AP Calculus BC

Parametric vs Cartesian Second Derivative

Q: How do I find where a parametric curve is concave up?

Compute \frac{d^2y}{dx^2} with the correct formula and find where it is positive. In practice that means comparing the sign of \frac{d}{dt}\left(\frac{dy}{dx}\right) with the sign of \frac{dx}{dt}, since concavity is up exactly when those two agree.

Q: What happens at a value of t where dx/dt is zero?

Both formulas break down there, since each divides by \frac{dx}{dt}. If \frac{dy}{dt} \ne 0 at that t, the curve has a vertical tangent and \frac{dy}{dx} genuinely is undefined. If \frac{dy}{dt} = 0 as well, the expression is 0/0 and the derivative may still exist: x = t^3, y = t^6 is just y = x^2, perfectly smooth at t = 0. So the point may be a cusp or an ordinary point, and you have to look at the limit to tell. Either way it splits the domain, so test concavity separately on each side.

The parametric second derivative is the t derivative of dy/dx, divided again by dx/dt. It is not the second t derivative of y over the second t derivative of x. In Cartesian form you differentiate twice in x and stop; in parametric form every step down costs another division by dx/dt.

Cartesian second derivative

Use when: The curve is given as y in terms of x, so differentiating the first derivative once more finishes the job.

Parametric second derivative

Use when: The curve is given by x and y as separate functions of a parameter, so concavity has to be built from the first derivative and divided by dx/dt.

Side by side

	Cartesian form	Parametric form
First derivative	$\frac{dy}{dx} = f'(x)$	$\frac{dy}{dx} = \frac{dy/dt}{dx/dt}$
Second derivative	$\frac{d^2y}{dx^2} = f''(x)$	$\frac{d^2y}{dx^2} = \dfrac{\frac{d}{dt}\left(\frac{dy}{dx}\right)}{\frac{dx}{dt}}$
What you differentiate at step two	The function $f'(x)$ , with respect to $x$	The quotient $\frac{dy/dt}{dx/dt}$ , with respect to $t$
Extra work	None beyond a second differentiation	One more division by $\frac{dx}{dt}$
Result for $x = t^2$ , $y = t^3$	$y = x^{3/2}$ , so $\frac{d^2y}{dx^2} = \frac{3}{4\sqrt{x}}$	$\frac{3/2}{2t} = \frac{3}{4t}$ , the same thing since $\sqrt{x} = t$

One rule carries the whole topic: differentiating with respect to $x$ means differentiating with respect to $t$ and dividing by $\frac{dx}{dt}$ . Apply it to $y$ and you get $\frac{dy}{dx} = \frac{dy/dt}{dx/dt}$ . Apply it a second time, now to the function $\frac{dy}{dx}$ rather than to $y$ , and you get concavity. The division by $\frac{dx}{dt}$ happens twice, once at each stage.

\frac{d^2y}{dx^2} = \frac{\dfrac{d}{dt}\!\left(\dfrac{dy}{dx}\right)}{\dfrac{dx}{dt}} \qquad \text{not} \qquad \frac{d^2y/dt^2}{d^2x/dt^2}

Test it on $x = t^2$ , $y = t^3$ with $t > 0$ . First, $\frac{dy}{dx} = \frac{3t^2}{2t} = \frac{3t}{2}$ . Differentiating that in $t$ gives $\frac{3}{2}$ , and dividing by $\frac{dx}{dt} = 2t$ gives $\frac{d^2y}{dx^2} = \frac{3}{4t}$ . Eliminating the parameter confirms it, since the curve is $y = x^{3/2}$ with second derivative $\frac{3}{4\sqrt{x}}$ and $\sqrt{x} = t$ . The tempting alternative would give $\frac{6t}{2} = 3t$ , which reports $3$ at $t = 1$ where the answer is $\frac{3}{4}$ .

The version that looks right and is not

$\frac{d^2y}{dx^2} \ne \frac{d^2y/dt^2}{d^2x/dt^2}$ . The first derivative really is a quotient of $t$ derivatives, so the second one looks as though it should be one too, and that guess is the most common parametric slip on the BC exam. Differentiate $\frac{dy}{dx}$ with respect to $t$ , then divide by $\frac{dx}{dt}$ one more time.

Frequently asked questions

Why do I divide by dx/dt a second time?

Because the second derivative is still a rate with respect to $x$ . You have found $\frac{d}{dt}\left(\frac{dy}{dx}\right)$ , which measures change per unit of $t$ , and converting any $t$ rate into an $x$ rate means dividing by $\frac{dx}{dt}$ . The same conversion is what produced $\frac{dy}{dx}$ in the first place.

How do I find where a parametric curve is concave up?

Compute $\frac{d^2y}{dx^2}$ with the correct formula and find where it is positive. In practice that means comparing the sign of $\frac{d}{dt}\left(\frac{dy}{dx}\right)$ with the sign of $\frac{dx}{dt}$ , since concavity is up exactly when those two agree.

What happens at a value of t where dx/dt is zero?

Both formulas break down there, since each divides by $\frac{dx}{dt}$ . If $\frac{dy}{dt} \ne 0$ at that $t$ , the curve has a vertical tangent and $\frac{dy}{dx}$ genuinely is undefined. If $\frac{dy}{dt} = 0$ as well, the expression is $0/0$ and the derivative may still exist: $x = t^3$ , $y = t^6$ is just $y = x^2$ , perfectly smooth at $t = 0$ . So the point may be a cusp or an ordinary point, and you have to look at the limit to tell. Either way it splits the domain, so test concavity separately on each side.

In the CED: Unit 9: Parametric, Polar, and Vector (BC)