Parametric Equations Solver

In Cartesian form a curve is a relation between $x$ and $y$. In parametric form both coordinates are written as functions of a third variable, the parameter, usually $t$:

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

Each value of $t$ produces one point $(f(t), g(t))$, and as $t$ runs through its interval the point traces the curve. Think of $t$ as time and the curve as the path of a moving object.

This buys you three things Cartesian form cannot give:

• Curves that fail the vertical line test. A full circle is not a function of $x$, but $x = \cos t$, $y = \sin t$ describes it in one piece.
• Direction and speed. The parametrisation says which way the point moves and how fast — information the picture alone does not carry.
• A natural extension to 3D, where $x$, $y$ and $z$ are each functions of $t$.

Because the same set of points can be traced by many different parametrisations, a parametric description carries strictly more information than the picture alone: it records where the motion starts and how fast it passes each point.

Line through the point $(x_0, y_0)$ with direction vector $\langle a, b \rangle$:

$x = x_0 + at, \qquad y = y_0 + bt$

Circle of radius $r$ centred at $(h, k)$:

$x = h + r\cos t, \qquad y = k + r\sin t, \qquad 0 \le t < 2\pi$

Ellipse with semi-axes $a$ and $b$: $x = h + a\cos t$, $y = k + b\sin t$.

To convert to Cartesian, eliminate $t$. For polynomial equations, solve the simpler equation for $t$ and substitute into the other. For trigonometric ones, isolate $\cos t$ and $\sin t$ and use $\cos^2 t + \sin^2 t = 1$.

To differentiate, divide the two rates rather than trying to find $y(x)$:

$\frac{dy}{dx} = \frac{dy/dt}{dx/dt}, \qquad \frac{dx}{dt} \neq 0$

• Losing the domain. $x = t^2$, $y = t$ traces only the half of the parabola $x = y^2$ with $x \ge 0$ — eliminating $t$ silently adds points the original curve never visits. State the restriction.
• Treating the parametrisation as unique. $x = t, y = 2t$ and $x = 3t, y = 6t$ draw the same line at different speeds. Both are correct answers.
• Computing $dy/dx$ as $\frac{dy}{dt} \cdot \frac{dx}{dt}$. It is a quotient, not a product.
• Mixing $t$ into a Cartesian answer. Once eliminated, the parameter must not appear anywhere in the final equation.
• Ignoring where $dx/dt = 0$. Those are the points with a vertical tangent, and the quotient formula fails there — handle them separately.
• Forgetting the second derivative rule: $\frac{d^2y}{dx^2} = \frac{d}{dt}\!\left(\frac{dy}{dx}\right) \div \frac{dx}{dt}$, not $\frac{d^2y/dt^2}{d^2x/dt^2}$.