Parametric Derivatives — AP® Calculus BC Lesson

Part 1 · Hook

Why this matters

A drone's flight plan gives its east position and its north position as functions of time. At any moment you can ask how fast it moves east, how fast it moves north, and how steep its path is. The first two are ordinary derivatives with respect to

t

. The third, the slope

\frac{dy}{dx}

of the path, comes from dividing one rate by the other, and that one division is the heart of Unit 9.

Part 2 · Before you start

What this builds on

Part 3 · Prerequisite check

Quick check before you start

1. If $y$ is a function of $x$ and $x$ is a function of $t$ , the chain rule says $\frac{dy}{dt} =$

$\frac{dy}{dx}\cdot\frac{dx}{dt}$
$\frac{dy}{dx} + \frac{dx}{dt}$
$\frac{dy}{dx} \div \frac{dx}{dt}$
$\frac{dx}{dy}\cdot\frac{dx}{dt}$

Show the answer

The rates multiply: change in $y$ per unit of $x$ , times change in $x$ per unit of $t$ .

Correct: $\frac{dy}{dx}\cdot\frac{dx}{dt}$ :
$\frac{dy}{dx} + \frac{dx}{dt}$ :
$\frac{dy}{dx} \div \frac{dx}{dt}$ :
$\frac{dx}{dy}\cdot\frac{dx}{dt}$ :

2. The path $x = t + 1$ , $y = t^2$ passes through which point when $t = 2$ ?

$(3, 4)$
$(2, 4)$
$(4, 3)$
$(3, 2)$

Show the answer

Put $t = 2$ into both equations: $x = 3$ and $y = 4$ .

Correct: $(3, 4)$ :
$(2, 4)$ :
$(4, 3)$ :
$(3, 2)$ :

3. The line through $(2, 5)$ with slope $-3$ is

$y - 5 = -3(x - 2)$
$y - 2 = -3(x - 5)$
$y + 5 = -3(x + 2)$
$y = -3x + 5$

Show the answer

Point-slope form: $y - y_1 = m(x - x_1)$ with $(x_1, y_1) = (2, 5)$ .

Correct: $y - 5 = -3(x - 2)$ :
$y - 2 = -3(x - 5)$ :
$y + 5 = -3(x + 2)$ :
$y = -3x + 5$ :

Part 4 · See it

See it first

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. — The curve $x = t^2$ , $y = t^3 - 3t$ passes through $(3, 0)$ twice, with slope $\sqrt{3}$ at $t = \sqrt{3}$ and $-\sqrt{3}$ at $t = -\sqrt{3}$ . Its tangent is horizontal at $t = \pm 1$ , where $\frac{dy}{dt} = 0$ , and vertical at $t = 0$ , where $\frac{dx}{dt} = 0$ . LevlPrep original diagram.

Part 5 · Step by step

How it works, step by step

$x$ and $y$ are both functions of $t$ $\frac{dx}{dt}$ and $\frac{dy}{dt}$ are the horizontal and vertical rates of the moving point
The chain rule gives $\frac{dy}{dt} = \frac{dy}{dx}\cdot\frac{dx}{dt}$ so $\frac{dy}{dx} = \frac{dy/dt}{dx/dt}$ wherever $\frac{dx}{dt} \ne 0$
$\frac{dy}{dt} = 0$ while $\frac{dx}{dt} \ne 0$ the point moves straight across for an instant: a horizontal tangent
$\frac{dx}{dt} = 0$ while $\frac{dy}{dt} \ne 0$ the point moves straight up or down for an instant: a vertical tangent

Part 6 · Key ideas

Key ideas

The slope of a parametric curve is $\frac{dy}{dx} = \frac{dy/dt}{dx/dt}$ , provided $\frac{dx}{dt} \ne 0$ . It is a function of $t$ : evaluate it at the time, not at the point.
A tangent line needs a point and a slope, both at the same $t$ : $y - y(t_0)$ $= \left.\frac{dy}{dx}\right|_{t=t_0}(x - x(t_0))$ .
Signs give direction: $\frac{dx}{dt} > 0$ moves right, $< 0$ left; $\frac{dy}{dt} > 0$ moves up, $< 0$ down.
Horizontal tangent: $\frac{dy}{dt} = 0$ and $\frac{dx}{dt} \ne 0$ . Vertical tangent: $\frac{dx}{dt} = 0$ and $\frac{dy}{dt} \ne 0$ . Both zero: look closer.

Part 7 · Misconception

A common mistake

The wrong idea: At $t = 1$ , $\frac{dx}{dt} = 0$ and $\frac{dy}{dt} = 5$ , so the tangent line is horizontal.

What actually happens: $\frac{dx}{dt} = 0$ means no horizontal motion: the point moves straight up, so the tangent line is vertical. A horizontal tangent needs $\frac{dy}{dt} = 0$ with $\frac{dx}{dt} \ne 0$ , because $\frac{dy}{dx} = \frac{dy/dt}{dx/dt}$ is 0 only when its numerator is.

Part 8 · Check yourself

Check yourself

Exam-style questions. Anything you miss goes into your review queue.

Graph

A path with direction arrows and two marked times

A point moves along the path

x(t) = t^3 - 3t

,

y(t) = 4 - t^2

for

-2 \le t \le 2

. The graph shows the path; the arrows give the direction of increasing

t

, and the points where

t = -1

and

t = 2

are marked.

The path x(t) = t cubed minus 3t, y(t) = 4 minus t squared for t from -2 to 2. It starts at (-2, 0) when t = -2, rises to the right, turns back to the left at (2, 3), passes over the top at (0, 4), turns again at (-2, 3) and comes down to (2, 0) when t = 2, crossing its earlier path at (0, 1). Arrows show the direction of increasing t. The point where t = -1, (2, 3), is marked in blue and the point where t = 2, (2, 0), in vermillion. — LevlPrep original diagram.

1. Find $\frac{dy}{dx}$ at the point where $t = 2$ .

No calculator. Type the exact value (a fraction, π, e or a root is fine).

Show the answer

$\frac{dy}{dt} = -2t = -4$ and $\frac{dx}{dt} = 3t^2 - 3 = 9$ at $t = 2$ , so $\frac{dy}{dx} = \frac{-4}{9} = -\frac{4}{9}$ .

Answer: $\frac{-4}{9}$ ≈ -0.444

2. Write an equation of the line tangent to the path at the point where $t = 2$ .

Type an expression for $y =$ .

Show the answer

The point is $(x(2), y(2))$ $= (8 - 6, 4 - 4)$ $= (2, 0)$ and the slope is $-\frac{4}{9}$ , so $y - 0 = -\frac{4}{9}(x - 2)$ .

Answer: $y =$ $\frac{-4}{9} (x - 2)$

3. Which statement describes the tangent line at the point where $t = -1$ ?

It is horizontal, because $\frac{dx}{dt} = 0$ there
It has slope 2, because $\frac{dy}{dt} = 2$ there
It is vertical, because $\frac{dx}{dt} = 0$ and $\frac{dy}{dt} \ne 0$ there
It does not exist, because $\frac{dy}{dx}$ is undefined there

Show the answer

At $t = -1$ , $\frac{dx}{dt} = 3(1) - 3 = 0$ and $\frac{dy}{dt} = 2 \ne 0$ . The point moves straight up for an instant, so the tangent line is vertical: the path turns from moving right to moving left at $(2, 3)$ .

It is horizontal, because $\frac{dx}{dt} = 0$ there: $\frac{dx}{dt} = 0$ makes the horizontal motion stop, which gives a vertical tangent. A horizontal tangent needs $\frac{dy}{dt} = 0$ .
It has slope 2, because $\frac{dy}{dt} = 2$ there: $\frac{dy}{dt} = 2$ is a rate per unit of time, not the slope; the slope divides by $\frac{dx}{dt}$ , which is 0 here.
Correct: It is vertical, because $\frac{dx}{dt} = 0$ and $\frac{dy}{dt} \ne 0$ there: Right: no horizontal motion and nonzero vertical motion give a vertical tangent line.
It does not exist, because $\frac{dy}{dx}$ is undefined there: An undefined slope is the slope of a vertical line; the tangent line exists and is vertical.

4. At which point is the tangent line to the path horizontal?

$(2, 3)$ and $(-2, 3)$ , where $t = -1$ and $t = 1$
$(0, 1)$ , where the path crosses itself
$(-2, 0)$ and $(2, 0)$ , where $y = 0$
$(0, 4)$ , where $t = 0$

Show the answer

Horizontal tangent: $\frac{dy}{dt} = -2t = 0$ with $\frac{dx}{dt} \ne 0$ . That is $t = 0$ , where $\frac{dx}{dt} = -3$ , at the point $(0, 4)$ .

$(2, 3)$ and $(-2, 3)$ , where $t = -1$ and $t = 1$ : There $\frac{dx}{dt} = 0$ and $\frac{dy}{dt} \ne 0$ : those tangents are vertical.
$(0, 1)$ , where the path crosses itself: The crossing point has two tangent lines, one for each pass, and neither is horizontal.
$(-2, 0)$ and $(2, 0)$ , where $y = 0$ : $y = 0$ is where the path meets the $x$ -axis; the slope there is $\frac{dy/dt}{dx/dt} = \pm\frac{4}{9}$ , not 0.
Correct: $(0, 4)$ , where $t = 0$ : Right: $\frac{dy}{dt} = 0$ and $\frac{dx}{dt} = -3 \ne 0$ at $t = 0$ .

5. A curve is given by $x = e^t$ , $y = t^2 + 1$ . Find $\frac{dy}{dx}$ at $t = 1$ .

No calculator. Type the exact value (a fraction, π, e or a root is fine).

Show the answer

$\frac{dy}{dt} = 2t$ and $\frac{dx}{dt} = e^t$ , so $\frac{dy}{dx} = \frac{2t}{e^t}$ , which is $\frac{2}{e}$ at $t = 1$ .

Answer: $\frac{2}{e}$ ≈ 0.736

6. A curve is given by $x = x(t)$ , $y = y(t)$ . Which expression gives the slope of the tangent line at a point where $\frac{dx}{dt} \ne 0$ ?

$\frac{dx/dt}{dy/dt}$
$\frac{dy}{dt}$
$\frac{dy}{dt} \cdot \frac{dx}{dt}$
$\frac{dy/dt}{dx/dt}$

Show the answer

By the chain rule $\frac{dy}{dt} = \frac{dy}{dx}\cdot\frac{dx}{dt}$ , so $\frac{dy}{dx} = \frac{dy/dt}{dx/dt}$ : rise per unit of time over run per unit of time.

$\frac{dx/dt}{dy/dt}$ : This is the reciprocal of the slope, run over rise.
$\frac{dy}{dt}$ : This is how fast $y$ changes per unit of time, not per unit of $x$ .
$\frac{dy}{dt} \cdot \frac{dx}{dt}$ : Multiplying the rates has no meaning here; the chain rule gives a quotient.
Correct: $\frac{dy/dt}{dx/dt}$ : Right: the chain rule, solved for $\frac{dy}{dx}$ .

7. The curve $x = t^2$ , $y = t^3 - 3t$ passes through $(3, 0)$ twice, at $t = -\sqrt{3}$ and at $t = \sqrt{3}$ . What can be said about the slope of the curve at $(3, 0)$ ?

It is 0, because $y = 0$ at $(3, 0)$
There are two tangent lines, with slopes $-\sqrt{3}$ and $\sqrt{3}$
There is one slope, found by putting $x = 3$ into a formula for $\frac{dy}{dx}$
It does not exist, because the curve crosses itself there

Show the answer

$\frac{dy}{dx} = \frac{3t^2 - 3}{2t}$ . At $t = \sqrt{3}$ : $\frac{6}{2\sqrt{3}} = \sqrt{3}$ . At $t = -\sqrt{3}$ : $-\sqrt{3}$ . Each pass through the point has its own tangent line, which is why slopes are found from $t$ , not from the point.

It is 0, because $y = 0$ at $(3, 0)$ : The height of a point says nothing about its slope; compute $\frac{dy/dt}{dx/dt}$ at each $t$ .
Correct: There are two tangent lines, with slopes $-\sqrt{3}$ and $\sqrt{3}$ : Right: one tangent line for each time the curve passes through the point.
There is one slope, found by putting $x = 3$ into a formula for $\frac{dy}{dx}$ : $\frac{dy}{dx}$ here is a formula in $t$ , and two values of $t$ give $x = 3$ , with different slopes.
It does not exist, because the curve crosses itself there: Each pass has a well-defined tangent line; there are simply two of them.

Part 9 · Summary

Summary

For a curve

x = x(t)

,

y = y(t)

, the derivatives

\frac{dx}{dt}

and

\frac{dy}{dt}

are the horizontal and vertical rates, and their signs give the direction of motion. The slope of the curve is

\frac{dy}{dx} = \frac{dy/dt}{dx/dt}

where

\frac{dx}{dt} \ne 0

, evaluated at a time. Horizontal tangents need

\frac{dy}{dt} = 0

, vertical tangents

\frac{dx}{dt} = 0

, with the other rate nonzero.

Part 10 · Up next

What comes next

Part 11 · Connections