Exam practice: transformations, areas and volumes Problem set A One question at a time

The graph of \(y=-x^3\) is transformed onto the graph of \(y=15-0.2x^3\) by a vertical translation of a units and a stretch parallel to the x-axis with scale factor b.

1. (a) Write down the value of a.

   • The graph moves up 15: a = 15

2. (b) Find the value of b.

   • \(y=-\left(\frac{x}{b}\right)^3=-\frac{x^3}{b^3}\), so \(\frac1{b^3}=0.2\)
   • \(b^3=5.000\), b = 1.71

The curve \(y=\sqrt{x}\) passes through the point \((0.04,\ 0.2)\).

1. (a) Find \(\frac{dy}{dx}\).

   • \(y=x^{1/2}\), so \(\frac{dy}{dx}=\frac12x^{-1/2}=\frac{1}{2\sqrt{x}}\)

2. (b) Hence show that the equation of the tangent at this point is \(y=2.5x+0.1\).

   • Gradient at x = 0.04: \(\frac{1}{2\times 0.2}=2.5\)
   • \(y-0.2=2.5(x-0.04)\)
   • \(y=2.5x+0.1\)

The point P has coordinates \((0.9,\ 1.8)\). It is transformed by a stretch with scale factor \(3\) in the x-direction, then a stretch with scale factor \(\frac{1}{2}\) in the y-direction, then a translation of 1 unit to the left and 2 units up.

1. (a) Find the coordinates of the image after the two stretches.

   • x-direction: 0.9 × 3 = 2.7
   • y-direction: 1.8 × 0.5 = 0.9

2. (b) Find the coordinates of the image after the translation as well.

   • Translate: x = 2.7 − 1 = 1.7
   • y = 0.9 + 2 = 2.9

A planetarium has an outer shell in the shape of a hemisphere of diameter 30 m, standing on vertical walls of height 9 m. Inside, the ceiling is a dome. The graph of \(y=-x^3\) is transformed onto the graph of \(y=22-0.04x^3\) by a vertical translation of a units and a stretch parallel to the x-axis with scale factor b.

1. (a) Write down the value of a.

   • The vertical translation is up by 22: a = 22

2. (b) Find the value of b.

   • Stretch parallel to the x-axis, scale factor b: \(y=-\left(\frac{x}{b}\right)^3=-\frac{x^3}{b^3}\)
   • Compare with \(y=22-0.04x^3\): \(\frac1{b^3}=0.04\), so \(b^3=25.00\)
   • b = 2.92

3. (c) The ceiling is modelled by rotating the curve \(y=22-0.04x^3\) through 360° about the y-axis, between y = 0 and y = 22. The units are metres. Find the volume of the space between the outer shell and the ceiling.

   • Outer shell: hemisphere plus cylinder, radius 15, height of cylinder 9: \(\frac23\pi(15)^3+\pi(15)^2(9)=13430.3\)
   • Rearrange for x: \(x^3=\frac{22-y}{0.04}\), so \(x^2=\left(\frac{22-y}{0.04}\right)^{2/3}\)
   • From the calculator: \(\pi\int_0^{22}\left(\frac{22-y}{0.04}\right)^{2/3}dy=2783.76\)
   • Volume between = 13430.3 − 2783.76 = 10600 m³

The shape of a decorative stainless-steel panel is modelled by the region bounded by the graphs of f and g, the x-axis and the line segment [AB]. Units are metres. The function f is defined by \(f(x)=\sqrt{x}\) for \(0\le x\le 0.64\), and \(f(x)=0.625x+0.4\) for \(0.64<x\le 1.2\). The graph of g is obtained from the graph of f by a stretch with scale factor \(\frac{1}{2}\) in the x-direction, followed by a stretch with scale factor \(\frac{4}{5}\) in the y-direction, followed by a translation of 0.3 units to the right. Point A on f has coordinates \((1.2,\ 1.15)\), and B is the image of A.

1. (a) Show that the tangent to \(y=\sqrt{x}\) at the point \((0.64,\ 0.8)\) is \(y=0.625x+0.4\).

   • \(y=\sqrt{x},\quad \frac{dy}{dx}=\frac{1}{2\sqrt{x}}\)
   • At x = 0.64: gradient \(\frac{1}{2\times 0.8}=0.625\)
   • \(y-0.8=0.625(x-0.64)\), so \(y=0.625x+0.4\)

2. (b) Find the coordinates of B, which are (p, q).

   • x-direction, factor 0.5: 1.2 → 0.6; y-direction, factor 0.8: 1.15 → 0.92
   • Translate 0.3 units: x = 0.6 + 0.3 = 0.9
   • p = 0.9, q = 0.92

3. (c) The function g is given by \(g(x)=h(x)\) for \(0.3\le x\le a\) and \(g(x)=x+b\) for \(a<x\le p\). Find an expression for h(x), the value of a and the value of b.

   • \(g(x)=\frac{4}{5}\,f\left(\frac{x-0.3}{\frac{1}{2}}\right)\)
   • First piece: \(h(x)=0.8\sqrt{2(x-0.3)}\)
   • The break in f is at x = 0.64: \(a=0.3+0.5\times 0.64=0.62\)
   • Second piece: \(\frac{4}{5}\left[0.625\times\frac{x-0.3}{\frac{1}{2}}+0.4\right]=x+0.02\)
   • b = 0.02

4. (d) Find the area enclosed by the graph of f, the x-axis and the line \(x=1.2\).

   • Two pieces: \(\int_0^{0.64}\sqrt{x}\,dx+\int_{0.64}^{1.2}(0.625x+0.4)\,dx\)
   • \(=0.341333+0.546000=0.887333\)
   • Area = 0.887 m²

5. (e) The area enclosed by the graph of g, the x-axis and the line x = p is 0.354933 m² correct to six significant figures. Find the area of the region bounded by f, g, the x-axis and [AB], that is, the area under f minus the area under g and minus the area below AB.

   • Trapezium below AB, width \(1.2-0.9=0.3\): \(\frac12(0.92+1.15)\times 0.3=0.310500\)
   • \(0.887333-0.354933-0.310500=0.221900\)
   • Area = 0.222 m²

The roof of a sports arena has an outer shell in the shape of a hemisphere of radius 14 m, standing on vertical walls of height 10 m. The inner ceiling is a dome.

1. (a) The ceiling is modelled by rotating the curve \(y=20-0.2x^2\) through 360° about the y-axis, between y = 0 and y = 20. The units are metres. Find the volume of the space between the outer shell and the ceiling.

   • Outer shell: \(\frac23\pi(14)^3+\pi(14)^2(10)=11904.5\)
   • Rearrange for x²: \(x^2=\frac{20-y}{0.2}\)
   • From the calculator: \(\pi\int_0^{20}\frac{20-y}{0.2}\,dy=3141.59\)
   • Volume between = 11904.5 − 3141.59 = 8760 m³

2. (b) The space is filled with foam costing $34 per cubic metre. Find the cost, to the nearest dollar.

   • 8762.95 × 34 = 297 940
   • The cost is $297 940

A pattern for a sail is drawn using a piecewise function. Units are metres. The function f is defined by \(f(x)=x^2\) for \(0\le x\le 1\) and \(f(x)=2x-1\) for \(1<x\le 3\). The point A has coordinates \((3,\ 5)\). The graph of g is obtained from the graph of f by a stretch with scale factor \(2\) in the x-direction, then a stretch with scale factor \(3\) in the y-direction, then a translation of 1 units to the right. B is the image of A.

1. (a) Find the coordinates of B.

   • x: 3 → 6 → 7
   • y: 5 → 15
   • B = (7, 15)

2. (b) Write down g(x) on each of its two pieces, with the interval for each.

   • \(g(x)=3f\left(\frac{x-1}{2}\right)\)
   • First piece: \(\frac{3}{4}(x-1)^2\), from x = 1 to x = 1 + 2×1 = 3
   • Second piece: \(3\left[2\times\frac{x-1}{2}-1\right]=3x−6\), from x = 3 to x = 7

3. (c) Find the area enclosed by the graph of f, the x-axis and the line \(x=3\).

   • \(\int_0^{1}x^2\,dx+\int_{1}^{3}(2x-1)\,dx=\frac{1^3}{3}+6\)
   • Area = 6.33

4. (d) Hence find the area enclosed by the graph of g and the x-axis.

   • The stretches multiply the area by 2 × 3 = 6. The translation changes nothing.
   • \(6\times 6.333=38\)