Second Derivatives of Parametric Equations

AP Calculus BC · Parametric, Polar & Vector-Valued Functions

Second Derivatives of Parametric Equations

Differentiate the slope with respect to t — then divide by dx/dt one more time.

key idea$\frac{d^{2}y}{dx^{2}}=\frac{\dfrac{d}{dt}\!\left(\dfrac{dy}{dx}\right)}{\dfrac{dx}{dt}}$

The rule

The second derivative is not the ratio of the second derivatives. It is the first derivative of $\dfrac{dy}{dx}$ — and since that expression is a function of $t$, differentiating it produces a $t$-derivative that must then be converted back by dividing by $\dfrac{dx}{dt}$ one more time.

Expression	Correct?	Why
$\dfrac{\frac{d}{dt}\left(\frac{dy}{dx}\right)}{\frac{dx}{dt}}$	yes	the definition, converted back to $x$
$\dfrac{d^{2}y/dt^{2}}{d^{2}x/dt^{2}}$	no	the pattern does not repeat at second order
$\dfrac{d}{dt}\left(\dfrac{dy}{dx}\right)$	no	this is $\dfrac{d}{dt}$, not $\dfrac{d}{dx}$ — one division missing
$\dfrac{d}{dx}\left(\dfrac{dy}{dx}\right)$	yes	the same thing, written before the conversion

Row 2 is the error worth naming out loud: the ratio trick works once and only once. At second order you must differentiate the slope expression and divide by $\dfrac{dx}{dt}$ again.

Step 1find $\dfrac{dy}{dx}$ as a function of $t$

Step 2differentiate it with respect to $t$

Step 3divide by $\dfrac{dx}{dt}$ again

$\dfrac{dy}{dx}$The slope, expressed as a function of the parameter — not of $x$.

$\dfrac{d}{dt}\left(\dfrac{dy}{dx}\right)$How fast the slope is changing per unit of $t$. An intermediate quantity, never the final answer.

$\dfrac{d^{2}y}{dx^{2}}$How fast the slope changes per unit of $x$ — this is what determines concavity.

Concave up$\dfrac{d^{2}y}{dx^{2}}\gt 0$. The sign of the intermediate $t$-derivative alone does not settle this.

Pattern recognition

four habits

Compute $\dfrac{dy}{dx}$ first and simplify it before differentiating — the algebra afterwards is much lighter.
The quotient rule usually appears in step 2, because $\dfrac{dy}{dx}$ is itself a fraction in $t$.
Divide by $\dfrac{dx}{dt}$, not by $\dfrac{d^{2}x}{dt^{2}}$.
Concavity questions are answered by the sign of the finished $\dfrac{d^{2}y}{dx^{2}}$, at the relevant $t$.

Worked examples

EX 1

The three-step computation

$x=t^{2}$, $y=t^{3}$

For $x=t^{2}$ and $y=t^{3}$, find $\dfrac{d^{2}y}{dx^{2}}$.

First derivative

$\dfrac{dx}{dt}=2t$, $\dfrac{dy}{dt}=3t^{2}$, so $\dfrac{dy}{dx}=\dfrac{3t^{2}}{2t}=\dfrac{3t}{2}$.

Simplify before differentiating

$\dfrac{3t}{2}$ is already simple — this is why step 1 should always be reduced first.

Differentiate with respect to $t$

$\dfrac{d}{dt}\left(\dfrac{3t}{2}\right)=\dfrac32$.

Divide by $\dfrac{dx}{dt}$

$\dfrac{3/2}{2t}=\dfrac{3}{4t}$.

Read the concavity

For $t\gt 0$ this is positive, so the curve is concave up there; for $t\lt 0$ it is concave down.

EX 2

The quotient rule appears

$x=t^{2}+1$, $y=t^{3}-t$

Find $\dfrac{d^{2}y}{dx^{2}}$ for $x=t^{2}+1$, $y=t^{3}-t$.

First derivative

$\dfrac{dy}{dx}=\dfrac{3t^{2}-1}{2t}$.

Rewrite to avoid the quotient rule

$=\dfrac{3t}{2}-\dfrac{1}{2t}=\dfrac32t-\dfrac12t^{-1}$.

Differentiate with respect to $t$

$\dfrac32+\dfrac12t^{-2}=\dfrac{3}{2}+\dfrac{1}{2t^{2}}$.

Divide by $\dfrac{dx}{dt}=2t$

$\dfrac{1}{2t}\left(\dfrac32+\dfrac{1}{2t^{2}}\right)=\dfrac{3}{4t}+\dfrac{1}{4t^{3}}$.

Note the tactic

Splitting the fraction in step 2 replaced the quotient rule with the power rule — less work and fewer sign errors.

EX 3

Concavity at a specific point

evaluate, then read the sign

For $x=\cos t$, $y=\sin t$ on $\left(0,\pi\right)$, determine the concavity at $t=\dfrac{\pi}{4}$.

First derivative

$\dfrac{dx}{dt}=-\sin t$, $\dfrac{dy}{dt}=\cos t$, so $\dfrac{dy}{dx}=-\cot t$.

Differentiate with respect to $t$

$\dfrac{d}{dt}\left(-\cot t\right)=\csc^{2}t$.

Divide by $\dfrac{dx}{dt}$

$\dfrac{\csc^{2}t}{-\sin t}=-\csc^{3}t$.

Evaluate at $t=\dfrac{\pi}{4}$

$\csc\dfrac{\pi}{4}=\sqrt2$, so the value is $-2\sqrt2\lt 0$.

Conclude

Concave down — as expected for the upper half of the unit circle.

EX 4

Diagnosing the classic error

why the shortcut fails

Show that $\dfrac{d^{2}y/dt^{2}}{d^{2}x/dt^{2}}$ gives the wrong answer for $x=t^{2}$, $y=t^{3}$.

Compute the second $t$-derivatives

$\dfrac{d^{2}x}{dt^{2}}=2$ and $\dfrac{d^{2}y}{dt^{2}}=6t$.

Form the tempting ratio

$\dfrac{6t}{2}=3t$.

Compare with the correct value

The true answer is $\dfrac{3}{4t}$ — not merely a different constant, but a completely different function.

Test one value

At $t=1$ the shortcut gives $3$ and the correct rule gives $\dfrac34$. At $t=2$: $6$ against $\dfrac38$. They even move in opposite directions.

Explain why

The first derivative absorbed one factor of $\dfrac{dx}{dt}$; the second derivative must absorb another. The naive ratio absorbs none.

Watch out

Divide by $\dfrac{dx}{dt}$, never by $\dfrac{d^{2}x}{dt^{2}}$

The conversion factor is the first derivative of $x$, at every order. This is the error the topic exists to prevent.

Simplify $\dfrac{dy}{dx}$ before differentiating

A reduced expression turns a quotient-rule computation into a power-rule one and removes most of the opportunity for error.

The intermediate $t$-derivative is not the answer

$\dfrac{d}{dt}\left(\dfrac{dy}{dx}\right)$ has the wrong units and the wrong sign behaviour. One more division is required.

Concavity comes from the finished expression

Read the sign of $\dfrac{d^{2}y}{dx^{2}}$ after the final division. The sign can flip at that step, because $\dfrac{dx}{dt}$ may be negative.

Course map. Topic 9.2 · LO CHA-3.G — calculate derivatives of parametric functions. BC-only.