Mastering Parametric Equations for Mathematics Examinations

What the examiner is testing

The core skill assessed is converting between parametric and Cartesian forms, often involving differentiation or integration. Marks are typically awarded for correct elimination of the parameter, accurate differentiation with respect to the parameter, and then using the chain rule to find derivatives with respect to \(x\) or \(y\).

The method

1. Identify the parameter: This is usually \(t\) or \(\theta\), defining \(x\) and \(y\) coordinates.
2. Eliminate the parameter (if converting to Cartesian):
   • Rearrange one parametric equation to make the parameter the subject.
   • Substitute this expression for the parameter into the other parametric equation.
   • Simplify the resulting Cartesian equation, ensuring it's in the required form (e.g., \(y = f(x)\) or an implicit form).
3. Differentiate with respect to the parameter: Calculate \(\frac{\mathrm{d}x}{\mathrm{d}t}\) and \(\frac{\mathrm{d}y}{\mathrm{d}t}\) (or \(\frac{\mathrm{d}x}{\mathrm{d}\theta}\) and \(\frac{\mathrm{d}y}{\mathrm{d}\theta}\)).
4. Apply the chain rule for derivatives: Use \(\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{\mathrm{d}y/\mathrm{d}t}{\mathrm{d}x/\mathrm{d}t}\) (or \(\frac{\mathrm{d}y/\mathrm{d}\theta}{\mathrm{d}x/\mathrm{d}\theta}\)).
5. Substitute the parameter value (if finding a specific gradient): If a specific point or value of the parameter is given, substitute it into the derivative expression.

Worked example

A curve is defined by the parametric equations \(x = 2t^2\) and \(y = t^3 - 3t\). Find the Cartesian equation of the curve.

$$ \begin{aligned} x &= 2t^2 \quad &(1) \\ y &= t^3 - 3t \quad &(2) \end{aligned} $$

From (1), we can make \(t^2\) the subject:
$$ t^2 = \frac{x}{2} $$
This implies \(t = \pm\sqrt{\frac{x}{2}}\).

Substitute \(t\) into (2):
$$ y = \left(\pm\sqrt{\frac{x}{2}}\right)^3 - 3\left(\pm\sqrt{\frac{x}{2}}\right) $$
$$ y = \pm\left(\frac{x}{2}\right)^{3/2} \mp 3\left(\frac{x}{2}\right)^{1/2} $$
$$ y = \pm\sqrt{\frac{x}{2}}\left(\frac{x}{2} - 3\right) $$
To eliminate the \(\pm\) sign, we can square both sides, but this might introduce extraneous solutions. A more common approach is to express \(t\) in terms of \(x\) and substitute.

Let's try a different approach: express \(t\) in terms of \(x\) from \(x = 2t^2\).
If \(t = \sqrt{\frac{x}{2}}\), then:
$$ y = \left(\sqrt{\frac{x}{2}}\right)^3 - 3\left(\sqrt{\frac{x}{2}}\right) $$
$$ y = \frac{x\sqrt{x}}{2\sqrt{2}} - \frac{3\sqrt{x}}{\sqrt{2}} $$
$$ y = \frac{x\sqrt{2x}}{4} - \frac{3\sqrt{2x}}{2} $$
$$ y = \frac{\sqrt{2x}}{4}(x - 6) $$
If \(t = -\sqrt{\frac{x}{2}}\), then:
$$ y = \left(-\sqrt{\frac{x}{2}}\right)^3 - 3\left(-\sqrt{\frac{x}{2}}\right) $$
$$ y = -\frac{x\sqrt{x}}{2\sqrt{2}} + \frac{3\sqrt{x}}{\sqrt{2}} $$
$$ y = -\frac{\sqrt{2x}}{4}(x - 6) $$
So, \(y^2 = \frac{2x}{16}(x-6)^2 = \frac{x}{8}(x-6)^2\).

Sanity check: If \(t=2\), \(x = 2(2^2) = 8\) and \(y = 2^3 - 3(2) = 8 - 6 = 2\).
Using the Cartesian equation \(y^2 = \frac{x}{8}(x-6)^2\):
\(2^2 = 4\).
\(\frac{8}{8}(8-6)^2 = 1 \times 2^2 = 4\). The values match.

Worked example: a harder one

A curve has parametric equations \(x = \sin t\) and \(y = \sin 2t\), for \(0 \le t \le \frac{\pi}{2}\).
Find the Cartesian equation of the curve and determine the range of possible values for \(y\).

$$ \begin{aligned} x &= \sin t \quad &(1) \\ y &= \sin 2t \quad &(2) \end{aligned} $$

The obvious first move is to substitute \(t = \arcsin x\) into the second equation.
$$ y = \sin(2 \arcsin x) $$
This is a valid Cartesian equation, but it's not in a simplified or easily usable form. It's hard to differentiate or find the range directly from this. This approach fails to give a simple algebraic relationship.

Instead, use trigonometric identities.
From (2), we know \(\sin 2t = 2 \sin t \cos t\).
So, \(y = 2 \sin t \cos t\).
From (1), \(x = \sin t\).
We also know that \(\cos^2 t + \sin^2 t = 1\), so \(\cos t = \pm\sqrt{1 - \sin^2 t}\).
Since \(0 \le t \le \frac{\pi}{2}\), \(\cos t \ge 0\), so \(\cos t = \sqrt{1 - \sin^2 t}\).
Substitute \(\sin t = x\) and \(\cos t = \sqrt{1 - x^2}\) into the expression for \(y\):
$$ y = 2x\sqrt{1 - x^2} $$
This is the Cartesian equation.

Now, determine the range of possible values for \(y\).
Given \(0 \le t \le \frac{\pi}{2}\):
For \(x = \sin t\): as \(t\) goes from \(0\) to \(\frac{\pi}{2}\), \(\sin t\) goes from \(0\) to \(1\). So, \(0 \le x \le 1\).
For \(y = \sin 2t\): as \(t\) goes from \(0\) to \(\frac{\pi}{2}\), \(2t\) goes from \(0\) to \(\pi\).
The function \(\sin(2t)\) for \(0 \le 2t \le \pi\) starts at \(0\), increases to a maximum of \(1\) (when \(2t = \frac{\pi}{2}\), so \(t = \frac{\pi}{4}\)), and then decreases back to \(0\).
Therefore, the range of \(y\) is \(0 \le y \le 1\).

Practice

1. A curve is given by the parametric equations \(x = 3t - 1\) and \(y = t^2 + 2\). Find the Cartesian equation of the curve.
2. Find \(\frac{\mathrm{d}y}{\mathrm{d}x}\) in terms of \(t\) for the curve given by \(x = \cos t\) and \(y = \sin t\).
3. A curve has parametric equations \(x = 2e^t\) and \(y = e^{2t} - 1\). Find the gradient of the tangent to the curve at the point where \(t = \ln 3\).
4. The parametric equations of a curve are \(x = \theta - \sin \theta\) and \(y = 1 - \cos \theta\) for \(0 \le \theta \le 2\pi\). Find the equation of the normal to the curve at the point where \(\theta = \frac{\pi}{2}\).

Answers:
1. \(y = \left(\frac{x+1}{3}\right)^2 + 2\) or \(9y = (x+1)^2 + 18\)
2. \(\frac{\mathrm{d}y}{\mathrm{d}x} = -\cot t\)
3. \(9\)
4. Working for Q4:
$$ x = \theta - \sin \theta \implies \frac{\mathrm{d}x}{\mathrm{d}\theta} = 1 - \cos \theta $$
$$ y = 1 - \cos \theta \implies \frac{\mathrm{d}y}{\mathrm{d}\theta} = \sin \theta $$
$$ \frac{\mathrm{d}y}{\mathrm{d}x} = \frac{\mathrm{d}y/\mathrm{d}\theta}{\mathrm{d}x/\mathrm{d}\theta} = \frac{\sin \theta}{1 - \cos \theta} $$
At \(\theta = \frac{\pi}{2}\):
$$ \frac{\mathrm{d}y}{\mathrm{d}x}\Big|_{\theta=\frac{\pi}{2}} = \frac{\sin(\frac{\pi}{2})}{1 - \cos(\frac{\pi}{2})} = \frac{1}{1 - 0} = 1 $$
The gradient of the tangent at \(\theta = \frac{\pi}{2}\) is \(1\).
The gradient of the normal is \(-\frac{1}{1} = -1\).

Find the coordinates of the point at \(\theta = \frac{\pi}{2}\):
$$
    x = \frac{\pi}{2} - \sin(\frac{\pi}{2}) = \frac{\pi}{2} - 1
    $$
$$
    y = 1 - \cos(\frac{\pi}{2}) = 1 - 0 = 1
    $$
The point is \((\frac{\pi}{2} - 1, 1)\).

Equation of the normal \(y - y_1 = m(x - x_1)\):
$$
    y - 1 = -1\left(x - \left(\frac{\pi}{2} - 1\right)\right)
    $$
$$
    y - 1 = -x + \frac{\pi}{2} - 1
    $$
$$
    y = -x + \frac{\pi}{2}
    $$

The three mistakes that lose marks

1. Incorrectly applying the chain rule for \(\frac{\mathrm{d}^2y}{\mathrm{d}x^2}\): Students often differentiate \(\frac{\mathrm{d}y}{\mathrm{d}x}\) with respect to \(t\) and call it \(\frac{\mathrm{d}^2y}{\mathrm{d}x^2}\).
   Wrong: \(\frac{\mathrm{d}^2y}{\mathrm{d}x^2} = \frac{\mathrm{d}}{\mathrm{d}t}\left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)\).
   Correct: \(\frac{\mathrm{d}^2y}{\mathrm{d}x^2} = \frac{\mathrm{d}}{\mathrm{d}x}\left(\frac{\mathrm{d}y}{\mathrm{d}x}\right) = \frac{\mathrm{d}}{\mathrm{d}t}\left(\frac{\mathrm{d}y}{\mathrm{d}x}\right) \times \frac{\mathrm{d}t}{\mathrm{d}x}\). Remember \(\frac{\mathrm{d}t}{\mathrm{d}x} = \frac{1}{\mathrm{d}x/\mathrm{d}t}\).
2. Forgetting domain restrictions when eliminating the parameter: When using substitutions like \(\cos t = \sqrt{1 - \sin^2 t}\), the sign of the square root depends on the given range of \(t\).
   Wrong: Always assuming \(\cos t = \sqrt{1 - \sin^2 t}\) without considering the quadrant of \(t\).
   Example: If \(t\) is in the range \(\frac{\pi}{2} < t < \pi\), then \(\cos t\) is negative, so \(\cos t = -\sqrt{1 - \sin^2 t}\).
3. Algebraic errors when simplifying after substitution: Especially common when dealing with fractions or powers.
   Wrong: \(y = (x^{1/2} - 2)^2 \implies y = x - 4\).
   Correct: \(y = (x^{1/2} - 2)^2 \implies y = (x^{1/2})^2 - 2(x^{1/2})(2) + 2^2 = x - 4x^{1/2} + 4\).

30-second recap

Parametric equations define \(x\) and \(y\) in terms of a third variable, the parameter. To convert to Cartesian, eliminate the parameter. For calculus, use the chain rule: \(\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{\mathrm{d}y/\mathrm{d}t}{\mathrm{d}x/\mathrm{d}t}\). Be careful with domain restrictions and the second derivative.