9.1 - Defining and Differentiating Parametric Equations

What Are Parametric Equations?

Unlike standard functions where

y = f(x)

, parametric equations use a third variable (usually

t

) to define

x

and

y

separately.

Definition: if

f

and

g

are continuous functions of

t

on an interval, then

x = f(t) \quad \text{and} \quad y = g(t)

are called parametric equations, and

t

is the parameter, which often represents time.

Key difference: you have two separate equations instead of one relationship between

x

and

y

Graphing Parametric Equations

Method 1: Table of Values

To graph parametric equations, build a table by choosing values of

t

and calculating the matching

(x, y)

coordinates.

Example 1: Basic Parametric Curve

Given:

x = 3t

and

y = t^2 + 1

for

t \in [-2, 2]

$t$	$x = 3t$	$y = t^2 + 1$	Point $(x, y)$
$-2$	$-6$	$5$	$(-6, 5)$
$-1$	$-3$	$2$	$(-3, 2)$
$0$	$0$	$1$	$(0, 1)$
$1$	$3$	$2$	$(3, 2)$
$2$	$6$	$5$	$(6, 5)$

9.1 - Defining and Differentiating Parametric Equations

Important: the arrows show the direction of motion as

t

increases. This matters, because parametric curves can cross themselves or reverse direction.

Calculator: Parametric Mode

On a TI-84

Step 1: Switch to parametric mode.

Press MODE

Scroll down and highlight PAR

Press ENTER

Step 2: Enter the equations.

Press Y=

You’ll see X₁T = and Y₁T =

Enter your equations, using the X,T,θ,n button for $t$

Step 3: Set the window.

Press WINDOW

Set Tmin, Tmax, and Tstep

Tmin and Tmax are the range of $t$ -values
Tstep is the interval between calculated points, so smaller means a smoother curve

Set Xmin, Xmax, Ymin, and Ymax as usual

Step 4: Press GRAPH.

Eliminating the Parameter

Eliminating the parameter means converting parametric equations back into a single equation in

x

and

y

, which is called rectangular form.

Example 2: Algebraic Elimination

Given:

x = 3t

and

y = t^2 + 1

Step 1: Solve one equation for $t$ .

x = 3t \quad \Rightarrow \quad t = \frac{x}{3}

Step 2: Substitute into the other equation.

y = t^2 + 1 = \left(\frac{x}{3}\right)^2 + 1 = \frac{x^2}{9} + 1

Result: the parametric equations describe the parabola

y = \frac{x^2}{9} + 1

Trigonometric Parametric Equations

When the parametric equations involve sine and cosine, use the Pythagorean identity:

\boxed{\sin^2 t + \cos^2 t = 1}

Example 3: Circular Motion

Given:

x = 3\cos t

and

y = 3\sin t

Step 1: Isolate sine and cosine.

\cos t = \frac{x}{3} \qquad \sin t = \frac{y}{3}

Step 2: Apply the Pythagorean identity.

\sin^2 t + \cos^2 t = 1

\left(\frac{y}{3}\right)^2 + \left(\frac{x}{3}\right)^2 = 1

\frac{y^2}{9} + \frac{x^2}{9} = 1

Step 3: Simplify. Multiply both sides by 9:

x^2 + y^2 = 9

Result: a circle centered at the origin with radius 3.

Derivatives of Parametric Equations

To find

\frac{dy}{dx}

for parametric equations, use

\boxed{\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{y'(t)}{x'(t)}}

provided that

\frac{dx}{dt} \neq 0

Alternative notation:

\frac{dy}{dx} = \frac{dy/dt}{dx/dt}

Important: this is the slope of the tangent line at any point on the parametric curve.

Example 4: Finding the Derivative

Given:

x = t^2 + 1

and

y = 2t^3 - 4t

. Find

\frac{dy}{dx}

Step 1: Find $\frac{dx}{dt}$ .

\frac{dx}{dt} = 2t

Step 2: Find $\frac{dy}{dt}$ .

\frac{dy}{dt} = 6t^2 - 4

Step 3: Apply the formula.

\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{6t^2 - 4}{2t}

Step 4: Simplify.

\frac{dy}{dx} = \frac{2(3t^2 - 2)}{2t} = \frac{3t^2 - 2}{t}

Answer:

\frac{dy}{dx} = \frac{3t^2 - 2}{t}

Example 5: Derivative with Exponentials

Given:

x = e^t

and

y = t^2 - 2t

. Find

\frac{dy}{dx}

Step 1: Find the derivatives with respect to $t$ .

\frac{dx}{dt} = e^t \qquad \frac{dy}{dt} = 2t - 2

Step 2: Apply the formula.

\frac{dy}{dx} = \frac{2t - 2}{e^t} = \frac{2(t - 1)}{e^t}

Finding Tangent Lines to Parametric Curves

To find the equation of a tangent line at a specific value

t = t_0

, you need three things:

The $x$ -coordinate: $x(t_0)$

The $y$ -coordinate: $y(t_0)$

The slope: $\frac{dy}{dx}\big|_{t=t_0}$

Then use point-slope form:

y - y_0 = m(x - x_0)

Example 6: Equation of a Tangent Line

Given:

x = 2t - 1

and

y = t^2 + 3t

. Find the tangent line when

t = 2

Step 1: Find the point at $t = 2$ .

x(2) = 2(2) - 1 = 3

y(2) = (2)^2 + 3(2) = 4 + 6 = 10

Point:

(3, 10)

Step 2: Find $\frac{dy}{dx}$ .

\frac{dx}{dt} = 2 \qquad \frac{dy}{dt} = 2t + 3 \qquad \frac{dy}{dx} = \frac{2t + 3}{2}

Step 3: Evaluate the slope at $t = 2$ .

\frac{dy}{dx}\bigg|_{t=2} = \frac{2(2) + 3}{2} = \frac{7}{2}

Step 4: Write the equation in point-slope form.

y - 10 = \frac{7}{2}(x - 3)

Step 5: Solve for $y$ .

y - 10 = \frac{7}{2}x - \frac{21}{2}

y = \frac{7}{2}x - \frac{21}{2} + \frac{20}{2}

y = \frac{7}{2}x - \frac{1}{2}

Answer:

y = \frac{7}{2}x - \frac{1}{2}

Example 7: Tangent Line with Trig Functions

Given:

x = e^{t}

and

y = \sin t

. Find the tangent line when

t = \frac{\pi}{4}

Step 1: Find the point.

x\left(\frac{\pi}{4}\right) = e^{\pi/4} \qquad y\left(\frac{\pi}{4}\right) = \sin\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2}

Step 2: Find $\frac{dy}{dx}$ .

\frac{dx}{dt} = e^t \qquad \frac{dy}{dt} = \cos t \qquad \frac{dy}{dx} = \frac{\cos t}{e^t}

Step 3: Evaluate at $t = \frac{\pi}{4}$ .

\frac{dy}{dx}\bigg|_{t=\pi/4} = \frac{\cos(\pi/4)}{e^{\pi/4}} = \frac{\sqrt{2}/2}{e^{\pi/4}} = \frac{\sqrt{2}}{2e^{\pi/4}}

Step 4: Write the equation.

y - \frac{\sqrt{2}}{2} = \frac{\sqrt{2}}{2e^{\pi/4}}\left(x - e^{\pi/4}\right)

Simplified:

y = \frac{\sqrt{2}}{2e^{\pi/4}}x - \frac{\sqrt{2}}{2e^{\pi/4}} \cdot e^{\pi/4} + \frac{\sqrt{2}}{2}

y = \frac{\sqrt{2}}{2e^{\pi/4}}x - \frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2}

y = \frac{\sqrt{2}}{2e^{\pi/4}}x

Horizontal and Vertical Tangent Lines

Because

\frac{dy}{dx} = \frac{dy/dt}{dx/dt}

, look at the top and the bottom of the fraction separately:

Horizontal tangent: $\frac{dy}{dt} = 0$ and $\frac{dx}{dt} \neq 0$ . The slope is $0$ divided by a number that isn’t zero, so the slope is $0$ .

Vertical tangent: $\frac{dx}{dt} = 0$ and $\frac{dy}{dt} \neq 0$ . The slope is undefined, since you’d be dividing by zero.

Both equal to zero? Then the formula gives

\frac{0}{0}

, and you can’t conclude anything yet. The curve might come to a sharp point (a cusp) there, like

x = t^2

y = t^3

t = 0

. It takes more work, like a limit, to figure out what happens.

Whenever

\frac{dx}{dt} = 0

, the formula can’t give you a number for the slope. That’s why it comes with the condition

\frac{dx}{dt} \neq 0

Example 8: Horizontal and Vertical Tangents

Given:

x = t^2 - 2t

and

y = t^3 - 12t

. Find every point where the tangent line is horizontal or vertical.

Step 1: Find both derivatives.

\frac{dx}{dt} = 2t - 2 \qquad \frac{dy}{dt} = 3t^2 - 12

Step 2: Horizontal tangents. Set

\frac{dy}{dt} = 0

3t^2 - 12 = 0 \quad \Rightarrow \quad t^2 = 4 \quad \Rightarrow \quad t = \pm 2

Check the bottom:

\frac{dx}{dt} = 2

t = 2

and

-6

t = -2

. Neither is zero, so both are horizontal tangents.

t = 2: \ (x, y) = (0, -16) \qquad t = -2: \ (x, y) = (8, 16)

Step 3: Vertical tangents. Set

\frac{dx}{dt} = 0

2t - 2 = 0 \quad \Rightarrow \quad t = 1

Check the top:

\frac{dy}{dt} = 3(1)^2 - 12 = -9 \neq 0

. So there’s a vertical tangent at

t = 1

, at the point

(x, y) = (-1, -11)

Answer: horizontal tangents at

(0, -16)

and

(8, 16)

, and a vertical tangent at

(-1, -11)

Important Distinctions

$\frac{dy}{dx}$ vs. $\frac{dy}{dt}$

$\frac{dy}{dt}$ tells you how $y$ changes with respect to the parameter $t$

$\frac{dx}{dt}$ tells you how $x$ changes with respect to $t$

$\frac{dy}{dx}$ tells you the slope of the tangent line to the curve, meaning how $y$ changes with $x$

Remember:

\frac{dy}{dx} = \frac{dy/dt}{dx/dt}

Parametric Curves vs. Functions

Parametric curves are NOT necessarily functions of

x

. They can:

Loop back on themselves

Have vertical tangent lines

Pass through the same $x$ -value more than once

That’s exactly why parametric equations are more flexible than standard functions.

Common Mistakes to Avoid

Mixing up the order in the derivative formula.

Wrong:

\frac{dy}{dx} = \frac{dx/dt}{dy/dt}

Correct:

\frac{dy}{dx} = \frac{dy/dt}{dx/dt}

Think “dy on top, dx on the bottom.”

Forgetting the chain rule when finding $\frac{dx}{dt}$ or $\frac{dy}{dt}$ . If

x = e^{2t}

, then

\frac{dx}{dt} = 2e^{2t}

. Don’t lose the 2.

Not indicating direction on parametric graphs. Always show arrows for the direction as

t

increases.

Dividing by zero. The formula only works when

\frac{dx}{dt} \neq 0

Problem-Solving Strategy

To graph parametric equations

Choose several values of $t$ in the given interval

Calculate the matching $(x, y)$ points

Plot the points and connect them with a smooth curve

Add arrows showing the direction as $t$ increases

To eliminate the parameter

Solve one equation for $t$

Substitute into the other equation

Simplify to get a relationship between $x$ and $y$ alone

For trig functions, use

\sin^2 t + \cos^2 t = 1

To find $\frac{dy}{dx}$

Find $\frac{dx}{dt}$

Find $\frac{dy}{dt}$

Divide: $\frac{dy}{dx} = \frac{dy/dt}{dx/dt}$

Simplify if you can

To find a tangent line

Find the point $(x(t_0), y(t_0))$

Find the slope $\frac{dy}{dx}\big|_{t=t_0}$

Use point-slope form: $y - y_0 = m(x - x_0)$

Solve for $y$ if needed

Key Takeaways

Parametric equations: both $x$ and $y$ are functions of a parameter $t$ , so $x = f(t)$ and $y = g(t)$

The parameter $t$ usually represents time and determines your position along the curve

Direction matters. Parametric curves have an orientation

Eliminating the parameter: solve for $t$ in one equation and substitute into the other

Pythagorean identity: use $\sin^2 t + \cos^2 t = 1$ for trig parametric equations

Derivative formula: $\frac{dy}{dx} = \frac{dy/dt}{dx/dt}$ , not the other way around

Tangent lines need the point $(x(t_0), y(t_0))$ and the slope $\frac{dy}{dx}\big|_{t=t_0}$

Horizontal tangents: $\frac{dy}{dt} = 0$ with $\frac{dx}{dt} \neq 0$ . Vertical tangents: $\frac{dx}{dt} = 0$ with $\frac{dy}{dt} \neq 0$

Calculator mode: switch to PAR for graphing parametric equations

Not always functions. Parametric curves can loop, cross themselves, or have vertical tangents

Flexibility is the point: parametric equations can represent curves that aren’t functions of $x$

Back Next

9.1 - Defining and Differentiating Parametric Equations

What Are Parametric Equations?

Graphing Parametric Equations

Method 1: Table of Values

Example 1: Basic Parametric Curve

Calculator: Parametric Mode

On a TI-84

Eliminating the Parameter

Example 2: Algebraic Elimination

Trigonometric Parametric Equations

Example 3: Circular Motion

Derivatives of Parametric Equations

Example 4: Finding the Derivative

Example 5: Derivative with Exponentials

Finding Tangent Lines to Parametric Curves

Example 6: Equation of a Tangent Line

Example 7: Tangent Line with Trig Functions

Horizontal and Vertical Tangent Lines

Example 8: Horizontal and Vertical Tangents

Important Distinctions

dydx\frac{dy}{dx}dxdy​ vs. dydt\frac{dy}{dt}dtdy​

Parametric Curves vs. Functions

Common Mistakes to Avoid

Problem-Solving Strategy

To graph parametric equations

To eliminate the parameter

To find dydx\frac{dy}{dx}dxdy​

To find a tangent line

Key Takeaways

$\frac{dy}{dx}$ vs. $\frac{dy}{dt}$

To find $\frac{dy}{dx}$