Question 4
Solution
Solution
$\text{The curve } S \text{ is given by the equation } y = \frac{x^2}{4} + 2 \text{ where } x \geq 0, \text{ and the line } l \text{ is given by } 2y - x - 4 = 0 \text{ where } x \geq 0. \text{ These intersect at points } A \text{ and } B.$
(a) (i) Coordinates of Point $A$
$\text{To find the intersection points of } S \text{ and } l, \text{ solve their equations simultaneously:}$
$y = \frac{x^2}{4} + 2 \quad \text{and} \quad 2y - x - 4 = 0.$
$\text{Substitute } y = \frac{x^2}{4} + 2 \text{ into } 2y - x - 4 = 0:$
$2\left(\frac{x^2}{4} + 2\right) - x - 4 = 0.$
$\text{Simplify:}$
$\frac{x^2}{2} + 4 - x - 4 = 0 \implies \frac{x^2}{2} - x = 0.$
$\text{Factorise:}$
$\frac{x}{2}(x - 2) = 0.$
$\text{This gives } x = 0 \text{ or } x = 2.$
$\text{For } x = 0, \text{ substitute into } y = \frac{x^2}{4} + 2:$
$y = \frac{0^2}{4} + 2 = 2.$
$\text{Thus, } A = (0, 2).$
(a) (ii) Coordinates of Point $B$
$\text{For } x = 2, \text{ substitute into } y = \frac{x^2}{4} + 2:$
$y = \frac{2^2}{4} + 2 = \frac{4}{4} + 2 = 1 + 2 = 3.$
$\text{Thus, } B = (2, 3).$
(b) Volume of the Solid Generated
$\text{The volume of the solid generated when the region between } S \text{ and } l \text{ is rotated about the } y\text{-axis is given by the formula:}$
$V = \pi \int_{y_1}^{y_2} \left( R(y)^2 - r(y)^2 \right) \, dy,$
$\text{where } R(y) \text{ and } r(y) \text{ are the outer and inner radii as functions of } y, \text{ and } y_1 \text{ and } y_2 \text{ are the bounds of } y.$
$\text{From the equations:}$
- $\text{The line } l \text{ can be rearranged as } x = 2y - 4, \text{ giving } r(y) = 2y - 4.$
- $\text{The curve } S \text{ is rearranged as } x = \sqrt{4(y - 2)}, \text{ giving } R(y) = \sqrt{4(y - 2)}.$
- $y_1 = 2 \text{ and } y_2 = 3.$
$\text{Substitute into the volume formula:}$
$V = \pi \int_{2}^{3} \left[ \left(\sqrt{4(y - 2)}\right)^2 - (2y - 4)^2 \right] \, dy.$
$\text{Simplify:}$
$V = \pi \int_{2}^{3} \left[ 4(y - 2) - (2y - 4)^2 \right] \, dy.$
$\text{Expand } (2y - 4)^2:$
$(2y - 4)^2 = 4y^2 - 16y + 16.$
$\text{Thus:}$
$V = -\pi \int_{2}^{3} \left( 4y^2 - 16y + 16 - 4y + 8 \right) \, dy.$
$\text{Combine terms:}$
$V = -\pi \int_{2}^{3} \left( 4y^2 - 20y + 24 \right) \, dy.$
$\text{Integrate term by term:}$
$\int 4y^2 \, dy = \frac{4y^3}{3}, \quad \int -20y \, dy = -10y^2, \quad \int 24 \, dy = 24y.$
$\text{Thus:}$
$V = -\pi \left[ \frac{4y^3}{3} - 10y^2 + 24y \right]_{2}^{3}.$
$\text{Evaluate at the bounds:}$
$\text{At } y = 3: \quad \frac{4(3)^3}{3} - 10(3)^2 + 24(3) = \frac{108}{3} - 90 + 72 = 36 - 90 + 72 = 18.$
$\text{At } y = 2: \quad \frac{4(2)^3}{3} - 10(2)^2 + 24(2) = \frac{32}{3} - 40 + 48 = \frac{32}{3} + 8 = \frac{56}{3}.$
$\text{Subtract:}$
$V = -\pi \left( 18 - \frac{56}{3} \right) = \pi \left( \frac{54}{3} - \frac{56}{3} \right) = \pi \left( \frac{2}{3} \right).$
$\text{Thus:}$
$V = \frac{2\pi}{3}.$
Question 5
Solution
(a) Draw the lines
$\text{The equations of the lines to be drawn are:}$
- $y = 2x + 5$
- $4y = x - 8,\; y = \frac{x - 8}{4}$
- $5y + 3x = 30,\; y = \frac{30 - 3x}{5}$
(b) Shade the region $R$
$y \leq 2x + 5, \quad 4y \geq x - 8, \quad 5y + 3x \leq 30.$
$\text{Shade below } y = 2x + 5,\; \text{above } y = \frac{x - 8}{4},\; \text{and below } y = \frac{30 - 3x}{5}.$
(c) Find the least value of $P = 2x - 5y$
$\text{Intersection of } y = 2x + 5 \text{ and } y = \frac{x - 8}{4}:$
$2x + 5 = \frac{x - 8}{4} \implies 8x + 20 = x - 8 \implies 7x = -28 \implies x = -4,\; y = -3.$
$\text{Intersection of } y = 2x + 5 \text{ and } y = \frac{30 - 3x}{5}:$
$2x + 5 = \frac{30 - 3x}{5} \implies 10x + 25 = 30 - 3x \implies 13x = 5$
$\implies x = \frac{5}{13},\; y = 2\left(\frac{5}{13}\right) + 5 = \frac{75}{13}.$
$\text{Intersection of } y = \frac{x - 8}{4} \text{ and } y = \frac{30 - 3x}{5}:$
$\frac{x - 8}{4} = \frac{30 - 3x}{5} \implies 5(x - 8) = 4(30 - 3x) \implies 5x - 40 = 120 - 12x$
$\implies 17x = 160 \implies x = \frac{160}{17},\; y = \frac{6}{17}.$
Evaluate $P = 2x - 5y$:
$P = 2(-4) - 5(-3) = 7.$
$P = 2\left(\frac{5}{13}\right) - 5\left(\frac{75}{13}\right) = \frac{-365}{13} \approx -28.07.$
$P = 2\left(\frac{160}{17}\right) - 5\left(\frac{6}{17}\right) = \frac{290}{17} \approx 17.05.$
Final Answer:
$P_{\min} = -28.1 \text{ at } \left(\frac{5}{13}, \frac{75}{13}\right).$
Question 6
Solution
(a) Show that the volume of the prism is $ \frac{5\sqrt{3}}{4} r^2 \, \mathrm{cm}^3 $
$\text{The cross-section } ABC \text{ is an isosceles triangle with:}$
$AB = AC = r \, \mathrm{cm}, \quad \angle CAB = \frac{\pi}{3} \, \text{radians}.$
$\text{The area of triangle } ABC \text{ is given by:}$
$\text{Area of } \triangle ABC = \frac{1}{2} \cdot AB \cdot AC \cdot \sin(\angle CAB).$
$\text{Substitute the values:}$
$\text{Area of } \triangle ABC = \frac{1}{2} \cdot r \cdot r \cdot \sin\left(\frac{\pi}{3}\right).$
$\text{Since } \sin\left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2}, \text{ the area becomes:}$
$\text{Area of } \triangle ABC = \frac{1}{2} \cdot r^2 \cdot \frac{\sqrt{3}}{2} = \frac{\sqrt{3}}{4} r^2 \, \mathrm{cm}^2.$
$\text{The volume of the prism is:}$
$\text{Volume} = \text{Area of cross-section } \times \text{height}.$
$\text{The height of the prism is } AE = 5 \, \mathrm{cm}, \text{ so:}$
$\text{Volume} = \frac{\sqrt{3}}{4} r^2 \cdot 5 = \frac{5\sqrt{3}}{4} r^2 \, \mathrm{cm}^3.$
$\text{Thus, the volume of the prism is:}$
$\boxed{\frac{5\sqrt{3}}{4} r^2 \, \mathrm{cm}^3}.$
(b) Rate of increase of the volume of the prism
$\text{The volume of the prism is:}$
$V = \frac{5\sqrt{3}}{4} r^2.$
$\text{Differentiate } V \text{ with respect to time } t:$
$\frac{dV}{dt} = \frac{5\sqrt{3}}{4} \cdot 2r \cdot \frac{dr}{dt}.$
$\text{Simplify:}$
$\frac{dV}{dt} = \frac{5\sqrt{3}}{2} r \cdot \frac{dr}{dt}.$
$\text{We are given that:}$
$\frac{dr}{dt} = 0.2 \, \mathrm{cm/s}.$
$\text{When the area of the rectangular face } BCDF \text{ is } 60 \, \mathrm{cm}^2,$
$\text{Area of } BCDF = BC \cdot AE = r \cdot 5.$
$\text{Substitute } 60:$
$r \cdot 5 = 60 \implies r = 12 \, \mathrm{cm}.$
$\text{Substitute into } \frac{dV}{dt}:$
$\frac{dV}{dt} = \frac{5\sqrt{3}}{2} \cdot 12 \cdot 0.2.$
$\text{Simplify:}$
$\frac{dV}{dt} = \frac{5\sqrt{3}}{2} \cdot 2.4 = 6\sqrt{3} \, \mathrm{cm}^3/\mathrm{s}.$
$\text{Thus, the rate of increase is:}$
$\boxed{6\sqrt{3} \, \mathrm{cm}^3/\mathrm{s}}.$
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