Parametric Derivatives — AP® Calculus BC Notes

A curve given by parametric equations $x = x(t)$ , $y = y(t)$ is the path of a moving point. Calculus asks three questions about it at each instant: how fast is $x$ changing, how fast is $y$ changing, and how steep is the path? The first two are derivatives you already know how to take. The third needs one new idea.

Two rates and a direction

Differentiate each coordinate with respect to $t$ . $\frac{dx}{dt}$ is the horizontal rate and $\frac{dy}{dt}$ the vertical rate of the point. Their signs tell you which way it moves:

Sign	Meaning
$\frac{dx}{dt} > 0$	$x$ is increasing: the point moves right
$\frac{dx}{dt} < 0$	$x$ is decreasing: the point moves left
$\frac{dy}{dt} > 0$	$y$ is increasing: the point moves up
$\frac{dy}{dt} < 0$	$y$ is decreasing: the point moves down

Every rule from Units 2 and 3 applies, one coordinate at a time: the power, product, quotient and chain rules all work exactly as before, with $t$ as the variable.

The slope of the curve

If $y$ can be thought of as a function of $x$ near a point, the chain rule says

$\frac{dy}{dt} = \frac{dy}{dx}\cdot\frac{dx}{dt}.$

Solve for $\frac{dy}{dx}$ :

$\frac{dy}{dx} = \frac{dy/dt}{dx/dt}, \qquad \text{provided } \frac{dx}{dt} \ne 0.$

In words: the slope is the vertical rate over the horizontal rate. If the point rises 6 units per second while it moves right 2 units per second, the path rises 3 units per unit of $x$ . Notice that $\frac{dy}{dx}$ comes out as a function of $t$ . You evaluate it at a time, not at a point.

Worked example 1. A curve is given by $x = 2t + 1$ , $y = t^2 - 4t$ . Find the line tangent to the curve at $t = 3$ .

Rates: $\frac{dx}{dt} = 2$ and $\frac{dy}{dt} = 2t - 4$ .

Slope: $\frac{dy}{dx}$ $= \frac{2t - 4}{2}$ $= t - 2$ , which is $1$ at $t = 3$ .

Point: $(x(3), y(3))$ $= (7, 9 - 12)$ $= (7, -3)$ . Use the same $t$ for the point and the slope.

Tangent line: $y + 3 = 1(x - 7)$ , that is, $y = x - 10$ .

Check by eliminating $t$ : $t = \frac{x - 1}{2}$ , so $y$ $= \frac{(x - 1)^2}{4} - 2(x - 1)$ and $\frac{dy}{dx} = \frac{x - 1}{2} - 2$ . At $x = 7$ this is $3 - 2 = 1$ . The parametric route was faster and needed no elimination.

Horizontal and vertical tangents

A fraction is 0 when its numerator is 0, and it is undefined when its denominator is 0. So:

Horizontal tangent: $\frac{dy}{dt} = 0$ and $\frac{dx}{dt} \ne 0$ . For an instant the point moves straight across.
Vertical tangent: $\frac{dx}{dt} = 0$ and $\frac{dy}{dt} \ne 0$ . For an instant the point moves straight up or down.

In worked example 1, $\frac{dy}{dt} = 2t - 4 = 0$ at $t = 2$ , while $\frac{dx}{dt} = 2$ . The tangent is horizontal at $(5, -4)$ , the lowest point of the curve.

When both rates are 0 at the same time, the point stops for an instant and the quotient tells you nothing. The curve may have a sharp corner there or a perfectly smooth tangent; you have to look more closely, so state that the test does not apply.

The curve x = t squared, y = t cubed minus 3t for t from about -2 to 2, a loop that crosses itself at (3, 0), with arrows for the direction of increasing t. At (3, 0) two dashed tangent lines are drawn: slope square root of 3 when t = square root of 3, and slope minus square root of 3 when t = minus square root of 3. Short horizontal tangent segments mark (1, 2) at t = -1 and (1, -2) at t = 1, and a vertical tangent marks the origin at t = 0. — **Figure 1.** One point, two slopes: why parametric slopes are found from $t$ . LevlPrep original diagram.

The figure shows why "evaluate at a time" matters. The curve $x = t^2$ , $y = t^3 - 3t$ passes through $(3, 0)$ twice. Its slope is $\frac{3t^2 - 3}{2t}$ , which is $\sqrt{3}$ at $t = \sqrt{3}$ and $-\sqrt{3}$ at $t = -\sqrt{3}$ : two tangent lines at one point. A formula in $x$ could never give both.

Curves given by a table

On the exam, $x(t)$ and $y(t)$ are often known only through a table of values and derivatives. The same rules apply to the numbers.

Worked example 2. At $t = 4$ , a curve has $x(4) = 2$ , $y(4) = -1$ , $x'(4) = -5$ and $y'(4) = 10$ . Find the tangent line and the direction of motion at $t = 4$ .

Slope: $\frac{y'(4)}{x'(4)}$ $= \frac{10}{-5}$ $= -2$ . Tangent line: $y + 1 = -2(x - 2)$ .

Direction: $x'(4) < 0$ , so the point moves left; $y'(4) > 0$ , so it moves up.

Common slips

Slip	Fix
Using $\frac{dy}{dt}$ as the slope	Divide by $\frac{dx}{dt}$ : the slope is change in $y$ per unit of $x$
Dividing upside down	$\frac{dy/dt}{dx/dt}$ : the $y$ rate on top, as in $\frac{dy}{dx}$
Using $t$ as the $x$ -coordinate	The point is $(x(t_0), y(t_0))$ ; compute both
Calling $\frac{dx}{dt} = 0$ horizontal	No horizontal motion means a vertical tangent

On calculator questions, store $x'(t_0)$ and $y'(t_0)$ before dividing, and give the slope to three decimal places, with the quotient written out as your setup.