Parametric differentiation

A parametric curve defines both coordinates using a third variable, usually $t$:

$x=f(t), \qquad y=g(t).$

As $t$ changes, the point $(x,y)$ moves along the curve. Parametric differentiation finds how $y$ changes with $x$ without first eliminating $t$.

If you need to review how parameters describe curves, begin with parametric equations.

The parametric derivative

Suppose $x$ and $y$ are differentiable functions of $t$. By the chain rule,

$\frac{dy}{dt}=\frac{dy}{dx}\frac{dx}{dt}.$

Therefore, provided $\dfrac{dx}{dt}\ne 0$,

$\boxed{\frac{dy}{dx}=\frac{dy/dt}{dx/dt}}.$

This is a quotient of two derivatives. It is not $\dfrac{dy}{dt}\div\dfrac{dt}{dx}$, and it is not found by differentiating $y$ with respect to $x$ directly.

Example 1: find the gradient

The curve is defined by

$x=t^2+1, \qquad y=t^3-2t.$

Find $\dfrac{dy}{dx}$ and the gradient when $t=2$.

Differentiate both coordinates with respect to $t$:

$\frac{dx}{dt}=2t, \qquad \frac{dy}{dt}=3t^2-2.$

Divide in the correct order:

$\frac{dy}{dx}=\frac{3t^2-2}{2t}.$

At $t=2$,

$\frac{dy}{dx}=\frac{3(2)^2-2}{2(2)}=\frac{10}{4}=\frac52.$

The parameter value identifies the point as well as its gradient:

$x=2^2+1=5, \qquad y=2^3-2(2)=4.$

So the curve has gradient $\dfrac52$ at $(5,4)$.

Tangents and normals

At $t=a$, first calculate the point

$\bigl(f(a),g(a)\bigr)$

and the tangent gradient

$m=\left.\frac{dy}{dx}\right|_{t=a}.$

The tangent through $(x_1,y_1)$ is

$y-y_1=m(x-x_1).$

If $m\ne 0$, the normal gradient is $-1/m$, since perpendicular gradients multiply to $-1$.

Example 2: equations of a tangent and normal

For

$x=2t+1, \qquad y=t^2+3,$

find the tangent and normal when $t=2$.

The point is

$x=5, \qquad y=7,$

and

$\frac{dx}{dt}=2, \qquad \frac{dy}{dt}=2t.$

Hence

$\frac{dy}{dx}=\frac{2t}{2}=t,$

so the tangent gradient at $t=2$ is $2$. Its equation is

$\boxed{y-7=2(x-5)}.$

The normal gradient is $-\dfrac12$, giving

$\boxed{y-7=-\frac12(x-5)}.$

Keep the answer in point-gradient form unless the question asks for another form. It displays both the point and gradient clearly.

Stationary points

A stationary point has a horizontal tangent, so

$\frac{dy}{dx}=0.$

For a regular parametric curve, this occurs when

$\frac{dy}{dt}=0 \qquad \text{and} \qquad \frac{dx}{dt}\ne 0.$

Solving only $\dfrac{dy}{dt}=0$ is not enough. You must check $\dfrac{dx}{dt}$ at every candidate parameter value.

Example 3: locate stationary points

Consider

$x=t^2+t, \qquad y=t^3-3t.$

Differentiate:

$\frac{dx}{dt}=2t+1, \qquad \frac{dy}{dt}=3t^2-3=3(t-1)(t+1).$

Candidates occur at $t=1$ and $t=-1$.

At $t=1$, $\dfrac{dx}{dt}=3\ne0$, and the point is

$x=2, \qquad y=-2.$

At $t=-1$, $\dfrac{dx}{dt}=-1\ne0$, and the point is

$x=0, \qquad y=2.$

Therefore the stationary points are

$\boxed{(2,-2) \text{ and } (0,2)}.$

Both checks matter because the same coordinate can sometimes be reached using different parameter values, and a zero denominator changes the geometry.

What if a derivative is zero?

The formula must be interpreted carefully.

Condition at $t=a$	Likely geometric meaning
$dy/dt=0$ and $dx/dt\ne0$	Horizontal tangent and a stationary point
$dx/dt=0$ and $dy/dt\ne0$	Vertical tangent, so $dy/dx$ is undefined
$dx/dt=0$ and $dy/dt=0$	Indeterminate form requiring further analysis

A vertical tangent is not a stationary point of $y$ as a function of $x$. Its gradient is not zero; it is undefined.

Example 4: distinguish horizontal and vertical tangents

Let

$x=t^2, \qquad y=t^3-3t.$

Then

$\frac{dx}{dt}=2t, \qquad \frac{dy}{dt}=3t^2-3.$

At $t=0$,

$\frac{dx}{dt}=0, \qquad \frac{dy}{dt}=-3.$

The point is $(0,0)$ and the tangent is vertical. Its equation is

$\boxed{x=0}.$

Horizontal tangents occur when $3t^2-3=0$, so $t=\pm1$. Since $2t\ne0$ at both values, both give genuine stationary points.

When both derivatives are zero

If $dx/dt$ and $dy/dt$ are both zero, writing $0/0$ does not determine the gradient. Simplify the expression for $dy/dx$, if possible, and examine its limit as $t$ approaches the parameter value.

For example, if

$x=t^2, \qquad y=t^4,$

then for $t\ne0$,

$\frac{dy}{dx}=\frac{4t^3}{2t}=2t^2.$

Although the original quotient is $0/0$ at $t=0$, the limiting gradient is

$\lim_{t\to0}2t^2=0.$

In fact, eliminating $t$ gives $y=x^2$ for $x\ge0$, confirming a horizontal tangent at the origin.

Do not automatically cancel factors and substitute without considering the curve. In harder cases, $0/0$ can indicate a cusp, a self-intersection or a parameterisation that momentarily stops. A limit, an alternative parameterisation or the Cartesian equation may be needed.

Second derivatives

The second derivative measures how the gradient changes with $x$. First find $\dfrac{dy}{dx}$ as a function of $t$, differentiate that expression with respect to $t$, then divide by $dx/dt$ again:

$\boxed{\frac{d^2y}{dx^2}=\frac{\dfrac{d}{dt}\left(\dfrac{dy}{dx}\right)}{dx/dt}}.$

Do not simply differentiate $dy/dx$ with respect to $t$ and stop. That gives the rate of change of the gradient with $t$, not with $x$.

Example 5: classify stationary points

For

$x=t+1, \qquad y=t^3-3t,$

we have

$\frac{dx}{dt}=1, \qquad \frac{dy}{dt}=3t^2-3,$

so

$\frac{dy}{dx}=3t^2-3.$

The stationary values are $t=\pm1$. Now

$\frac{d}{dt}\left(\frac{dy}{dx}\right)=6t,$

and therefore

$\frac{d^2y}{dx^2}=\frac{6t}{1}=6t.$

At $t=-1$, $d^2y/dx^2=-6<0$, so the point is a local maximum. At $t=1$, $d^2y/dx^2=6>0$, so the point is a local minimum.

Their coordinates are

$t=-1: (x,y)=(0,2),$

$t=1: (x,y)=(2,-2).$

A trigonometric example

The ellipse

$x=4\cos t, \qquad y=3\sin t$

has

$\frac{dx}{dt}=-4\sin t, \qquad \frac{dy}{dt}=3\cos t.$

Therefore

$\frac{dy}{dx}=-\frac{3\cos t}{4\sin t}.$

At $t=\dfrac{\pi}{4}$, the point is

$\left(2\sqrt2,\frac{3\sqrt2}{2}\right),$

and the gradient is

$-\frac{3(\sqrt2/2)}{4(\sqrt2/2)}=-\frac34.$

The tangent is consequently

$y-\frac{3\sqrt2}{2}=-\frac34\left(x-2\sqrt2\right).$

Notice that $t$ is an angle here, but parametric differentiation works in exactly the same way as for polynomial functions.

A reliable exam method

1. Differentiate $x$ and $y$ separately with respect to $t$.
2. Form $\dfrac{dy}{dx}=\dfrac{dy/dt}{dx/dt}$ and simplify.
3. Use the given parameter value or solve the required gradient condition.
4. Check whether $dx/dt$ is zero.
5. Substitute into the original equations to find the point.
6. Write the tangent or normal using point-gradient form.
7. For classification, calculate $d^2y/dx^2$ and evaluate its sign.

Common mistakes

• Reversing the quotient and calculating $(dx/dt)/(dy/dt)$.
• Substituting a value of $t$ before differentiating, which turns the coordinates into constants.
• Using the derivative expressions to calculate the point instead of the original equations.
• Declaring $dy/dt=0$ sufficient for a stationary point without checking $dx/dt$.
• Calling a vertical tangent stationary because $dx/dt=0$.
• Taking the negative reciprocal before evaluating the tangent gradient.
• Writing $d^2y/dx^2=d/dt(dy/dx)$ and forgetting to divide by $dx/dt$.

Check your understanding

For

$x=t^2+2t, \qquad y=t^3-3t,$

answer the following.

1. Find $dy/dx$ in terms of $t$.
2. Find the equation of the tangent at $t=1$.
3. Find all stationary points and verify that each is genuine.
4. Find $d^2y/dx^2$ in terms of $t$.

Answers

First,

$\frac{dx}{dt}=2t+2, \qquad \frac{dy}{dt}=3t^2-3,$

so

$\frac{dy}{dx}=\frac{3t^2-3}{2t+2}=\frac{3(t-1)(t+1)}{2(t+1)}=\frac32(t-1), \qquad t\ne-1.$

At $t=1$, the point is $(3,-2)$ and the gradient is $0$, so the tangent is

$\boxed{y=-2}.$

The numerator of the unsimplified derivative is zero at $t=1$ and $t=-1$. At $t=1$, $dx/dt=4\ne0$, so $(3,-2)$ is stationary. At $t=-1$, both derivatives are zero, so it cannot be accepted using the regular stationary-point test. The simplified gradient tends to $-3$ as $t\to-1$, so the point $(-1,2)$ is not stationary.

Finally,

$\frac{d}{dt}\left(\frac{dy}{dx}\right)=\frac32,$

and

$\boxed{\frac{d^2y}{dx^2}=\frac{3/2}{2t+2}=\frac{3}{4(t+1)}}$

where this expression is defined.

What to learn next

Review tangents and normals if forming line equations is not yet automatic. Then study stationary points and curve sketching to connect derivatives with the wider shape of a curve, or parametric modelling to see why parameterised curves are useful in context.