Second Derivatives of Parametric Equations
The second derivative of a parametric curve
- We found $\tfrac{dy}{dx}$ for a parametric curve. Now: how curved is it? That needs $\tfrac{d^2y}{dx^2}$.
- The catch: you can't just divide the second derivatives in $t$.
- Instead, differentiate the first slope $\tfrac{dy}{dx}$ with respect to $t$, then divide by $\tfrac{dx}{dt}$ again.
- It's the parametric slope rule, applied a second time to $\tfrac{dy}{dx}$.
The correct formula
- Treat the first derivative $\tfrac{dy}{dx}$ as a new function of $t$, and take its parametric slope:
-
$$\frac{d^2y}{dx^2}=\frac{\dfrac{d}{dt}\!\left(\dfrac{dy}{dx}\right)}{\dfrac{dx}{dt}}$$
- Differentiate $\tfrac{dy}{dx}$ with respect to $t$, then divide by $\tfrac{dx}{dt}$ (same denominator as before).
- The concavity's sign comes from this, just like an ordinary $f''$.
Concavity of a parametric curve
y = ax³ + bx
The parametric second derivative tells concavity — computed by differentiating $\tfrac{dy}{dx}$ in $t$, then dividing by $\tfrac{dx}{dt}$.
The parametric second derivative $\dfrac{d^2y}{dx^2}$ equals...
Differentiate $\tfrac{dy}{dx}$ in $t$, then divide by $\tfrac{dx}{dt}$.
Order the steps to find the parametric second derivative.
Slope, differentiate in $t$, divide by $\tfrac{dx}{dt}$.
Why it isn't the obvious ratio
- It is not $\dfrac{d^2y/dt^2}{d^2x/dt^2}$ — that shortcut is wrong.
- $\tfrac{dy}{dx}$ is already a quotient of $t$-derivatives, so differentiating it needs the quotient rule (in $t$), not a naive second-derivative ratio.
- The right process always goes through $\tfrac{d}{dt}\big(\tfrac{dy}{dx}\big)$ first.
- Skipping that gives the classic wrong answer.
The parametric second derivative equals $\dfrac{d^2y/dt^2}{d^2x/dt^2}$.
That naive ratio is wrong.
The two-step routine
- 1. Find $\tfrac{dy}{dx}=\dfrac{dy/dt}{dx/dt}$ and simplify.
- 2. Differentiate that with respect to $t$, then divide by $\tfrac{dx}{dt}$.
- Concave up where $\tfrac{d^2y}{dx^2}>0$, concave down where $<0$ — same as always.
- Keep the same denominator $\tfrac{dx}{dt}$ in both steps.
For $x=t^2$, $y=t^3$ ($\tfrac{dy}{dx}=\tfrac{3t}{2}$), $\dfrac{d^2y}{dx^2}=$
$\frac{d}{dt}(\tfrac{3t}{2})=\tfrac32$; divide by $2t$: $\tfrac{3}{4t}$.
After differentiating $\tfrac{dy}{dx}$ in $t$, you divide by $\dfrac{dx}{dt}$, the ____ denominator as the first derivative.
Both steps divide by $\tfrac{dx}{dt}$.
The sign of the parametric second derivative tells you the curve's...
Positive → concave up, negative → concave down.
The second derivative is not $\dfrac{d^2y/dt^2}{d^2x/dt^2}$. You must differentiate the first derivative $\tfrac{dy}{dx}$ with respect to $t$ and then divide by $\tfrac{dx}{dt}$ again. Using the naive ratio of second derivatives is the single most common parametric mistake.
For $x=t^2,\ y=t^3$ (so $\tfrac{dy}{dx}=\tfrac{3t}{2}$), find $\tfrac{d^2y}{dx^2}$.
- $\dfrac{d}{dt}\!\left(\dfrac{3t}{2}\right)=\dfrac{3}{2}$.
- Divide by $\dfrac{dx}{dt}=2t$: $\dfrac{d^2y}{dx^2}=\dfrac{3/2}{2t}=\dfrac{3}{4t}$.
- (Not $\tfrac{6t}{2}=3t$, which the wrong ratio would give.)
The parametric second derivative is $\dfrac{d^2y}{dx^2}=\dfrac{\frac{d}{dt}\left(\frac{dy}{dx}\right)}{\frac{dx}{dt}}$ — differentiate the first slope $\tfrac{dy}{dx}$ in $t$, then divide by $\tfrac{dx}{dt}$ again. It is not $\frac{d^2y/dt^2}{d^2x/dt^2}$. Its sign gives concavity as usual.