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IB Mathematics: Applications and Interpretation HL · AHL 4: Vectors · AHL 4.30

Vectors: exam review Exam review: second paper One question at a time

Try each question on paper first. Open the working a line at a time, then say whether you had it.

Section A: short questions

  1. A hiker’s beacon is at the point P and a rescue helicopter is at the point Q. Relative to an origin O, the position vectors are \(\overrightarrow{OP} = -2\mathbf{i} + 5\mathbf{j} + \mathbf{k}\) and \(\overrightarrow{OQ} = 3\mathbf{i} + \mathbf{j} - 3\mathbf{k}\), in metres. [6]

    1. (a) Write down \(\overrightarrow{PQ}\) as a column vector. [2]

      • M1 for \(\overrightarrow{OQ} - \overrightarrow{OP}\) (end point minus start point)
      • A1 for \(\begin{pmatrix} 5 \\ -4 \\ -4 \end{pmatrix}\)

    2. (b) Find the exact distance between P and Q. [2]

      • M1 for \(\sqrt{(5)^2 + (-4)^2 + (-4)^2}\)
      • A1 for \(\sqrt{57}\)

    3. (c) Find the vector of magnitude 12 in the same direction as \(\overrightarrow{PQ}\). Give the components to 3 significant figures. [2]

      • M1 for dividing \(\overrightarrow{PQ}\) by its magnitude and multiplying by 12, \(\dfrac{12}{\sqrt{57}}\begin{pmatrix} 5 \\ -4 \\ -4 \end{pmatrix}\)
      • A1 for \(\begin{pmatrix} 7.95 \\ -6.36 \\ -6.36 \end{pmatrix}\)

  2. In triangle OCD, \(\overrightarrow{OC} = \mathbf{c}\) and \(\overrightarrow{OD} = \mathbf{d}\). The point N is the midpoint of OD. The point R lies on CD with CR:RD = 2:1. The point S is such that OCSD is a parallelogram. [6]

    OCDSNR

    1. (a) Find \(\overrightarrow{CD}\) in terms of \(\mathbf{c}\) and \(\mathbf{d}\). [1]

      • A1 for \(\mathbf{d} - \mathbf{c}\)

    2. (b) Show that \(\overrightarrow{OR} = \dfrac{1}{3}(\mathbf{c} + 2\mathbf{d})\). [2]

      • M1 for \(\overrightarrow{OR} = \overrightarrow{OC} + \dfrac{2}{3}\overrightarrow{CD}\)
      • A1 for \(\mathbf{c} + \dfrac{2}{3}(\mathbf{d} - \mathbf{c}) = \dfrac{1}{3}(\mathbf{c} + 2\mathbf{d})\) (answer given, so the working must show the step)

    3. (c) Show that N, R and S lie on a straight line. [3]

      • M1 for \(\overrightarrow{NR} = \overrightarrow{OR} - \overrightarrow{ON} = \dfrac{1}{6}(2\mathbf{c} + \mathbf{d})\)
      • M1 for \(\overrightarrow{NS} = \overrightarrow{OS} - \overrightarrow{ON} = \mathbf{c} + \mathbf{d} - \overrightarrow{ON} = \dfrac{1}{2}(2\mathbf{c} + \mathbf{d})\)
      • R1 for \(\overrightarrow{NS} = 3\overrightarrow{NR}\), and N is a common point, so N, R, S are collinear

  3. Let \(\mathbf{u} = \begin{pmatrix} p \\ 9 \\ -6 \end{pmatrix}\), \(\mathbf{v} = \begin{pmatrix} -2 \\ q \\ 4 \end{pmatrix}\) and \(\mathbf{w} = \begin{pmatrix} k + 1 \\ 2k \\ 3 \end{pmatrix}\), where \(p\), \(q\) and \(k\) are constants. [6]

    1. (a) Given that \(\mathbf{u}\) is parallel to \(\mathbf{v}\), find the value of \(p\) and the value of \(q\). [3]

      • M1 for \(\mathbf{u} = \lambda\mathbf{v}\) (or equal ratios of components)
      • M1 for \(-6 = \lambda(4)\) and so \(\lambda = -\frac{3}{2}\)
      • A1 for \(p = 3\) and \(q = -6\)

    2. (b) Given that \(|\mathbf{w}| = 8\), find the possible values of \(k\). Give your answers to 3 significant figures. [3]

      • M1 for \((k + 1)^2 + (2 k)^2 + (3)^2 = 64\)
      • A1 for the equation \(5 k^{2} + 2 k - 54 = 0\)
      • A1 for \(k = -3.49\) and \(k = 3.09\)

  4. The line \(l\) passes through the points P\((-3, 4, 1)\) and Q\((1, 2, 6)\). The point R has coordinates \((5, 0, 10)\). [6]

    1. (a) Find a vector equation of \(l\). Give your answer in the form \(\mathbf{r} = \mathbf{a} + \lambda\mathbf{b}\). [2]

      • M1 for \(\overrightarrow{PQ} = \begin{pmatrix} 4 \\ -2 \\ 5 \end{pmatrix}\) (or \(\overrightarrow{QP}\)) as the direction
      • A1 for \(\mathbf{r} = \begin{pmatrix} -3 \\ 4 \\ 1 \end{pmatrix} + \lambda\begin{pmatrix} 4 \\ -2 \\ 5 \end{pmatrix}\) (any point of l and any multiple of the direction)

    2. (b) Determine whether R lies on \(l\). [2]

      • M1 for equating \(\begin{pmatrix} -3 \\ 4 \\ 1 \end{pmatrix} + \lambda\begin{pmatrix} 4 \\ -2 \\ 5 \end{pmatrix} = \begin{pmatrix} 5 \\ 0 \\ 10 \end{pmatrix}\) and solving one component: \(5 = 4 \lambda - 3 \Rightarrow \lambda = 2\)
      • R1 for a component that does not agree: with \(\lambda = 2\) the other components give \(5, 0, 11\), which is not \((5,0,10)\), so R is not on \(l\)

    3. (c) Write down parametric equations of \(l\). [2]

      • A1 for two correct equations, \(x = 4 \lambda - 3,\ y = 4 - 2 \lambda\)
      • A1 for all three, with \(z = 5 \lambda + 1\)

  5. A yacht is sailing in a race with constant velocity. At time \(t\) hours after the start its position vector is \(\mathbf{r} = \begin{pmatrix} 6 \\ -2 \end{pmatrix} + t\begin{pmatrix} -7 \\ 5 \end{pmatrix}\), in km relative to the race marker M. The first component is the distance east and the second is the distance north. [7]

    1. (a) Find the speed of the yacht. [2]

      • M1 for \(\sqrt{(-7)^2 + (5)^2}\)
      • A1 for \(\sqrt{74}\) = 8.60

    2. (b) Find the bearing on which the yacht is moving. Give your answer to the nearest degree. [2]

      • M1 for a sketch or the angle with the north-south line: \(\tan\theta = \dfrac{7}{5}\), 54.5°
      • A1 for the bearing 306° (360° − 54.5°)

    3. (c) Find the distance of the yacht from the race marker M when \(t = 3.5\). [3]

      • M1 for \(\mathbf{r} = \begin{pmatrix} 6 \\ -2 \end{pmatrix} + 3.5\begin{pmatrix} -7 \\ 5 \end{pmatrix}\)
      • A1 for \(\mathbf{r} = \begin{pmatrix} -18.5 \\ 15.5 \end{pmatrix}\)
      • A1 for \(\sqrt{(-18.5)^2 + (15.5)^2}\) = 24.1

  6. The scalar product (dot product) of two vectors is written \(\mathbf{a} \cdot \mathbf{b}\). [7]

    1. (a) The vectors \(\mathbf{a}\) and \(\mathbf{b}\) have \(|\mathbf{a}| = 5\) and \(|\mathbf{b}| = 9\). The angle between them is 110°. Find \(\mathbf{a} \cdot \mathbf{b}\). Give your answer to 3 significant figures. [2]

      • M1 for \(\mathbf{a} \cdot \mathbf{b} = 5 \times 9 \times \cos 110^\circ\)
      • A1 for \(-15.4\)

    2. (b) Let \(\mathbf{p} = \begin{pmatrix} 4 \\ 1 \\ -2 \end{pmatrix}\) and \(\mathbf{q} = \begin{pmatrix} -3 \\ 5 \\ 2 \end{pmatrix}\). Find the angle between \(\mathbf{p}\) and \(\mathbf{q}\). Give your answer to the nearest degree. [3]

      • M1 for \(\mathbf{p} \cdot \mathbf{q} = (4)(-3) + (1)(5) + (-2)(2) = -11\)
      • M1 for \(\cos\theta = \dfrac{-11}{\sqrt{21}\sqrt{38}}\)
      • A1 for 113° (the cosine is negative, so the angle is obtuse)

    3. (c) The vector \(\mathbf{s} = t\mathbf{i} + 3\mathbf{j} - 2\mathbf{k}\) is perpendicular to \(\mathbf{z} = 2\mathbf{i} - 3\mathbf{j} + 4\mathbf{k}\). Find the value of \(t\). [2]

      • M1 for \(\mathbf{s} \cdot \mathbf{z} = 0:\ 2 t - 17 = 0\)
      • A1 for \(t = \frac{17}{2}\)

  7. The vector product (cross product) of two vectors is written \(\mathbf{a} \times \mathbf{b}\). [7]

    1. (a) The vectors \(\mathbf{a}\) and \(\mathbf{b}\) have \(|\mathbf{a}| = 5\) and \(|\mathbf{b}| = 8\). The angle between them is 115°. Find \(|\mathbf{a} \times \mathbf{b}|\). Give your answer to 3 significant figures. [2]

      • M1 for \(|\mathbf{a} \times \mathbf{b}| = 5 \times 8 \times \sin 115^\circ\)
      • A1 for 36.3

    2. (b) Let \(\mathbf{p} = -\mathbf{i} + 2\mathbf{j} + 4\mathbf{k}\) and \(\mathbf{q} = 3\mathbf{i} + \mathbf{j} - 2\mathbf{k}\). Find \(\mathbf{p} \times \mathbf{q}\). [2]

      • M1 for the component form \(\begin{pmatrix} (2)(-2) - (4)(1) \\ (4)(3) - (-1)(-2) \\ (-1)(1) - (2)(3) \end{pmatrix}\)
      • A1 for \(\begin{pmatrix} -8 \\ 10 \\ -7 \end{pmatrix}\)

    3. (c) Find the component of \(\mathbf{p}\) in the direction of \(\mathbf{q}\). Give your answer to 3 significant figures. [3]

      • M1 for the scalar product \(\mathbf{p} \cdot \mathbf{q} = -9\) (direction goes with the scalar product)
      • M1 for dividing by \(|\mathbf{q}| = \sqrt{14}\)
      • A1 for -2.41

  8. A toy tram runs on a straight track. Its velocity at time \(t\) seconds is \(v = 9\sin(0.6t)\) m s⁻¹. A second toy tram runs on a parallel straight track. When its displacement from the station is \(s\) metres, its velocity is \(v = 3 + 0.2s^2\) m s⁻¹. [6]

    1. (a) Find the acceleration of the first tram when \(t = 2\). Give your answer to 3 significant figures. [2]

      • M1 for differentiating the velocity with respect to \(t\) (chain rule): \(a = 5.4\cos(0.6t)\)
      • A1 for \(a(2) = 1.96\) (calculator in radian mode)

    2. (b) Find the change in displacement of the first tram between \(t = 0\) and \(t = 4\). Give your answer to 3 significant figures. [2]

      • M1 for the definite integral \(\displaystyle\int_0^{4} 9\sin(0.6t)\,dt\)
      • A1 for \(26.1\) metres

    3. (c) Find the acceleration of the second tram when \(s = 3\). [2]

      • M1 for using \(a = v\dfrac{dv}{ds} = (3 + 0.2s^2)(0.4s)\)
      • A1 for \(a = 5.76\)

  9. A buggy moves across a flat field. Its velocity at time \(t\) seconds is \(\mathbf{v} = \begin{pmatrix} 6 t^{2} - 2 \\ 4 t - 7 \end{pmatrix}\) m s⁻¹, for \(t \geq 0\). When \(t = 0\) its position vector relative to a fixed origin O is \(\begin{pmatrix} -3 \\ 4 \end{pmatrix}\) metres. [7]

    1. (a) Find the acceleration of the buggy when \(t = 3\). [2]

      • M1 for differentiating each component: \(\mathbf{a} = \begin{pmatrix} 12 t \\ 4 \end{pmatrix}\)
      • A1 for \(\mathbf{a}(3) = \begin{pmatrix} 36 \\ 4 \end{pmatrix}\)

    2. (b) Find an expression for the position vector of the buggy at time \(t\). [3]

      • M1 for integrating each component of the velocity
      • A1 for the correct integrals, \(2 t^{3} - 2 t\) and \(2 t^{2} - 7 t\), plus constants
      • A1 for using \(\mathbf{r}(0) = \begin{pmatrix} -3 \\ 4 \end{pmatrix}\) to find the constants: \(\mathbf{r} = \begin{pmatrix} 2 t^{3} - 2 t - 3 \\ 2 t^{2} - 7 t + 4 \end{pmatrix}\)

    3. (c) Find the speed of the buggy when \(t = 2\). Give your answer to 3 significant figures. [2]

      • M1 for \(|\mathbf{v}(2)| = \sqrt{(22)^2 + (1)^2}\)
      • A1 for \(22.0\)

  10. A football is kicked from a point O on level ground. Its position vector relative to O at time \(t\) seconds is \(\mathbf{r} = \begin{pmatrix} 11t \\ 12t - 4.9t^2 \end{pmatrix}\) metres, where the first component is horizontal and the second is vertically upward. [6]

    1. (a) Find the velocity vector of the football at time \(t\). [2]

      • M1 for differentiating each component of \(\mathbf{r}\) with respect to \(t\)
      • A1 for \(\begin{pmatrix} 11 \\ 12 - 9.8t \end{pmatrix}\)

    2. (b) Find the greatest height of the football above the ground. Give your answer to 3 significant figures. [2]

      • M1 for setting the vertical velocity to zero, \(12 - 9.8t = 0\), so \(t = 1.22\)
      • A1 for \(12(1.22) - 4.9(1.22)^2 = 7.35\)

    3. (c) Find the horizontal distance from O to the point where the football lands. Give your answer to 3 significant figures. [2]

      • M1 for setting the vertical displacement to zero, \(12t - 4.9t^2 = 0\), so \(t = 2.45\) (reject \(t = 0\))
      • A1 for \(11(2.45) = 26.9\)

  11. A carriage runs round a circular track. Its position vector at time \(t\) seconds, relative to a fixed origin O, is \(\mathbf{r} = \begin{pmatrix} -2 + 3\sin(1.5t) \\ 5 + 3\cos(1.5t) \end{pmatrix}\) metres. A second carriage starts at the lowest point of the circle and goes round anticlockwise, taking 5 seconds for each complete turn, on the same circle. [8]

    1. (a) Show that the carriage moves on a circle, and write down the centre and the radius of the circle. [3]

      • M1 for \(\mathbf{r} - \begin{pmatrix} -2 \\ 5 \end{pmatrix}\) having components 3\sin(1.5t) and 3\cos(1.5t)
      • M1 for \(\sqrt{3^2\cos^2(1.5t) + 3^2\sin^2(1.5t)} = 3\) using \(\sin^2\theta + \cos^2\theta = 1\)
      • A1 for centre \((-2, 5)\) and radius \(3\) metres

    2. (b) State whether the carriage goes round the circle clockwise or anticlockwise, and find the time it takes to go round once. Give your answer to 3 significant figures. [2]

      • M1 for the direction of motion at \(t = 0\), from the position and velocity (\(\mathbf{r}(0) - \text{centre} = \begin{pmatrix} 0 \\ 3 \end{pmatrix}, \ \mathbf{v}(0) = \begin{pmatrix} 4.5 \\ 0 \end{pmatrix}\)): from the highest point of the circle the carriage moves to the right, so clockwise
      • A1 for clockwise and \(T = \dfrac{2\pi}{1.5} = 4.19\) seconds

    3. (c) Write down an expression for the position vector of the second carriage at time \(t\) seconds. [3]

      • M1 for angle speed \(\dfrac{2\pi}{5}\) so the argument is \(\dfrac{2\pi t}{5}\)
      • M1 for centre \(\begin{pmatrix} -2 \\ 5 \end{pmatrix}\) plus radius \(3\) with the correct start point and direction (at t=0 position (cx, cy-R) and velocity (Rw, 0) to the right: anticlockwise)
      • A1 for \(\mathbf{r} = \begin{pmatrix} -2 + 3\sin(\dfrac{2\pi t}{5}) \\ 5 - 3\cos(\dfrac{2\pi t}{5}) \end{pmatrix}\)

Section B: long questions

  1. Two straight pipes in a tunnel are modelled by lines. The lines \(l_1\) and \(l_2\) are given by \(l_1:\ \mathbf{r} = \begin{pmatrix} -2 \\ 3 \\ 1 \end{pmatrix} + \lambda\begin{pmatrix} 1 \\ 2 \\ -2 \end{pmatrix}\) and \(l_2:\ \mathbf{r} = -2\mathbf{i} + 8\mathbf{j} - 5\mathbf{k} + \mu\left(2\mathbf{i} - \mathbf{j} + 2\mathbf{k}\right)\). [14]

    1. (a) Show that \(l_1\) and \(l_2\) intersect, and find the coordinates of the point of intersection. [4]

      • M1 for equating the \(x\) and \(y\) components of the two lines: \(\lambda - 2 = 2 \mu - 2\) and \(2 \lambda + 3 = 8 - \mu\)
      • A1 for \(\lambda = 2\) and \(\mu = 1\)
      • R1 for checking the \(z\) component: \(-3 = -3\), so the lines intersect
      • A1 for \((0, 7, -3)\)

    2. (b) Find the acute angle between \(l_1\) and \(l_2\). [3]

      • M1 for the direction vectors \(\begin{pmatrix} 1 \\ 2 \\ -2 \end{pmatrix}\) and \(\begin{pmatrix} 2 \\ -1 \\ 2 \end{pmatrix}\) (not the position vectors)
      • M1 for \(\cos\theta = \dfrac{|-4|}{\sqrt{9}\sqrt{9}}\)
      • A1 for 63.6°

    3. (c) The point P has coordinates \((3, 6, -2)\). Find the shortest distance from P to \(l_2\). Give your answer to 3 significant figures. [4]

      • M1 for \(\overrightarrow{PB} = \begin{pmatrix} 2 \mu - 5 \\ 2 - \mu \\ 2 \mu - 3 \end{pmatrix}\) for a general point B on \(l_2\)
      • M1 for \(\overrightarrow{PB} \cdot \begin{pmatrix} 2 \\ -1 \\ 2 \end{pmatrix} = 0\)
      • A1 for \(\mu = 2\) and \(\overrightarrow{PB} = \begin{pmatrix} -1 \\ 0 \\ 1 \end{pmatrix}\)
      • A1 for \(\sqrt{2}\) = 1.41

    4. (d) A third line \(l_3\) has parametric equations \(x = 2 s + 1,\ y = s,\ z = 6 - s\). Show that \(l_1\) and \(l_3\) do not intersect. [3]

      • M1 for equating two components of \(l_1\) and \(l_3\): \(\lambda - 2 = 2 s + 1\) and \(2 \lambda + 3 = s\)
      • A1 for \(\lambda = -3\) and \(s = -3\)
      • R1 for the \(z\) components: \(7 \neq 9\), so the lines do not intersect

  2. Two search helicopters, H and K, fly over a forest. Positions are measured in metres from a fixed point O, with \(\mathbf{i}\) east, \(\mathbf{j}\) north and \(\mathbf{k}\) vertically upwards, and \(t\) is the time in seconds. At \(t = 0\) H is at \((-50, 30, 20)\) and moves with constant velocity \(\begin{pmatrix} 10 \\ -5 \\ 3 \end{pmatrix}\) m/s. At \(t = 0\) K is at \((-80, 60, 25)\) and moves with constant speed 14 m/s in the direction of \(\begin{pmatrix} 6 \\ -3 \\ 2 \end{pmatrix}\). [14]

    1. (a) Find the speed of H. Give your answer to 3 significant figures. [2]

      • M1 for \(\sqrt{(10)^2 + (-5)^2 + (3)^2}\)
      • A1 for 11.6

    2. (b) Show that the velocity of K is \(\begin{pmatrix} 12 \\ -6 \\ 4 \end{pmatrix}\). [3]

      • A1 for \(|\begin{pmatrix} 6 \\ -3 \\ 2 \end{pmatrix}| = 7\)
      • M1 for \(\mathbf{v} = 14 \times \dfrac{1}{7}\begin{pmatrix} 6 \\ -3 \\ 2 \end{pmatrix}\)
      • A1 for \(\begin{pmatrix} 12 \\ -6 \\ 4 \end{pmatrix}\) (answer given, so the scaling step must be seen)

    3. (c) Write down the position vector of K at time \(t\). [1]

      • A1 for \(\mathbf{r}_{K} = \begin{pmatrix} -80 \\ 60 \\ 25 \end{pmatrix} + t\begin{pmatrix} 12 \\ -6 \\ 4 \end{pmatrix}\)

    4. (d) Find the distance between H and K when \(t = 5\). Give your answer to 3 significant figures. [2]

      • M1 for finding both positions at the time, or the vector between them: \(\begin{pmatrix} -20 \\ 25 \\ 10 \end{pmatrix}\)
      • A1 for 33.5

    5. (e) Show that H and K do not collide. [3]

      • M1 for equating the \(x\) components: \(10 t - 50 = 12 t - 80\) gives \(t = 15\)
      • M1 for equating a second component: \(30 - 5 t = 60 - 6 t\) gives \(t = 30\)
      • R1 for the values of \(t\) being different, so the same \(t\) does not work in every component and they do not collide

    6. (f) Use your calculator to find the time at which they are closest together, and the distance between them at that time. Give your answers to 3 significant figures. [3]

      • M1 for an expression for the distance, \(d = |\mathbf{r}_{K} - \mathbf{r}_{H}| = \sqrt{6 t^{2} - 170 t + 1825}\)
      • A1 for \(t = 14.2\)
      • A1 for \(d = 24.9\)

  3. A triangular solar panel DEF is held up by three posts. The corners are D\((1, -2, 4)\), E\((4, 0, 3)\) and F\((2, 3, 7)\), in metres. [14]

    1. (a) Find \(\overrightarrow{DE}\) and \(\overrightarrow{DF}\) as column vectors. [2]

      • A1 for \(\overrightarrow{DE} = \begin{pmatrix} 3 \\ 2 \\ -1 \end{pmatrix}\)
      • A1 for \(\overrightarrow{DF} = \begin{pmatrix} 1 \\ 5 \\ 3 \end{pmatrix}\)

    2. (b) Find the angle EDF. Give your answer to the nearest degree. [3]

      • M1 for the scalar product \(\overrightarrow{DE} \cdot \overrightarrow{DF} = (3)(1) + (2)(5) + (-1)(3) = 10\)
      • M1 for \(\cos\theta = \dfrac{10}{\sqrt{14}\sqrt{35}}\)
      • A1 for 63°

    3. (c) Find \(\overrightarrow{DE} \times \overrightarrow{DF}\). [2]

      • M1 for the component form of the vector product, e.g. \((2)(3) - (-1)(5)\)
      • A1 for \(\begin{pmatrix} 11 \\ -10 \\ 13 \end{pmatrix}\)

    4. (d) Hence find the area of triangle DEF. Give your answer to 3 significant figures. [2]

      • M1 for \(\tfrac{1}{2}|\overrightarrow{DE} \times \overrightarrow{DF}|\)
      • A1 for \(\tfrac{1}{2}\sqrt{390}\) = 9.87

    5. (e) Find a unit vector that is perpendicular to both \(\overrightarrow{DE}\) and \(\overrightarrow{DF}\). [2]

      • M1 for dividing \(\begin{pmatrix} 11 \\ -10 \\ 13 \end{pmatrix}\) by its magnitude \(\sqrt{390}\)
      • A1 for \(\pm\dfrac{1}{\sqrt{390}}\begin{pmatrix} 11 \\ -10 \\ 13 \end{pmatrix}\) (either direction)

    6. (f) Find the shortest distance from F to the line DE. Give your answer to 3 significant figures. [3]

      • M1 for the perpendicular distance as the height of the triangle, \(\text{area} = \tfrac{1}{2}|\overrightarrow{DE}| \times h\), or the component \(\dfrac{|\overrightarrow{DE} \times \overrightarrow{DF}|}{|\overrightarrow{DE}|}\)
      • A1 for \(|\overrightarrow{DE}| = \sqrt{14}\)
      • A1 for 5.28

  4. P is a cabin on a circular track. Its position vector relative to the centre O of the track at time \(t\) seconds, where \(t \geq 0\), is \(\mathbf{r}_P = \begin{pmatrix} 6\cos(0.08t^2) \\ 6\sin(0.08t^2) \end{pmatrix}\) metres. A second cabin Q starts from the same point as P, 4 seconds after P, and moves in exactly the same way. [12]

    1. (a) Show that P moves on a circle with centre O, and write down the radius of the circle. [2]

      • M1 for \(|\mathbf{r}_P|^2 = 36\cos^2(0.08t^2) + 36\sin^2(0.08t^2) = 36\) using \(\sin^2\theta + \cos^2\theta = 1\)
      • A1 for \(|\mathbf{r}_P| = 6\), constant, so P moves on a circle of radius 6 metres

    2. (b) Find an expression for the velocity of P at time \(t\). [3]

      • M1 for differentiating \(6\cos(0.08t^2)\) with the chain rule
      • A1 for \(-0.96t\sin(0.08t^2)\)
      • A1 for \(0.96t\cos(0.08t^2)\)

    3. (c) Find the speed of P when \(t = 3\). [2]

      • M1 for \(|\mathbf{v}_P| = \sqrt{(0.96t)^2\sin^2(0.08t^2) + (0.96t)^2\cos^2(0.08t^2)} = 0.96t\), or evaluating the magnitude on the calculator at \(t = 3\)
      • A1 for \(2.88\)

    4. (d) Find the number of complete revolutions that P makes in the first 15 seconds. [2]

      • M1 for the angle turned, \(0.08(15)^2 = 18.0\) radians, divided by \(2\pi\) to give \(2.86\) revolutions
      • A1 for 2 complete revolutions (whole number only)

    5. (e) Write down an expression for the position vector \(\mathbf{r}_Q\) of Q at time \(t\), for \(t \geq 4\). [1]

      • A1 for replacing \(t\) by \((t - 4)\) in \(\mathbf{r}_P\)

    6. (f) Find the distance between P and Q when \(t = 8\). Give your answer to 3 significant figures. [2]

      • M1 for \(|\mathbf{r}_P - \mathbf{r}_Q|\) with \(t = 8\), using \(\mathbf{r}_P\) at angle \(0.08(8)^2 = 5.12\) and \(\mathbf{r}_Q\) at angle \(0.08(4)^2 = 1.28\)
      • A1 for 11.3

  5. A cyclist rides along a straight road. The velocity of the cyclist at time \(t\) seconds is \(v = t^{2} - 10 t + 16\) m s⁻¹, for \(t \geq 0\). [7]

    1. (a) Find the times when the cyclist is at rest, and for each time say whether the cyclist changes direction. [3]

      • M1 for solving \(t^{2} - 10 t + 16 = 0\): \(t = 2\), \(t = 8\)
      • A1 for both times
      • R1 for \(v\) changing sign (positive, then negative, then positive) at each of these times, so the direction changes at both

    2. (b) Find the total distance travelled by the cyclist in the first 10 seconds. Give your answer to 3 significant figures. [2]

      • M1 for \(\displaystyle\int_0^{10} |t^{2} - 10 t + 16|\,dt\), using the calculator
      • A1 for \(65.3\) metres (the displacement is \(-6.67\) metres, which is not the distance)

    3. (c) The cyclist starts at the point where \(s = 0\). Find the first time after it starts at which it returns to this point. Give your answer to 3 significant figures. [2]

      • M1 for \(s = \displaystyle\int_0^t v\,dt = \frac{t^{3}}{3} - 5 t^{2} + 16 t\), set equal to 0 and solved with technology
      • A1 for \(t = 4.63\) (reject \(t = 0\) and any negative time)

  6. A stone is thrown from a point 1.8 m above level ground, with speed 14 m s⁻¹ at an angle of 40° above the horizontal. Take O as the point on the ground directly below the point of release, with \(\mathbf{i}\) horizontal and \(\mathbf{j}\) vertically upward. Ignore air resistance and take \(g = 9.8\) m s⁻². [12]

    1. (a) Write down the position vector of the stone at time \(t\) seconds after it is thrown, in the form \(\mathbf{r} = \begin{pmatrix} x \\ y \end{pmatrix}\) where \(x\) and \(y\) are in terms of \(t\). [2]

      • M1 for horizontal component \(14\cos 40^\circ = 10.7\) and vertical component \(14\sin 40^\circ = 9.00\)
      • A1 for \(\mathbf{r} = \begin{pmatrix} 10.7t \\ 1.8 + 9.00t - 4.9t^2 \end{pmatrix}\) (with \(g = 9.8\))

    2. (b) Find the time that the stone takes to reach the ground, and the horizontal distance from O to the point where it lands. [3]

      • M1 for setting the vertical component equal to 0: \(1.8 + 9.00t - 4.9t^2 = 0\)
      • A1 for \(t = 2.02\) (the positive root)
      • A1 for \(x = 10.7 \times 2.02 = 21.6\) m

    3. (c) Find the greatest height of the stone above the ground. [2]

      • M1 for the top at \(9.00 - 9.8t = 0\), \(t = 0.918\), and substituting into \(y\)
      • A1 for \(5.93\) m

    4. (d) Find the speed of the stone when it lands, and the angle below the horizontal at which it lands. [3]

      • M1 for \(\mathbf{v}(t) = \begin{pmatrix} 10.7 \\ 9.00 - 9.8t \end{pmatrix}\) at \(t = 2.02\): \(\begin{pmatrix} 10.7 \\ -10.8 \end{pmatrix}\)
      • A1 for speed \(15.2\) m s⁻¹
      • A1 for \(\tan^{-1}\dfrac{10.8}{10.7} = 45.2^\circ\) below the horizontal

    5. (e) The stone passes a wall, of height 6.5 m, at a horizontal distance 10 m from O. Decide whether the stone clears the wall, and find the vertical distance between the stone and the top of the wall. [2]

      • M1 for \(t = \dfrac{10}{10.7} = 0.932\), then \(y = 5.93\) m at this time
      • A1 for comparing 5.93 with 6.5: the stone does not clear the wall, 0.569 m below the top

jmaths.xyz · AI HL · AHL 4.30 Unit review