2.4, 3.4 - Differentiating parametric curves | Cambridge A-Level Mathematics

A curve can be described by giving both coordinates in terms of a third variable. You will follow such a curve, turn its two parameter derivatives into a gradient, and find tangents and normals without first finding an equation of the form $y=f(x)$ .

A curve given by a parameter

How can one curve contain both $(4,2)$ and $(4,-2)$ , yet still be easy to describe? Let $x=t^2$ and $y=t$ . The inputs $t=2$ and $t=-2$ give those two points. We have described both branches using one moving parameter.

Parametric equations

Parametric equations give the coordinates of a point using a common parameter, such as $x=f(t)$ , $y=g(t)$ . Each permitted value of $t$ supplies one ordered pair $(x,y)$ . The parameter need not be time.

Use the same parameter value in both equations. Changing $t$ traces the curve, and the permitted parameter interval determines which part is traced.

Worked example 1: following a sideways parabola

Question. For $x=t^2$ , $y=t$ , find the points corresponding to $t=-2,0,2$ , and find a Cartesian equation.

Plan. Substitute each value into both coordinates, then eliminate $t$ using $t=y$ .

\begin{array}{c|rrr} t&-2&0&2\\\hline x&4&0&4\\ y&-2&0&2 \end{array}

Since $t=y$ , the equation $x=t^2$ becomes $x=y^2$ .

Answer. The points are $(4,-2)$ , $(0,0)$ , $(4,2)$ ; the curve is $x=y^2$ , with $x\ge0$ .

Check. Each point satisfies $x=y^2$ . As $t$ increases through these values, the point moves towards the vertex along the lower branch, then away along the upper branch.

Shift only the $y$ -equation up by $1$ . The parameter substitution is unchanged, but elimination now uses $t=y-1$ .

Worked example 2: a shifted curve with a restricted parameter

Question. For $x=t^2$ , $y=t+1$ , where $-2\le t\le2$ , find the endpoints and a Cartesian equation with the restriction that selects the traced arc.

Plan. Substitute the parameter endpoints, then express $t$ in terms of $y$ .

\begin{aligned} t=-2&\Longrightarrow(x,y)=(4,-1)\\ t=2&\Longrightarrow(x,y)=(4,3)\\ t=y-1&\Longrightarrow x=(y-1)^2 \end{aligned}

The parameter restriction becomes $-1\le y\le3$ .

Answer. Endpoints $(4,-1)$ , $(4,3)$ ; $x=(y-1)^2$ , with $-1\le y\le3$ .

Check. At $t=0$ , the point is $(0,1)$ , the shifted vertex. Both endpoints satisfy the Cartesian equation and the stated restriction.

The parameter interval matters: the Cartesian equation alone would include the rest of the parabola too.

[DIAGRAM: asset_name: A parameter follows a curve; asset_slug: parametric-parabola; caption: Increasing t traces both branches, and t equals zero gives a vertical tangent; description: Coordinate plot of x equals t squared and y equals t, from t equals minus two to two. Points are labelled t equals minus two at four comma minus two, t equals zero at the origin, t equals one at one comma one, and t equals two at four comma two. Arrows show increasing t. A vertical tangent at the origin and the tangent of gradient one half at one comma one are drawn.]
Increasing t traces both branches, and t equals zero gives a vertical tangent

Watch out

$x=t^2$ does not imply $t=\sqrt x$ for every real $t$ . Negative $t$ gives the other branch. Use the other coordinate or the parameter interval before choosing a square-root sign.

Follow a new upward shift and interval.

The parametric gradient

A gradient measures vertical change per horizontal change, rather than either change per unit parameter. If a small parameter change gives $\Delta x$ and $\Delta y$ , then $\Delta y/\Delta x=(\Delta y/\Delta t)/(\Delta x/\Delta t)$ . Taking limits gives the parametric gradient when the horizontal rate is nonzero.

The parametric gradient

For differentiable $x=f(t)$ , $y=g(t)$ , at a parameter value where $\mathrm dx/\mathrm dt\ne0$ ,

\frac{\mathrm dy}{\mathrm dx} =\frac{\mathrm dy/\mathrm dt}{\mathrm dx/\mathrm dt}.

In the formula booklet.

Differentiate both coordinates with respect to $t$ , then divide in the order “vertical rate over horizontal rate”. The result will usually still contain $t$ .

Worked example 3: a gradient without eliminating the parameter

Question. Find $\mathrm dy/\mathrm dx$ for $x=t^2$ , $y=t+1$ , and evaluate it at $t=1$ .

Plan. Find the two parameter derivatives, then take their ratio.

\begin{aligned} \frac{\mathrm dx}{\mathrm dt}&=2t,&\frac{\mathrm dy}{\mathrm dt}&=1\\ \frac{\mathrm dy}{\mathrm dx}&=\frac1{2t}&& (t\ne0) \end{aligned}

Answer. $\mathrm dy/\mathrm dx=1/(2t)$ , and the gradient at $t=1$ is $1/2$ .

Check. Near $t=1$ , the curve is $y=1+\sqrt x$ , whose derivative at $x=1$ is $1/(2\sqrt1)=1/2$ .

Keep $x=t^2$ , but replace the linear $y$ -coordinate with an exponential. The numerator is now a changing rate too.

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2.4, 3.4 - Differentiating parametric curves

A curve given by a parameter

The parametric gradient

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