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Parametric Equation Calculator

Last updated: March 28, 2026

Calculator

Results

x(t)

3.535527

y(t)

2.121324

dx/dt

-3.53554

dy/dt

2.121316

dy/dx (slope)

-0.599998

Results

x(t)

3.535527

y(t)

2.121324

dx/dt

-3.53554

dy/dt

2.121316

dy/dx (slope)

-0.599998

Parametric equations describe curves by expressing both $$x$$ and $$y$$ as separate functions of a third variable, the parameter $$t$$. Instead of writing $$y = f(x)$$, you write $$x = x(t)$$ and $$y = y(t)$$, which allows you to trace paths that a standard function cannot represent — such as loops, spirals, and multi-valued curves. The classic ellipse parametrization is:

$$x(t) = a\cos t + h, \quad y(t) = b\sin t + k$$

where $$a$$ and $$b$$ are the semi-axes and $$(h, k)$$ is the center. Our Parametric Equation Calculator evaluates these equations at any parameter value $$t$$, giving you the point coordinates $$(x, y)$$, the velocity components $$dx/dt$$ and $$dy/dt$$, and the slope of the tangent line $$dy/dx$$.

Parametric curves are fundamental in physics (projectile trajectories, planetary orbits), computer graphics (Bezier curves, animation paths), and engineering (cam profiles, robotic arm motion). The ability to compute derivatives parametrically — using $$\frac{dy}{dx} = \frac{dy/dt}{dx/dt}$$ — is a key skill tested in AP Calculus BC and university-level courses. This tool lets you verify your hand calculations and build intuition for how the parameter $$t$$ traces out the curve.

Visual Analysis

How It Works

Given the parametric equations for an ellipse centered at $$(h, k)$$:

$$x(t) = a\cos t + h$$

$$y(t) = b\sin t + k$$

Position: Substitute your value of $$t$$ to find the point $$(x, y)$$ on the ellipse.

Derivatives: Differentiate each equation with respect to $$t$$: $$\frac{dx}{dt} = -a\sin t, \quad \frac{dy}{dt} = b\cos t$$

Slope (dy/dx): By the chain rule: $$\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{b\cos t}{-a\sin t} = -\frac{b}{a}\cot t$$

The slope is undefined when $$dx/dt = 0$$ (i.e., at $$t = 0$$ and $$t = \pi$$), corresponding to the top and bottom of the ellipse where the tangent line is vertical.

Understanding Your Results

The outputs x(t) and y(t) give the exact point on the ellipse for your chosen parameter value. The dx/dt and dy/dt values represent the horizontal and vertical components of the velocity vector as the point traces the curve. Their ratio dy/dx gives the slope of the tangent line at that point. A very large dy/dx (displayed as 999999) indicates a near-vertical tangent. When $$a = b$$, the ellipse becomes a circle and the slope simplifies to $$-\cot t$$.

Worked Examples

Ellipse at t = π/4

Inputs

a5
b3
h0
k0
t0.7854

Results

x val3.535534
y val2.12132
dx dt-3.535534
dy dt2.12132
dy dx-0.6

At t = π/4: x = 5cos(π/4) ≈ 3.536, y = 3sin(π/4) ≈ 2.121. Slope = −(3/5)cot(π/4) = −0.6.

Shifted Circle at t = π/2

Inputs

a4
b4
h2
k-1
t1.5708

Results

x val2
y val3
dx dt-4
dy dt0
dy dx0

Circle of radius 4 centered at (2, −1). At t = π/2: x = 2, y = 3 (top of circle). Tangent is horizontal (dy/dx = 0).

Frequently Asked Questions

A parametric equation defines a curve using a parameter $$t$$ rather than expressing $$y$$ directly as a function of $$x$$. Both coordinates are given as functions: $$x = x(t)$$ and $$y = y(t)$$. As $$t$$ varies over an interval, the point $$(x(t), y(t))$$ traces out the curve. This approach can represent curves that are not functions, such as circles and figure-eights.

Use the chain rule: $$\frac{dy}{dx} = \frac{dy/dt}{dx/dt}$$. First differentiate $$y(t)$$ and $$x(t)$$ separately with respect to $$t$$, then divide. This gives the slope of the tangent line at the point corresponding to parameter $$t$$. The formula is undefined when $$dx/dt = 0$$, which corresponds to vertical tangent lines.

The parameter $$t$$ is an independent variable that controls position along the curve. In physics, $$t$$ often represents time, so $$(x(t), y(t))$$ gives the position of a moving object at time $$t$$. In the ellipse parametrization, $$t$$ represents the eccentric anomaly, an angle-like quantity that goes from $$0$$ to $$2\pi$$ for one complete traversal.

For the ellipse $$x = a\cos t + h, y = b\sin t + k$$: isolate the trig functions: $$\cos t = (x-h)/a$$ and $$\sin t = (y-k)/b$$. Then use the identity $$\cos^2 t + \sin^2 t = 1$$ to get: $$\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1$$. This is the standard Cartesian equation of an ellipse.

The tangent line is horizontal when $$dy/dt = 0$$ (and $$dx/dt \neq 0$$), which occurs at $$t = 0$$ and $$t = \pi$$ for the ellipse (the rightmost and leftmost points). The tangent is vertical when $$dx/dt = 0$$ (and $$dy/dt \neq 0$$), at $$t = \pi/2$$ and $$t = 3\pi/2$$ (the top and bottom).

This calculator is configured for the ellipse parametrization $$x = a\cos t + h, y = b\sin t + k$$, which includes circles as a special case ($$a = b$$). For other parametric curves like cycloids, Lissajous figures, or custom paths, you would need to modify the parametric equations. The derivative computation method ($$dy/dx = (dy/dt)/(dx/dt)$$) applies universally to all parametric curves.

Sources & Methodology

Stewart, J. Calculus: Early Transcendentals, 9th Edition (2020); Anton, H. Calculus, 11th Edition (2016); Khan Academy — Parametric Equations and Polar Coordinates (2024)
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