- Parametric equations describe a curve by giving \(x\) and \(y\) separately as functions of a third variable, the parameter, often time \(t\).
- To find the Cartesian equation, eliminate the parameter: make \(t\) the subject of one equation and substitute it into the other.
- The gradient is \(\frac{dy}{dx} = \frac{dy/dt}{dx/dt}\), and the second derivative is \(\frac{d^2y}{dx^2} = \frac{d}{dt}\left(\frac{dy}{dx}\right) \div \frac{dx}{dt}\).
- In our case, a ball kicked at 25 m/s (15 m/s across, 20 m/s up) lands after about 4.08 s, 61.2 m away, after peaking at 20.4 m, and is still 7.5 m high when it passes 55 m.
- Exam point: never divide by \(\frac{dx}{dt}\) when it is zero. Those points have vertical tangents.
The case: will a long kick clear the bar?
A player takes a long kick from 55 metres out. The ball leaves the ground at 25 m/s. Its velocity splits into a horizontal part of 15 m/s and a vertical part of 20 m/s, so the launch angle is about 53° above the horizontal. Ignoring air resistance and taking \(g = 9.8\) m/s², the position of the ball \(t\) seconds after the kick is
where \(x\) is the horizontal distance and \(y\) the height, both in metres. The bar is 3 m high and 55 m away. (The figures are illustrative.)
The coach wants to know several things. How high does the ball go, and when? Where does it land? Is it high enough when it reaches 55 m to clear a 3 m bar? And at what angle is it falling at that point? These are questions about a path in two dimensions that unfolds over time. The natural tool is a pair of parametric equations.
The concept: one parameter, two coordinates
Most curves you meet first are written as \(y = f(x)\): one equation linking \(x\) and \(y\) directly. Parametric equations do it differently. Each coordinate is given its own formula in terms of a third variable, called the parameter. As the parameter changes, the point \((x, y)\) moves and traces out the curve.
For moving objects, the parameter is usually time. This is powerful because it keeps information that a single \(y = f(x)\) equation throws away: not just where the ball goes, but when it gets there. It also handles curves that are not functions at all, such as circles and loops, which fail the vertical line test.[3]
Here \(f(t)\) and \(g(t)\) give the horizontal and vertical positions, and \(f'(t)\) and \(g'(t)\) are their rates of change with respect to \(t\). For motion, \(\frac{dx}{dt}\) is the horizontal velocity and \(\frac{dy}{dt}\) the vertical velocity. Their ratio \(\frac{dy}{dx}\) is the gradient of the path: how much height changes per metre travelled sideways. The formula comes from the chain rule, \(\frac{dy}{dt} = \frac{dy}{dx} \times \frac{dx}{dt}\). When \(\frac{dx}{dt} = 0\) the gradient is undefined and the tangent is vertical.
Mathematics Made Simple explains parametric equations and their derivatives with worked examples and graphs.Get the textbook · $12
Why this method fits the case
A projectile moves in two directions at once, and each direction obeys a simple law. Horizontally, with no air resistance, nothing speeds the ball up or slows it down, so \(x\) grows at a steady 15 m/s. Vertically, gravity pulls the ball down, so \(y\) follows a quadratic in \(t\). Writing the two motions separately, with time as the shared parameter, is the cleanest way to describe this.[3]
The parametric form also answers “when” questions directly. The time to the top, the time of flight and the time to reach the bar all come straight from the \(t\)-equations. If we only had \(y\) as a function of \(x\), we would have to work back to time using \(x = 15t\) anyway.
Parametric equations are not always the best form. If you only care about the shape of a path, and it is a simple function, the Cartesian equation is often easier to sketch and to integrate. If air resistance matters, the separate equations become coupled differential equations and need numerical methods. And for curves defined by distance and angle from a fixed point, such as spirals, polar coordinates are usually simpler.
Setting it up
Here are the quantities in the kick.
| Quantity | Symbol | Value | Meaning |
|---|---|---|---|
| Time since the kick | \(t\) | seconds, \(t \ge 0\) | The parameter |
| Horizontal velocity | \(\frac{dx}{dt}\) | 15 m/s (constant) | \(25\cos\theta\) |
| Initial vertical velocity | \(\frac{dy}{dt}\) at \(t=0\) | 20 m/s | \(25\sin\theta\) |
| Launch angle | \(\theta\) | \(\tan^{-1}\frac{4}{3} \approx 53.1°\) | Angle above the ground |
| Gravity | \(g\) | 9.8 m/s² | Gives the \(-4.9t^2\) term |
| Bar | — | 3 m high, 55 m away | Target to clear |
From the equations, \(\frac{dx}{dt} = 15\) and \(\frac{dy}{dt} = 20 – 9.8t\). So the gradient of the path at time \(t\) is
Solving it
Find the greatest height of the ball, the time of flight and the horizontal distance to where it lands.
The ball rises to about 20.4 m after 2.04 s and lands about 61.2 m away after 4.08 s.
Eliminate the parameter to find the Cartesian equation of the path. Does the ball clear the 3 m bar 55 m away? At what angle is it travelling there?
The ball clears the bar by about 4.5 m, falling at roughly 47° below the horizontal.
A different curve. An ellipse is given by \(x = 3\cos t\), \(y = 2\sin t\), \(0 \le t < 2\pi\). Find its Cartesian equation and the equation of the tangent at \(t = \frac{\pi}{4}\).
The ellipse is \(\frac{x^2}{9} + \frac{y^2}{4} = 1\), and the tangent at \(t = \frac{\pi}{4}\) is \(y = -\frac{2}{3}x + 2\sqrt{2}\).
Make parametric equations automatic with the Practice Questions book: 1,565 questions, all fully worked.Get 1,565 questions · $9
What the answer tells us
The kick works comfortably. At 55 m the ball is about 7.5 m up, so it clears a 3 m bar by more than 4 m, and it would carry about 61 m before landing. By then the ball has been in the air for 3.67 s and is past its peak, dropping steeply at about 47°.
Figure 1 plots the path from the parametric equations, with markers at one-second intervals. Because \(x\) grows at a constant rate, the markers are evenly spaced horizontally. Vertically they bunch together near the top, where the ball is moving slowly upwards.
Reasonableness check. The path should be symmetric about the peak, because gravity slows the rise and speeds the fall at the same rate. The peak is at \(x = 15 \times 2.04 \approx 30.6\) m, exactly half the range of 61.2 m. The landing speed should equal the launch speed, 25 m/s, since there is no air resistance: at \(t = 4.08\), the velocity is \((15, -20)\), whose size is \(\sqrt{15^2 + 20^2} = 25\). Both checks pass.
Sensitivity. The range of a projectile on level ground is \(\frac{2v_x v_y}{g}\). If the player kicked at 20 m/s at the same angle, so \(v_x = 12\) and \(v_y = 16\), the range would fall to about 39 m and the ball would not reach the bar at all. At 25 m/s, a 45° kick would go furthest on level ground, about 63.8 m. At 55 m the two kicks are almost equally high (about 7.6 m for 45° against 7.5 m for 53°), so either clears the bar comfortably; the steeper kick is much higher in mid-flight (20.4 m against 15.9 m at its peak), which would matter if a defender stood in the way.
For more on parametric curves, including polar and vector-valued functions, see the Mathematics Made Simple textbook, which shows each curve with a graph. To practise eliminating parameters, finding tangents and normals, and working with projectile paths, use the Mathematics Made Simple Practice Questions book.
Master parametric, polar and vector functions with both Mathematics Made Simple books for $16.Both books · $16
Practice questions
Try each question on paper first, then check the worked answer.
EasyQ1. A curve has parametric equations \(x = 2t + 1\), \(y = t^2\). Find its Cartesian equation. Show worked answer
Substitute: \(y = \left(\frac{x – 1}{2}\right)^2 = \frac{(x – 1)^2}{4}\).
A parabola with vertex at \((1, 0)\).
EasyQ2. For \(x = t^2\), \(y = t^3\), find \(\frac{dy}{dx}\) in terms of \(t\), and its value at \(t = 2\). Show worked answer
\(\frac{dy}{dx} = \frac{3t^2}{2t} = \frac{3t}{2}\) (for \(t \ne 0\)).
At \(t = 2\): \(\frac{dy}{dx} = 3\).
MediumQ3. A circle is given by \(x = 2 + 5\cos t\), \(y = -1 + 5\sin t\). Find its centre and radius, and the gradient of the tangent at \(t = \frac{\pi}{3}\). Show worked answer
\(\frac{dy}{dx} = \frac{5\cos t}{-5\sin t} = -\cot t\).
At \(t = \frac{\pi}{3}\): \(-\cot\frac{\pi}{3} = -\frac{1}{\sqrt{3}} \approx -0.577\).
MediumQ4. A ball is thrown with \(x = 12t\), \(y = 16t – 4.9t^2\) (metres, seconds). Find the time of flight, the range and the greatest height. Show worked answer
Range: \(12 \times 3.265 \approx 39.2\) m.
Top when \(16 – 9.8t = 0\), \(t \approx 1.633\) s; height \(= \frac{16^2}{2 \times 9.8} \approx 13.1\) m.
ExamQ5. A curve has parametric equations \(x = t^2 + 1\), \(y = 2t – t^3\), for \(t > 0\). (a) Find \(\frac{dy}{dx}\) in terms of \(t\). (b) Find the equation of the tangent at \(t = 1\). (c) Find \(\frac{d^2y}{dx^2}\) at \(t = 1\). [7 marks] Show worked answer
\(\frac{dy}{dx} = \frac{2 – 3t^2}{2t}\). [A1]
(b) At \(t = 1\): point \((2, 1)\), gradient \(\frac{2 – 3}{2} = -\frac{1}{2}\). [B1]
Tangent: \(y – 1 = -\frac{1}{2}(x – 2)\), so \(y = -\frac{1}{2}x + 2\). [M1 A1]
(c) Write \(\frac{dy}{dx} = \frac{1}{t} – \frac{3t}{2}\). Then \(\frac{d}{dt}\left(\frac{dy}{dx}\right) = -\frac{1}{t^2} – \frac{3}{2}\). [M1]
Divide by \(\frac{dx}{dt} = 2t\): at \(t = 1\), \(\frac{-1 – 1.5}{2} = -\frac{5}{4}\). [A1]
Frequently asked questions
Why use parametric equations instead of \(y = f(x)\)?
They keep track of the parameter, often time, so you can answer “when” as well as “where”. They also describe curves that are not functions, such as circles, ellipses and loops, and they make motion in two directions easy to model separately.
Can every parametric curve be written as a Cartesian equation?
Not always in a neat form. Some, like the projectile path or a circle, eliminate easily. Others, such as \(x = t + \sin t\), \(y = \cos t\), have no simple Cartesian equation. You can still find gradients, areas and arc lengths using the parametric formulas.
How do I find where a parametric curve has a horizontal or vertical tangent?
Horizontal tangents occur where \(\frac{dy}{dt} = 0\) and \(\frac{dx}{dt} \ne 0\). Vertical tangents occur where \(\frac{dx}{dt} = 0\) and \(\frac{dy}{dt} \ne 0\). If both are zero, the point needs closer study; the curve may have a cusp there.
How do I find the speed of an object moving along a parametric path?
Speed is the size of the velocity vector: \(\sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2}\). For the kick, the speed at launch is \(\sqrt{15^2 + 20^2} = 25\) m/s. Integrating speed over time gives the distance travelled along the curve.
References
- College Board. AP Calculus AB and BC Course and Exam Description. New York: College Board.
- International Baccalaureate Organization (2019). Mathematics: analysis and approaches guide. Cardiff: IBO.
- Strang, G. & Herman, E. (2016). Calculus Volume 2. Houston, TX: OpenStax. openstax.org
- Cambridge Assessment International Education. Cambridge International AS & A Level Mathematics 9709: Syllabus for examination in 2026 and 2027. cambridgeinternational.org
