(i) The rate of change of the height $h$ against time $t$ is represented by ${dh \over dt}$

\begin{align} h & = {t(10 - t) \over 4} \\ & = {10t - t^2 \over 4} \\ & = {10t \over 4} - {t^2 \over 4} \\ & = {5 \over 2}t - {1 \over 4}t^2 \\ \\ {dh \over dt} & = {5 \over 2} - {1 \over 4}(2)(t) \\ & = {5 \over 2} - {1 \over 2}t \\ \\ \text{When } & t = 2, \\ {dh \over dt} & = {5 \over 2} - {1 \over 2}(2) \\ & = 1.5 \text{ km/s} \end{align}

(ii)

\begin{align} \text{Substitute } & t = 6 \text{ into } {dh \over dt}, \\ {dh \over dt} & = {5 \over 2} - {1 \over 2}(6) \\ & = -0.5 \text{ km/s} \end{align}

(i) The rate of change of the length $l$ against time $t$ is represented by ${dl \over dt}$

\begin{align} l & = {t^3 \over 3} - 4t + 10 \\ \\ {dl \over dt} & = {(3)(t^2) \over 3} - 4(1) + 0 \\ & = t^2 - 4 \\ \\ \text{When } & {dl \over dt} = 5, \\ 5 & = t^2 - 4 \\ 9 & = t^2 \\ \pm\sqrt{9} & = t \\ \pm 3 & = t \\ \\ t & = 3 \phantom{00}\text{ or }\phantom{00} t = -3 \text{ (N.A.)} \end{align}

(ii) Since the length is decreasing at a rate of 4 mm/s, ${dl \over dt} = - 4$

\begin{align} \text{When } & {dl \over dt} = -4, \\ -4 & = t^2 - 4 \\ -4 + 4 & = t^2 \\ 0 & = t^2 \\ \pm\sqrt{0} & = t \\ 0 & = t \end{align}

(i)

\begin{align} {dr \over dt} & = 2t \\ \\ \text{When } & t = 2, \\ {dr \over dt} & = 2(2) \\ & = 4 \text{ cm/s} \end{align}

(ii) The graph is a quadratic curve (minimum curve $\cup$)

\begin{align} r & = t^2 + 2 \\ & = (t - 0)^2 + 2 \\ \\ \text{Since coefficient} &\text{ of } t^2 \text{ is positive, it is a minimum curve } (\cap) \\ \\ \text{Turni} & \text{ng point is } (0, 2) \\ \\ \text{When } & t = 0, \\ r & = (0)^2 + 2 \\ & = 2 \\ \\ \implies & \text{Vertical intercept is } (0, 2) \end{align}

$$ \therefore \text{As } t \text{ increases, } r \text{ increases at an increasing rate} $$

(i)

\begin{align} \text{When } & t = 0, \\ r & = 3 + {2 \over 1 + (0)} \\ & = 5 \text{ cm} \end{align}

(ii)

\begin{align} r & = 3 + {2 \over 1 + t} \\ & = 3 + 2(1 + t)^{-1} \\ \\ {dr \over dt} & = 2(-1)(1 + t)^{-2} . (1) \phantom{00000} [\text{Chain rule}] \\ & = -2(1 + t)^{-2} \\ & = -2\left[ {1 \over (1 + t)^2} \right] \\ & = -{2 \over (1 + t)^2} \\ \\ \text{When } & t = 3, \\ {dr \over dt} & = -{2 \over [1 + (3)]^2} \\ & = -{1 \over 8} \text{ cm/s} \\ \\ \text{The ball} & \text{oon is deflating at this instant} \end{align}

(i) Going from 10 am to 11 am, one hour (60 minutes) has elapsed

\begin{align} \text{When } & t = 60, \\ f(t) & = {1 \over 10 000 000}[6(60)^2 + 5]^2 \\ & = 46.6776025 \\ & \approx 46 \text{ people} \end{align}

(ii)

\begin{align} f'(t) & = {1 \over 10000000}(2)(6t^2 + 5). (12t) \phantom{00000} [\text{Chain rule}] \\ & = {1 \over 5000000}(6t^2 + 5)(12t) \\ & = {12t(6t^2 + 5) \over 5000000} \\ \\ \text{When } & t = 60, \\ f'(t) & = {12(60)[6(60)^2 + 5] \over 5000000} \\ & = 3.11112 \\ & \approx 3 \text{ people/min} \end{align}

(i)

\begin{align} {ds \over dv} & = {(1) \over 8} + {(2)(v) \over 80} \\ & = {1 \over 8} + {v \over 40} \\ \\ \text{When } & v = 60, \\ {ds \over dv} & = {1 \over 8} + {(60) \over 40} \\ & = 1.625 \end{align}

(ii)

When the speed of the van is 60 km/h, the rate of change of the stopping distance s with respect to the speed of the van v is 1.625 km per km/h.

(i)(a)

\begin{align} \text{When } & x = 3, \\ C(x) & = {19200 \over (3)} \\ & = \$6400 \end{align}

(i)(b)

\begin{align} \text{When } & x = 8, \\ C(x) & = {19200 \over (8)} \\ & = \$2400 \end{align}

(ii) Note the question is asking for average rate of change. Use the answers from (i).

\begin{align} \text{Average rate of change} & = {\text{Change in annual gasoline cost} \over \text{Change in } x} \\ & = {2400 - 6400 \over 8 - 3} \\ & = -800 \text{ dollars per km/l} \end{align}

(iii)

\begin{align} C(x) & = {19200 \over x} \\ & = 19200 x^{-1} \\ \\ C'(x) & = 19200(-1)(x^{-2}) \\ & = -19200(x^{-2}) \\ & = -19200\left(1 \over x^2 \right) \\ & = -{19200 \over x^2} \\ \\ \text{When } & x = 5, \\ C'(x) & = -{19200 \over (5)^2} \\ & = -768 \text{ dollars per km/l} \end{align}