(a)

\begin{align} y & = 2x^2 + {1 \over x} \\ & = 2x^2 + x^{-1} \\ \\ {dy \over dx} & = 2(2)(x) + (-1)(x^{-2}) \\ & = 4x - x^{-2} \\ & = 4x - {1 \over x^2} \\ \\ {dy \over dt} & = {dy \over dx} \times {dx \over dt} \\ & = \left(4x - {1 \over x^2}\right) \times 2 \\ & = 8x - {2 \over x^2} \\ \\ \text{When } & x = 1, \\ {dy \over dt} & = 8(1) - {2 \over (1)^2} \\ & = 6 \text{ units per second} \end{align}

(b)

\begin{align} y & = {3 \over (2x - 3)^3} \\ & = 3(2x - 3)^{-3} \\ \\ {dy \over dx} & = 3(-3)(2x - 3)^{-4} . (2) \phantom{000000} [\text{Chain rule}] \\ & = -18(2x - 3)^{-4} \\ & = -{18 \over (2x - 3)^4} \\ \\ {dy \over dt} & = {dy \over dx} \times {dx \over dt} \\ & = -{18 \over (2x - 3)^4} \times 2 \\ & = -{36 \over (2x - 3)^4} \\ \\ \text{When } & x = 2, \\ {dy \over dt} & = -{36 \over [2(2) - 3]^4} \\ & = -36 \text{ units per second} \end{align}

(c)

\begin{align} {dy \over dx} & = (3)(x^2) + 0 \\ & = 3x^2 \\ \\ {dy \over dt} & = {dy \over dx} \times {dx \over dt} \\ & = 3x^2 \times 2 \\ & = 6x^2 \\ \\ \text{Substitute } & y = 10 \text{ into } y = x^3 + 2, \\ 10 & = x^3 + 2 \\ 8 & = x^3 \\ \sqrt[3]{8} & = x \\ 2 & = x \\ \\ \text{Substitute } & x = 2 \text{ into } {dy \over dt}, \\ {dy \over dt} & = 6(2)^2 \\ & = 24 \text{ units per second} \end{align}

(d)

\begin{align} u & = x &&& v & = x + 1 \\ {du \over dx} & = 1 &&& {dv \over dx} & = 1 \end{align} \begin{align} {dy \over dx} & = {(x + 1)(1) - (x)(1) \over (x + 1)^2} \phantom{000000} [\text{Quotient rule}] \\ & = {x + 1 - x \over (x + 1)^2} \\ & = {1 \over (x + 1)^2} \\ \\ {dy \over dt} & = {dy \over dx} \times {dx \over dt} \\ & = {1 \over (x + 1)^2} \times 2 \\ & = {2 \over (x + 1)^2} \\ \\ \text{Substitute } & y = 2 \text{ into } y = {x \over x + 1}, \\ 2 & = {x \over x + 1} \\ 2(x + 1) & = x \\ 2x + 2 & = x \\ 2x - x & = -2 \\ x & = -2 \\ \\ \text{Substitute } & x = -2 \text{ into } {dy \over dt}, \\ {dy \over dt} & = {2 \over (-2 + 1)^2} \\ & = 2 \text{ units per second} \end{align}

(a)

\begin{align} {dy \over dx} & = (3)(x^2) - 2(2)(x) \\ & = 3x^2 - 4x \\ \\ {dy \over dt} & = {dy \over dx} \times {dx \over dt} \\ \\ {dx \over dt} & = { {dy \over dt} \over {dy \over dx} } \\ & = {4 \over 3x^2 - 4x} \\ \\ \text{When } & x = 3, \\ {dx \over dt} & = {4 \over 3(3)^2 - 4(3)} \\ & = {4 \over 15} \text{ units per second} \end{align}

(b)

\begin{align} u & = 3x^2 &&& v & = 1 + x \\ {du \over dx} & = 6x &&& {dv \over dx} & = 1 \end{align} \begin{align} {dy \over dx} & = { (1 + x)(6x) - (3x^2)(1) \over (1 + x)^2 } \phantom{000000} [\text{Quotient rule}] \\ & = {6x(1 + x) - 3x^2 \over (1 + x)^2} \\ & = { 6x + 6x^2 - 3x^2 \over (1 + x)^2 } \\ & = {6x + 3x^2 \over (1 + x)^2} \\ \\ {dy \over dt} & = {dy \over dx} \times {dx \over dt} \\ \\ {dx \over dt} & = { {dy \over dt} \over {dy \over dx} } \\ & = { 4 \over {6x + 3x^2 \over (1 + x)^2} } \\ & = {4(1 + x)^2 \over 6x + 3x^2} \\ \\ \text{When } & x = 2, \\ {dx \over dt} & = {4(1 + 2)^2 \over 6(2) + 3(2)^2} \\ & = {3 \over 2} \text{ units per second} \end{align}

(c)

\begin{align} y & = \sqrt{2x + 7} \\ & = (2x + 7)^{1 \over 2} \\ \\ {dy \over dx} & = {1 \over 2}(2x + 7)^{-{1 \over 2}} . (2) \phantom{000000} [\text{Chain rule}] \\ & = (2x + 7)^{-{1 \over 2}} \\ & = {1 \over \sqrt{2x + 7}} \\ \\ {dy \over dt} & = {dy \over dx} \times {dx \over dt} \\ \\ {dx \over dt} & = { {dy \over dt} \over {dy \over dx} } \\ & = {4 \over {1 \over \sqrt{2x + 7}} } \\ & = 4\sqrt{2x + 7} \\ \\ \text{Substitute } & y = 3 \text{ into } y = \sqrt{2x + 7}, \\ 3 & = \sqrt{2x + 7} \\ (3)^2 & = (\sqrt{2x + 7})^2 \\ 9 & = 2x + 7 \\ 2 & = 2x \\ {2 \over 2} & = x \\ 1 & = x \\ \\ \text{Substitute } & x = 1 \text{ into } {dx \over dt}, \\ {dx \over dt} & = 4\sqrt{2(1) + 7} \\ & = 12 \text{ units per second} \end{align}

(d)

\begin{align} y & = x(x - 4) \\ & = x^2 - 4x \\ \\ {dy \over dx} & = 2x - 4 \\ \\ {dy \over dt} & = {dy \over dx} \times {dx \over dt} \\ \\ {dx \over dt} & = { {dy \over dt} \over {dy \over dx} } \\ & = { 4 \over 2x - 4} \\ \\ \text{Substitute } & y = 5 \text{ into } y = x(x - 4), \\ 5 & = x(x - 4) \\ 5 & = x^2 - 4x \\ 0 & = x^2 - 4x - 5 \\ 0 & = (x - 5)(x + 1) \\ \\ x - 5 = 0 \phantom{000} & \text{or} \phantom{000} x + 1 = 0 \\ x = 5 \phantom{000} & \phantom{or000+1} x = - 1 \text{ (Reject, since } x > 0) \\ \\ \text{Substitute } & x = 5 \text{ into } {dx \over dt}, \\ {dx \over dt} & = {4 \over 2(5) - 4} \\ & = {2 \over 3} \text{ units per second} \end{align}

(i)

\begin{align} u & = 2x - 1 &&& v & = x + 1 \\ {du \over dx} & = 2 &&& {dv \over dx} & = 1 \end{align} \begin{align} {dy \over dx} & = {(x + 1)(2) - (2x - 1)(1) \over (x + 1)^2} \phantom{000000} [\text{Quotient rule}] \\ & = {2(x + 1) - (2x - 1) \over (x + 1)^2} \\ & = {2x + 2 - 2x + 1 \over (x + 1)^2} \\ & = {3 \over (x + 1)^2} \end{align}

(ii) The point A, where the curve cuts the y-axis, is the y-intercept (where x = 0)

\begin{align} \text{Substitute } & x = 0 \text{ into eqn of curve,} \\ y & = {2(0) - 1 \over (0) + 1} \\ & = {-1 \over 1} \\ & = - 1 \\ \\ \therefore & \phantom{.} A(0, -1) \\ \\ \text{Substitute } & x = 0 \text{ into } {dy \over dx}, \\ {dy \over dx} & = {3 \over (0 + 1)^2} \\ & = 3 \\ \\ \text{Gradient of tangent at } A & = 3 \\ \\ y & = mx + c \\ y & = 3x + c \\ \\ \text{Using } & A(0, -1), \\ -1 & = 3(0) + c \\ -1 & = c \\ \\ \text{Eqn of tangent: } & y = 3x - 1 \end{align}

(iii)

\begin{align} {dx \over dt} & = 0.03 \text{ units per second} \\ \\ {dy \over dx} & = 3 \phantom{00000} [\text{from (ii)]} \\ \\ {dy \over dt} & = {dy \over dx} \times {dx \over dt} \\ & = 3 \times 0.03 \\ & = 0.09 \text{ units per second} \end{align}

(i)

\begin{align} u & = 2x - 10 &&& v & = x + 1 \\ {du \over dx} & = 2 &&& {dv \over dx} & = 1 \end{align} \begin{align} {dy \over dx} & = {(x + 1)(2) - (2x - 10)(1) \over (x + 1)^2} \phantom{000000} [\text{Quotient rule}] \\ & = {2(x + 1) - (2x - 10) \over (x + 1)^2} \\ & = {2x + 2 - 2x + 10 \over (x + 1)^2} \\ & = {12 \over (x + 1)^2} \\ \\ \text{Since } (x + 1)^2 > 0 & \text{ for } x \ne -1, \\ {12 \over (x + 1)^2} & > 0 \\ \therefore {dy \over dx} & > 0 \end{align}

(ii)

\begin{align} {dy \over dt} & = {dy \over dx} \times {dx \over dt} \\ \\ \text{Since } & {dy \over dt} = 3{dx \over dt}, \\ {dy \over dx} & = 3 \\ {12 \over (x + 1)^2} & = 3 \\ 12 & = 3(x + 1)^2 \\ {12 \over 3} & = (x + 1)^2 \\ 4 & = (x + 1)^2 \\ \pm \sqrt{4} & = x + 1 \\ \pm 2 & = x + 1 \end{align} \begin{align} x + 1 & = 2 && \text{ or } & x + 1 & = -2 \\ x & = 1 &&& x & = - 3 \end{align}

(a)

\begin{align} \text{Let } A \text{ denote } & \text{the area of the circle} \\ \text{Let } r \text{ denote } & \text{the radius of the circle} \\ \\ {dA \over dt} & = 2\pi \text{ cm}^2 \text{/s} \\ \\ 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} } \\ & = { 2 \pi \over 2\pi r } \\ & = {1 \over r} \\ \\ \text{When } & r = 6, \\ {dr \over dt} & = {1 \over 6} \text{ cm/s} \end{align}

(b)

\begin{align} \text{Let } A \text{ denote } & \text{the area of the circle} \\ \text{Let } r \text{ denote } & \text{the radius of the circle} \\ \\ {dA \over dt} & = 10\pi \text{ m}^2 \text{/s} \\ \\ 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} } \\ & = { 10 \pi \over 2\pi r } \\ & = {5 \over r} \\ \\ \text{When } & r = 2, \\ {dr \over dt} & = {5 \over 2} \text{ m/s} \end{align}

(i)

\begin{align} {dV \over dt} & = 3 \text{ cm}^3 \text{/s} \\ \\ V & = x^2 + 3x \\ \\ {dV \over dx} & = (2)(x) + 3(1) \\ & = 2x + 3 \\ \\ {dV \over dt} & = {dV \over dx} \times {dx \over dt} \\ \\ {dx \over dt} & = { {dV \over dt} \over {dV \over dx} } \\ & = { 3 \over 2x + 3 } \end{align}

(ii)

\begin{align} \text{Substitute } & x = 3 \text{ into } {dx \over dt}, \\ {dx \over dt} & = {3 \over 2(3) + 3} \\ & = {1 \over 3} \text{ cm/s} \\ \\ \text{Substitute } & x = 6 \text{ into } {dx \over dt}, \\ {dx \over dt} & = {3 \over 2(6) + 3} \\ & = {1 \over 5} \text{ cm/s} \end{align}

(a)

\begin{align} \text{Let } A \text{ denote } & \text{the area of the square} \\ \text{Let } l \text{ denote } & \text{the length of the square} \\ \\ {dA \over dt} & = 10 \text{ cm}^2 \text{/s} \\ \\ A & = l \times l \\ A & = l^2 \\ \\ {dA \over dl} & = 2l \\ \\ {dA \over dt} & = {dA \over dl} \times {dl \over dt} \\ \\ {dl \over dt} & = { {dA \over dt} \over {dA \over dl} } \\ & = {10 \over 2l} \\ & = {5 \over l} \\ \\ \text{Substitute } & A = 4 \text{ into } A = l^2, \\ 4 & = l^2 \\ \pm \sqrt{4} & = l \\ \pm 2 & = l \\ \\ l & = 2 \text{ or } - 2 \text{ (Reject, since } l > 0) \\ \\ \text{Substitute } & l = 2 \text{ into } {dl \over dt}, \\ {dl \over dt} & = {5 \over 2} \text{ cm/s} \end{align}

(b)

\begin{align} \text{Let } A \text{ denote } & \text{the surface area of the cube} \\ \text{Let } l \text{ denote } & \text{the length of the cube} \\ \\ {dA \over dt} & = 0.2 \text{ cm}^2 \text{/s} \\ \\ A & = 6(l \times l) \\ & = 6l^2 \\ \\ {dA \over dl} & = 12l \\ \\ {dA \over dt} & = {dA \over dl} \times {dl \over dt} \\ \\ {dl \over dt} & = { {dA \over dt} \over {dA \over dl} } \\ & = {0.2 \over 12l} \\ & = {1 \over 60l} \\ \\ \text{When } & l = 1, \\ {dl \over dt} & = {1 \over 60(1)} \\ & = {1 \over 60} \text{ cm/s} \\ \\ \\ \text{Let } V \text{ denote } & \text{the volume of the cube} \\ \\ V & = l \times l \times l \\ & = l^3 \\ \\ {dV \over dl} & = 3l^2 \\ \\ {dV \over dt} & = {dV \over dl} \times {dl \over dt} \\ & = 3l^2 \times {1 \over 60} \\ & = {3l^2 \over 60} \\ & = {l^2 \over 20} \\ \\ \text{When } & l = 1, \\ {dV \over dt} & = {(1)^2 \over 20} \\ & = {1 \over 20} \text{ cm}^3 \text{/s} \end{align}

(i)

\begin{align} \text{Let } x \text{ denote } & \text{the breadth of the rectangle} \\ \\ {dx \over dt} & = 4 \text{ cm/s} \\ \\ \text{Length} & = 2 \times \text{Breadth} \\ y & = 2x \\ \\ {dy \over dx} & = 2 \\ \\ {dy \over dt} & = {dy \over dx} \times {dx \over dt} \\ & = 2 \times 4 \\ & = 8 \text{ cm/s} \end{align}

(ii)

\begin{align} \text{Let } A \text{ denote } & \text{the area of the rectangle} \\ \\ A & = \text{Length} \times \text{Breadth} \\ & = 2x \times x \\ & = 2x^2 \\ \\ {dA \over dx} & = 4x \\ \\ {dA \over dt} & = {dA \over dx} \times {dx \over dt} \\ & = 4x \times 4 \\ & = 16x \\ \\ \text{When } & x = 8, \\ {dA \over dt} & = 16(8) \\ & = 128 \text{ cm}^2 \text{/s} \end{align}

(i) Since PQ is parallel to the y-axis, the x-coordinate of Q is p.

\begin{align} y & = x^2 \\ \\ \text{When } & x = p, \\ y & = (p)^2 \\ & = p^2 \\ \\ \therefore & \phantom{.} Q(p, p^2) \\ \\ \text{Area of trapezium, } A & = {1 \over 2} \times \text{Sum of parallel sides} \times \text{Height} \\ & = {1 \over 2} \times (OR + PQ) \times OP \\ & = {1 \over 2} \times (1 + p^2) \times p \\ & = {1 \over 2}p (1 + p^2) \end{align}

(ii)

\begin{align} {dp \over dt} & = 0.5 \text{ units per second} \\ \\ A & = {1 \over 2}p (1 + p^2) \\ & = {1 \over 2}p + {1 \over 2}p^3 \\ \\ {dA \over dp} & = {1 \over 2} + {1 \over 2}(3)p^2 \\ & = {1 \over 2} + {3 \over 2}p^2 \\ \\ {dA \over dt} & = {dA \over dp} \times {dp \over dt} \\ & = \left({1 \over 2} + {3 \over 2}p^2\right) \times 0.5 \\ & = {1 \over 4} + {3 \over 4}p^2 \\ \\ \text{When } & p = 3, \\ {dA \over dt} & = {1 \over 4} + {3 \over 4}(3)^2 \\ & = 7 \text{ square units per second} \end{align}

(i) The formula to find the area of triangle can be found in the provided formula sheet

\begin{align} \text{Let } y \text{ denote } & \text{the area of triangle } ABC \\ \\ \text{Area of triangle} & = {1 \over 2}ab\sin c \\ y & = {1 \over 2}(AB)(AC)(\sin \angle A) \\ & = {1 \over 2}(2x)(x + 1)(\sin 150^\circ) \\ & = (x)(x + 1)\left( 1 \over 2 \right) \\ & = {1 \over 2}x (x + 1)\text{ cm}^2 \phantom{00} \text{ (Shown)} \end{align}

(ii)

\begin{align} {dx \over dt} & = 2 \text{ cm/s} \\ \\ y & = {1 \over 2}x(x + 1) \\ & = {1 \over 2}x^2 + {1 \over 2}x \\ \\ {dy \over dx} & = {1 \over 2}(2)x + {1 \over 2}(1) \\ & = x + {1 \over 2} \\ \\ {dy \over dt} & = {dy \over dx} \times {dx \over dt} \\ & = \left(x + {1 \over 2}\right) \times 2 \\ & = 2x + 1 \\ \\ \text{When } & x = 4, \\ {dy \over dt} & = 2(4) + 1 \\ & = 9 \text{ cm}^2 \text{/s} \end{align}

(i)

\begin{align} \text{Substitute } & x = 2x \text{ into } y = 2x + 1, \\ y & = 2(2x) + 1 \\ & = 4x + 1 \\ \\ \therefore & \phantom{.} Q(2x, 4x + 1) \\ \therefore & \phantom{.} R(x, 4x + 1) \\ \\ {dx \over dt} & = 1.2 \text{ units per second} \\ \\ \text{Let } A \text{ denote } & \text{the area of the parallelogram } OPQR \\ \\ A & = \text{Base} \times \text{Height} \\ & = OP \times PR \\ & = x \times (4x + 1) \\ & = (4x^2 + x) \text{ units}^2 \\ \\ {dA \over dx} & = 4(2)x + 1 \\ & = 8x + 1 \\ \\ {dA \over dt} & = {dA \over dx} \times {dx \over dt} \\ & = (8x + 1) \times 1.2 \\ & = 1.2(8x + 1) \\ & = 9.6x + 1.2 \\ \\ \text{When } & x = 1.5, \\ {dA \over dt} & = 9.6(1.5) + 1.2 \\ & = 15.6 \text{ square units per second} \end{align}

(ii) Recall that the distance between two points can be found by $\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$

\begin{align} \text{Let } L \text{ denote } & \text{the length of the diagonal } OQ \\ \\ L & = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \\ & = \sqrt{(2x - 0)^2 + (4x + 1 - 0)^2} \\ & = \sqrt{(2x)^2 + (4x + 1)^2} \\ & = \sqrt{4x^2 + 16x^2 + 8x + 1} \\ & = \sqrt{20x^2 + 8x + 1} \\ & = (20x^2 + 8x + 1)^{1 \over 2} \\ \\ {dL \over dx} & = {1 \over 2}(20x^2 + 8x + 1)^{-{1 \over 2}} . (40x + 8) \phantom{000000} [\text{Chain rule}] \\ & = {1 \over 2}(40x + 8) (20x^2 + 8x + 1)^{-{1 \over 2}} \\ & = (20x + 4)\left( 1 \over \sqrt{20x^2 + 8x + 1} \right) \\ & = {20x + 4 \over \sqrt{20x^2 + 8x + 1}} \\ \\ {dL \over dt} & = {dL \over dx} \times {dx \over dt} \\ & = {20x + 4 \over \sqrt{20x^2 + 8x + 1}} \times 1.2 \\ & = {20x + 4 \over \sqrt{20x^2 + 8x + 1}} \times {6 \over 5} \\ & = {6(20x + 4) \over 5\sqrt{20x^2 + 8x + 1}} \\ \\ \text{When } & x = 1.5, \\ {dL \over dt} & = {6[20(1.5) + 4] \over 5\sqrt{20(1.5)^2 + 8(1.5) + 1}} \\ & \approx 5.36 \text{ units per second} \end{align}

\begin{align} {dI \over dt} & = 2 \text{ A/s} \\ \\ P & = I^2 R \\ \\ \text{When } & R = 400, \\ P & = I^2 (400) \\ & = 400I^2 \\ \\ {dP \over dI} & = 400(2)I \\ & = 800I \\ \\ {dP \over dt} & = {dP \over dI} \times {dI \over dt} \\ & = 800I \times 2 \\ & = 1600I \\ \\ \text{When } & I = 3, \\ {dP \over dt} & = 1600(3) \\ & = 4800 \text{ W/s} \end{align}

Note: The length of PQ is always 6 m

\begin{align} \text{Let } y \text{ denote the } & \text{distance between } Q \text{ and the bottom of the wall} \\ \text{Let } x \text{ denote the } & \text{distance between } P \text{ and the bottom of the wall} \\ \\ {dy \over dt} & = 0.6 \text{ m/s} \\ \\ \text{By Py} & \text{thagoras theorem,} \\ 6^2 & = x^2 + y^2 \\ 36 & = x^2 + y^2 \\ 36 - x^2 & = y^2 \\ \\ y^2 & = 36 - x^2 \\ y & = \pm\sqrt{36-x^2} \\ \\ y & = \sqrt{36-x^2} \phantom{00}\text{ or }\phantom{00} y = -\sqrt{36-x^2}\text{ (N.A.)} \\ & = (36 - x^2)^{1 \over 2} \\ \\ {dy \over dx} & = \left(1 \over 2\right) (36 - x^2)^{-{1 \over 2}} . (-2x) \phantom{000000} [\text{Chain rule}] \\ & = (-x)(36 - x^2)^{-{1 \over 2}} \\ & = (-x)\left(1 \over \sqrt{36 -x^2}\right) \\ & = -{x \over \sqrt{36 - x^2}} \\ \\ {dy \over dt} & = {dy \over dx} \times {dx \over dt} \\ \\ {dx \over dt} & = { {dy \over dt} \over {dy \over dx} } \\ & = { 0.6 \over -{x \over \sqrt{36 - x^2}} } \\ & = -{0.6\sqrt{36 - x^2} \over x} \\ \\ \text{When } & x = 4.8, \\ {dx \over dt} & = -{0.6\sqrt{36 - (4.8)^2} \over 4.8} \\ & = -0.45 \text{ m/s} \\ \\ \therefore \text{Top of ladder } & \text{is sliding down 0.45 m/s} \end{align}

Note: x is decreasing!

\begin{align} {dx \over dt} & = -50 \text{ m/s} \\ \\ \text{Let } y \text{ denote } & \text{the distance } HL \\ \\ \text{By Py} & \text{thagoras theorem,} \\ y^2 & = x^2 + 270^2 \\ y^2 & = x^2 + 72 \phantom{.} 900 \\ y & = \pm \sqrt{x^2 + 72 \phantom{.} 900} \\ \\ y & = \sqrt{x^2 + 72 \phantom{.} 900} \phantom{0}\text{ or }\phantom{0} y = -\sqrt{x^2 + 72 \phantom{.} 900} \text{ (N.A.)} \\ & = (x^2 + 72 \phantom{.} 900)^{1 \over 2} \\ \\ {dy \over dx} & = {1 \over 2} (x^2 + 72 \phantom{.} 900)^{-{1 \over 2}} . (2x) \phantom{000000} [\text{Chain rule}] \\ & = (x)(x^2 + 72 \phantom{.} 900)^{-{1 \over 2}} \\ & = (x)\left( 1 \over \sqrt{x^2 + 72 \phantom{.} 900} \right) \\ & = {x \over \sqrt{x^2 + 72 \phantom{.} 900}} \\ \\ {dy \over dt} & = {dy \over dx} \times {dx \over dt} \\ & = {x \over \sqrt{x^2 + 72 \phantom{.} 900}} \times -50 \\ & = -{50x \over \sqrt{x^2 + 72 \phantom{.} 900}} \\ \\ \text{When } & x = 144, \\ {dy \over dt} & = -{50(144) \over \sqrt{(144)^2 + 72 \phantom{.} 900}} \\ & = -23{9 \over 17} \text{ m/s} \end{align}

(i) Note: The top of the man's shadow is at the leftmost vertex of the triangle

The distance between the top of the man's shadow and the street lamp is x m.

Since the distance between the man and the street lamp is y m, the distance between the top of the man's shadow and the man is (x - y) m.

With this information, we can form two similar right-angled triangles as shown below:

\begin{align} \text{By simi} & \text{lar triangles,} \\ {7 \over 2} & = {x \over x - y} \\ 7(x - y) & = 2(x) \\ 7x - 7y & = 2x \\ 7x - 2x & = 7y \\ 5x & = 7y \\ {5 \over 7}x & = y \end{align}

(ii)

\begin{align} {dy \over dt} & = {5 \over 3} \text{ m/s} \\ \\ y & = {5 \over 7}x \\ \\ {dy \over dx} & = {5 \over 7} \\ \\ {dy \over dt} & = {dy \over dx} \times {dx \over dt} \\ \\ {dx \over dt} & = { {dy \over dt} \over {dy \over dx} } \\ & = { {5 \over 3} \over {5 \over 7} } \\ & = {7 \over 3} \text{ m/s} \end{align}

(i) Motorist A's path can modelled as the x-axis while motorist B's path can modelled as the y-axis. The junction P can be modelled as the origin (0, 0).

\begin{align} \text{Distance} & = \text{Speed} \times \text{Time} \\ \\ \text{Distance travelled by } & A \text{ after } t \text{ hours} = 75t \\ \\ \text{Distance travelled by } & B \text{ after } t \text{ hours} = 60t \end{align}

Note: Since A reaches junction P half an hour after B, B is 60 x ¹/₂ = 30 km ahead

\begin{align} \text{Distance, } D & = \sqrt{(-75t - 0)^2 + [(0 - (30 + 60t)]^2} \\ & = \sqrt{ (-75t)^2 + (-30 - 60t)^2 } \\ & = \sqrt{ 5625 t^2 + 900 + 3600t + 3600t^2 } \\ & = \sqrt{ 9225 t^2 + 3600t + 900 } \\ & = (9225t^2 + 3600t + 900)^{1 \over 2} \\ \\ {dD \over dt} & = {1 \over 2}(9225t^2 + 3600t + 900)^{-{1 \over 2}} . (18 \phantom{.} 450t + 3600) \\ \\ \text{When } & t = {1 \over 3}, \\ {dD \over dt} & = {1 \over 2}\left[9225\left(1 \over 3\right)^2 + 3600\left(1 \over 3\right) + 900 \right]^{-{1 \over 2}} . \left[ 18 \phantom{.} 450\left(1 \over 3\right) + 3600 \right] \\ & = 87.206 \\ & \approx 87.2 \text{ km/h} \end{align}

(ii) The diagram in (i) denotes A at junction P, with B half an hour behind.

\begin{align} t & = -{1 \over 2} + {1 \over 3} \\ & = -{1 \over 6} \\ \\ \text{When } & t = -{1 \over 6}, \\ {dD \over dt} & = {1 \over 2}\left[9225\left(-{1 \over 6}\right)^2 + 3600\left(-{1 \over 6}\right) + 900 \right]^{-{1 \over 2}} . \left[ 18 \phantom{.} 450\left(-{1 \over 6}\right) + 3600 \right] \\ & = 11.129 \\ & \approx 11.1 \text{ km/h} \end{align}

(a)

\begin{align} \text{If radius increases by a small amount,} & \text{ increase in area is equal to circumference of circle} \\ \\ \text{Change in area} & = 2 \pi r \\ \\ \therefore {dA \over dr} & = 2 \pi r \end{align}

(b)

(Analogy - Applying a very thin layer of paint around a ball causes the volume of the ball to increase by approximately the curved surface area of the ball)

\begin{align} \text{If radius increases by a small amount,} & \text{ increase in volume is equal to curved surface area of sphere} \\ \\ \therefore {dV \over dr} & = 4 \pi r^2 \end{align}