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AI HL · AHL 3

Further Integration

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AHL 3.13

Exam practice: transformations, areas and volumes

AHL 3.13

Exam practice: transformations, areas and volumes

Target: I can read the translation and the stretch factor from a transformed graph such as y = a − kx³. (+4 more)

By the end of this section

Key points

Transformations

  • Do the transformations in the order the question gives. A stretch changes the distance from the axis, so it changes a translation made before it.
  • A stretch with scale factor s in the x-direction replaces x by \(\frac xs\). Not by sx. Compare \((3x)^2\) with \(9x^2\) and take care which axis the stretch is parallel to.
  • For \(y=-x^3\) transformed onto \(y=a-kx^3\): a is the vertical translation, and \(\frac1{b^3}=k\), so \(b=\frac{1}{\sqrt[3]{k}}\).
  • For a piecewise function, change every piece and every limit. The breakpoint moves like any other x-value.

Volumes of revolution about the y-axis

  • About the y-axis the volume is \(V=\pi\int_c^d x^2\,dy\). Write x² in terms of y, and use y-limits.
  • A hemisphere has volume \(\frac23\pi r^3\) and a cylinder \(\pi r^2h\). For the space between two solids, find the volume of each and subtract.
  • Keep the unrounded values until the last line. Then give 3 significant figures and the unit.

Tangents and areas

  • The derivative of \(\sqrt x\) is \(\frac{1}{2\sqrt x}\). For a "show that" question, end on the line you were given.
  • The area under a piecewise function is one integral for each piece, each with its own limits.
  • A trapezium with parallel sides p and q and width w has area \(\frac12(p+q)w\).
  • A stretch with scale factor s in the x-direction and t in the y-direction multiplies the area by st. A translation does not change it.

On the calculator

A volume of revolution about the y-axis

  • Write x² in terms of y first.
  • Enter π × (x² in terms of y) as the function, with the y-limits.
  • Read the integral. Do not round until you subtract.

Screenshot and video to come

Area under a piecewise function

  • Integrate each piece over its own interval.
  • Add the results. The breakpoint is the upper limit of one piece and the lower limit of the next.

Screenshot and video to come

Checking a transformed graph

  • Graph f and g on the same axes.
  • Check that B is on the graph of g, and that g meets the x-axis where you expect.
  • Check the area of g with the calculator's integral.

Screenshot and video to come

Worked examples

Problems like the ones in class. Pick one, try it, then open the working.
Reading a transformed cubic

The graph of \(y=-x^3\) is transformed onto the graph of \(y=18-0.05x^3\) by a vertical translation of a units and a stretch parallel to the x-axis with scale factor b. Find a and b.

Working
  1. The translation

    \(a=18\)

    The graph moves up 18, so the constant is 18.

  2. The stretch

    \(y=-\left(\frac{x}{b}\right)^3=-\frac{x^3}{b^3}\)

    A stretch with scale factor b parallel to the x-axis replaces x by x/b. Not by bx.

  3. Compare

    \(\frac{1}{b^3}=0.05\ \Rightarrow\ b^3=20\)

    Match the coefficient of x³.

  4. Solve

    \(b=\sqrt[3]{20}=2.71\)

    Cube root, 3 significant figures.

Video walkthrough coming
The space between two domes

The outer shell of an observatory is a hemisphere of diameter 24 m, standing on vertical walls 8 m high. The inner ceiling is modelled by rotating the curve \(y=18-0.05x^3\) through 360° about the y-axis, between y = 0 and y = 18. Units are metres. Find the volume of the space between the outer shell and the ceiling.

−15−10−5510155101520xy188
Working
  1. Outer shell

    \(\frac23\pi(12)^3+\pi(12)^2(8)=7238.23\)

    Half a sphere is \(\frac23\pi r^3\). The walls are a cylinder, \(\pi r^2h\).

  2. x in terms of y

    \(x^3=\frac{18-y}{0.05}\ \Rightarrow\ x^2=\left(\frac{18-y}{0.05}\right)^{2/3}\)

    Rotating about the y-axis, the volume is \(\pi\int x^2\,dy\), so the integrand must be in y.

  3. Inner volume

    From the calculator: \(\pi\int_0^{18}\left(\frac{18-y}{0.05}\right)^{2/3}dy=1717.02\)

    The limits are the y-values, 0 to 18.

  4. Subtract

    7238.23 − 1717.02 = 5520 m³

    Keep the unrounded values until the end.

Video walkthrough coming
The tangent to a square root

(a) Find \(\frac{dy}{dx}\) for \(y=\sqrt{x}\). (b) Hence show that the tangent to the curve at the point \((0.25,\ 0.5)\) is \(y=x+0.25\).

Working
  1. Differentiate

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

    Write the root as a power.

  2. Gradient

    \(\frac{1}{2\sqrt{0.25}}=\frac{1}{2\times0.5}=1\)

    Substitute x = 0.25.

  3. Tangent

    \(y-0.5=1(x-0.25)\), so \(y=x+0.25\)

    That is the line you were asked to show. End on it.

Video walkthrough coming
The image of a point

The shape of a metal bracket is modelled using the function f defined by \(f(x)=\sqrt{x}\) for \(0\le x\le 0.25\), and \(f(x)=x+0.25\) for \(0.25<x\le 1\). Units are metres. The point A is (1, 1.25). The graph of g is obtained from the graph of f by a stretch with scale factor 1/2 in the x-direction, then a stretch with scale factor 1/2 in the y-direction, then a translation of 0.25 units to the right. B is the image of A. Find the coordinates of B.

Working
  1. Stretch in x

    \((1,\ 1.25)\to(0.5,\ 1.25)\)

    Scale factor 1/2: halve the x-coordinate.

  2. Stretch in y

    \((0.5,\ 1.25)\to(0.5,\ 0.625)\)

    Halve the y-coordinate.

  3. Translate

    \((0.5,\ 0.625)\to(0.75,\ 0.625)\)

    Add 0.25 to x. The translation is last, so it is not stretched.

  4. State

    \(B=(0.75,\ 0.625)\)

    p = 0.75 and q = 0.625.

Video walkthrough coming
The new function, piece by piece

The shape of a metal bracket is modelled using the function f defined by \(f(x)=\sqrt{x}\) for \(0\le x\le 0.25\), and \(f(x)=x+0.25\) for \(0.25<x\le 1\). Units are metres. The point A is (1, 1.25). The graph of g is obtained from the graph of f by a stretch with scale factor 1/2 in the x-direction, then a stretch with scale factor 1/2 in the y-direction, then a translation of 0.25 units to the right. B is the image of A. The function g is given by \(g(x)=h(x)\) for \(0.25\le x\le a\) and \(g(x)=x+b\) for \(a<x\le 0.75\). Find the value of b.

0.20.40.60.810.20.40.60.811.21.4xyAB
Working
  1. Write g in terms of f

    \(g(x)=\frac12 f\bigl(2(x-0.25)\bigr)\)

    Stretch factor 1/2 in x: replace x by 2x. Translate right 0.25: replace x by x − 0.25. Vertical factor 1/2 in front.

  2. The first piece

    \(h(x)=\frac12\sqrt{2(x-0.25)}\)

    Use the square root piece of f.

  3. The break

    \(a=0.25+0.5\times0.25=0.375\)

    The break at x = 0.25 on f moves to x = 0.375.

  4. The second piece

    \(\frac12\bigl[2(x-0.25)+0.25\bigr]=x-0.125\)

    Use the line piece of f. So b = −0.125.

Video walkthrough coming
Area by subtracting

The shape of a metal bracket is modelled using the function f defined by \(f(x)=\sqrt{x}\) for \(0\le x\le 0.25\), and \(f(x)=x+0.25\) for \(0.25<x\le 1\). Units are metres. The point A is (1, 1.25). The graph of g is obtained from the graph of f by a stretch with scale factor 1/2 in the x-direction, then a stretch with scale factor 1/2 in the y-direction, then a translation of 0.25 units to the right. B is the image of A. The area enclosed by the graph of g, the x-axis and the line x = 0.75 is 0.184896 m² (6 significant figures). Find the area of the region bounded by f, g, the x-axis and the line segment [AB]; that is, the area under f, minus the area under g, minus the area below AB.

0.20.40.60.810.20.40.60.811.21.4xyAB
Working
  1. Area under f

    \(\int_0^{0.25}\sqrt{x}\,dx+\int_{0.25}^{1}(x+0.25)\,dx=0.0833333+0.656250=0.739583\)

    Two pieces, so two integrals. Each uses its own limits.

  2. The trapezium

    \(\frac12(0.625+1.25)\times0.25=0.234375\)

    Parallel sides are the heights at B and A; the width is 1 − 0.75 = 0.25.

  3. Subtract

    0.739583 − 0.184896 − 0.234375 = 0.320312

    Area under f, minus area under g, minus the trapezium.

  4. State

    Area = 0.320 m²

    3 significant figures.

Video walkthrough coming