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

The graph of \(y=-x^3\) is transformed onto the graph of \(y=9-0.3x^3\) 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 9: a = 9

2. (b) Find the value of b.

   • \(y=-\left(\frac{x}{b}\right)^3=-\frac{x^3}{b^3}\), so \(\frac1{b^3}=0.3\)
   • \(b^3=3.333\), b = 1.49

The curve \(y=\sqrt{x}\) passes through the point \((0.01,\ 0.1)\).

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 \(y=5x+0.05\).

   • Gradient at x = 0.01: \(\frac{1}{2\times 0.1}=5\)
   • \(y-0.1=5(x-0.01)\)
   • \(y=5x+0.05\)

The point P has coordinates \((1.2,\ 0.8)\). It is transformed by a stretch with scale factor \(\frac{1}{4}\) in the x-direction, then a stretch with scale factor \(5\) in the y-direction, then a translation of 2 units to the right and 3 units down.

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

   • x-direction: 1.2 × 0.25 = 0.3
   • y-direction: 0.8 × 5 = 4

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

   • Translate: x = 0.3 + 2 = 2.3
   • y = 4 − 3 = 1

A glasshouse has an outer roof in the shape of a hemisphere of diameter 20 m, standing on vertical walls of height 6 m. Inside, there is a ceiling of glass panels shaped as a dome. The graph of \(y=-x^3\) is transformed onto the graph of \(y=14-0.06x^3\) 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 14: a = 14

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 \(y=14-0.06x^3\): \(\frac1{b^3}=0.06\), so \(b^3=16.67\)
   • b = 2.55

3. (c) The ceiling is modelled by rotating the curve \(y=14-0.06x^3\) through 360° about the y-axis, between y = 0 and y = 14. The units are metres. Find the volume of the space between the outer shell and the ceiling.

   • Outer shell: hemisphere plus cylinder, radius 10, height of cylinder 6: \(\frac23\pi(10)^3+\pi(10)^2(6)=3979.35\)
   • Rearrange for x: \(x^3=\frac{14-y}{0.06}\), so \(x^2=\left(\frac{14-y}{0.06}\right)^{2/3}\)
   • From the calculator: \(\pi\int_0^{14}\left(\frac{14-y}{0.06}\right)^{2/3}dy=1000.19\)
   • Volume between = 3979.35 − 1000.19 = 2980 m³

The shape of a ceramic tile in a mosaic 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 \(0\le x\le 0.0625\), and \(f(x)=2x+0.125\) for \(0.0625<x\le 0.25\). 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 \(\frac{1}{2}\) in the y-direction, followed by a translation of 0.05 units to the right. Point A on f has coordinates \((0.25,\ 0.625)\), and B is the image of A.

1. (a) Show that the tangent to \(y=\sqrt{x}\) at the point \((0.0625,\ 0.25)\) is \(y=2x+0.125\).

   • \(y=\sqrt{x},\quad \frac{dy}{dx}=\frac{1}{2\sqrt{x}}\)
   • At x = 0.0625: gradient \(\frac{1}{2\times 0.25}=2\)
   • \(y-0.25=2(x-0.0625)\), so \(y=2x+0.125\)

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

   • x-direction, factor 0.5: 0.25 → 0.125; y-direction, factor 0.5: 0.625 → 0.3125
   • Translate 0.05 units: x = 0.125 + 0.05 = 0.175
   • p = 0.175, q = 0.3125

3. (c) The function g is given by \(g(x)=h(x)\) for \(0.05\le x\le a\) and \(g(x)=2x+b\) for \(a<x\le p\). Find an expression for h(x), the value of a and the value of b.

   • \(g(x)=\frac{1}{2}\,f\left(\frac{x-0.05}{\frac{1}{2}}\right)\)
   • First piece: \(h(x)=0.5\sqrt{2(x-0.05)}\)
   • The break in f is at x = 0.0625: \(a=0.05+0.5\times 0.0625=0.08125\)
   • Second piece: \(\frac{1}{2}\left[2\times\frac{x-0.05}{\frac{1}{2}}+0.125\right]=2x-0.0375\)
   • b = −0.0375

4. (d) Find the area enclosed by the graph of f, the x-axis and the line \(x=0.25\).

   • Two pieces: \(\int_0^{0.0625}\sqrt{x}\,dx+\int_{0.0625}^{0.25}(2x+0.125)\,dx\)
   • \(=0.0104167+0.0820312=0.0924479\)
   • Area = 0.0924 m²

5. (e) The area enclosed by the graph of g, the x-axis and the line x = p is 0.0231120 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 \(0.25-0.175=0.075\): \(\frac12(0.3125+0.625)\times 0.075=0.0351562\)
   • \(0.0924479-0.0231120-0.0351562=0.0341797\)
   • Area = 0.0342 m²

A swimming-pool hall has an outer shell in the shape of a hemisphere of radius 9 m, standing on vertical walls of height 7 m. The inner ceiling is a dome.

1. (a) The ceiling is modelled by rotating the curve \(y=12-0.2x^2\) through 360° about the y-axis, between y = 0 and y = 12. The units are metres. Find the volume of the space between the outer shell and the ceiling.

   • Outer shell: \(\frac23\pi(9)^3+\pi(9)^2(7)=3308.10\)
   • Rearrange for x²: \(x^2=\frac{12-y}{0.2}\)
   • From the calculator: \(\pi\int_0^{12}\frac{12-y}{0.2}\,dy=1130.97\)
   • Volume between = 3308.10 − 1130.97 = 2180 m³

2. (b) The space is filled with insulation costing $26 per cubic metre. Find the cost, to the nearest dollar.

   • 2177.12 × 26 = 56 605
   • The cost is $56 605

A design for a garden path edging is drawn using a piecewise function. Units are metres. The function f is defined by \(f(x)=x^2\) for \(0\le x\le 2\) and \(f(x)=4x-4\) for \(2<x\le 3\). The point A has coordinates \((3,\ 8)\). 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 \(4\) in the y-direction, then a translation of 2 units to the left. B is the image of A.

1. (a) Find the coordinates of B.

   • x: 3 → 6 → 4
   • y: 8 → 32
   • B = (4, 32)

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

   • \(g(x)=4f\left(\frac{x+2}{2}\right)\)
   • First piece: \((x+2)^2\), from x = −2 to x = −2 + 2×2 = 2
   • Second piece: \(4\left[4\times\frac{x+2}{2}-4\right]=8x\), from x = 2 to x = 4

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

   • \(\int_0^{2}x^2\,dx+\int_{2}^{3}(4x-4)\,dx=\frac{2^3}{3}+6\)
   • Area = 8.67

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

   • The stretches multiply the area by 2 × 4 = 8. The translation changes nothing.
   • \(8\times 8.667=69.3\)