Parametric equations – Year 13 A level maths revision notes to download as PDF

Worked example 1

Find the Cartesian equation of (a) \(x=2t-1\), \(y=t^2+t\); (b) \(x=3t\), \(y=\dfrac2t\).

(a) \(t=\dfrac{x+1}2\), so \(y=\dfrac{(x+1)^2}{4}+\dfrac{x+1}{2}=\dfrac{x^2+4x+3}{4}\). (b) Multiplying gives \(xy=6\).

Worked example 2

Find the Cartesian equation of (a) \(x=2\cos t\), \(y=3\sin t\); (b) \(x=1+2\cos t\), \(y=-3+2\sin t\); (c) \(x=\sin t\), \(y=\cos2t\).

(a) \(\dfrac{x^2}{4}+\dfrac{y^2}{9}=1\), an ellipse. (b) \((x-1)^2+(y+3)^2=4\), a circle with centre \((1,-3)\) and radius 2. (c) \(\cos2t=1-2\sin^2t\) gives \(y=1-2x^2\). Also \(x=\sec t\), \(y=\tan t\) gives \(x^2-y^2=1\), a hyperbola.

Worked example 3

Describe the part of the curve traced by (a) \(x=\sin t\), \(y=\cos2t\); (b) \(x=e^t\), \(y=e^{2t}+1\); (c) \(x=\dfrac1{1+t}\), \(y=\dfrac t{1+t}\).

(a) \(-1\le x\le1\): only the part of \(y=1-2x^2\) between \(x=-1\) and \(x=1\). (b) \(y=x^2+1\), but \(e^t>0\), so only \(x>0\). (c) \(x+y=1\), but \(x\) is never 0, so the point \((0,1)\) is missing.

Worked example 4

Find where \(x=t^2\), \(y=2t\) meets \(y=x-3\).

\(2t=t^2-3\), so \(t^2-2t-3=0\), \((t-3)(t+1)=0\) and \(t=3\) or \(t=-1\). The points are \((9,6)\) and \((1,-2)\).

Worked example 5

A ball is thrown from height 2 m: \(x=12t\), \(y=2+10t-4.9t^2\). Find the greatest height and where it lands.

The vertical velocity is zero when \(10-9.8t=0\), \(t=1.02\) s, and \(y=2+\dfrac{100}{19.6}=7.10\) m. It lands when \(y=0\): \(4.9t^2-10t-2=0\), so \(t=\dfrac{10+\sqrt{139.2}}{9.8}=2.22\) s and \(x=26.7\) m. The model ignores air resistance, treats the ball as a particle and assumes constant gravity.

Worked example 6

Find the minimum distance from the origin to the curve \(x=t+1\), \(y=t-1\).

The distance is \(\sqrt{(t+1)^2+(t-1)^2}=\sqrt{2t^2+2}\), with minimum \(\sqrt2\) at \(t=0\).

Worked example 7: Converting with a hyperbola

Find the Cartesian equation of \(x=\sec t\), \(y=\tan t\) and state the part of the curve traced for \(0\le t\lt \dfrac\pi2\).

Use \(1+\tan^2t=\sec^2t\): \(1+y^2=x^2\), so \(x^2-y^2=1\), a hyperbola. For \(0\le t\lt \dfrac\pi2\): \(x=\sec t\ge1\) and \(y=\tan t\ge0\), so only the upper branch on the right (\(x\ge1\), \(y\ge0\)) is traced.

Worked example 8: Parametric equations for a given circle

Write parametric equations for the circle \((x-2)^2+(y+1)^2=9\).

The centre is \((2,-1)\) and the radius is 3. Use \(x=2+3\cos t\) and \(y=-1+3\sin t\). Check: \((3\cos t)^2+(3\sin t)^2=9\). For \(0\le t<2\pi\) the point makes one anticlockwise circuit starting at \((5,-1)\). Starting elsewhere, or going clockwise, is possible by changing the parameter, so a parametric description is not unique.

Worked example 9: Finding where a curve meets an axis

Find where \(x=t^2-3t\), \(y=t+2\) meets the \(x\)-axis and the \(y\)-axis.

\(x\)-axis: \(y=0\) gives \(t=-2\) and \(x=4+6=10\): the point \((10,0)\). \(y\)-axis: \(x=0\) gives \(t(t-3)=0\), so \(t=0\) or \(t=3\): points \((0,2)\) and \((0,5)\). Always substitute back to find both coordinates; the value of \(t\) alone is not the answer.

Worked example 10: Area under a parametric curve

Find the area between the curve \(x=t^2\), \(y=t+1\) for \(0\le t\le2\), the \(x\)-axis and the lines \(x=0\) and \(x=4\).

\(\displaystyle\int y\,dx=\int_0^2(t+1)(2t)\,dt=\int_0^2(2t^2+2t)\,dt=\left[\dfrac{2t^3}3+t^2\right]_0^2=\dfrac{16}3+4=\dfrac{28}3\). The limits are values of \(t\) (\(t=0\) gives \(x=0\), \(t=2\) gives \(x=4\)), so we convert the integral to a \(t\)-integral using \(dx=\dfrac{dx}{dt}dt\).

Worked example 11: Gradient from parametric equations

Find the gradient of the curve \(x=t^2+1\), \(y=t^3-t\) at the point where \(t=2\).

\(\dfrac{dx}{dt}=2t\) and \(\dfrac{dy}{dt}=3t^2-1\). So \(\dfrac{dy}{dx}=\dfrac{3t^2-1}{2t}\). At \(t=2\): \(\dfrac{11}4=2.75\). The point is \((5,6)\), so the tangent is \(y-6=2.75(x-5)\).

Worked example 12: A projectile modelled parametrically

A stone is thrown from the top of a cliff: \(x=8t\), \(y=20+6t-4.9t^2\). Find when it hits the sea (\(y=0\)) and how far from the base of the cliff it lands.

\(4.9t^2-6t-20=0\) gives \(t=\dfrac{6+\sqrt{36+392}}{9.8}=\dfrac{6+20.69}{9.8}=2.72\) s. Then \(x=8\times2.72=21.8\) m. The negative root is rejected because the stone is thrown at \(t=0\).

Worked example 13: A curve with a fractional parameter

Find the Cartesian equation of \(x=\dfrac{t}{t-1}\), \(y=\dfrac{1}{t+2}\) and state any values that are never reached.

From the second equation, \(t+2=\dfrac1y\), so \(t=\dfrac1y-2=\dfrac{1-2y}{y}\). Then \(t-1=\dfrac{1-3y}{y}\) and \(x=\dfrac{(1-2y)/y}{(1-3y)/y}=\dfrac{1-2y}{1-3y}\). Rearranging: \(x-3xy=1-2y\), so \(y(2-3x)=1-x\) and \(y=\dfrac{1-x}{2-3x}\). The equation \(\dfrac{t}{t-1}=1\) has no solution, so \(x=1\) is never reached, and \(\dfrac1{t+2}\) is never zero, so \(y=0\) is never reached. Also \(t=1\) and \(t=-2\) are not allowed. The Cartesian equation is therefore valid only for \(x\ne1\) and \(y\ne0\).

Worked example 14: Direction of travel and speed

A particle moves on the path \(x=4t\), \(y=3t^2\). Find its speed at \(t=1\) and describe its direction of travel.

The velocity components are \(\dfrac{dx}{dt}=4\) and \(\dfrac{dy}{dt}=6t\). At \(t=1\) they are 4 and 6, so the speed is \(\sqrt{16+36}=\sqrt{52}=7.21\) m/s. The direction makes angle \(\tan^{-1}\dfrac64=56.3^\circ\) with the positive \(x\)-axis. The path is a parabola \(y=\dfrac{3x^2}{16}\) and the particle moves to the right as \(t\) increases.

Worked example 15: Sketching from a table

Sketch the curve \(x=t^2-2t\), \(y=t+1\) for \(-1\le t\le3\) by plotting points, then find its Cartesian equation.

Table: \(t=-1\): \((3,0)\); \(t=0\): \((0,1)\); \(t=1\): \((-1,2)\); \(t=2\): \((0,3)\); \(t=3\): \((3,4)\). Plotting these gives a parabola opening to the right with vertex at \((-1,2)\), traced from bottom to top as \(t\) increases. Cartesian equation: \(t=y-1\), so \(x=(y-1)^2-2(y-1)=y^2-4y+3\). Check with \(t=3\): \(y=4\), \(x=16-16+3=3\). ✓

Worked example 16: Comparing two moving objects

Two drones have positions \(\mathbf r_1\): \(x=2t\), \(y=10-t\) and \(\mathbf r_2\): \(x=14-t\), \(y=3t-2\). Do the paths cross, and do the drones collide?

Paths: eliminate \(t\). For drone 1: \(t=x/2\), so \(y=10-x/2\). For drone 2: \(t=14-x\), so \(y=3(14-x)-2=40-3x\). Equate: \(10-\dfrac x2=40-3x\), so \(\dfrac{5x}2=30\) and \(x=12\), \(y=4\). The paths cross at \((12,4)\). Drone 1 is there when \(2t=12\), \(t=6\). Drone 2 is there when \(14-t=12\), \(t=2\). The times differ, so the drones do not collide, even though their paths cross.