 marks

Since $R = 4$ , $L = 3$ , $C = 0.75$ , we check $\frac{4L}{R^{2}} = \frac{4(3)}{16} = 0.75$ , consistent with (ii), and

k = \frac{R}{2L} = \frac{4}{6} = \frac{2}{3}.

So $I = \text{Ate}^{- \frac{2}{3}t}$ , with derivative (from (ii), with this $k$ )

\frac{dI}{dt} = Ae^{- \frac{2}{3}t}\left( 1 - \frac{2}{3}t \right).

At a stationary value, $\frac{dI}{dt} = 0$ . Since $Ae^{- \frac{2}{3}t} \neq 0$ (as $A > 0$ ),

1 - \frac{2}{3}t = 0

t = \frac{3}{2}.

The corresponding value of $I$ is

I = A\left( \frac{3}{2} \right)e^{- \frac{2}{3}\left( \frac{3}{2} \right)}

= \frac{3A}{2}e^{- 1}.

To confirm this is a maximum, we use the second derivative found in (ii), $\frac{d^{2}I}{dt^{2}} = Ae^{- kt}\left( k^{2}t - 2k \right)$ , evaluated at $t = \frac{3}{2} = \frac{1}{k}$ :

\left. \ \frac{d^{2}I}{dt^{2}} \right|_{t = \frac{3}{2}} = Ae^{- 1}\left( k^{2} \cdot \frac{1}{k} - 2k \right)

= Ae^{- 1}(k - 2k)

= - \text{Ake}^{- 1}

= - \frac{2A}{3e}.

Since $A > 0$ , this is negative, so $t = \frac{3}{2}$ gives a maximum.

$\therefore$ the maximum value of $I$ is $\frac{3A}{2e}$ , attained at $t = \frac{3}{2}$ .

Part (iv)

2 marks

Since $t$ is time after the power is switched on, we sketch $I = \text{Ate}^{- \frac{2}{3}t}$ for $t \geq 0$ . The curve starts at the origin ( $I = 0$ when $t = 0$ ), rises to the maximum turning point $\left( \frac{3}{2},\ \frac{3A}{2e} \right)$ found in (iii), then decreases and approaches $I = 0$ as $t \rightarrow \infty$ without ever reaching it (since $e^{- \frac{2}{3}t} > 0$ for all finite $t$ ).

2018 H2 Maths Paper 1, Question 10

Part (i)

Part (ii)

Part (iii)

Part (iv)

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