Defining and Differentiating Parametric Equations

3. Visual Connection

5. Concept Development

BC-only topic. Parametric equations describe both coordinates using a shared parameter:

Each parameter value selects the point \((x(t),y(t))\). As \(t\) increases, the ordered points trace a curve with a particular starting point, ending point, direction, and possibly repeated points. A rectangular equation alone may describe the same geometric set but usually does not preserve that tracing information.

Reading a parametric curve

• Make a short table of \(t\), \(x(t)\), and \(y(t)\) to locate points.
• Use increasing \(t\)-values to determine orientation.
• The signs of \(x'(t)\) and \(y'(t)\) indicate instantaneous left/right and down/up motion.
• Different parameter intervals can trace only part of a curve or trace the same part more than once.

Eliminating the parameter can reveal the familiar shape. For example, \(x=2t+3\), \(y=3t-4\) gives \(y=\tfrac32x-\tfrac{17}{2}\), but the specified \(t\)-interval is still needed to identify the segment and its direction.

First derivative

If the parameter can locally be viewed as producing \(y=F(x)\), then \(y(t)=F(x(t))\). The chain rule gives

provided \(x'(t)\ne0\). The component derivative \(y'(t)\) is vertical change per unit \(t\); it is not by itself the slope in the \(xy\)-plane. The quotient compares vertical and horizontal change.

Tangent line at \(t=t_0\)

1. Find the point \((x_0,y_0)=(x(t_0),y(t_0))\).
2. Compute \(x'(t_0)\) and \(y'(t_0)\).
3. If \(x'(t_0)\ne0\), calculate \(m=y'(t_0)/x'(t_0)\).
4. Write \(y-y_0=m(x-x_0)\).

A parameter value is not an \(x\)-coordinate unless the problem explicitly makes \(x(t)=t\).

Horizontal and vertical tangents

After finding candidate \(t\)-values, substitute them into both original equations to report points. If \(x'(t)=y'(t)=0\), the quotient is \(0/0\) and these tests are inconclusive. Simplifying the slope for nearby \(t\), eliminating the parameter, or examining a limit may reveal a tangent, cusp, or other singular behavior.

Repeated points

Different parameter values can produce the same \((x,y)\). Evaluate the slope at each parameter value separately: a self-intersection can have two distinct tangent lines at one geometric point.

Scope boundary

This topic concerns the first derivative and tangent behavior. The second parametric derivative is developed in Topic 9.2, parametric arc length in Topic 9.3, and vector motion quantities in Topics 9.4-9.6.

6. Detailed Worked Example and Error Check

Example 1: Shape, interval, and orientation

Let \(x=2t+3\), \(y=3t-4\), \(-2\le t\le3\). Solving \(t=(x-3)/2\) gives

At \(t=-2\) the point is \((-1,-10)\); at \(t=3\) it is \((9,5)\). The equations therefore trace that line segment from lower left to upper right. Also, \(x'=2\), \(y'=3\), so \(dy/dx=3/2\), matching the rectangular slope.

Example 2: Tangent line at a parameter value

Let \(x=t^2+1\) and \(y=t^3-t\). Then

At \(t=1\), the point is \((2,0)\) and the slope is \(1\). Thus the tangent line is

Example 3: A tangent to a circle

For \(x=4\cos t\), \(y=4\sin t\), the curve is the circle \(x^2+y^2=16\), traced counterclockwise. At \(t=\pi/4\),

The tangent line is \(y-2\sqrt2=-(x-2\sqrt2)\), or \(y=-x+4\sqrt2\).

Example 4: Horizontal and vertical tangents

Consider \(x=t^2-1\), \(y=t^3-3t\). Since

horizontal tangents occur at \(t=\pm1\), where \(x'\ne0\), giving the points \((0,-2)\) and \((0,2)\). A vertical tangent occurs at \(t=0\), where \(y'=-3\ne0\), giving the point \((-1,0)\).

Example 5: Two tangents at one point

For \(x=t^2-1\), \(y=t^3-t\), both \(t=1\) and \(t=-1\) produce \((0,0)\). The slope is

At \(t=1\) the slope is \(1\), while at \(t=-1\) it is \(-1\). The curve crosses itself at the origin with tangent lines \(y=x\) and \(y=-x\).

Example 6: Projectile path

A projectile has \(x=20t\) and \(y=5+24t-16t^2\). Its path slope is

At \(t=1/2\), the point is \((10,13)\), the slope is \(2/5\), and the tangent line is \(y-13=\tfrac25(x-10)\). The path has a horizontal tangent when \(24-32t=0\), so \(t=3/4\).

Example 7: Derivative information from a table

Suppose a table gives \(x(2)=3\), \(y(2)=-1\), \(x'(2)=-2\), and \(y'(2)=5\). The point is \((3,-1)\), and

The tangent line is \(y+1=-\tfrac52(x-3)\). Because \(x'<0\) and \(y'>0\), the point is moving left and up as \(t\) increases.

Example 8: When both component derivatives vanish

For \(x=t^2\), \(y=t^3\), both derivatives are zero at \(t=0\). The standard horizontal and vertical tests do not apply directly. For \(t\ne0\),

The curve \(y^2=x^3\) has a cusp at the origin with horizontal tangent line \(y=0\). The conclusion comes from the limiting geometry, not from declaring \(0/0=0\).

Common errors

• Treating \(dy/dt\) as \(dy/dx\).
• Reversing the quotient and computing \(x'(t)/y'(t)\).
• Writing a tangent line through \((t_0,y(t_0))\) instead of \((x(t_0),y(t_0))\).
• Calling \(y'(t)=0\) horizontal without checking \(x'(t)\ne0\).
• Calling \(x'(t)=0\) vertical without checking \(y'(t)\ne0\).
• Assuming \(0/0\) determines a tangent.
• Eliminating the parameter and forgetting the interval or orientation.
• Reporting a parameter value when the question asks for a point.