FPM(2023) June Paper 02 Question 4/5/6 and solution

Question 4

The curve $S$ with equation $y=\frac{x^2}{4}+2$ where $x \geqslant 0$ and the line $l$ with equation $2 y-x-4=0$ where $x \geqslant 0$ intersect at the points $A$ and $B$, as shown in Figure 2.

(a) (i) Show that the coordinates of point $A$ are $(0,2)$

(ii) Find the coordinates of the point $B$ [4 Marks]

The finite region bounded by $S$ and $l$, shown shaded in Figure 2, is rotated through $2 \pi$ radians about the $y$-axis.

(b) Use algebraic integration to find the volume of the solid generated.

Give your answer in terms of $\pi$ [ 4 Marks]

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

(a) On the grid opposite draw the line with equation

(i) $y=2 x+5$

(ii) $4 y=x-8$

(iii) $5 y+3 x=30$ [3  Marks]

(b) Show, by shading, the region $R$ defined by the inequalities

$$

y \leqslant 2 x+5 \quad 4 y \geqslant x-8 \quad 5 y+3 x \leqslant 30

$$

[ 1 Mark]

For all points in $R$ with coordinates $(x, y)$

$$

P=2 x-5 y

$$

(c) Using your graph, find the least value of $P$

[ 3 Marks]

Solution

(a) Draw the lines

$\text{The equations of the lines to be drawn are:}$

1. $y = 2x + 5$
2. $4y = x - 8,\; y = \frac{x - 8}{4}$
3. $5y + 3x = 30,\; y = \frac{30 - 3x}{5}$

(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

Figure 3 shows a right triangular prism $A B C D E F$. A cross section $A B C$ of the prism is a triangle in which $A B=A C=r \mathrm{~cm}$ and $\angle C A B=\frac{\pi}{3}$ radians.

In the prism

$$

A E=B F=C D=5 \mathrm{~cm} \quad E D=E F=r \mathrm{~cm} \text { and } \angle D E F=\frac{\pi}{3} \text { radians }

$$

(a) Show that the volume of the prism is $\frac{5 \sqrt{3}}{4} r^2 \mathrm{~cm}^3$ [ 1 Mark]

The volume of the prism is increasing in such a way that the size of $\angle C A B$ and the size of $\angle D E F$ remain constant and the length of $A E$, the length of $B F$ and the length of $C D$ remain constant.

The lengths of $A B, A C, E D$ and $E F$ are each increasing at a constant rate of $0.2 \mathrm{~cm} / \mathrm{s}$

(b) Find the exact rate of increase, in $\mathrm{cm}^3 / \mathrm{s}$, of the volume of the prism when the area of the rectangular face $B C D F$ is $60 \mathrm{~cm}^2$ [ 5 Marks]

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

$\text{The volume of the prism is:}$

$V = \frac{5\sqrt{3}}{4} r^2.$

$\text{Differentiate } V \text{ with respect to time } t:$

$\frac{dV}{dt} = \frac{5\sqrt{3}}{4} \cdot 2r \cdot \frac{dr}{dt}.$

$\text{Simplify:}$

$\frac{dV}{dt} = \frac{5\sqrt{3}}{2} r \cdot \frac{dr}{dt}.$

$\text{We are given that:}$

$\frac{dr}{dt} = 0.2 \, \mathrm{cm/s}.$

$\text{When the area of the rectangular face } BCDF \text{ is } 60 \, \mathrm{cm}^2,$

$\text{Area of } BCDF = BC \cdot AE = r \cdot 5.$

$\text{Substitute } 60:$

$r \cdot 5 = 60 \implies r = 12 \, \mathrm{cm}.$

$\text{Substitute into } \frac{dV}{dt}:$

$\frac{dV}{dt} = \frac{5\sqrt{3}}{2} \cdot 12 \cdot 0.2.$

$\text{Simplify:}$

$\frac{dV}{dt} = \frac{5\sqrt{3}}{2} \cdot 2.4 = 6\sqrt{3} \, \mathrm{cm}^3/\mathrm{s}.$

$\text{Thus, the rate of increase is:}$

$\boxed{6\sqrt{3} \, \mathrm{cm}^3/\mathrm{s}}.$