Second Derivatives of
Parametric Equations
Master the technique for finding concavity and curvature of parametrically defined curves using \(\frac{d^2y}{dx^2}\).
AP Calculus BC
What Are Parametric Equations?
Instead of defining \( y \) directly as a function of \( x \), parametric equations define both coordinates in terms of a third variable called the parameter, typically \( t \):
This representation is powerful for modeling motion, projectile paths, and curves that fail the vertical line test.
Review: First Derivative \(\frac{dy}{dx}\)
Before tackling the second derivative, recall the chain rule approach to the first derivative of a parametric curve:
This gives the slope of the tangent line at any point on the curve. The second derivative extends this to measure how that slope changes.
Why Does \(\frac{d^2y}{dx^2}\) Matter?
Concavity
Determines whether the curve bends upward or downward at a given point.
Inflection Points
Identifies where the curve changes concavity: where \(\frac{d^2y}{dx^2} = 0\) or is undefined.
Motion Analysis
In physics, this relates to the acceleration component perpendicular to the velocity vector.
The Core Formula
In words: take the derivative with respect to \( t \) of the first derivative \(\frac{dy}{dx}\), then divide by \(\frac{dx}{dt}\). This is the chain rule applied a second time.
Deriving the Formula
1
Start with the first derivative: \(\frac{dy}{dx} = \frac{y'(t)}{x'(t)}\), which is itself a function of \( t \).
2
Apply the definition: \(\frac{d^2y}{dx^2} = \frac{d}{dx}\!\left(\frac{dy}{dx}\right)\). Since everything depends on \( t \), use the chain rule: \(\frac{d}{dx} = \frac{1}{dx/dt} \cdot \frac{d}{dt}\).
3
Combine: \(\frac{d^2y}{dx^2} = \frac{\frac{d}{dt}\!\left(\frac{dy}{dx}\right)}{\frac{dx}{dt}}\). This is the formula.
Visual Interpretation
Right vs. Wrong Formula
| ✓ Correct | ✕ Incorrect | |
|---|---|---|
| Formula | \(\dfrac{\frac{d}{dt}\!\left(\frac{dy}{dx}\right)}{\frac{dx}{dt}}\) | \(\dfrac{\frac{d^2y}{dt^2}}{\frac{d^2x}{dt^2}}\) |
| Meaning | Rate of change of slope w.r.t. \(x\) | Ratio of separate second derivatives |
| Uses chain rule? | Yes | No |
| Gives concavity? | Yes | No |
How to Apply the Formula
1
Find \(\frac{dx}{dt}\) and \(\frac{dy}{dt}\): Differentiate both parametric equations with respect to \( t \).
2
Compute \(\frac{dy}{dx}\): Form the ratio \(\frac{dy/dt}{dx/dt}\). Simplify.
3
Differentiate \(\frac{dy}{dx}\) with respect to \( t \): Use quotient rule or simplify first, then differentiate.
4
Divide by \(\frac{dx}{dt}\): This final division completes the formula: \(\frac{d^2y}{dx^2} = \frac{\frac{d}{dt}(dy/dx)}{dx/dt}\).
5
Interpret the sign: Positive means concave up, negative means concave down.
Example 1: Basic Polynomial
1
Derivatives: \(\frac{dx}{dt} = 2t\), \(\frac{dy}{dt} = 3t^2\)
2
First derivative: \(\frac{dy}{dx} = \frac{3t^2}{2t} = \frac{3t}{2}\)
3
Differentiate w.r.t. \(t\): \(\frac{d}{dt}\!\left(\frac{3t}{2}\right) = \frac{3}{2}\)
4
Divide by \(\frac{dx}{dt}\): \(\frac{d^2y}{dx^2} = \frac{3/2}{2t} = \frac{3}{4t}\)
Example 2: Trigonometric Curve
1
Derivatives: \(\frac{dx}{dt} = -\sin t\), \(\frac{dy}{dt} = \cos t\)
2
First derivative: \(\frac{dy}{dx} = \frac{\cos t}{-\sin t} = -\cot t\)
3
Differentiate w.r.t. \(t\): \(\frac{d}{dt}(-\cot t) = \csc^2 t\)
4
Divide by \(\frac{dx}{dt}\): \(\frac{d^2y}{dx^2} = \frac{\csc^2 t}{-\sin t} = -\csc^3 t\)
Unit Circle: Concavity Analysis
Example 3: Quotient Rule Required
1
Derivatives: \(\frac{dx}{dt} = 1 + \frac{1}{t} = \frac{t+1}{t}\), \(\frac{dy}{dt} = 1 - \frac{1}{t} = \frac{t-1}{t}\)
2
First derivative: \(\frac{dy}{dx} = \frac{(t-1)/t}{(t+1)/t} = \frac{t-1}{t+1}\)
3
Quotient rule on \(\frac{t-1}{t+1}\): \(\frac{d}{dt}\!\left(\frac{t-1}{t+1}\right) = \frac{(t+1)(1) - (t-1)(1)}{(t+1)^2} = \frac{2}{(t+1)^2}\)
4
Divide: \(\frac{d^2y}{dx^2} = \frac{2/(t+1)^2}{(t+1)/t} = \frac{2t}{(t+1)^3}\)
Finding Inflection Points
An inflection point occurs where the concavity changes. For parametric curves, this means finding values of \( t \) where \(\frac{d^2y}{dx^2}\) changes sign.
1
Set the numerator of \(\frac{d^2y}{dx^2}\) equal to zero and solve for \( t \).
2
Check for sign changes on either side of each candidate \( t \)-value.
3
Plug the \( t \)-value back into \( x(t) \) and \( y(t) \) to get the actual point on the curve.
Common Mistakes to Avoid
✕ Wrong Ratio
Using \(\frac{d^2y/dt^2}{d^2x/dt^2}\) instead of the correct chain rule formula. This gives a completely different (and wrong) answer.
✕ Forgetting the Denominator
Computing \(\frac{d}{dt}(\frac{dy}{dx})\) but forgetting to divide by \(\frac{dx}{dt}\). The result has the wrong units and meaning.
✕ Quotient Rule Errors
Making algebraic errors in the quotient rule when differentiating \(\frac{dy}{dx}\) with respect to \( t \). Always double-check sign and order.
Summary of Derivatives
| Quantity | Formula | Tells You |
|---|---|---|
| \(\frac{dy}{dx}\) | \(\dfrac{dy/dt}{dx/dt}\) | Slope of tangent line |
| \(\frac{d^2y}{dx^2}\) | \(\dfrac{\frac{d}{dt}(dy/dx)}{dx/dt}\) | Concavity of curve |
| Concave up | \(\frac{d^2y}{dx^2} > 0\) | Curve bends upward |
| Concave down | \(\frac{d^2y}{dx^2} < 0\) | Curve bends downward |
| Inflection point | \(\frac{d^2y}{dx^2}\) changes sign | Concavity reversal |
Question 1 of 5
Given \( x = t^2 \) and \( y = t^3 \), what is \(\frac{d^2y}{dx^2}\)?
We found \(\frac{dy}{dx} = \frac{3t}{2}\), so \(\frac{d}{dt}\!\left(\frac{3t}{2}\right) = \frac{3}{2}\). Dividing by \(\frac{dx}{dt} = 2t\) gives \(\frac{3}{4t}\). Choice (a) forgets to divide by \(2t\), choice (d) incorrectly uses \(\frac{d^2y/dt^2}{d^2x/dt^2}\).
Question 2 of 5
For the unit circle \( x = \cos t, \; y = \sin t \), which expression equals \(\frac{d^2y}{dx^2}\)?
\(\frac{dy}{dx} = -\cot t\). Then \(\frac{d}{dt}(-\cot t) = \csc^2 t\). Dividing by \(\frac{dx}{dt} = -\sin t\) gives \(\frac{\csc^2 t}{-\sin t} = -\csc^3 t\). Choice (a) is the first derivative, and (b) forgets the final division.
Question 3 of 5
A student claims: "To find \(\frac{d^2y}{dx^2}\) for parametric equations, just compute \(\frac{d^2y/dt^2}{d^2x/dt^2}\)." This is:
The ratio \(\frac{d^2y/dt^2}{d^2x/dt^2}\) does not equal \(\frac{d^2y}{dx^2}\) in general. The correct formula requires the chain rule: differentiate \(\frac{dy}{dx}\) with respect to \(t\), then divide by \(\frac{dx}{dt}\). Choice (d) is tempting, but when \(\frac{dx}{dt}\) is constant, \(\frac{d^2x}{dt^2} = 0\), making that ratio undefined.
Question 4 of 5
If \( x = e^t \) and \( y = e^{2t} \), what is \(\frac{d^2y}{dx^2}\)?
\(\frac{dx}{dt} = e^t\), \(\frac{dy}{dt} = 2e^{2t}\), so \(\frac{dy}{dx} = \frac{2e^{2t}}{e^t} = 2e^t\). Then \(\frac{d}{dt}(2e^t) = 2e^t\). Finally, \(\frac{d^2y}{dx^2} = \frac{2e^t}{e^t} = 2\). This makes sense: since \(y = x^2\), the second derivative \(\frac{d^2y}{dx^2} = 2\) is constant.
Question 5 of 5
For \( x = e^t \) and \( y = t^2 \), what is \(\frac{d^2y}{dx^2}\)?
\(\frac{dx}{dt} = e^t\), \(\frac{dy}{dt} = 2t\), so \(\frac{dy}{dx} = \frac{2t}{e^t}\). Using the quotient rule: \(\frac{d}{dt}\!\left(\frac{2t}{e^t}\right) = \frac{2e^t - 2te^t}{e^{2t}} = \frac{2(1-t)}{e^t}\). Dividing by \(\frac{dx}{dt} = e^t\): \(\frac{d^2y}{dx^2} = \frac{2(1-t)}{e^{2t}}\). Choice (a) forgets the \((1-t)\) factor from the quotient rule. Choice (c) is the first derivative, not the second.