Defining and Differentiating Parametric Equations: Study Guide (Calculus BC 9.1)

BC-only material. Parametric equations are part of Calculus BC only. Calculus AB students do not need this topic.

Before you start: prerequisites

You need three earlier ideas. If any is shaky, revisit it from the Calculus BC roadmap first.

Prerequisite	Topic	What you use it for here
Basic derivatives (powers, trig, eˣ, ln x)	2.5–2.7	Finding dx/dt and dy/dt
Chain rule	3.1	Explains why dy/dx = (dy/dt) ÷ (dx/dt), and is needed inside many dx/dt and dy/dt
Tangent line from a point and a slope	2.1, 4.6	Writing y − y₀ = m(x − x₀)

Notation on this page: x′(t) means dx/dt and y′(t) means dy/dt. Angles are in radians.

What a parametric curve is

So far, most curves you have met were written as y = f(x): one output y for each input x. A parametric curve describes x and y separately, each as a function of a third variable, the parameter, usually t:

x = f(t), y = g(t)

Each value of t gives one point (f(t), g(t)). As t increases, the point moves and traces the curve. It often helps to think of t as time and the curve as the path of a moving object.

Three things change compared with y = f(x):

• The curve has a direction. As t increases, the point travels one way along the curve. This is called the orientation. Graphs usually show it with arrows.
• The curve can turn back or cross itself. It does not have to pass the vertical line test. The same point (x, y) can be reached at two different values of t.
• Different equations can trace the same path. x = cos t, y = sin t and x = cos 2t, y = sin 2t both give the unit circle; the second goes round twice as fast.

A table of values is the simplest way to see a curve. Take x = t² + 1, y = t³ − 4t:

t	−2	−1	0	1	2
x = t² + 1	5	2	1	2	5
y = t³ − 4t	0	3	0	−3	0

The point (5, 0) appears twice, at t = −2 and at t = 2. So this curve crosses itself there. Figure 1 shows the whole picture.

Background: eliminating the parameter. Sometimes you can solve one equation for t and substitute into the other. For x = 2t + 1, y = t²: t = (x − 1)/2, so y = (x − 1)²/4, a parabola. This is useful for checking work, but it is often messy or impossible, and you never need it to find dy/dx.

Deriving dy/dx with the chain rule

Suppose that, near some point, the curve can be written as y as a function of x, and x and y both depend on t. The chain rule says

dy/dt = (dy/dx) · (dx/dt)

If dx/dt ≠ 0, divide both sides by dx/dt:

dy/dx = (dy/dt) ÷ (dx/dt), provided dx/dt ≠ 0

The value of dy/dx at a point is the slope of the line tangent to the curve at that point. Leibniz notation makes the formula easy to remember: the dt’s “cancel”. But it is the chain rule that makes it true, so you must still check that dx/dt is not zero.

Quick check with the parabola above. x = 2t + 1, y = t² gives dx/dt = 2 and dy/dt = 2t, so dy/dx = 2t/2 = t. From y = (x − 1)²/4, dy/dx = (x − 1)/2, and (x − 1)/2 = t. ✓ Both methods agree.

Notice that the answer dy/dx = t is written in terms of t, not x. This is normal for parametric curves. To find a slope, you substitute a value of t.

Horizontal and vertical tangents

The fraction (dy/dt) ÷ (dx/dt) tells you where the tangent line is flat or upright.

Condition	Tangent line	Reason
dy/dt = 0 and dx/dt ≠ 0	Horizontal	Slope = 0 ÷ (non-zero) = 0
dx/dt = 0 and dy/dt ≠ 0	Vertical	Slope = (non-zero) ÷ 0 is undefined; the point moves straight up or down for an instant
dx/dt = 0 and dy/dt = 0	No conclusion yet	0 ÷ 0: the formula gives no answer

Always check the other derivative. Solving dy/dt = 0 finds only candidates for horizontal tangents. Each one must also have dx/dt ≠ 0.

Why the both-zero case needs care. Take x = t³, y = t⁶. At t = 0, dx/dt = 3t² = 0 and dy/dt = 6t⁵ = 0. But y = (t³)² = x², a parabola with a perfectly ordinary horizontal tangent at the origin. In other curves, both derivatives being zero gives a sharp point (a cusp). So when both are zero, simplify dy/dx first or look at the graph; do not just say “horizontal” or “vertical”.

Worked example 1: tangent lines on a self-crossing curve

Question. For the curve x = t² + 1, y = t³ − 4t (Figure 1), without a calculator: (a) find dy/dx in terms of t; (b) find the equation of the tangent line at t = 1; (c) find the slope of the curve at the point (5, 0); (d) find the points where the tangent line is horizontal or vertical.

(a) Differentiate each coordinate, then divide.

1. dx/dt = 2t and dy/dt = 3t² − 4.
2. dy/dx = (3t² − 4) / (2t), for t ≠ 0.

(b) Tangent line at t = 1.

1. Point: x = 1 + 1 = 2 and y = 1 − 4 = −3, so the point is (2, −3).
2. Slope: dy/dx = (3 − 4)/2 = −½.
3. Line: y − (−3) = −½(x − 2), which simplifies to y = −½x − 2.

(c) Slope at (5, 0). You are given a point, not a value of t, so find t first.

1. x = 5 gives t² + 1 = 5, so t = 2 or t = −2.
2. y = 0 at both: 2³ − 8 = 0 and (−2)³ + 8 = 0. So the curve passes through (5, 0) twice.
3. At t = 2: dy/dx = (12 − 4)/4 = 2. At t = −2: dy/dx = 8/(−4) = −2.

The curve has two tangent lines at (5, 0): y = 2x − 10 and y = −2x + 10. A question that asks for “the slope at (5, 0)” here must also say which value of t it means.

(d) Horizontal and vertical tangents.

1. Horizontal: dy/dt = 3t² − 4 = 0 gives t = ±2/√3 (about ±1.155). At these values dx/dt = 2t ≠ 0. ✓ The points are (7/3, 16√3/9) and (7/3, −16√3/9), about (2.33, 3.08) and (2.33, −3.08).
2. Vertical: dx/dt = 2t = 0 gives t = 0. There dy/dt = −4 ≠ 0. ✓ The point is (1, 0).

Checks. Figure 1 shows a highest and a lowest point near x = 2.33, and the curve is upright at its leftmost point (1, 0). The signs also match: at t = 1 the point moves right (dx/dt = 2 > 0) and down (dy/dt = −1 < 0), so the slope should be negative. ✓

Worked example 2: the path of a drone, in context

Context. A survey drone flies in a vertical plane. Its horizontal distance east of a mast is x(t) = 9t − t³/3 metres, and its height above the ground is y(t) = 20 + 8t − 2t² metres, where t is the time in seconds, for 0 ≤ t ≤ 4.

(a) Find dy/dx in terms of t. (b) When is the path horizontal, and when is it vertical? Interpret each. (c) Find the slope of the path at t = 4. The slope is positive. Is the drone climbing?

(a) dx/dt = 9 − t² and dy/dt = 8 − 4t, so dy/dx = (8 − 4t)/(9 − t²), for t ≠ 3.

(b) Horizontal and vertical.

1. dy/dt = 8 − 4t = 0 at t = 2. There dx/dt = 9 − 4 = 5 ≠ 0, so the path is horizontal at t = 2, at the point (46/3, 28), about (15.33, 28). The drone has stopped rising and starts to descend: this is its highest point, 28 m.
2. dx/dt = 9 − t² = 0 at t = 3 (t = −3 is outside the interval). There dy/dt = 8 − 12 = −4 ≠ 0, so the path is vertical at t = 3, at (18, 26). The drone has stopped moving east and starts to move back west: this is its furthest point from the mast, 18 m.

(c) Slope at t = 4.

1. dx/dt = 9 − 16 = −7 and dy/dt = 8 − 16 = −8.
2. dy/dx = (−8)/(−7) = 8/7 ≈ 1.14.
3. Interpretation. The path is rising to the right at the drone’s position (14.67, 20). But the drone is not climbing: dy/dt = −8 < 0, so its height is falling at 8 m/s, and dx/dt = −7 < 0, so it is moving west. It is travelling down and to the left along a line that slopes up to the right.

Units. dy/dx is metres of height per metre of horizontal distance, so it has no units. It describes the shape of the path, not how fast the drone moves.

Checks. At t = 1 the slope is 4/8 = ½, positive, and the drone really is moving up and right (dx/dt = 8, dy/dt = 4), which matches Figure 2. ✓ dx/dt and dy/dt are never both zero on 0 ≤ t ≤ 4 (one is zero only at t = 3, the other only at t = 2). ✓

Common misconceptions

• “dy/dx = (dx/dt) ÷ (dy/dt).” The formula is upside down. Think “dy over dx”: the y-derivative goes on top.
• Substituting x into a formula in t. If dy/dx = (3t² − 4)/(2t), the slope “at x = 5” is not found by putting t = 5. Find the value (or values) of t that give the point first.
• Assuming one point means one value of t. A parametric curve can pass through the same point twice with different slopes, as at (5, 0) in Worked example 1.
• Stopping at dy/dt = 0. That gives candidates only. A horizontal tangent also needs dx/dt ≠ 0; if both are 0, the formula gives no answer.
• Mixing up horizontal and vertical. dx/dt = 0 means no sideways movement at that instant, so the tangent is vertical, not horizontal.
• Reading dy/dx as “going up”. A positive slope only says the path rises to the right. The object can still be moving down and to the left, as at t = 4 in Worked example 2.
• Treating dy/dx as a speed. Speed and velocity need both dx/dt and dy/dt; they come later in Unit 9.

Where this leads

Topic 9.2 differentiates again to find d²y/dx², which tells you the concavity of a parametric curve. Topic 9.3 uses dx/dt and dy/dt to find the length of a parametric curve, building on the arc length work in Topic 8.13. Topics 9.4 to 9.6 turn the parameter into time and study velocity, speed and motion in the plane. Next: Second Derivatives of Parametric Equations (9.2).

Try the practice questions now, then use the revision notes and the checklist to consolidate.