For parametric equations $x = f(t)$ and $y = g(t)$, we can find $\frac{dy}{dx}$ using the Chain Rule. This allows us to find slopes of tangent lines, determine where the curve has horizontal or vertical tangents, and compute concavity.
🎯 In this section you will learn
- The formula for $\frac{dy}{dx}$ for parametric curves: $\frac{dy}{dx} = \frac{dy/dt}{dx/dt}$
- How to find horizontal tangents ($dy/dt = 0$, $dx/dt \neq 0$)
- How to find vertical tangents ($dx/dt = 0$, $dy/dt \neq 0$)
- The second derivative formula: $\frac{d^2y}{dx^2} = \frac{d/dt(dy/dx)}{dx/dt}$
📌 First Derivative
$$ \frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{g'(t)}{f'(t)} \quad \text{provided } f'(t) \neq 0 $$
The slope of the tangent line at parameter value $t$
💡 Derivation
By the Chain Rule: $\frac{dy}{dt} = \frac{dy}{dx} \cdot \frac{dx}{dt}$, so $\frac{dy}{dx} = \frac{dy/dt}{dx/dt}$.
Given $x = t^2$, $y = t^3$, find $\frac{dy}{dx}$.
①
Find $dx/dt$ and $dy/dt$
$\frac{dx}{dt} = 2t$, $\frac{dy}{dt} = 3t^2$
②
Apply the formula
$\frac{dy}{dx} = \frac{3t^2}{2t} = \frac{3t}{2}$, for $t \neq 0$
For $x = t^2$, $y = t^3$, find the tangent line at $t = 2$.
①
Find the point
$x = 2^2 = 4$, $y = 2^3 = 8$ → $(4, 8)$
②
Find the slope at $t=2$
$\frac{dy}{dx} = \frac{3t}{2}$, so at $t=2$, slope $= \frac{3(2)}{2} = 3$
③
Write the tangent line
$y - 8 = 3(x - 4)$ → $y = 3x - 4$
📌 Horizontal and Vertical Tangents
| Tangent Type | Condition | Formula |
|---|
| Horizontal Tangent | $\frac{dy}{dx} = 0$ | $\frac{dy}{dt} = 0$ and $\frac{dx}{dt} \neq 0$ |
| Vertical Tangent | $\frac{dy}{dx}$ is infinite | $\frac{dx}{dt} = 0$ and $\frac{dy}{dt} \neq 0$ |
For $x = t^2 - 4$, $y = t^3 - 3t$, find where the curve has horizontal and vertical tangents.
①
Find derivatives
$\frac{dx}{dt} = 2t$, $\frac{dy}{dt} = 3t^2 - 3 = 3(t^2 - 1)$
②
Horizontal tangents: $\frac{dy}{dt}=0$
$3(t^2 - 1)=0$ → $t = \pm 1$
$t=1$: $(x,y)=(-3, -2)$
$t=-1$: $(x,y)=(-3, 2)$
③
Vertical tangents: $\frac{dx}{dt}=0$
$2t = 0$ → $t = 0$
$t=0$: $(x,y)=(-4, 0)$
📌 Second Derivative
$$ \frac{d^2y}{dx^2} = \frac{\frac{d}{dt}\left(\frac{dy}{dx}\right)}{\frac{dx}{dt}} $$
Used to find concavity of parametric curves
For $x = t^2$, $y = t^3$, find $\frac{d^2y}{dx^2}$.
①
Find $\frac{dy}{dx}$
$\frac{dy}{dx} = \frac{3t^2}{2t} = \frac{3t}{2}$
②
Differentiate with respect to $t$
$\frac{d}{dt}\left(\frac{dy}{dx}\right) = \frac{d}{dt}\left(\frac{3t}{2}\right) = \frac{3}{2}$
③
Divide by $\frac{dx}{dt} = 2t$
$\frac{d^2y}{dx^2} = \frac{3/2}{2t} = \frac{3}{4t}$ (for $t \neq 0$)
⚠️ Common Mistakes
- Forgetting the Chain Rule: $\frac{dy}{dx} = \frac{dy/dt}{dx/dt}$, not $\frac{dy}{dt} \cdot \frac{dx}{dt}$.
- Horizontal vs. Vertical tangents: Horizontal when $dy/dt = 0$, vertical when $dx/dt = 0$ (provided the other is nonzero).
- Second derivative formula: $\frac{d^2y}{dx^2} = \frac{d/dt(dy/dx)}{dx/dt}$, not $\frac{d^2y/dt^2}{d^2x/dt^2}$.
🔍 Key Takeaways
- $\frac{dy}{dx} = \frac{dy/dt}{dx/dt}$.
- Horizontal tangents occur when $\frac{dy}{dt} = 0$ and $\frac{dx}{dt} \neq 0$.
- Vertical tangents occur when $\frac{dx}{dt} = 0$ and $\frac{dy}{dt} \neq 0$.
- Second derivative: $\frac{d^2y}{dx^2} = \frac{d/dt(dy/dx)}{dx/dt}$.
← Back to Module Page