Parametric Equations Calculator

Graph parametric curves and compute derivatives, speed and rectangular form step by step
x = t^2, y = t^3 - 3t, find dy/dx at t = 2
second derivative of x = t^2, y = t^3 - 3t
eliminate the parameter for x = 3cos(t), y = 2sin(t)
speed of x = t - sin(t), y = 1 - cos(t) at t = pi/2

What Parametric Equations Are

A parametric curve describes both coordinates through a third variable, the parameter tt:

x=f(t),y=g(t),t[a,b]x = f(t), \qquad y = g(t), \qquad t \in [a, b]

As tt runs over its interval the point (f(t),g(t))(f(t), g(t)) traces a path. Unlike y=F(x)y = F(x), this can double back, cross itself or close into a loop — a circle needs two half-functions in rectangular form but only one parametrisation:

x=rcost,y=rsint,t[0,2π]x = r\cos t, \quad y = r\sin t, \quad t \in [0, 2\pi]

A line through (x0,y0)(x_0, y_0) with direction a,b\langle a, b \rangle is x=x0+atx = x_0 + at, y=y0+bty = y_0 + bt, the component form of the vector equation r(t)=r0+tv\mathbf{r}(t) = \mathbf{r}_0 + t\mathbf{v}.

The parameter also carries orientation and timing, which rectangular form throws away.

Calculus on Parametric Curves

First derivative

dydx=dy/dtdx/dt,dxdt0\frac{dy}{dx} = \frac{dy/dt}{dx/dt}, \qquad \frac{dx}{dt} \neq 0

This is the chain rule rearranged. Where dx/dt=0dx/dt = 0 but dy/dt0dy/dt \neq 0 the tangent is vertical; where both vanish the point may be a cusp and the formula says nothing.

Second derivative

d2ydx2=ddt ⁣(dydx)dx/dt\frac{d^2y}{dx^2} = \frac{\dfrac{d}{dt}\!\left(\dfrac{dy}{dx}\right)}{dx/dt}

Differentiate the first derivative with respect to tt, then divide by dx/dtdx/dt again. It is emphatically not d2y/dt2d2x/dt2\frac{d^2y/dt^2}{d^2x/dt^2}.

Speed and arc length

speed=(dxdt)2+(dydt)2,L=ab(dxdt)2+(dydt)2dt\text{speed} = \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2}, \qquad L = \int_a^b \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2}\,dt

The arc-length integral counts every retraced piece, so it equals the geometric length only when the curve is traversed once.

Eliminating the parameter

Solve one equation for tt and substitute, or use an identity such as cos2t+sin2t=1\cos^2 t + \sin^2 t = 1 for trigonometric pairs. Always carry the domain restriction across.

Common Mistakes to Avoid

  • Dividing the second derivatives: d2ydx2\frac{d^2y}{dx^2} requires differentiating dy/dxdy/dx with respect to tt and dividing by dx/dtdx/dt once more.
  • Losing the domain when eliminating tt: x=t2x = t^2, y=ty = t gives x=y2x = y^2, but only the branch with x0x \ge 0 — and every restriction on tt transfers.
  • Ignoring dx/dt=0dx/dt = 0: those parameter values are exactly where vertical tangents and cusps hide.
  • Confusing speed with dy/dxdy/dx: dy/dxdy/dx is the tangent slope, a shape property; speed measures how fast the point moves and depends on the parametrisation.
  • Assuming the parametrisation is unique: the same curve has infinitely many, differing in orientation and rate.

Examples

Step 1: Differentiate each component: dxdt=2t\dfrac{dx}{dt} = 2t and dydt=3t23\dfrac{dy}{dt} = 3t^2 - 3
Step 2: Form the quotient: dydx=3t232t\dfrac{dy}{dx} = \dfrac{3t^2 - 3}{2t} (valid for t0t \neq 0)
Step 3: Substitute t=2t = 2: numerator 3(4)3=93(4) - 3 = 9, denominator 2(2)=42(2) = 4
Step 4: dydx=94\dfrac{dy}{dx} = \dfrac{9}{4}
Answer: dydx=94=2.25\dfrac{dy}{dx} = \dfrac{9}{4} = 2.25

Step 1: From above, dydx=3t232t=32t32t1\dfrac{dy}{dx} = \dfrac{3t^2-3}{2t} = \dfrac{3}{2}t - \dfrac{3}{2}t^{-1}
Step 2: Differentiate with respect to tt: ddt ⁣(dydx)=32+32t2\dfrac{d}{dt}\!\left(\dfrac{dy}{dx}\right) = \dfrac{3}{2} + \dfrac{3}{2}t^{-2}
Step 3: Divide by dxdt=2t\dfrac{dx}{dt} = 2t: d2ydx2=32+32t22t\dfrac{d^2y}{dx^2} = \dfrac{\frac{3}{2} + \frac{3}{2t^2}}{2t}
Step 4: Simplify by multiplying numerator and denominator by 2t22t^2: 3t2+34t3\dfrac{3t^2 + 3}{4t^3}
Step 5: At t=2t = 2 this is 3(4)+34(8)=1532\dfrac{3(4)+3}{4(8)} = \dfrac{15}{32}, positive, so the curve is concave up there
Answer: d2ydx2=3(t2+1)4t3\dfrac{d^2y}{dx^2} = \dfrac{3(t^2+1)}{4t^3}

Step 1: Solve each for the trig function: cost=x3\cos t = \dfrac{x}{3} and sint=y2\sin t = \dfrac{y}{2}
Step 2: Apply cos2t+sin2t=1\cos^2 t + \sin^2 t = 1
Step 3: (x3)2+(y2)2=1\left(\dfrac{x}{3}\right)^2 + \left(\dfrac{y}{2}\right)^2 = 1
Step 4: This is an ellipse with semi-axes 33 and 22, traced counter-clockwise once for t[0,2π)t \in [0, 2\pi)
Answer: x29+y24=1\dfrac{x^2}{9} + \dfrac{y^2}{4} = 1

Frequently Asked Questions

It is a pair of equations x = f(t) and y = g(t) that give both coordinates in terms of a third variable. The parameter t sweeps through an interval and the point (x, y) traces out a curve, which lets you describe paths that fail the vertical line test, such as circles and loops.

Differentiate x and y separately with respect to t, then divide: dy/dx = (dy/dt)/(dx/dt). The formula needs dx/dt to be non-zero; where dx/dt = 0 and dy/dt is non-zero the tangent line is vertical.

Because d2y/dx2 means the rate of change of the slope dy/dx with respect to x, not a ratio of two second derivatives. You differentiate dy/dx with respect to t and then divide by dx/dt once more, which is the chain rule applied a second time.

Solve one equation for t and substitute into the other, or use a Pythagorean identity when sine and cosine are involved. Then restrict the resulting equation to the range of x and y that the parameter actually produces, or you will describe more of the curve than exists.

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