Parametric equations and gradients

What you’ll learn

• Convert suitable parametric equations into Cartesian form.
• Differentiate parametric curves and find tangents and normals.
• Track restrictions imposed by the parameter.

Understand the idea

Parametric equations describe both coordinates using a third variable. Eliminate the parameter to obtain a Cartesian relationship, but retain restrictions from the original parameter domain. The Cartesian equation alone may describe more of a curve than the given parametrisation reaches.

The chain rule gives $dy/dx=(dy/dt)/(dx/dt)$ whenever $dx/dt\ne0$. Evaluate the coordinates and this ratio at the requested parameter value before forming a tangent or normal. A zero denominator requires separate analysis rather than ordinary division.

Trigonometric parametrisations can be eliminated using identities: $x=a\cos t$ and $y=b\sin t$ give an ellipse relationship. Polynomial parametrisations often allow direct substitution. Choose the simplest elimination that preserves any sign or domain information.

Key formulae

$$x=f(t),\quad y=g(t)\quad\Rightarrow\quad\frac{dy}{dx}=\frac{g'(t)}{f'(t)}$$

$$y-g(t_0)=\frac{g'(t_0)}{f'(t_0)}[x-f(t_0)]$$

$$x=a\cos t,\ y=b\sin t\Rightarrow\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$$

Worked example

For $x=t^2+1$, $y=2t^3$, find the tangent at $t=1$.

1. At $t=1$, the point is $(2,2)$.
2. $dx/dt=2t$ and $dy/dt=6t^2$, so $dy/dx=3t$ for $t\ne0$.
3. The tangent gradient at $t=1$ is $3$, so $y-2=3(x-2)$.

Answer: $y=3x-4$.