Topic 9.1 Notes – Defining and Differentiating Parametric Equations

1. What a Parametric Curve Is

A parametric curve is defined by two equations:

$$x = x(t), \quad y = y(t)$$

Each value of $t$ gives a point $(x(t), y(t))$ in the xy-plane. As $t$ changes, that point moves along a path.

So even though everything depends on $t$, the graph is still drawn in the regular coordinate plane. You never graph $t$.

Here’s the key shift in thinking:

• In regular functions, $y$ depends directly on $x$.
• In parametrics, both $x$ and $y$ depend on $t$.

That lets curves:

• Move left and right
• Loop back on themselves
• Pass through the same point more than once
• Have motion with direction

To visualize that motion idea, look at this example curve traced as $t$ increases.

Parametric curve for $x(t)=t^2-1$, $y(t)=t^3-3t$, $-2 \le t \le 2$

Notice the arrows. Direction matters in parametrics. Two curves can look identical but be traced differently depending on how $t$ changes.

You already know how to differentiate expressions like $x(t)$ and $y(t)$. We’re just going to use those same derivative rules again.

2. The Derivative of a Parametric Curve

We still want the slope of the tangent line, which is

$$\frac{dy}{dx}$$

But we don’t have $y$ written as a function of $x$. Both depend on $t$. So we use the chain rule idea:

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

This only works when $\frac{dx}{dt} \ne 0$.

Why this makes sense:

• $\frac{dy}{dt}$ tells how fast $y$ changes.
• $\frac{dx}{dt}$ tells how fast $x$ changes.
• Their ratio tells how fast $y$ changes with respect to $x$.

That ratio is the slope of the tangent line.

On quizzes and FRQs, this formula is the entire point of the question. If you write only $dy/dt$, you won’t get credit. They want the ratio.

3. How to Find the Slope at a Specific Value of t

Here’s the clean process.

1. Differentiate $x(t)$ to find $dx/dt$.
2. Differentiate $y(t)$ to find $dy/dt$.
3. Form the ratio $\frac{dy/dt}{dx/dt}$.
4. Simplify.
5. Plug in the given $t$-value (if one is provided).

Example:

$$x(t) = 3t^2 + 1, \quad y(t) = t^3 - 4t$$

Differentiate:

$$\frac{dx}{dt} = 6t$$ $$\frac{dy}{dt} = 3t^2 - 4$$

Form the ratio:

$$\frac{dy}{dx} = \frac{3t^2 - 4}{6t}$$

If asked for the slope at $t = 2$:

$$\frac{dy}{dx}\Big|_{t=2} = \frac{3(4) - 4}{12} = \frac{8}{12} = \frac{2}{3}$$

If they ask for the equation of the tangent line, also find the point:

$$x(2)=13, \quad y(2)=0$$

Then use point-slope form:

$$y - 0 = \frac{2}{3}(x - 13)$$

On FRQs, don’t skip writing the actual point. They often award a separate point for it.

If no value of $t$ is given, leave your answer in terms of $t$.

4. Special Cases and What They Mean

Horizontal Tangent

$$\frac{dy}{dt} = 0 \quad \text{and} \quad \frac{dx}{dt} \ne 0$$

Then

$$\frac{dy}{dx} = 0$$

The curve is momentarily flat.

Vertical Tangent

$$\frac{dx}{dt} = 0 \quad \text{and} \quad \frac{dy}{dt} \ne 0$$

The ratio is undefined. The tangent line is vertical.

Both Equal Zero

$$\frac{dx}{dt} = 0 \quad \text{and} \quad \frac{dy}{dt} = 0$$

The slope is indeterminate. In this course, you usually just recognize this situation. Further analysis is beyond what you’re expected to do in this specific topic.