Second Derivatives Parametric
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
The first derivative $dy/dx$ gives the slope of the tangent to a parametric curve. The second derivative $d^2y/dx^2$ gives information about concavity. But for a parametric curve, $d^2y/dx^2$ is not simply $\dfrac{d^2y/dt^2}{d^2x/dt^2}$ -- that is the most common error in this topic. The correct formula differentiates $dy/dx$ (itself a function of $t$) with respect to $x$, applying the chain rule a second time.
Quick Reference
$$\frac{d^2y}{dx^2} = \frac{\dfrac{d}{dt}\!\left(\dfrac{dy}{dx}\right)}{\dfrac{dx}{dt}}.$$
Step-by-step:
- Compute $\dfrac{dy}{dx} = \dfrac{dy/dt}{dx/dt}$ (this is a function of $t$).
- Differentiate it with respect to $t$: $\dfrac{d}{dt}\!\left(\dfrac{dy}{dx}\right)$.
- Divide by $\dfrac{dx}{dt}$.
Concavity: The parametric curve is concave up when $d^2y/dx^2 > 0$, concave down when $d^2y/dx^2 < 0$.
Key Concepts
1. Why Not $(d^2y/dt^2)/(d^2x/dt^2)$?
The notation $d^2y/dx^2$ means “differentiate $dy/dx$ with respect to $x$.” Using the chain rule: $$\frac{d^2y}{dx^2} = \frac{d}{dx}\!\left(\frac{dy}{dx}\right) = \frac{d}{dt}\!\left(\frac{dy}{dx}\right)\cdot\frac{dt}{dx} = \frac{\frac{d}{dt}(dy/dx)}{dx/dt}.$$
The ratio $(d^2y/dt^2)/(d^2x/dt^2)$ is a different quantity and is generally wrong for the second derivative of $y$ with respect to $x$.
2. Procedure in Practice
Let $p(t) = \dfrac{dy}{dx} = \dfrac{g'(t)}{f'(t)}$ (already computed from the first derivative step). Then: $$\frac{d^2y}{dx^2} = \frac{p'(t)}{f'(t)}.$$
Differentiate $p(t)$ with respect to $t$ using the quotient rule: $$p'(t) = \frac{g''(t)f'(t) - g'(t)f''(t)}{[f'(t)]^2}.$$
So: $$\frac{d^2y}{dx^2} = \frac{g''f' - g'f''}{(f')^3}.$$
Worked Example
For $x = t^2 - 1$, $y = t^3 - 3t$, find $d^2y/dx^2$ and determine where the curve is concave up.
Step 1 -- First derivative (from previous example): $$\frac{dy}{dx} = \frac{3(t^2-1)}{2t} = \frac{3t^2 - 3}{2t}.$$
Step 2 -- Differentiate with respect to $t$:
Let $p(t) = \dfrac{3t^2-3}{2t}$. Use the quotient rule: $$p'(t) = \frac{6t\cdot 2t - (3t^2-3)\cdot 2}{4t^2} = \frac{12t^2 - 6t^2 + 6}{4t^2} = \frac{6t^2+6}{4t^2} = \frac{3(t^2+1)}{2t^2}.$$
Step 3 -- Divide by $dx/dt = 2t$:
$$\frac{d^2y}{dx^2} = \frac{p'(t)}{2t} = \frac{3(t^2+1)}{2t^2\cdot 2t} = \frac{3(t^2+1)}{4t^3}.$$
Concavity: $d^2y/dx^2 > 0$ iff $\dfrac{3(t^2+1)}{4t^3} > 0$ iff $t^3 > 0$ iff $t > 0$.
The curve is concave up for $t > 0$ (which corresponds to the right half of the curve in $x$).
Boxed answers:
- $\dfrac{d^2y}{dx^2} = \dfrac{3(t^2+1)}{4t^3}$
- Concave up when $t > 0$.
Common Errors Summary
| Error | Example | Correction |
|---|---|---|
| Using $(d^2y/dt^2)/(d^2x/dt^2)$ | Computing $d^2y/dx^2 = 6t / 2 = 3t$ | The correct formula is $(d/dt)(dy/dx)$ divided by $dx/dt$; the second $t$-derivatives go in the numerator of $d/dt(dy/dx)$, not directly |
| Forgetting to divide by $dx/dt$ | Reporting $p'(t)$ as the second derivative | $d^2y/dx^2 = p'(t)/(dx/dt)$; the extra division by $dx/dt$ is required |
| Sign errors in the quotient rule for $p'(t)$ | Dropping a term in $p'(t)$ | Write out the quotient rule carefully: $(uv' - vu')/v^2$ where $u = dy/dt$, $v = dx/dt$ |
Common Misconceptions
$d^2y/dx^2 = (d^2y/dt^2)/(d^2x/dt^2)$.
This is the composition-is-not-chaining error at the second level. The notation $d^2y/dx^2$ means “differentiate $dy/dx$ with respect to $x$,” not “take the ratio of second $t$-derivatives.” Applying the chain rule correctly gives $d^2y/dx^2 = \frac{d}{dt}(dy/dx)/(dx/dt)$, where the numerator differentiates the first-derivative expression with respect to $t$. For $x = t^2$, $y = t^3$, the first derivative is $3t/2$ and the correct second derivative is $\frac{d}{dt}(3t/2)/(2t) = (3/2)/(2t) = 3/(4t)$, not $6t/(2) = 3t$.
Leveled Practice
Level 1 -- Compute the Second Derivative
Problem 1. For $x = t$, $y = t^2$, compute $d^2y/dx^2$ directly. Verify using the $y = x^2$ formula.
Show answer
$dx/dt = 1$, $dy/dt = 2t$.
$dy/dx = 2t/1 = 2t$.
$d/dt(dy/dx) = 2$.
$d^2y/dx^2 = 2/1 = 2$.
Verification: $y = x^2 \Rightarrow dy/dx = 2x = 2t \Rightarrow d^2y/dx^2 = 2$. Confirmed.
Level 2 -- Concavity Analysis
Problem 2. For $x = \cos t$, $y = \sin t$, compute $d^2y/dx^2$ and interpret.
Show answer
$dx/dt = -\sin t$, $dy/dt = \cos t$.
$dy/dx = \cos t / (-\sin t) = -\cot t$.
$d/dt(-\cot t) = \csc^2 t$.
$d^2y/dx^2 = \csc^2 t / (-\sin t) = -\csc^3 t / 1 = -1/\sin^3 t$.
Since $\csc^3 t > 0$ for $\sin t > 0$ (upper half of circle) and $\csc^3 t < 0$ for $\sin t < 0$ (lower half), we get $d^2y/dx^2 < 0$ always. The unit circle is always concave toward the origin (concave down as a function of $x$). Makes sense: the circle curves away from the tangent line on both the upper and lower arcs.
Mastery Checklist
Mental Model
The second derivative $d^2y/dx^2$ measures how the slope $dy/dx$ changes as you move along the curve (in $x$). For a parametric curve, both slope and horizontal position change with $t$. The slope changes at rate $p'(t)$ per unit $t$, while $x$ changes at rate $dx/dt$ per unit $t$. The rate of change of slope with respect to $x$ is $p'(t)/(dx/dt)$ -- the chain rule applied a second time.
Connections
Looking back
- Parametric derivatives (Section 10.2): $dy/dx$ computed there is the input to this formula.
- Concavity (Section 3.3): $d^2y/dx^2 > 0$ means concave up, just as for ordinary functions.