Defining and Differentiating Parametric Equations

Use one parameter for point, direction, and tangent rates, checking both tangent conditions.

College Board Topic 9.1 · CED p. 166 (PDF p. 171) · CED effective Fall 2020; project course outline 2026–27.

A parametric curve gives two coordinates from one input: $x = x (t)$ and $y = y (t)$ . Think of t as a clock, even when it is just a parameter. The ordered pair tells you where the point is; increasing t tells you which way it travels. Eliminating t can reveal a familiar curve, but it can lose direction and the allowed part of that curve.

The two rates $\frac{d x}{d t}$ and $\frac{d y}{d t}$ measure coordinate change per unit of t. A tangent slope measures vertical change per horizontal change. The chain rule gives $\frac{d y}{d x} = \frac{d y / d t}{d x / d t}$ , provided $\frac{d x}{d t} \neq 0$ . Evaluate both derivatives at the same t. You cannot divide a derivative at one time by a derivative at another.

A horizontal tangent needs $d y / d t = 0$ and $d x / d t \neq 0$ . A vertical tangent occurs when $d x / d t = 0$ and $d y / d t \neq 0$ . If both vanish, the quotient rule for slope says nothing: inspect a limit or a local parameterization before deciding. Do not call every zero-over-zero point a cusp.

For $x = 2 \cos t$ , $y = 3 \sin t$ , the point at $t = π / 2$ is $(0, 3)$ . Its rates are $(- 2, 0)$ , so it has a horizontal tangent and is moving left. The same ellipse is traced again if you extend t from 2π to 4π; a graph alone hides that.

A point moves around the ellipse; the parameter records its direction as well as its location.

Let's work through it

For $x = t^{2} + 1$ , $y = t^{3} - 3 t$ , at $t = 2$ find $(x, y) = (5, 2)$ . Why: Use the same parameter value in both coordinates.
Differentiate: $x' = 2 t$ , $y' = 3 t^{2} - 3$ . Why: The derivative gives change with respect to t.
Compute $d y / d x = 9 / 4$ , then $y - 2 = (9 / 4) (x - 5)$ . Why: Divide the coordinate rates because x′(2)=4 is nonzero; use point-slope form.

Try it from memory

Pause & retrieve

Try it before you peek.

Write your answer and the reason from memory. Then compare with the worked answer. Your self-assessment is a study decision, not an automatically graded score.

Question 1 of 20 self-assessed

Question 1

For

x = t

y = t^{2}

, what is the slope at

t = 3

Your answer and reasoning

JavaScript enables the written-work controls. You can still try the questions below on paper and then open their reasoning.

Question 1: prompt and worked reasoning

For

x = t

y = t^{2}

, what is the slope at

t = 3

6

Divide 2t by 1, then evaluate at 3.

Question 2: prompt and worked reasoning

x' = 0

and

y' = 4

at a point, describe its tangent.

Vertical.

The vertical coordinate changes while the horizontal coordinate has zero instantaneous change.

See every recall question with its answer

These references preserve each prompt and its answer for direct evidence links. Try the interactive questions above before opening the worked reasoning.

Retrieval 1

For

x = t

y = t^{2}

, what is the slope at

t = 3

Worked answer and reason

6

Divide 2t by 1, then evaluate at 3.

Retrieval 2

x' = 0

and

y' = 4

at a point, describe its tangent.

Worked answer and reason

Vertical.

The vertical coordinate changes while the horizontal coordinate has zero instantaneous change.

Catch the tempting mistake

dy/dt is the tangent slope.: It is a rate with respect to the parameter. Divide by dx/dt when that denominator is nonzero.
Eliminating the parameter preserves everything.: It can erase orientation, parameter restrictions, and repeated tracing.

Make the picture move

BC: parametric and polar calculus

Track position, direction, and speed separately.

Try first. Compare t = 0 and t = π/2. Which point has a vertical tangent, and which has the greater speed?

x = 2 cos t y = sin t v = \sqrt{{x'}^{2} + {y'}^{2}}

Choose a scenario Parameter t 0.7854

Scroll sideways to see the full graph on a small screen. The table below gives the numerical values.

Parametric ellipse
Short segment: direction of increasing parameter

A closed ellipse with the selected position marked. A short dark segment points in the direction of increasing parameter when velocity is nonzero. At t = 0.7854, position is (1.4142, 0.7071) and speed is 1.5811. Slope is -0.5.

Position (x, y)(1.4142, 0.7071)

Speed per time unit1.5811

Slope dy/dx-0.5

At t = 0.7854, position is (1.4142, 0.7071) and speed is 1.5811. Slope is -0.5.

Compare the numbers

Computed values for this scenario (rounded for display)
Time t	x	y	Speed	dy/dx
0.7854	1.4142	0.7071	1.5811	-0.5
0	2	0	1	undefined
0.7854	1.4142	0.7071	1.5811	-0.5
1.5708	0	1	2	0
3.1416	-2	0	1	undefined

What’s going on? The velocity is (−2 sin t, cos t). Speed is its nonnegative length; dy/dx is the ratio of its components when dx/dt ≠ 0. At t = 0 the tangent is vertical and speed is 1; at t = π/2 the tangent is horizontal and speed is 2.

Keep in mind. The short segment is a scaled velocity direction. Its screen length is not a distance traveled. These formulas describe this parametrization; changing the parametrization can change speed without changing the curve.

Calculator workspace and limits

For calculus, set the angle selector to Radians. This scientific calculator evaluates arithmetic and functions. It does not graph arbitrary functions, find numerical roots, evaluate numerical derivatives, or compute definite integrals. Those tasks need an approved graphing tool; the May 2027 exam permits approved handheld graphing calculators and Bluebook’s built-in Desmos on calculator-required parts. A graph or decimal estimate helps you check a conclusion; your setup, hypotheses, units and justification still do the explaining.

Scientific calculator

This calculator requires JavaScript. Use the physics formula sheets and show your working manually.

Calculation is unavailable without JavaScript.

+ − * / ^ ! · pi · e · sqrt · sin, cos, tan · asin, acos, atan · atan2(y,x) · ln, log · abs · exp · hypot(x,y). Use * between factors. Keep units in your written working.

Put it to work

Unit 9 problem 1 · No calculator · analytic
Unit 9 problem 2 · No calculator · analytic
Unit 9 problem 3 · No calculator · multiple
Unit 9 problem 12 · No calculator · numerical

For the teacher

Ask learners to sketch the ordered pairs before showing the ellipse. Require both tangent conditions, not only the zero numerator.

Unit planning and formative guidance →

Glossary and deeper reading · 4 refreshers

These editorial refreshers supplement the reading. Definitions and examples are added guidance; original source wording and recorded answers remain in their original context.

Chain rule
The chain rule differentiates a composition by multiplying the outer derivative at the inner output by the inner derivative. For differentiable functions, (f(g(x)))′=f′(g(x))g′(x). For example, the derivative of sin(x²) is cos(x²)·2x. Nested compositions require a factor for each successive layer.
Read more: 3.1 · Chain rule →
Derivative
A derivative gives the instantaneous rate of change of a function with respect to its input. At a point, it is the finite limit of the difference quotient when that limit exists. For example, the derivative of x² is 2x, so the tangent slope at x=3 is 6. A derivative function collects those rates across the inputs where the function is differentiable.
Read more: 2.2 · Derivative →
Vertical tangent
A vertical tangent is a tangent line parallel to the vertical axis, with no finite slope. A graph can have a vertical tangent even though its derivative does not exist as a finite real number at that input. For example, y=∛x has the vertical tangent x=0 at the origin as its nearby slopes tend to positive infinity. This case is separate from the finite-slope tangent approximation available at a differentiable point.
Read more: 2.4 · Differentiability and vertical tangents →
Quotient rule
The quotient rule differentiates a quotient where the denominator is nonzero and both functions are differentiable. It gives (f/g)′=(f′g−fg′)/g². For example, differentiating x/(x+1) produces 1/(x+1)² for x≠−1. The order of the subtraction and the denominator square both matter.
Read more: 2.9 · Quotient rule →

Let's work through it

Try it from memory

Try it before you peek.

Question 1

Retrieval 1

Retrieval 2

Catch the tempting mistake

Make the picture move

Track position, direction, and speed separately.

Put it to work

Chain rule

Derivative

Vertical tangent

Quotient rule

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