P4.2 Parametric equations
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A2 content for Pure Mathematics 4 (specification sections 3.1, 5.1 (parametric part) and 6.5). Implicit differentiation and volumes of revolution have their own P4 pages.
What this page covers
7 points
- What a parameter is; plotting a curve from a table of \(t\)-values and reading its direction
- Eliminating the parameter algebraically to get a Cartesian equation
- Eliminating the parameter with trigonometric identities: \(\sin^2 + \cos^2 = 1\), double-angle formulae, \(\sec^2 = 1 + \tan^2\), \(\csc^2 = 1 + \cot^2\)
- The domain and range of the Cartesian equation when the parameter is restricted
- \(\frac{dy}{dx} = \frac{dy}{dt} \div \frac{dx}{dt}\); horizontal and vertical tangents
- Equations of tangents and normals to a parametric curve
- The area under a parametric curve, \(\int y\frac{dx}{dt}\,dt\), with the limits changed to values of \(t\)
In one minute
6 points
- A parametric curve gives both coordinates in terms of a third variable, the parameter (usually \(t\) or \(\theta\)), as \(x = f(t)\), \(y = g(t)\), so each value of \(t\) gives one point and the curve has a direction.
- To get a Cartesian equation, eliminate \(t\): make \(t\) the subject of one equation and substitute, or (with trigonometric functions) use an identity that links the two expressions.
- Domain and range: the Cartesian equation may describe more points than the curve. The \(x\)-values and \(y\)-values the curve actually reaches come from the allowed values of \(t\).
- Gradient: \(\frac{dy}{dx} = \frac{dy}{dt} \div \frac{dx}{dt}\). Tangent horizontal where \(\frac{dy}{dt} = 0\); vertical where \(\frac{dx}{dt} = 0\).
- Area: \(\int y\,dx = \int y\frac{dx}{dt}\,dt\), with the \(x\)-limits replaced by the matching \(t\)-limits.
- None of these results is in the formulae list. Learn them.
Contents
14 sections
- What are parametric equations?
- How do you turn parametric equations into a Cartesian equation?
- How do you eliminate a parameter inside trigonometric functions?
- Which values of x and y does the curve actually reach?
- How do you find dy/dx for a parametric curve?
- How do you find the tangent and normal to a parametric curve?
- How do you find the area under a parametric curve?
- How do the pieces fit together in a full question?
- Exam tips for P4.2
- Common misconceptions
- How this topic connects
- Key formulae
- Check your understanding
- Specification checklist
What are parametric equations?
Instead of linking \(y\) directly to \(x\), both coordinates are given as functions of a third variable, the parameter: \(x = f(t)\), \(y = g(t)\). Each value of \(t\) gives exactly one point \((f(t), g(t))\). As \(t\) increases the point moves along the curve, so a parametric curve has a direction.
A parametric curve can do things that \(y = f(x)\) cannot: turn back on itself, have vertical tangents, or cross itself. To sketch a curve, make a table of \(t\), \(x\) and \(y\), plot the points in order of \(t\) and join them smoothly.
| \(t\) | −2 | −1 | 0 | 1 | 2 |
|---|---|---|---|---|---|
| \(x = t^2\) | 4 | 1 | 0 | 1 | 4 |
| \(y = t^3 - 3t\) | −2 | 2 | 0 | −2 | 2 |
The most familiar parametric curve is the path of a projectile. Launched with horizontal speed 12 m s⁻¹ and vertical speed 9 m s⁻¹, its position after \(t\) seconds is \(x = 12t\), \(y = 9t - 4.9t^2\). Time is the parameter: it tells you when the projectile is at each point, which a Cartesian equation cannot. Eliminating \(t = \frac{x}{12}\) gives the path \(y = \frac34 x - \frac{49}{1440}x^2\).
Joining the points in order of \(x\) instead of in order of \(t\). In the table above \(t = -1\) and \(t = 1\) both give \(x = 1\), but they are different points, \((1, 2)\) and \((1, -2)\), on different parts of the curve.
How do you turn parametric equations into a Cartesian equation?
A Cartesian equation links \(x\) and \(y\) only. To find it, you eliminate the parameter. When one equation is easy to rearrange (often a linear one such as \(x = 2t - 1\)), make \(t\) the subject and substitute into the other equation, then simplify into the form asked for. A fraction inside a fraction is cleared by multiplying top and bottom by the same thing.
Sometimes it is quicker to combine the equations directly. For example, if \(x = t^2\) and \(y = t^3 - 3t = t(t^2 - 3)\), then squaring gives \(y^2 = t^2(t^2 - 3)^2 = x(x - 3)^2\).
A curve \(C\) has parametric equations \(x = 2t - 1\), \(y = \frac{t}{t+2}\), \(t \ne -2\). (a) Show that a Cartesian equation of \(C\) is \(y = \frac{x+1}{x+5}\). (b) Find the coordinates of the point where \(C\) crosses the \(y\)-axis.
(a) Make \(t\) the subject of the simpler equation: \(x = 2t - 1 \Rightarrow t = \frac{x+1}{2}\). Substitute, then multiply top and bottom by 2: \(y = \frac{(x+1)/2}{(x+1)/2 + 2} = \frac{x+1}{x+1+4}\). \(y = \frac{x+1}{x+5}\) as required.
(b) On the \(y\)-axis \(x = 0\), so \(t = \frac12\) and \(y = \frac{1/2}{1/2 + 2} = \frac15\). \(\left(0, \frac15\right)\) Check with the Cartesian equation: \(\frac{0+1}{0+5} = \frac15\).
How do you eliminate a parameter inside trigonometric functions?
You cannot sensibly make \(t\) the subject of \(x = 3\sin t\) (you would get inverse functions). Instead, find an identity that connects the two trigonometric expressions. First isolate the trigonometric function in each equation, for example \(x = 2 + 3\cos\theta \Rightarrow \cos\theta = \frac{x-2}{3}\); then substitute into the identity. Pick the identity by looking at what appears:
| The equations contain… | Use the identity | Typical result |
|---|---|---|
| \(\sin t\) and \(\cos t\) | \(\sin^2 t + \cos^2 t = 1\) | a circle or an ellipse |
| \(\sin t\) (or \(\cos t\)) and \(\cos 2t\) | \(\cos 2t = 1 - 2\sin^2 t = 2\cos^2 t - 1\) | a parabola |
| \(\sin t\) or \(\cos t\) and \(\sin 2t\) | \(\sin 2t = 2\sin t\cos t\), then \(\sin^2 + \cos^2 = 1\) | often \(y^2 = \ldots\) |
| \(\tan t\) and \(\sec t\) | \(\sec^2 t = 1 + \tan^2 t\) | a hyperbola-like curve |
| \(\cot t\) and \(\csc t\) | \(\csc^2 t = 1 + \cot^2 t\) | a hyperbola-like curve |
(a) Find a Cartesian equation of the curve \(x = 2 + 3\cos\theta\), \(y = 3\sin\theta - 1\), \(0 \le \theta < 2\pi\), and describe the curve. (b) A curve has parametric equations \(x = 3\sin t\), \(y = 4\cos 2t\), \(-\frac{\pi}{2} \le t \le \frac{\pi}{2}\). Show that its Cartesian equation is \(y = 4 - \frac89 x^2\).
(a) Isolate: \(\cos\theta = \frac{x-2}{3}\), \(\sin\theta = \frac{y+1}{3}\). Use \(\sin^2\theta + \cos^2\theta = 1\) and multiply by 9. \((x-2)^2 + (y+1)^2 = 9\): a circle with centre \((2, -1)\) and radius 3
(b) \(\cos 2t\) is linked to \(\sin t\) by a double-angle formula: \(y = 4(1 - 2\sin^2 t) = 4 - 8\sin^2 t\). Substitute \(\sin t = \frac{x}{3}\): \(y = 4 - 8 \times \frac{x^2}{9}\). \(y = 4 - \frac89 x^2\) as required. Choose \(\cos 2t = 1 - 2\sin^2 t\) because \(x\) involves \(\sin t\); the version with \(\cos^2 t\) would need a second step.
Squaring and adding before isolating the trigonometric functions. From \(x = 2 + 3\cos\theta\) and \(y = 3\sin\theta - 1\), writing \(x^2 + y^2 = 9\) is wrong: the constants 2 and −1 must be moved to the other side first.
Which values of x and y does the curve actually reach?
The Cartesian equation can describe more than the parametric curve. In worked example 2(b), \(y = 4 - \frac89 x^2\) is a whole parabola, but the curve is only the part with \(-3 \le x \le 3\). So a complete answer gives the Cartesian equation together with its domain, the set of \(x\)-values the curve reaches. The range is the set of \(y\)-values it reaches.
To find them, work out the values that \(x = f(t)\) and \(y = g(t)\) take over the allowed values of \(t\). Use what you know about each function: \(-1 \le \sin t \le 1\), \(e^t > 0\), \(t^2 \ge 0\), \(\sec t \ge 1\) for \(0 \le t < \frac{\pi}{2}\), and so on. The restriction on \(t\) can also decide a sign: if \(y^2 = \ldots\) and \(y\) is known to be positive, take the positive square root.
A curve \(C\) has parametric equations \(x = \tan t + 1\), \(y = 3\sec t\), \(0 \le t < \frac{\pi}{2}\). (a) Find a Cartesian equation of \(C\) in the form \(y = \sqrt{f(x)}\), where \(f(x)\) is a quadratic. (b) State the set of values of \(x\) (the domain) and of \(y\) (the range) on \(C\).
(a) Isolate: \(\tan t = x - 1\), \(\sec t = \frac{y}{3}\). Use \(\sec^2 t = 1 + \tan^2 t\): \(\frac{y^2}{9} = 1 + (x-1)^2 \Rightarrow y^2 = 9x^2 - 18x + 18\). For \(0 \le t < \frac{\pi}{2}\), \(\sec t > 0\), so \(y > 0\). \(y = \sqrt{9x^2 - 18x + 18}\)
(b) As \(t\) runs from 0 towards \(\frac{\pi}{2}\), \(\tan t\) increases from 0 without limit and \(\sec t\) increases from 1 without limit. Domain \(x \ge 1\); range \(y \ge 3\)
When a question says "state the domain of f" after you have found \(y = f(x)\), the answer comes from the parametric equation for \(x\), not from the formula for \(f\). A formula like \(\sqrt{9x^2 - 18x + 18}\) is defined for every \(x\), but the curve only has \(x \ge 1\).
How do you find dy/dx for a parametric curve?
- To find the gradient at a point you need its value of \(t\). If you are given the coordinates, find the \(t\) that gives both of them.
- Horizontal tangent (gradient 0): \(\frac{dy}{dt} = 0\) with \(\frac{dx}{dt} \ne 0\).
- Vertical tangent (gradient undefined): \(\frac{dx}{dt} = 0\) with \(\frac{dy}{dt} \ne 0\).
- Simplify the quotient, especially with trigonometric functions: \(-\frac{2\cos\theta}{4\sin\theta} = -\frac12\cot\theta\).
The curve \(C\) has parametric equations \(x = t^2\), \(y = t^3 - 3t\). (a) Find \(\frac{dy}{dx}\) in terms of \(t\). (b) Find the gradient of \(C\) at the point \((4, 2)\). (c) Find the coordinates of the points where the tangent to \(C\) is horizontal, and of the point where it is vertical.
(a) \(\frac{dx}{dt} = 2t\), \(\frac{dy}{dt} = 3t^2 - 3\). \(\frac{dy}{dx} = \frac{3t^2 - 3}{2t}\)
(b) Find \(t\) at \((4, 2)\) using both coordinates: \(t^2 = 4\) gives \(t = \pm2\); \(y = 2\) only for \(t = 2\) (\(t = -2\) gives \(y = -2\)). Gradient \(= \frac{3\times4 - 3}{2\times2}\). \(\frac94\)
(c) Horizontal where \(\frac{dy}{dt} = 0\): \(3t^2 - 3 = 0 \Rightarrow t = \pm1\), and \(\frac{dx}{dt} = \pm2 \ne 0\) there. \((1, -2)\) and \((1, 2)\) Vertical where \(\frac{dx}{dt} = 0\): \(2t = 0 \Rightarrow t = 0\), with \(\frac{dy}{dt} = -3 \ne 0\). \((0, 0)\)
At the crossing point \((3, 0)\) the two branches have gradients \(\frac{3\times3 - 3}{\pm2\sqrt3} = \pm\sqrt3\): two different tangents through one point, which no curve \(y = f(x)\) can have.
How do you find the tangent and normal to a parametric curve?
The method is the one from P1; only the gradient is found differently.
- Find the point: substitute the value of \(t\) into both parametric equations.
- Find the gradient \(m\): substitute the same \(t\) into \(\frac{dy}{dx}\). For the normal use \(-\frac1m\).
- Use \(y - y_1 = m(x - x_1)\), with \(x\) and \(y\), never \(t\), in the final equation.
The ellipse \(E\) has parametric equations \(x = 4\cos\theta\), \(y = 2\sin\theta\), \(0 \le \theta < 2\pi\). The point \(P\) on \(E\) has parameter \(\theta = \frac{\pi}{4}\). (a) Show that \(\frac{dy}{dx} = -\frac12\cot\theta\). (b) Find an equation of the tangent to \(E\) at \(P\), in the form \(x + ay = b\), where \(a\) and \(b\) are exact constants. (c) Find an equation of the normal to \(E\) at \(P\). (d) The tangent meets the \(x\)-axis at \(A\) and the \(y\)-axis at \(B\). Find the area of triangle \(OAB\).
(a) \(\frac{dx}{d\theta} = -4\sin\theta\), \(\frac{dy}{d\theta} = 2\cos\theta\), so \(\frac{dy}{dx} = \frac{2\cos\theta}{-4\sin\theta}\). \(-\frac12\cot\theta\) as required.
(b) At \(\theta = \frac{\pi}{4}\): \(P = \left(4\times\frac{\sqrt2}{2}, 2\times\frac{\sqrt2}{2}\right) = (2\sqrt2, \sqrt2)\) and \(m = -\frac12\cot\frac{\pi}{4} = -\frac12\). Then \(y - \sqrt2 = -\frac12(x - 2\sqrt2) \Rightarrow 2y - 2\sqrt2 = -x + 2\sqrt2\). \(x + 2y = 4\sqrt2\)
(c) Normal gradient \(-1 \div \left(-\frac12\right) = 2\): \(y - \sqrt2 = 2(x - 2\sqrt2)\). \(y = 2x - 3\sqrt2\)
(d) Tangent intercepts: \(y = 0\) gives \(x = 4\sqrt2\), so \(A(4\sqrt2, 0)\); \(x = 0\) gives \(y = 2\sqrt2\), so \(B(0, 2\sqrt2)\). Area \(= \frac12 \times 4\sqrt2 \times 2\sqrt2 = \frac12 \times 16\). 8
Leaving \(\theta\) (or \(t\)) in the equation of a tangent. The gradient \(-\frac12\cot\theta\) is a formula; the tangent at \(P\) uses the number \(-\frac12\). A final answer such as \(y - \sqrt2 = -\frac12\cot\theta\,(x - 2\sqrt2)\) is not a straight line.
How do you find the area under a parametric curve?
The curve must be above the \(x\)-axis for the result to be the area. Keep the limits in the order "\(t\) at the left end, \(x = a\)" to "\(t\) at the right end, \(x = b\)", even if the first is the larger number. The method:
- find the \(t\)-values at the ends of the region (from the \(x\)-values given, or from where the curve meets an axis);
- differentiate \(x\) with respect to \(t\);
- multiply \(y\) by \(\frac{dx}{dt}\) and simplify (this is often the step a "show that" part is testing);
- integrate with respect to \(t\) between the \(t\)-limits.
If \(x\) decreases as \(t\) increases, \(\frac{dx}{dt}\) is negative and the left-hand end has the larger \(t\). Writing the limits in the correct order makes the answer positive; you can then swap them and change the sign of the integrand, as in worked example 7.
The curve \(C\) has parametric equations \(x = t^2 + 2t\), \(y = 3 - t\), \(0 \le t \le 3\). Find the area of the region bounded by \(C\) and the coordinate axes.
Ends of the region. \(t = 0\) gives \((0, 3)\), on the \(y\)-axis; \(t = 3\) gives \((15, 0)\), on the \(x\)-axis. \(x\) increases from 0 to 15 as \(t\) goes from 0 to 3.
Set up. \(\frac{dx}{dt} = 2t + 2\), so Area \(= \int_0^3 (3 - t)(2t + 2)\,dt = \int_0^3 (6 + 4t - 2t^2)\,dt\).
Integrate. \(\left[6t + 2t^2 - \tfrac23 t^3\right]_0^3 = 18 + 18 - 18 - 0\). 18
The curve \(C\) has parametric equations \(x = 4\cos t\), \(y = \sin 2t\), \(0 \le t \le \frac{\pi}{2}\). The region \(R\) is bounded by \(C\), the \(x\)-axis and the line \(x = 2\). (a) Show that the area of \(R\) is given by \(\int_{\pi/3}^{\pi/2} 8\sin^2 t\cos t\,dt\). (b) Hence find the exact area of \(R\).
(a) R runs from \(x = 0\) to \(x = 2\). \(x = 0\): \(\cos t = 0\), \(t = \frac{\pi}{2}\); \(x = 2\): \(\cos t = \frac12\), \(t = \frac{\pi}{3}\). With the limits in \(x\)-order and \(\sin 2t = 2\sin t\cos t\): Area \(= \int_{\pi/2}^{\pi/3} \sin 2t \times (-4\sin t)\,dt = \int_{\pi/2}^{\pi/3} -8\sin^2 t\cos t\,dt\). Swap the limits and change the sign: Area \(= \int_{\pi/3}^{\pi/2} 8\sin^2 t\cos t\,dt\) as required.
(b) Integrate by recognition: the derivative of \(\sin^3 t\) is \(3\sin^2 t\cos t\). \(\left[\tfrac83\sin^3 t\right]_{\pi/3}^{\pi/2} = \frac83\left(1 - \frac{3\sqrt3}{8}\right)\), using \(\sin^3\frac{\pi}{3} = \frac{3\sqrt3}{8}\). \(\frac83 - \sqrt3\) (≈ 0.935)
Putting the \(x\)-values 0 and 2 in as the limits of a \(t\)-integral. Once the integrand is in terms of \(t\), the limits must be values of \(t\) (here \(\frac{\pi}{2}\) and \(\frac{\pi}{3}\)). Mixing them spoils an otherwise correct integral.
How do the pieces fit together in a full question?
Longer questions on this topic usually combine a Cartesian equation (with its domain), a tangent or normal, and an area. The last example does all three.
The curve \(C\) has parametric equations \(x = \tan t\), \(y = \sin 2t\), \(0 \le t \le \frac{\pi}{4}\).
(a) Show that a Cartesian equation of \(C\) is \(y = \frac{2x}{1+x^2}\), and state the domain of \(x\) on \(C\).
(b) Find \(\frac{dy}{dx}\) in terms of \(t\).
(c) The point \(P\) on \(C\) has parameter \(t = \frac{\pi}{6}\). Find an equation of the tangent to \(C\) at \(P\), in the form \(y = mx + c\), where \(m\) and \(c\) are exact constants.
(d) Find the exact area of the region bounded by \(C\), the \(x\)-axis and the line \(x = 1\).
(a) Write \(\sin 2t\) in terms of \(\tan t\): \(\sin 2t = 2\sin t\cos t = 2\frac{\sin t}{\cos t}\cos^2 t = \frac{2\tan t}{\sec^2 t} = \frac{2\tan t}{1 + \tan^2 t}\). So \(y = \frac{2x}{1+x^2}\); \(\tan t\) increases from 0 to 1 on \(0 \le t \le \frac{\pi}{4}\), so the domain is \(0 \le x \le 1\).
(b) \(\frac{dx}{dt} = \sec^2 t\), \(\frac{dy}{dt} = 2\cos 2t\), so \(\frac{dy}{dx} = \frac{2\cos 2t}{\sec^2 t}\). \(2\cos 2t\cos^2 t\)
(c) \(P = \left(\tan\frac{\pi}{6}, \sin\frac{\pi}{3}\right) = \left(\frac{\sqrt3}{3}, \frac{\sqrt3}{2}\right)\) and \(m = 2 \times \frac12 \times \left(\frac{\sqrt3}{2}\right)^2 = \frac34\). Then \(y - \frac{\sqrt3}{2} = \frac34\left(x - \frac{\sqrt3}{3}\right) \Rightarrow y = \frac34 x - \frac{\sqrt3}{4} + \frac{\sqrt3}{2}\). \(y = \frac34 x + \frac{\sqrt3}{4}\)
(d) \(x = 0\) at \(t = 0\) and \(x = 1\) at \(t = \frac{\pi}{4}\). Area \(= \int_0^{\pi/4} \sin 2t\sec^2 t\,dt = \int_0^{\pi/4} \frac{2\sin t\cos t}{\cos^2 t}\,dt = \int_0^{\pi/4} 2\tan t\,dt = \big[2\ln\sec t\big]_0^{\pi/4} = 2\ln\sqrt2 - 2\ln 1\). \(\ln 2\) As a check, \(\int_0^1 \frac{2x}{1+x^2}\,dx = \big[\ln(1+x^2)\big]_0^1 = \ln 2\) as well.
Exam tips for P4.2
Write the change of limits as a separate line: "\(x = 0 \Rightarrow t = \frac{\pi}{2}\), \(x = 2 \Rightarrow t = \frac{\pi}{3}\)". It shows the step clearly on its own, and it stops you putting \(x\)-values into a \(t\)-integral.
If the target is an integral such as \(\int 8\sin^2 t\cos t\,dt\), the working must show \(y \times \frac{dx}{dt}\) written out, any identity used (\(\sin 2t = 2\sin t\cos t\)), and how the limits and sign were sorted out. Jumping straight to the printed answer proves nothing.
When you are given a point rather than a parameter, solve one equation for \(t\) and check each possible value in the other. With \(x = t^2\), both \(t = 2\) and \(t = -2\) give \(x = 4\), but only one of them gives the right \(y\).
"Find \(\frac{dy}{dx}\) in terms of \(t\)" is complete once the quotient is simplified (\(-\frac12\cot\theta\), \(2\cos 2t\cos^2 t\)). Do not try to convert it into \(x\); that wastes time and invites errors.
Common misconceptions
| Misconception | Why it is wrong | Correct idea |
|---|---|---|
| \(\frac{dy}{dx} = \frac{dx}{dt} \div \frac{dy}{dt}\). | That is \(\frac{dx}{dy}\), the reciprocal of the gradient. | \(\frac{dy}{dx} = \frac{dy}{dt} \div \frac{dx}{dt}\): the variable you want on top goes on top. |
| The area under a parametric curve is \(\int y\,dt\). | The area is \(\int y\,dx\); changing the variable to \(t\) changes \(dx\) as well. | Area \(= \int y\frac{dx}{dt}\,dt\). |
| The limits of the \(t\)-integral are the \(x\)-values at the ends of the region. | After substitution everything is in terms of \(t\), including the limits. | Convert each \(x\)-limit to its value of \(t\) first. |
| The Cartesian equation describes exactly the same points as the parametric equations. | The parameter's range can restrict the curve (\(-1 \le \sin t \le 1\) keeps \(|x| \le 3\) when \(x = 3\sin t\)). | Give the Cartesian equation with the domain the parametric curve actually covers. |
| A vertical tangent occurs where \(\frac{dy}{dt} = 0\). | \(\frac{dy}{dt} = 0\) makes the gradient zero, so the tangent is horizontal. | Vertical tangent: \(\frac{dx}{dt} = 0\) (with \(\frac{dy}{dt} \ne 0\)). |
| If \(x = 2 + 3\cos\theta\) and \(y = 3\sin\theta\), then \(x^2 + y^2 = 9\). | The identity applies to \(\cos\theta\) and \(\sin\theta\), not to \(x\) and \(y\) when constants have been added. | Isolate first: \(\cos\theta = \frac{x-2}{3}\), so \((x-2)^2 + y^2 = 9\). |
How this topic connects
P3 Trigonometry and differentiation
Eliminating a parameter relies on the P3 identities (double-angle formulae, \(\sec^2 = 1 + \tan^2\), \(\csc^2 = 1 + \cot^2\)), and \(\frac{dy}{dx} = \frac{dy}{dt} \div \frac{dx}{dt}\) is the chain rule. Tangents and normals use the straight-line work of P1; circles from P2 reappear as parametric curves.
P4.4 Implicit differentiation and P4.6 Integration
Implicit differentiation is the other way to differentiate a curve that is not written as \(y = f(x)\). The area formula is integration by substitution, and volumes of revolution use the same idea: \(\pi\int y^2\frac{dx}{dt}\,dt\).
Mechanics and IAL Physics
A projectile's position is a parametric curve with time as the parameter: \(x = (u\cos\alpha)t\), \(y = (u\sin\alpha)t - \frac12 gt^2\). Eliminating \(t\) gives the equation of the path, and \(\frac{dy}{dx}\) gives the direction of motion. Position vectors such as \(\mathbf{r} = f(t)\mathbf{i} + g(t)\mathbf{j}\) are the same idea, and projectile motion in IAL Physics Unit 1 treats horizontal and vertical motion separately in the same way.
Key formulae
| Formula | Use | In the formulae list? |
|---|---|---|
| \(\frac{dy}{dx} = \frac{dy}{dt} \div \frac{dx}{dt}\) | gradient of a parametric curve | No — learn it |
| Area \(= \int y\frac{dx}{dt}\,dt\), between \(t\)-limits | area between a parametric curve and the \(x\)-axis | No — learn it |
| \(\sin^2 t + \cos^2 t = 1\) | eliminating \(t\) from sin and cos (circles, ellipses) | No — learn it |
| \(\sin 2t = 2\sin t\cos t\); \(\cos 2t = 1 - 2\sin^2 t = 2\cos^2 t - 1\) | eliminating \(t\) when a double angle appears | No — learn it |
| \(\sec^2 t = 1 + \tan^2 t\); \(\csc^2 t = 1 + \cot^2 t\) | eliminating \(t\) from tan and sec, or cot and cosec | No — derive it from sin² + cos² = 1 |
| \(y - y_1 = m(x - x_1)\); \(m_{\text{normal}} = -\frac1m\) | equations of tangents and normals | No — learn it |
| \(\int \tan t\,dt = \ln|\sec t| + c\) | a standard integral that is useful in parametric areas | Yes |
Check your understanding
1. Find a Cartesian equation of the curve \(x = 3t - 1\), \(y = \frac{2}{t}\), \(t \ne 0\).
\(t = \frac{x+1}{3}\), so \(y = 2 \div \frac{x+1}{3} = \frac{6}{x+1}\), \(x \ne -1\).
2. Show that the curve \(x = 1 + 2\cos t\), \(y = 2\sin t - 3\) is a circle, and state its centre and radius.
\(\cos t = \frac{x-1}{2}\) and \(\sin t = \frac{y+3}{2}\). Then \(\sin^2 + \cos^2 = 1\) gives \((x-1)^2 + (y+3)^2 = 4\): centre \((1, -3)\), radius 2.
3. A curve has \(x = e^t\), \(y = e^{2t} + 1\), \(t \in \mathbb{R}\). Find its Cartesian equation, its domain and its range.
\(e^{2t} = (e^t)^2 = x^2\), so \(y = x^2 + 1\). Since \(e^t > 0\), the domain is \(x > 0\), and then the range is \(y > 1\). It is only the right half of the parabola, without its vertex.
4. For \(x = t^3\), \(y = t^2 + t\), find the gradient of the curve at the point where \(t = 1\).
\(\frac{dy}{dx} = \frac{2t+1}{3t^2} = \frac33 = 1\) at \(t = 1\) (the point \((1, 2)\)).
5. A student finds the area under a parametric curve and gets a negative answer, although the curve is above the \(x\)-axis. What has probably gone wrong?
\(x\) decreases as \(t\) increases, so \(\frac{dx}{dt} < 0\), and the student has written the smaller \(t\) as the lower limit. The lower limit should be the value of \(t\) at the left-hand end of the region (the smaller \(x\)). With the limits in that order, the answer comes out positive.
6. For the ellipse \(x = 4\cos\theta\), \(y = 2\sin\theta\), find the points where the tangent is horizontal.
\(\frac{dy}{d\theta} = 2\cos\theta = 0\) when \(\theta = \frac{\pi}{2}\) or \(\frac{3\pi}{2}\), and \(\frac{dx}{d\theta} = -4\sin\theta \ne 0\) there. The points are \((0, 2)\) and \((0, -2)\), the top and bottom of the ellipse.
Specification checklist
- (P4 3.1) I can plot a curve from its parametric equations, using a table of parameter values, and show its direction.
- (P4 3.1) I can eliminate the parameter algebraically to find a Cartesian equation in a required form.
- (P4 3.1) I can eliminate a trigonometric parameter using \(\sin^2 + \cos^2 = 1\), double-angle formulae, \(\sec^2 = 1 + \tan^2\) or \(\csc^2 = 1 + \cot^2\).
- (P4 3.1) I can find the domain and range of the Cartesian form from the allowed values of the parameter, and choose the correct sign of a square root.
- (P4 5.1) I can find \(\frac{dy}{dx}\) for a parametric curve as \(\frac{dy}{dt} \div \frac{dx}{dt}\), in simplified form in terms of the parameter.
- (P4 5.1) I can find the equations of tangents and normals to a parametric curve, and the points where the tangent is horizontal or vertical.
- (P4 6.5) I can write the area under a parametric curve as \(\int y\frac{dx}{dt}\,dt\) with the limits changed to values of the parameter.
- (P4 6.5) I can evaluate such integrals exactly, using identities and standard integrals, and deal correctly with \(x\) decreasing as the parameter increases.