Parametric equations – Year 13 A level maths worksheet answers to download as PDF

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1. Answer to Question 1

   1. \(t=-2\): \((-3,1)\); \(t=-1\): \((-1,-2)\); \(t=0\): \((1,-3)\); \(t=1\): \((3,-2)\); \(t=2\): \((5,1)\).
   2. \(2t+1=7\) gives \(t=3\). Check: \(y=9-3=6\). Yes, the point is \((7,6)\) when \(t=3\).
   3. \(x=0\) needs \(t=-0.5\), which gives \(y=0.25-3=-2.75\ne-3\). So \((0,-3)\) is not on the curve.

2. Answer to Question 2

   Method: to find a Cartesian equation, eliminate the parameter. If one equation is linear in \(t\), make \(t\) the subject and substitute. If the equations involve \(\sin t\) and \(\cos t\), isolate them and use \(\sin^2t+\cos^2t=1\) or a double angle formula.

   1. The \(x\) equation is linear, so make \(t\) the subject and substitute into \(y\). \(t=\dfrac{x+2}{3}\), so \(y=\dfrac{(x+2)^2}{9}+1=\dfrac{x^2+4x+13}{9}\). The curve is a parabola.
   2. Make \(t\) the subject of \(x=2t\) and substitute. \(t=\dfrac x2\), so \(y=\dfrac{8}{x/2}=\dfrac{16}{x}\), i.e. \(xy=16\). The curve is a rectangular hyperbola; \(t=0\) is not allowed, so \(x\ne0\).

3. Answer to Question 3

   Method: to find a Cartesian equation, eliminate the parameter. If one equation is linear in \(t\), make \(t\) the subject and substitute. If the equations involve \(\sin t\) and \(\cos t\), isolate them and use \(\sin^2t+\cos^2t=1\) or a double angle formula.

   1. Isolate \(\cos t\) and \(\sin t\). \(\cos t=\dfrac x4\), \(\sin t=\dfrac y3\). Using \(\cos^2t+\sin^2t=1\): \(\dfrac{x^2}{16}+\dfrac{y^2}{9}=1\), an ellipse. It has semi-axes 4 along the \(x\)-axis and 3 along the \(y\)-axis.
   2. Isolate \(\cos t\) and \(\sin t\) after moving the constants. \(\cos t=\dfrac{x-2}{3}\), \(\sin t=\dfrac{y+1}{3}\), so \((x-2)^2+(y+1)^2=9\): a circle with centre \((2,-1)\) and radius 3. The constants in the parametric equations give the centre.

4. Answer to Question 4

   1. \(\cos2t=1-2\sin^2t\), so \(y=1-2x^2\).
   2. \(-1\le\sin t\le1\) so \(-1\le x\le1\). Then \(x^2\) lies between 0 and 1, so \(y\) lies between \(-1\) and \(1\): \(-1\le y\le1\). The curve is only the part of the parabola with \(|x|\le1\).

5. Answer to Question 5

   Method: to find where a parametric curve meets a line or an axis, substitute the expressions for \(x\) and \(y\) into the equation of the line, solve for \(t\), then substitute each value of \(t\) back to get the coordinates.

   1. Replace \(y\) by \(2t\) and \(x\) by \(t^2+1\) in \(y=x-4\). Substitute: \(2t=t^2+1-4\), so \(t^2-2t-3=0\). This is the required equation.
   2. Factorise the quadratic. \((t-3)(t+1)=0\), so \(t=3\) or \(t=-1\). Points: \(t=3\): \((10,6)\); \(t=-1\): \((2,-2)\). Check: \(6=10-4\) and \(-2=2-4\), so both points lie on the line.

6. Answer to Question 6

   1. \(y=0\): \(t(t^2-4)=0\), so \(t=0\) or \(t=\pm2\). \(t=0\): \((-4,0)\); \(t=\pm2\): \((0,0)\).
   2. \(x=0\): \(t=\pm2\), giving \(y=0\) both times: the point \((0,0)\).
   3. The origin is reached for \(t=2\) and \(t=-2\): the curve crosses itself at the origin, so it has a loop.

7. Answer to Question 7

   1. Vertical velocity \(=12-9.8t=0\) when \(t=1.2245\) s. \(y=1+12(1.2245)-4.9(1.2245)^2=8.35\) m.
   2. \(y=0\): \(4.9t^2-12t-1=0\), \(t=\dfrac{12+\sqrt{144+19.6}}{9.8}=2.530\) s, so \(x=15\times2.530=37.9\) m.
   3. At \(x=9\), \(t=0.6\) and \(y=1+7.2-1.764=6.44\) m, which is above 2.2 m: the ball clears the wall.
   4. The ball is a particle and there is no air resistance (or spin); gravity is constant at \(9.8\) m s\(^{-2}\).

8. Answer to Question 8

   Method: in a parametric model, \(t\) is usually time. Answer each question by deciding first whether you need a value of \(t\) (from a condition on \(x\) or \(y\)) or a position (by substituting \(t\)).

   1. Isolate the sine and cosine, square and add. \(x^2+(y-12)^2=100(\sin^2+\cos^2)=100\): a circle with centre \((0,12)\) and radius 10 m. The centre is 12 m above the foot of the wheel.
   2. The height \(y\) is smallest when the cosine is 1 and largest when it is \(-1\). Lowest \(12-10=2\) m; highest \(12+10=22\) m. So the car moves between 2 m and 22 m above the ground.
   3. Substitute \(t=5\): the angle is \(\dfrac{5\pi}{20}=\dfrac\pi4\). \(y=12-10\cos(\pi/4)=12-7.07=4.93\) m. So after 5 seconds the car is about 4.93 m above the ground.
   4. Solve the inequality for the angle, then convert to time. \(12-10\cos\theta>17\) means \(\cos\theta<-0.5\), so \(\theta=\dfrac{\pi t}{20}\) between \(\dfrac{2\pi}3\) and \(\dfrac{4\pi}3\), i.e. \(t\) between \(\dfrac{40}3\) and \(\dfrac{80}3\), a time of \(\dfrac{40}3=13.3\) s. So the car is above 17 m for about 13.3 seconds of each 40-second turn, which is one third of the time.

9. Answer to Question 9

   Method: to find a Cartesian equation, eliminate the parameter. If one equation is linear in \(t\), make \(t\) the subject and substitute. If the equations involve \(\sin t\) and \(\cos t\), isolate them and use \(\sin^2t+\cos^2t=1\) or a double angle formula.

   1. Use \(\cos t=\dfrac x{40}\), \(\sin t=\dfrac y{25}\) and \(\cos^2t+\sin^2t=1\). \(\dfrac{x^2}{1600}+\dfrac{y^2}{625}=1\). This is an ellipse with semi-axes 40 m and 25 m.
   2. Use Pythagoras: \(d^2=x^2+y^2\). \(d^2=1600\cos^2t+625\sin^2t=625+975\cos^2t\) (using \(\sin^2t=1-\cos^2t\)). This is the required result.
   3. Find the extreme values of \(d^2\), then take square roots. \(\cos^2t\) ranges from 0 to 1, so \(d^2\) from 625 to 1600: least distance 25 m, greatest 40 m. These are the semi-axes of the ellipse.
   4. Substitute \(t=\tfrac\pi3\) (that is \(60^\circ\)). \(x=40\cos60^\circ=20\) m, \(y=25\sin60^\circ=21.65\) m: \((20,21.65)\). So the drone is 20 m east and about 21.65 m north of the tower.

10. Answer to Question 10

    Method: in a parametric model, \(t\) is usually time. Answer each question by deciding first whether you need a value of \(t\) (from a condition on \(x\) or \(y\)) or a position (by substituting \(t\)).

    1. Make \(t\) the subject of the linear equation and substitute. \(t=\dfrac x5\) so \(y=12x-\dfrac{4.9x^2}{25}=12x-0.196x^2\). The path is a parabola.
    2. The rocket stops rising when \(\dfrac{dy}{dt}=0\). \(60-9.8t=0\) gives \(t=6.12\) s and \(y=\dfrac{3600}{19.6}=183.7\) m. So the rocket reaches about 184 m after about 6.12 s.
    3. Set \(y=0\) and ignore the solution \(t=0\), which is the launch. \(y=0\): \(t=\dfrac{60}{4.9}=12.24\) s, so \(x=5\times12.24=61.2\) m. So it would land about 61.2 m away.
    4. Substitute the time of the highest point into \(x\). At \(t=6.12\) s, \(x=30.6\) m. This is half of the range, by the symmetry of the parabola.

11. Answer to Question 11

    1. \(t=0\): \((0,0)\). \(t=\pi/2\): \(x=0.35(1.5708-1)=0.20\), \(y=0.35\). \(t=\pi\): \(x=0.35\pi=1.10\), \(y=0.70\). \(t=3\pi/2\): \(x=0.35(4.7124+1)=2.00\), \(y=0.35\). \(t=2\pi\): \(x=2.20\), \(y=0\).
    2. Greatest height \(0.70\) m (the diameter of the wheel). One turn moves the bicycle \(0.35\times2\pi=2.20\) m.
    3. At the lowest point the reflector touches the road, and a rolling wheel has no slipping, so that point of the wheel has zero velocity relative to the ground.

12. Answer to Question 12

    Method: in a parametric model, \(t\) is usually time. Answer each question by deciding first whether you need a value of \(t\) (from a condition on \(x\) or \(y\)) or a position (by substituting \(t\)).

    1. \(t=\dfrac x{20}\) gives \(y=30-5\dfrac{x^2}{400}=30-\dfrac{x^2}{80}\).
    2. \(y=0\): \(t=\sqrt6=2.449\) s, \(x=20\sqrt6=49.0\) m.
    3. \(y=30-\dfrac{900}{80}=18.75\) m.
    4. A sloping landing area falling away would be reached later and further, because the ground is lower below the path.

13. Answer to Question 13

    Method: to find where a parametric curve meets a line or an axis, substitute the expressions for \(x\) and \(y\) into the equation of the line, solve for \(t\), then substitute each value of \(t\) back to get the coordinates.

    1. On the line \(y=x\), so set \(3\sin t=4\cos t\) and divide by \(\cos t\). Substitute: \(3\sin t=4\cos t\), so \(\tan t=\dfrac43\). This is the required condition.
    2. Tangent is positive in the first and third quadrants, so there are two values of \(t\) in one turn. \(t=0.9273\) (or \(0.9273+\pi\)). \(\cos t=0.6\), \(\sin t=0.8\), so \(x=2.4\), \(y=2.4\): \((2.4,2.4)\). The other point is \((-2.4,-2.4)\). The two points are symmetric about the origin.
    3. Use Pythagoras on the differences of the coordinates. The distance is \(\sqrt{4.8^2+4.8^2}=4.8\sqrt2=6.79\) thousand km. So the two points are about 6790 km apart.

14. Answer to Question 14

    Method: in a parametric model, \(t\) is usually time. Answer each question by deciding first whether you need a value of \(t\) (from a condition on \(x\) or \(y\)) or a position (by substituting \(t\)).

    1. Use \(d^2=x^2+y^2\) and expand both brackets. \(d^2=(3t-6)^2+(4t-5)^2=9t^2-36t+36+16t^2-40t+25=25t^2-76t+61\). This is the required expression.
    2. Minimising \(d^2\) is the same as minimising \(d\), so differentiate \(d^2\) with respect to \(t\). \(\dfrac{d(d^2)}{dt}=50t-76=0\), so \(t=1.52\) hours. \(d^2=25(1.52)^2-76(1.52)+61=3.24\), so \(d=1.8\) km. So the boat is closest after 1.52 hours (about 1 hour 31 minutes), at a distance of 1.8 km.
    3. Compare the minimum distance with the range of the light. \(1.8<2\), so yes. So the sailors can see the light at their closest approach, for a short time.

15. Answer to Question 15

    1. \(y=x^2+1\). Since \(e^t>0\), \(x>0\), and then \(y>1\). Only the part with \(x>0\) is traced.
    2. \(x+y=\dfrac{1+t}{1+t}=1\), so \(y=1-x\). Since \(x=\dfrac1{1+t}\) is never zero, the point \((0,1)\) is missing.
    3. \(y=x^2-2\). \(x=t^2\) ranges over \(0\le x\le4\) for \(-1\le t\le2\) (with \(x=0\) at \(t=0\), \(x=1\) at \(t=\pm1\)). So only \(0\le x\le4\) is traced, \(-2\le y\le14\).

16. Answer to Question 16

    1. \(t=1\): \((0,0)\); \(t=-1\): \((0,0)\). The curve crosses itself at the origin.
    2. \(y=0\) when \(t=0,\pm1\): points \((-1,0)\) and \((0,0)\).
    3. Area \(=\left|\int_{-1}^{1}y\dfrac{dx}{dt}dt\right|=\left|\int_{-1}^{1}(t^3-t)(2t)dt\right|=\left|\int_{-1}^{1}(2t^4-2t^2)dt\right|=\left|\left[\dfrac{2t^5}5-\dfrac{2t^3}3\right]_{-1}^{1}\right|=\left|2\left(\dfrac25-\dfrac23\right)\right|=\dfrac8{15}\).
    4. Replacing \(t\) by \(-t\) changes \(y\) to \(-y\) but leaves \(x\) unchanged, so the curve is symmetrical about the \(x\)-axis.

17. Answer to Question 17

    1. \(y^2=t^6=(t^2)^3=x^3\).
    2. \(\dfrac{dy}{dx}=\dfrac{dy/dt}{dx/dt}=\dfrac{3t^2}{2t}=\dfrac{3t}2\). At \(t=2\) the gradient is 3.
    3. The point is \((4,8)\), so \(y-8=3(x-4)\), i.e. \(y=3x-4\).
    4. Substitute into \(y^2=x^3\): \((3x-4)^2=x^3\), i.e. \(x^3-9x^2+24x-16=0\), which factorises as \((x-4)^2(x-1)=0\). The repeated root \(x=4\) is the tangent point; \(x=1\) gives \(y=3-4=-1\), i.e. \((1,-1)\).

18. Answer to Question 18

    1. \(x=2+3t\) matches 2, 5, 8, 11, 14. \(y\): \(t=1\): \(1+4-0.5=4.5\); \(t=2\): \(1+8-2=7\); \(t=3\): \(1+12-4.5=8.5\); \(t=4\): \(1+16-8=9\). All match.
    2. \(t=\dfrac{x-2}3\), so \(y=1+\dfrac{4(x-2)}3-\dfrac{(x-2)^2}{18}\).
    3. \(t=8\): \(x=26\), \(y=1+32-32=1\). The drone would be back at its starting height after 8 s, having climbed to 9 m at \(t=4\) (maximum), which is plausible for a parabolic flight but not for a drone that can hover or change direction, so the model should only be trusted within the range of the data.

19. Answer to Question 19

    1. \(5.625t=4.5\) gives \(t=0.8\) s.
    2. \(y=2+5.25(0.8)-4.9(0.64)=2+4.2-3.136=3.064\) m.
    3. The ball is at 3.064 m, only 0.014 m above the hoop centre, so it passes through the hoop (the ring is 0.45 m wide). The shot is successful.
    4. \(5.25-9.8t=0\) gives \(t=0.536\) s, \(y=2+\dfrac{5.25^2}{19.6}=3.41\) m.
    5. The ball is treated as a particle with no air resistance or spin; the ring is treated as a point.

20. Answer to Question 20

    Method: to find a Cartesian equation, eliminate the parameter. If one equation is linear in \(t\), make \(t\) the subject and substitute. If the equations involve \(\sin t\) and \(\cos t\), isolate them and use \(\sin^2t+\cos^2t=1\) or a double angle formula.

    1. \(x^2+y^2=\dfrac{(1-t^2)^2+4t^2}{(1+t^2)^2}=\dfrac{(1+t^2)^2}{(1+t^2)^2}=1\).
    2. \(t=0\): \((1,0)\); \(t=1\): \((0,1)\); \(t=-1\): \((0,-1)\).
    3. \(x=-1\) needs \(1-t^2=-1-t^2\), i.e. \(1=-1\), which is impossible. As \(t\to\pm\infty\) the point approaches \((-1,0)\) but never reaches it.

    So this parametrisation describes the whole unit circle except the single point \((-1,0)\).

21. Answer to Question 21

    1. \(y=2\sin t\cos t\), so \(y^2=4\sin^2t\cos^2t=4x^2(1-x^2)\).
    2. \(\sin t=0\) and \(\sin2t=0\) for \(t=0,\pi,2\pi\). At \(t=0\) and \(t=\pi\) the curve passes through the origin with different directions, so it crosses itself: a figure of eight.
    3. \(y=\sin2t\) is greatest, 1, when \(t=\dfrac\pi4\); then \(x=\sin\dfrac\pi4=0.707\).