Parametric Equations of Curves

A pair of equations

$$x = f(t), \quad y = g(t), \quad a \le t \le b$$

are parametric equations for a curve. You graph the curve by plugging values of t into x and y, then plotting the points as usual.


Example. The parametric equations

$$x = \cos t, \quad y = \sin t, \quad 0 \le t \le 2\pi$$

represent a circle of radius 1 centered at the origin.

$$\hbox{\epsfysize=1.75in \epsffile{parametric-equations1.eps}}$$

You can sometimes recover the x-y equation of a parametric curve by eliminating t from the parametric equations. In this case,

$$x^2 + y^2 = (\cos t)^2 + (\sin t)^2 = 1.$$

Notice that the graph of a circle is not the graph of a function. Parametric equations can represent more general curves than function graphs can.


Example. The parametric equations

$$x = t\cos t, \quad y = t\sin t,$$

represent a spiral.

$$\hbox{\epsfysize=1.75in \epsffile{parametric-equations2.eps}}$$

This is not the graph of a function $y = f(x)$ .


Example. Find the x-y equation for

$$x = 5 + 2\cos t, \quad y = -3 + \sin t.$$

Notice that

$$\cos t = \dfrac{x - 5}{2}, \quad \sin t = y + 3.$$

So

$$\left(\dfrac{x - 5}{2}\right)^2 + (y + 3)^2 = (\cos t)^2 + (\sin t)^2 = 1.$$

This is the standard form for the equation of an ellipse.


Example. The parametric equations

$$x = 3\cos t + \cos 3t, \quad y = 3\sin t - \sin 3t, \quad 0 \le t \le 2\pi$$

represent a hypocycloid of four cusps.

$$\hbox{\epsfysize=1.75in \epsffile{parametric-equations3.eps}}$$

In this case, it would be difficult to eliminate t to obtain an x-y equation.


Example. To parametrize a curve means to obtain parametric equations for the curve.

If you have x-y equations in which x or y is solved for, it's easy. For example, to parametrize $y = x^2$ , set $x = t$ . Then $y
   = x^2 = t^2$ . A parametrization is given by

$$x = t, \quad y = t^2.$$

There are infinitely many ways to parametrize a curve. For example,

$$x = t + 42, \quad y = (t + 42)^2,$$

is another parametrization of $y = x^2$ .

To parametrize $x = 3y -
   y^2$ , set $y = t$ . Then $x
   = 3y - y^2 = 3t - t^2$ , so

$$x = 3t - t^2, \quad y = t$$

is a parametrization of $x
   = 3y - y^2$ . (This is how you can graph x-in-terms-of-y equations on your calculator.)


Example. If $(a,b)$ and $(c,d)$ are points, the segment from $(a,b)$ to $(c,d)$ may be parametrized by

$$x = a + t(c - a), \quad y = b + t(d - b), \quad 0 \le t \le 1.$$

Notice that when $t = 0$ , $(x,y) = (a,b)$ , and when $t = 1$ , $(x,y) =
   (c,d)$ .

If you allow t to range from $-\infty$ to $+\infty$ , this gives the line through $(a,b)$ and $(c,d)$ .

For example, the line through $(3,-6)$ and $(5,2)$ is

$$x = 3 + 2t, \quad y = -6 + 8t.$$

An analogous result holds for lines in 3 dimensions (or in any number of dimensions).


Example. Find parametric equations for the Folium of Descartes:

$$x^3 + y^3 = 3xy.$$

Set $y = xt$ . Then

$$x^3 + x^3t^3 = 3x^2t, \quad x + xt^3 = 3t, \quad x(1 + t^3) = 3t, \quad x = \dfrac{3t}{1 + t^3}.$$

Therefore, $y = xt =
   \dfrac{3t^2}{1 + t^3}$ . A parametrization is given by

$$x = \dfrac{3t}{1 + t^3}, \quad y = \dfrac{3t^2}{1 + t^3}.$$

$$\hbox{\epsfysize=1.75in \epsffile{parametric-equations4.eps}}\quad\halmos$$


The first and second derivatives give information about the shape of a curve. Here's how to find the derivatives for a parametric curve.

First,

$$\der x t\cdot \der y x = \der y t,$$

by the Chain Rule. Solving for $\der y x$ gives

$$\der y x = \dfrac{\der y t}{\der x t}.$$

To find the second derivative, I differentiate the first derivative.

$$\dfrac{d^2y}{dx^2} = \der {} x \left(\der y x\right).$$

Since $\der y x$ will come out in terms of t, I want to be sure to differentiate $\der y x$ with respect to t. Use the Chain Rule again:

$$\der {} x \left(\der y x\right) = \der t x\cdot \left[\der {} t \left(\der y x\right)\right] = \dfrac{\der {} t \left(\der y x\right)}{\der x t}.$$

That is,

$$\dfrac{d^2y}{dx^2} = \dfrac{\der {} t \left(\der y x\right)}{\der x t}.$$


Example. Find $\der y x$ and $\dfrac{d^2y}{dx^2}$ for

$$x = t^2 + t + 2, \quad y = 2t^3 - t + 5.$$

First,

$$\der x t = 2t + 1, \quad \der y t = 6t^2 - 1.$$

So

$$\der y x = \dfrac{\der y t}{\der x t} = \dfrac{6t^2 - 1}{2t + 1}.$$

Next,

$$\der {} t \left(\der y x\right) = \der {} t \dfrac{6t^2 - 1}{2t + 1} = \dfrac{(2t + 1)(12t) - (6t^2 - 1)(2)}{(2t + 1)^2}.$$

So

$$\dfrac{d^2y}{dx^2} = \dfrac{\der {} t \left(\der y x\right)}{\der x t} = \dfrac{\dfrac{(2t + 1)(12t) - (6t^2 - 1)(2)}{(2t + 1)^2}}{2t + 1} = \dfrac{(2t + 1)(12t) - (6t^2 - 1)(2)}{(2t + 1)^3}.\quad\halmos$$


Example. At what points on the curve

$$x = t^3 + t + 2, \quad y = 2t^3 - 3t^2 - 12t + 5$$

is the tangent line horizontal?

Find the derivative:

$$\der y x = \dfrac{6t^2 - 6t - 12}{3t^2 + 1} = \dfrac{6(t - 2)(t + 1)}{3t^2 + 1}.$$

$\der y x = 0$ for $t = 2$ and for $t = -1$ .

When $t = 2$ , $x = 12$ and $y = -15$ . When $t
   = -1$ , $x = 0$ and $y =
   12$ . There are horizontal tangents are $(12,-15)$ and at $(0,12)$ .

$$\hbox{\epsfysize=1.75in \epsffile{parametric-equations5.eps}}\quad\halmos$$


Example. Find the equation of the tangent line to the curve

$$x = e^t + t^2, \quad y = e^{2t} + 3t$$

at the point corresponding to $t = 1$ .

$$\der y x = \dfrac{2e^{2t} + 3}{e^t + 2t}, \quad\hbox{so}\quad \left.\der y x\right|_{t=1} = \dfrac{2e^2 + 3}{e + 2}.$$

When $t = 1$ , $x = e + 1$ and $y = e^2 + 3$ . The tangent line is

$$y - (e^2 + 3) = \left(\dfrac{2e^2 + 3}{e + 2}\right) (x - (e + 1)).\quad\halmos$$


Send comments about this page to: Bruce.Ikenaga@millersville.edu.

Bruce Ikenaga's Home Page

Copyright 2013 by Bruce Ikenaga