Derivatives of Parametric Curves
Textbook Reference
| Primary source | OpenStax Calculus Volume 2, Section 7.2: “Calculus of Parametric Curves” |
| Direct link | https://openstax.org/books/calculus-volume-2/pages/7-2-calculus-of-parametric-curves |
| Textbook used in class | Stewart, Calculus, Section 10.2: “Calculus with Parametric Curves” |
Opening Scenario
A particle moves along the curve $x = t^2$, $y = t^3 - 3t$. What is the slope of the tangent to its path at $t = 1$? There is no explicit equation $y = f(x)$ to differentiate, but the chain rule gives: $$\frac{dy}{dx} = \frac{dy/dt}{dx/dt},$$ which requires only the derivatives of $x$ and $y$ with respect to $t$. The slope $dy/dx$ is a ratio of two rates.
Quick Reference
For a parametric curve $x = f(t)$, $y = g(t)$ with $f'(t) \neq 0$: $$\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{g'(t)}{f'(t)}.$$
Tangent line at $t = t_0$: Slope $= \dfrac{g'(t_0)}{f'(t_0)}$, passes through $(f(t_0), g(t_0))$.
Horizontal tangent: $dy/dt = 0$ and $dx/dt \neq 0$ at that $t$.
Vertical tangent: $dx/dt = 0$ and $dy/dt \neq 0$ at that $t$.
Key Concepts
1. Deriving the Formula
Think of $y$ as a function of $x$, and $x$ as a function of $t$. By the chain rule: $$\frac{dy}{dt} = \frac{dy}{dx}\cdot\frac{dx}{dt}.$$
Solving for $dy/dx$ (when $dx/dt \neq 0$): $$\frac{dy}{dx} = \frac{dy/dt}{dx/dt}.$$
This formula requires $dx/dt \neq 0$ (the curve is not moving purely vertically at that point).
2. Horizontal and Vertical Tangents
The tangent line is horizontal when $dy/dx = 0$, which requires $dy/dt = 0$ (numerator zero) and $dx/dt \neq 0$ (denominator nonzero).
The tangent line is vertical when $dy/dx$ is undefined due to $dx/dt = 0$ with $dy/dt \neq 0$.
When both $dx/dt = 0$ and $dy/dt = 0$ at the same $t$, the situation requires more analysis (it is a singular point).
3. Writing the Tangent Line
At $t = t_0$, the tangent line passes through $(x_0, y_0) = (f(t_0), g(t_0))$ with slope $m = g'(t_0)/f'(t_0)$: $$y - y_0 = m(x - x_0).$$
This is the point-slope form; substitute $x_0$, $y_0$, $m$ directly.
Worked Example
For $x = t^2 - 1$, $y = t^3 - 3t$, find (a) $dy/dx$; (b) the tangent line at $t = 2$; (c) the values of $t$ where the tangent is horizontal or vertical.
(a) $dx/dt = 2t$, $dy/dt = 3t^2 - 3 = 3(t^2-1)$.
$$\frac{dy}{dx} = \frac{3(t^2-1)}{2t}.$$
(b) Tangent at $t = 2$:
Point: $x_0 = 4-1 = 3$, $y_0 = 8-6 = 2$.
Slope: $m = 3(4-1)/(2\cdot 2) = 9/4$.
Tangent line: $y - 2 = \dfrac{9}{4}(x-3)$, i.e., $y = \dfrac{9}{4}x - \dfrac{19}{4}$.
(c) Horizontal tangents: $dy/dt = 3(t^2-1) = 0 \Rightarrow t = \pm 1$.
At $t = 1$: $dx/dt = 2 \neq 0$. $\checkmark$ Point: $(0, -2)$. At $t = -1$: $dx/dt = -2 \neq 0$. $\checkmark$ Point: $(0, 2)$.
Vertical tangents: $dx/dt = 2t = 0 \Rightarrow t = 0$.
At $t = 0$: $dy/dt = 3(0-1) = -3 \neq 0$. $\checkmark$ Point: $(-1, 0)$.
Boxed answers:
- $\dfrac{dy}{dx} = \dfrac{3(t^2-1)}{2t}$
- Tangent at $t=2$: $y = \frac{9}{4}x - \frac{19}{4}$
- Horizontal tangents at $(0,-2)$ and $(0,2)$; vertical tangent at $(-1,0)$.
Common Errors Summary
| Error | Example | Correction |
|---|---|---|
| Computing $dx/dy$ instead of $dy/dx$ | Writing $\frac{dy}{dx} = \frac{dx/dt}{dy/dt}$ | Slope is $\dfrac{dy}{dx} = \dfrac{dy/dt}{dx/dt}$; $y$-derivative on top, $x$-derivative on bottom |
| Using $t = 0$ when $dx/dt = 0$ to declare a horizontal tangent | Calling $t=0$ a horizontal tangent | When $dx/dt = 0$, the tangent is vertical (or singular), not horizontal |
| Writing the tangent line using $t$ as the variable | $y - y_0 = m(t - t_0)$ | The tangent line is in $x$ and $y$: $y - y_0 = m(x - x_0)$ |
Common Misconceptions
$dy/dx = (dy/dt) \cdot (dx/dt)$, multiplying rather than dividing.
This is the composition-is-not-chaining error. The chain rule gives $dy/dt = (dy/dx)(dx/dt)$, so solving for the slope yields $dy/dx = (dy/dt)/(dx/dt)$, a quotient. Multiplying the two rates instead of dividing conflates the chain rule with a product and produces a quantity with the wrong dimensions. For $x = t^2$, $y = t^3$, the correct slope is $(3t^2)/(2t) = 3t/2$, not $(3t^2)(2t) = 6t^3$.
$dx/dt = 0$ signals a horizontal tangent.
This is the height-vs-slope error applied parametrically. When $dx/dt = 0$ and $dy/dt \ne 0$, the curve is moving purely vertically and the tangent line is vertical, not horizontal. A horizontal tangent requires $dy/dt = 0$ with $dx/dt \ne 0$.
Leveled Practice
Level 1 -- Compute the Slope
Problem 1. For $x = 3\cos t$, $y = 2\sin t$, find $dy/dx$ and evaluate it at $t = \pi/4$.
Show answer
$dx/dt = -3\sin t$, $dy/dt = 2\cos t$.
$\dfrac{dy}{dx} = \dfrac{2\cos t}{-3\sin t} = -\dfrac{2}{3}\cot t$.
At $t = \pi/4$: $\cot(\pi/4) = 1$. Slope $= -2/3$.
Level 2 -- Full Tangent Line
Problem 2. For $x = t - t^3$, $y = 1 + t^2$, find the tangent line at $t = 1$.
Show answer
$dx/dt = 1-3t^2$, $dy/dt = 2t$.
At $t=1$: $dx/dt = -2$, $dy/dt = 2$. Slope $= 2/(-2) = -1$.
Point: $x = 1-1 = 0$, $y = 1+1 = 2$. So $(0,2)$.
Tangent line: $y - 2 = -1(x-0)$, i.e., $y = -x+2$.
Mastery Checklist
Mental Model
The slope $dy/dx$ is the ratio of vertical rate to horizontal rate. The parameter $t$ drives both $x$ and $y$. The horizontal rate is $dx/dt$ (how fast $x$ changes per unit $t$), and the vertical rate is $dy/dt$. The slope $dy/dx$ divides these: vertical rate divided by horizontal rate. When the horizontal rate is zero (the curve is moving straight up or down), the slope is undefined and the tangent is vertical.
Connections
Looking back
- Chain rule (Chapter 2): $dy/dt = (dy/dx)(dx/dt)$ rearranged to $dy/dx = (dy/dt)/(dx/dt)$.
Looking ahead
- Second derivative of parametric curves (Section 10.2): Iterate the formula to find $d^2y/dx^2$.
- Polar tangent lines (Section 10.4): Convert polar to parametric, then apply this formula.
Back to Direction and Orientation | Next: Second Derivatives