(i)

\begin{align} \text{Substitute } & x = 3 \text{ into equation of curve,} \\ y & = {10 \over (3)} - (3) \\ & = {1 \over 3} \\ \\ \text{Coordinates} & \text{ is } \left( 3, {1 \over 3} \right) \\ \\ y & = {10 \over x} - x \\ & = 10x^{-1} - x \\ \\ {dy \over dx} & = 10(-1)(x^{-2}) - (1) \\ & = -10 \left( 1 \over x^2 \right) - 1 \\ & = -{10 \over x^2} - 1 \\ \\ \text{When } & x = 3, \\ {dy \over dx} & = -{10 \over (3)^2} - 1 \\ & = -{19 \over 9} \\ \\ \text{Gradient of normal} \times -{19 \over 9} & = -1 \\ \text{Gradient of normal} & = -1 \div -{19 \over 9} \\ & = {9 \over 19} \\ \\ y & = mx + c \\ y & = {9 \over 19}x + c \\ \\ \text{Using } & \left(3, {1 \over 3}\right), \\ {1 \over 3} & = {9 \over 19}(3) + c \\ {1 \over 3} & = {27 \over 19} + c \\ -{62 \over 57} & = c \\ \\ \text{Eqn of normal: } y & = {9 \over 19}x - {62 \over 57} \\ 57y & = 27x - 62 \end{align}

(ii)

\begin{align} {dy \over dx} & = -{10 \over x^2} - 1 \\ \\ \text{When } & {dy \over dx} = -{7 \over 2}, \\ -{7 \over 2} & = -{10 \over x^2} - 1 \\ -{7 \over 2} + 1 & = -{10 \over x^2} \\ -{5 \over 2} & = -{10 \over x^2} \\ {5 \over 2} & = {10 \over x^2} \\ (x^2)5 & = 2(10) \\ 5x^2 & = 20 \\ x^2 & = {20 \over 5} \\ x^2 & = 4 \\ x & = \pm\sqrt{4} \\ x & = \pm 2 \end{align} \begin{align} x & = 2 && \text{ or } & x & = -2 \\ \\ \text{Substitute } & \text{into eqn of curve,} &&& \text{Substitute } & \text{into eqn of curve,} \\ y & = {10 \over (2)} - (2) &&& y & = {10 \over (-2)} - (-2) \\ & = 3 &&& & -3 \\ \\ \therefore & \phantom{.} (2, 3) &&& \therefore & \phantom{.} (-2, -3) \end{align}

(i)

\begin{align} 8y & = x^2 - kx + 17 \\ y & = {1 \over 8}x^2 - {k \over 8}x + {17 \over 8} \\ \\ {dy \over dx} & = {1 \over 8}(2)(x) - {k \over 8}(1) + 0 \\ & = {1 \over 4}x - {k \over 8} \end{align} \begin{align} \text{When } & x = -3, &&& \text{When } & x = 5, \\ {dy \over dx} & = {1 \over 4}(-3) - {k \over 8} &&& {dy \over dx} & = {1 \over 4}(5) - {k \over 8} \\ & = -{3 \over 4} - {k \over 8} &&& & = {5 \over 4} - {k \over 8} \\ & = -{1 \over 8} (6 + k) &&& & = {1 \over 8} (10 - k) \end{align}

(ii) Recall that for perpendicular lines, $m_1 \times m_2 = -1$

\begin{align} m_1 \times m_2 & = -1 \\ -{1 \over 8}(6 + k) \times {1 \over 8}(10 - k) & = -1 \\ -{1 \over 8} \left(1 \over 8\right)(6 + k)(10 - k) & = -1 \\ -{1 \over 64}(6 + k)(10 - k) & = - 1 \\ {1 \over 64}(6 + k)(10 - k) & = 1 \\ (6 + k)(10 - k) & = 64 \\ 60 - 6k + 10k - k^2 & = 64 \\ -k^2 + 4k + 60 - 64 & = 0 \\ -k^2 + 4k - 4 & = 0 \\ k^2 - 4k + 4 & = 0 \\ k^2 - 2(k)(2) + 2^2 & = 0 \\ (k - 2)^2 & = 0 \\ k - 2 & = \sqrt{0} \\ k - 2 & = 0 \\ k & = 2 \end{align}

(i)

\begin{align} {dy \over dx} & = 3(3)(x^2) - 5(1) + 0 \\ & = 9x^2 - 5 \\ \\ \text{When } & x = 1, \\ {dy \over dx} & = 9(1)^2 - 5 \\ & = 4 \\ \\ \text{Gradient of tangent} & = 4 \\ \\ y & = mx + c \\ y & = 4x + c \\ \\ \text{Using } & A(1, 2), \\ 2 & = 4(1) + c \\ 2 & = 4 + c \\ -2 & = c \\ \\ \text{Eqn of tangent:} & \phantom{0} y = 4x - 2 \end{align}

(ii) Recall that parallel lines have the same gradient

\begin{align} {dy \over dx} & = 9x^2 - 5 \\ \\ \text{When } & {dy \over dx} = 4, \\ 4 & = 9x^2 - 5 \\ 9 & = 9x^2 \\ 1 & = x^2 \\ \pm \sqrt{1} & = x \\ \pm 1 & = x \\ \\ x = 1 & \text{ (Point } A) \phantom{00} \text{ or } \phantom{00} x = -1 \\ \\ \text{Substitute } & x = -1 \text{ into eqn of curve,} \\ y & = 3(-1)^3 - 5(-1) + 4 \\ & = 6 \\ \\ \text{Coordinates} & \text{ of other point is } (-1, 6) \end{align}

(i)

\begin{align} y & = ax + {b \over x} \\ \\ \text{Using } & P(2, 7), \\ (7) & = a(2) + {b \over (2)} \\ 7 & = 2a + {b \over 2} \\ 14 & = 4a + b \\ 14 - 4a & = b \\ \\ b & = 14 - 4a \phantom{0}\text{ --- (1)} \\ \\ \\ \text{Eqn of normal: } y + 2x & = 11 \\ y & = -2x + 11 \\ \\ \text{Gradient of normal} & = -2 \\ \\ \text{Gradient of tangent} \times -2 & = -1 \\ \text{Gradient of tangent} & = {-1 \over -2} \\ & = {1 \over 2} \\ \\ \text{Eqn of curve: } y & = ax + {b \over x} \\ & = ax + b x^{-1} \\ \\ {dy \over dx} & = a(1) + b(-1)(x^{-2}) \\ & = a - b x^{-2} \\ & = a - {b \over x^2} \\ \\ \text{When } & x = 2, \\ {dy \over dx} & = a - {b \over 2^2} \\ & = a - {b \over 4} \\ \\ \text{When } & {dy \over dx} = {1 \over 2}, \\ {1 \over 2} & = a - {b \over 4} \\ 2 & = 4a - b \phantom{000} \text{--- (2)} \\ \\ \text{Substitute } & \text{(1) into (2),} \\ 2 & = 4a - (14 - 4a) \\ 2 & = 4a - 14 + 4a \\ 2 + 14 & = 8a \\ 16 & = 8a \\ {16 \over 8} & = a \\ 2 & = a \\ \\ \text{Substitute } & a = 2 \text{ into (1),} \\ b & = 14 - 4(2) \\ & = 6 \\ \\ \therefore a & = 2, b = 6 \end{align}

(ii)

\begin{align} \text{Eqn of curve: } y & = ax + {b \over x} \\ y & = 2x + {6 \over x} \phantom{000} \text{--- (3)} \\ \\ \text{Eqn of normal: } y & = -2x +11 \phantom{000} \text{--- (4)} \\ \\ \text{Substitute} & \text{ (4) into (3),} \\ -2x + 11 & = 2x + {6 \over x} \\ -2x^2 + 11x & = 2x^2 + 6 \\ -4x^2 + 11x - 6 & = 0 \\ 4x^2 - 11x + 6 & = 0 \\ (4x - 3)(x - 2) & = 0 \end{align} \begin{align} 4x - 3 & = 0 && \text{ or } & x - 2 & = 0 \\ 4x & = 3 &&& x & = 2 \phantom{0} (\text{Point } P) \\ x & = {3 \over 4} \\ \\ \text{Substitute } & \text{into (4),} \\ y & = -2 \left(3 \over 4\right) + 11 \\ & = {19 \over 2} \\ \\ \therefore & \phantom{.} Q \left({3 \over 4}, {19 \over 2}\right) \end{align}

\begin{align} u & = 2x^2 &&& v & = x - 1 \\ {du \over dx} & = 4x &&& {dv \over dx} & = 1 \end{align} \begin{align} {dy \over dx} & = {(x - 1)(4x) - (2x^2)(1) \over (x - 1)^2} \phantom{00000} [\text{Quotient rule}] \\ & = {4x(x - 1) - 2x^2 \over (x - 1)^2} \\ & = {4x^2 - 4x - 2x^2 \over (x - 1)^2} \\ & = {2x^2 - 4x \over (x - 1)^2} \\ & = {2x(x - 2) \over (x - 1)^2} \\ \\ \text{For } x > 2, & \phantom{.} x - 2 > 0 \text{ and } (x - 1)^2 > 0 \\ \\ \therefore {2x(x - 2) \over (x - 1)^2} & > 0 \\ {dy \over dx} & > 0 \\ \\ \therefore \text{Function is } & \text{increasing for } x > 2 \end{align}

(i)

\begin{align} V & = {60 \over p} \\ & = 60 p^{-1} \\ \\ {dV \over dp} & = 60(-1)(p^{-2}) \\ & = -60\left(1 \over p^2\right) \\ & = -{60 \over p^2} \end{align}

(ii)

\begin{align} {dp \over dt} & = 3 \text{ units per second} \\ \\ {dV \over dt} & = {dV \over dp} \times {dp \over dt} \\ & = -{60 \over p^2} \times 3 \\ & = -{180 \over p^2} \\ \\ \text{When } & p = 20, \\ {dV \over dt} & = -{180 \over (20)^2} \\ & = -0.45 \text{ cubic units per second} \end{align}

(i) Note: The area of the circular patch increases at a steady rate of 4 cm²/s.

\begin{align} \text{Let } A \text{ denote } & \text{the area of the circular patch} \\ \\ \text{After } 16 & \text{ seconds,} \\ A & = 4 \times 16 \\ & = 64 \text{ cm}^2 \\ \\ \text{Let } r \text{ denote } & \text{the radius of the circular patch} \\ \\ A & = \pi r^2 \\ \\ \text{When } & A = 64, \\ 64 & = \pi r^2 \\ {64 \over \pi} & = r^2 \\ {64 \over \pi} & = r^2 \\ \pm\sqrt{64 \over \pi} & = r \\ \pm {\sqrt{64} \over \sqrt{\pi}} & = r \\ \pm{8 \over \sqrt{\pi}} & = r \\ \\ r = {8 \over \sqrt{\pi}} \phantom{0}&\text{ or }\phantom{0} r = -{8 \over \sqrt{\pi}} \text{ (N.A.)} \end{align}

(ii)

\begin{align} A & = \pi r^2 \\ \\ {dA \over dr} & = \pi(2)(r) \\ & = 2\pi r \\ \\ {dA \over dt} & = {dA \over dr} \times {dr \over dt} \\ \\ {dr \over dt} & = { {dA \over dt} \over {dA \over dr} } \\ & = {4 \over 2\pi r} \\ & = {2 \over \pi r} \\ \\ \text{When } & r = {8 \over \sqrt{\pi}}, \\ {dr \over dt} & = {2 \over \pi \left(8 \over \sqrt{\pi}\right)} \\ & = {2 \over {8\pi \over \sqrt{\pi}} } \\ & = {2 \over 8 \sqrt{\pi} } \\ & = {1 \over 4\sqrt{\pi}} \text{ cm/s} \end{align}

(i) Formula for arc length (in radian) is $r \theta$

\begin{align} s & = r \theta \\ & = 8 \theta \\ \\ {ds \over d\theta} & = 8 \\ \\ {ds \over dt} & = {ds \over d\theta} \times {d\theta \over dt} \\ & = 8 \times {\pi \over 2} \\ & = {8\pi \over 2} \\ & = 4\pi \text{ cm/s} \end{align}

(ii) Formula for sector area (in radian) is ${1 \over 2} r^2 \theta$

\begin{align} A & = {1 \over 2} r^2 \theta \\ & = {1 \over 2} (8)^2 \theta \\ & = 32 \theta \\ \\ {dA \over d\theta} & = 32 \\ \\ {dA \over dt} & = {dA \over d\theta} \times {d\theta \over dt} \\ & = 32 \times {\pi \over 2} \\ & = {32\pi \over 2} \\ & = 16\pi \text{ cm}^2 \text{/s} \end{align}

(i)

\begin{align} u & = x &&& v & = \sqrt{x - 2} \\ & &&& & = (x - 2)^{1 \over 2} \\ {du \over dx} & = 1 &&& {dv \over dx} & = {1 \over 2}(x - 2)^{-{1 \over 2}} . (1) \phantom{00000} [\text{Chain rule}] \\ & &&& & = {1 \over 2}(x - 2)^{-{1 \over 2}} \\ & &&& & = {1 \over 2\sqrt{x - 2}} \end{align} \begin{align} {dy \over dx} & = { (\sqrt{x - 2})(1) - (x)\left(1 \over 2\sqrt{x - 2}\right) \over (\sqrt{x - 2})^2} \phantom{000000} [\text{Quotient rule}] \\ & = {\sqrt{x - 2} - {x \over 2\sqrt{x - 2}} \over x - 2} \\ & = {\sqrt{x - 2} - {x \over 2\sqrt{x - 2}} \over x - 2} \times {2\sqrt{x - 2} \over 2\sqrt{x - 2}} \\ & = {\sqrt{x - 2}(2\sqrt{x - 2}) - x \over (x - 2)(2\sqrt{x - 2})} \\ & = {2(x - 2) - x \over 2(x - 2)\sqrt{x - 2}} \\ & = {2x - 4 - x \over 2(x - 2)(x - 2)^{1 \over 2}} \\ & = {x - 4 \over 2(x - 2)^{3 \over 2}} \\ & = {x - 4 \over 2 \left(\sqrt{x - 2}\right)^3} \phantom{0} \text{ (Shown)} \end{align}

(ii)

\begin{align} {dy \over dx} & = {x - 4 \over 2 \left(\sqrt{x - 2}\right)^3} \\ \\ \text{When } & x = 4, \\ {dy \over dx} & = {(4) - 4 \over 2(\sqrt{(4) - 2})^3} \\ & = 0 \\ \\ \implies \text{Tangent } & \text{at } x = 4 \text{ is a horizontal line} \\ \implies \text{Normal } & \text{at } x = 4 \text{ is a vertical line} \\ \\ \therefore \text{Eqn of nor} & \text{mal: } x = 4 \end{align}

(i) Recall that parallel lines have the same gradient

\begin{align} \text{Curve 1: }y & = ax^2 + bx + 2 \\ \\ \text{Using } & \left(1 , {1 \over 2}\right), \\ {1 \over 2} & = a(1)^2 + b(1) + 2 \\ {1 \over 2} & = a + b + 2 \\ {1 \over 2} - 2 & = a + b \\ -{3 \over 2} & = a + b \\ \\ a & = -b - {3 \over 2} \phantom{0} \text{ --- (1)} \\ \\ \\ \text{Curve 1: } {dy \over dx} & = a(2)x + b(1) \\ & = 2ax + b \\ \\ \text{When } & x = 1, \\ {dy \over dx} & = 2a(1) + b \\ & = 2a + b \\ \\ \\ \text{Curve 2: } y & = x^2 + 6x + 4 \\ \\ {dy \over dx} & = 2x + 6 \\ \\ \text{When } & x = -2, \\ {dy \over dx} & = 2(-2) + 6 \\ & = 2 \\ \\ \text{Gradient of normal} & = {-1 \over 2} \\ & = -{1 \over 2} \\ \\ \therefore 2a + b & = -{1 \over 2} \phantom{000} \text{--- (2)} \\ \\ \text{Substitute } & \text{(1) into (2),} \\ 2\left(-b - {3 \over 2}\right) + b & = -{1 \over 2} \\ -2b - 3 + b & = -{1 \over 2} \\ -2b + b & = -{1 \over 2} + 3 \\ -b & = {5 \over 2} \\ b & = -{5 \over 2} \\ \\ \text{Substitute } & b = -{5 \over 2} \text{ into (1),} \\ a & = -\left(-{5 \over 2}\right) - {3 \over 2} \\ & = 1 \\ \\ \therefore a & = 1, b = -{5 \over 2} \end{align}

(ii)

\begin{align} \text{Substitute } & x = 2 \text{ into eqn of tangent}, \\ y & = 2(2) - 16 \\ & = - 12 \\ \\ \text{Coordinates} & \text{ of point is } (2, -12) \\ \\ \text{Eqn of curve: } & y = ax^3 + bx \\ \\ \text{Using } & (2, -12), \\ -12 & = a(2)^3 + b(2) \\ -12 & = 8a + 2b \\ -6 & = 4a + b \phantom{0} \text{ --- (1)} \\ \\ \\ \text{Eqn of tangent: } & y = 2x - 16 \\ \\ \text{Gradient of tangent} & = 2 \\ \\ \text{Eqn of curve: } & y = ax^3 + bx \\ \\ {dy \over dx} & = a(3)x^2 + b(1) \\ & = 3ax^2 + b \\ \\ \text{When } & x = 2 \text{ and } {dy \over dx} = 2, \\ 2 & = 3a(2)^2 + b \\ 2 & = 12a + b \\ 2 - 12a & = b \\ \\ b & = 2 - 12a \phantom{0} \text{ --- (2)} \\ \\ \text{Substitute } & \text{(2) into (1),} \\ -6 & = 4a + (2 - 12a) \\ -6 & = 4a + 2 - 12a \\ -6 - 2 & = 4a - 12a \\ -8 & = -8a \\ {-8 \over -8} & = a \\ 1 & = a \\ \\ \text{Substitute } & a = 1 \text{ into (2),} \\ b & = 2 - 12(1) \\ & = -10 \\ \\ \therefore a & = 1, b = -10 \end{align}

Recall that for perpendicular lines, $m_1 \times m_2 = -1$

\begin{align} 4y & = x + 3 \\ y & = {1 \over 4}x + {3 \over 4} \\ \\ \text{Gradient of line} & = {1 \over 4} \\ \\ \text{Gradient of tangent} \times {1 \over 4} & = -1 \phantom{00000} [\text{Perpendicular lines}] \\ \text{Gradient of tangent} & = -1 \div {1 \over 4} \\ & = -4 \\ \\ \text{Eqn of curve: } & y = (x + 1)^2 \\ \\ {dy \over dx} & = 2(x + 1). (1) \phantom{00000} [\text{Chain rule}] \\ & = 2(x + 1) \\ \\ \text{When } & {dy \over dx} = -4, \\ -4 & = 2(x + 1) \\ {-4 \over 2} & = x + 1 \\ -2 & = x + 1 \\ -3 & = x \\ \\ \text{Substitute } & x = -3 \text{ into eqn of curve,} \\ y & = (-3 + 1)^2 \\ & = 4 \\ \\ \text{Coordinates} & \text{ of point is } (-3, 4) \\ \\ \text{Gradient of tangent at } (-3, 4) & = -4 \\ \\ y & = mx + c \\ y & = -4x + c \\ \\ \text{Using } & (-3, 4), \\ 4 & = -4(-3) + c \\ 4 & = 12 + c \\ -8 & = c \\ \\ \text{Eqn of tangent: } y & = -4x - 8 \\ \\ 4x + y + 8 & = 0 \end{align}

\begin{align} \require{cancel} y & = {ab \over x} \\ & = ab x^{-1} \\ \\ {dy \over dx} & = ab(-1)(x^{-2}) \\ & = -ab\left(1 \over x^2\right) \\ & = -{ab \over x^2} \\ \\ \text{When } & x = a, \\ {dy \over dx} & = -{ab \over a^2} \\ & = -{b \over a} \\ \\ \text{Gradient of tangent} & = -{b \over a} \\ \\ y & = mx + c \\ y & = -{b \over a}x + c \\ \\ \text{Using } & P(a, b), \\ b & = -{b \over a}(a) + c \\ b & = -b + c \\ 2b & = c \\ \\ \text{Eqn of tangent: } & y = -{b \over a}x + 2b \\ \\ \text{When } & y = 0, \\ 0 & = -{b \over a}x + 2b \\ {b \over a}x & = 2b \\ \\ x & = 2b \div {b \over a} \\ & = 2b \times {a \over b} \\ & = {2\cancel{b}a \over \cancel{b}} \\ & = 2a \\ \\ \therefore & \phantom{0} Q(2a, 0) \\ \\ \text{When } & x = 0, \\ y & = -{b \over a}(0) + 2b \\ & = 2b \\ \\ \therefore & \phantom{0} R(0, 2b) \\ \\ \text{Distance between two points} & = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \\ \\ PQ & = \sqrt{ (a - 2a)^2 + (b - 0)^2 } \\ & = \sqrt{ (-a)^2 + (b)^2} \\ & = \sqrt{a^2 + b^2} \\ \\ RP & = \sqrt{ (0 - a)^2 + (2b - b)^2 } \\ & = \sqrt{ (-a)^2 + (b)^2} \\ & = \sqrt{a^2 + b^2} \\ \\ \therefore PQ & = RP \text{ (Shown)} \end{align}

(i)

\begin{align} s & = {v \over 4} + {v^2 \over 60} \\ \\ {ds \over dv} & = {(1) \over 4} + {(2)(v) \over 60} \\ & = {1 \over 4} + {v \over 30} \\ \\ \text{When } & v = 45, \\ {ds \over dv} & = {1 \over 4} + {(45) \over 30} \\ & = 1.75 \text{ km per km/h} \end{align}

(ii)

When the vehicle is traveling at 45 km/h, for every 1 km/h increase (or decrease) in the speed of the vehicle, the stopping distance increases (or decreases) by 1.75 km.

(i)

\begin{align} y & = \sqrt{{22 \over x} - x} \\ & = \sqrt{22x^{-1} - x} \\ & = (22x^{-1} - x)^{1 \over 2} \\ \\ {dy \over dx} & = {1 \over 2}(22x^{-1} - x)^{-{1 \over 2}} . [22(-1)(x^{-2}) - 1] \phantom{00000} [\text{Chain rule}] \\ & = {1 \over 2}\left(1 \over \sqrt{22x^{-1} - x}\right)(-22x^{-2} - 1) \\ & = {1 \over 2\sqrt{22x^{-1} - x}}\left[-22\left(1 \over x^2\right) - 1\right] \\ & = {1 \over 2\sqrt{22x^{-1} - x}}\left(-{22 \over x^2} - 1\right) \\ & = {1 \over 2\sqrt{22x^{-1} - x}}(-1)\left({22 \over x^2} + 1\right) \\ & = {1 \over 2\sqrt{22x^{-1} - x}}(-1)\left({22 \over x^2} + {x^2 \over x^2}\right) \\ & = {1 \over 2\sqrt{22x^{-1} - x}}(-1)\left({22 + x^2 \over x^2}\right) \\ & = (-1)\left( 22 + x^2 \over 2x^2\sqrt{22x^{-1} - x} \right) \\ & = -{22 + x^2 \over 2x^2\sqrt{22x^{-1} - x}} \\ & = -{22 + x^2 \over 2x^2\sqrt{{22 \over x} - x}} \\ & = -{22 + x^2 \over 2x^2\sqrt{{1 \over x} (22 - x^2)}} \\ & = -{22 + x^2 \over 2x^2 \sqrt{1 \over x}\sqrt{22 - x^2}} \\ & = -{22 + x^2 \over 2x^2 (x^{-{1 \over 2}})\sqrt{22 - x^2}} \\ & = -{22 + x^2 \over 2x^{3 \over 2}\sqrt{22 - x^2}} \end{align}

(ii)

\begin{align} {dy \over dt} & = {dy \over dx} \times {dx \over dt} \\ & = -{22 + x^2 \over 2x^{3 \over 2}\sqrt{22 - x^2}} \times 2 \\ & = -{22 + x^2 \over x^{3 \over 2}\sqrt{22 - x^2}} \\ \\ \text{When } & x = 2, \\ {dy \over dt} & = -{22 + (2)^2 \over (2)^{3 \over 2} \sqrt{22 - (2)^2}} \\ & = -{13 \over 6} \text{ cm/s} \end{align}

(i)

\begin{align} RP & = p - (-1) \\ & = (p + 1) \text{ units} \\ \\ \text{Substitute } & x = p \text{ into eqn of curve,} \\ y & = (p)^2 \\ & = p^2 \\ \\ \therefore & \phantom{.} Q(p, p^2) \\ \\ PQ & = p^2 \text{ units} \\ \\ \text{Area of triangle }PQR, A & = {1 \over 2} \times \text{base} \times \text{height} \\ & = {1 \over 2} \times RP \times PQ \\ & = {1 \over 2} \times (p + 1) \times (p^2) \\ & = {1 \over 2}(p + 1)(p^2) \\ & = {1 \over 2}(p^3 + p^2) \text{ square units} \end{align}

(ii)

\begin{align} A & = {1 \over 2}(p^3 + p^2) \\ & = {1 \over 2}p^3 + {1 \over 2} p^2 \\ \\ {dA \over dp} & = {1 \over 2}(3)(p^2) + {1 \over 2}(2)(p) \\ & = {3 \over 2}p^2 + p \\ \\ {dA \over dt} & = {dA \over dp} \times {dp \over dt} \\ & = \left( {3 \over 2}p^2 + p \right) \times 3 \\ & = {9 \over 2} p^2 + 3p \\ \\ \text{When } & p = 5, \\ {dA \over dt} & = {9 \over 2} (5)^2 + 3(5) \\ & = 127.5 \text{ square units per second} \end{align}

(i)

\begin{align} \text{Let } x \text{ denote } & \text{the distance between the man and the lamp post} \\ \\ {dx \over dt} & = 2 \text{ m/s} \\ \\ \text{Let } y \text{ denote } & \text{the length of the man's shadow} \end{align}

Two similar triangles can be formed:

\begin{align} \text{By si} & \text{milar triangles,} \\ {5 \over 1.5} & = {x + y \over y} \\ 5y & = 1.5(x + y) \\ 5y & = 1.5x + 1.5y \\ 5y - 1.5y & = 1.5x \\ 3.5y & = 1.5x \\ y & = 1.5x \div 3.5 \\ y & = {3 \over 7}x \\ \\ {dy \over dx} & = {3 \over 7}(1) \\ & = {3 \over 7} \\ \\ {dy \over dt} & = {dy \over dx} \times {dx \over dt} \\ & = \left(3 \over 7 \right) \times (2) \\ & = {6 \over 7} \text{ m/s} \end{align}

(ii)

\begin{align} \text{Let } z \text{ denote } & \text{the distance between the top of his shadow and the lamp post} \end{align}

Two similar triangles can be formed:

\begin{align} \text{By si} & \text{milar triangles,} \\ {5 \over 1.5} & = {z \over y} \\ 5y & = 1.5z \\ \\ z & = 5y \div 1.5 \\ & = {10 \over 3}y \\ \\ {dz \over dy} & = {10 \over 3}(1) \\ & = {10 \over 3} \\ \\ {dz \over dt} & = {dz \over dy} \times {dy \over dt} \\ & = \left(10 \over 3\right) \times \left(6 \over 7 \right) \\ & = {20 \over 7} \text{ m/s} \end{align}