Parametric Differentiation

§ 01What Does "Parametric" Actually Mean?

Most curves you have met so far are described by a single equation linking x and y directly, like y = x² or x² + y² = 4. Parametric equations do something different: they describe x and y each as separate functions of a third variable called a parameter.

Think of it like tracking a moving object. Instead of saying "the object's path satisfies y = x²", you say "at time t, the object is at position x = t² and y = t³." The parameter t drives both coordinates simultaneously. As t changes, the point (x, y) traces out a curve.

Parametric form

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

The parameter t does not have to represent time, it is just a convenient label for whatever is driving the two coordinates. For circles it is often an angle. For projectile paths it is time. For many mathematical curves it is simply a number with no physical meaning at all.

A concrete example — the circle

The unit circle x² + y² = 1 can be written parametrically as:

Unit circle in parametric form

x = cos(t)
y = sin(t) 0 ≤ t ≤ 2π

As t goes from 0 to 2π, the point (cos t, sin t) travels once around the unit circle. The parameter t here is the angle. You can verify this is the same circle: substitute and check that cos²(t) + sin²(t) = 1 ✓.

Why use parametric equations at all? Some curves — like cycloids, spirals, and Lissajous figures — are very awkward to write as a single equation y = f(x) but are beautifully simple in parametric form. Parametric equations also make it natural to track direction of motion, which is important in mechanics and geometry.

§ 02Deriving dy/dx for Parametric Curves

The formula dy/dx = (dy/dt) ÷ (dx/dt) looks like a neat trick, but it comes directly from the chain rule. Let us see exactly where it comes from.

We want dy/dx: the gradient of the curve at a point. Both x and y depend on t, so we cannot differentiate y directly with respect to x. But we can use the chain rule, which tells us how rates of change relate to each other.

Derivation of dy/dx for parametric equations

We want dy/dx. Write the chain rule connecting the rate of change of y with respect to t and with respect to x:

dy/dt = (dy/dx) · (dx/dt) This is the chain rule: "the rate of change of y with respect to t equals the rate of change of y with respect to x, multiplied by the rate of change of x with respect to t."

We want dy/dx, not dy/dt. Divide both sides by dx/dt (provided dx/dt ≠ 0):

dy/dx = (dy/dt) ÷ (dx/dt) This only works when dx/dt ≠ 0. When dx/dt = 0, the tangent is vertical and dy/dx is undefined at that point.

In practice: differentiate y with respect to t to get dy/dt. Differentiate x with respect to t to get dx/dt. Divide them:

dy/dx = (dy/dt) / (dx/dt) = ẏ / ẋ The dot notation ẏ and ẋ (pronounced "y-dot" and "x-dot") is shorthand for dy/dt and dx/dt. You will see both notations in textbooks.

The Parametric Differentiation Formula

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

This formula is the entire foundation of parametric differentiation. Everything else (second derivatives, tangents, normals) builds on it.

§ 03The Four-Step Method

Every parametric differentiation problem follows the same routine. Learn these four steps and you can handle any curve.

Step 01

Find dx/dt

Differentiate the x equation with respect to t. Use whatever rules are needed, power, trig, chain, product.

Step 02

Find dy/dt

Differentiate the y equation with respect to t in exactly the same way.

Step 03

Divide

Form dy/dx = (dy/dt) ÷ (dx/dt). Simplify if possible. The result is the gradient in terms of t.

Step 04

Substitute t

If a specific gradient or point is needed, substitute the value of t at that point into dy/dx.

Worked example — the parabola x = t², y = 2t

This is a common parametric form. Let us find dy/dx and the gradient at the point where t = 3.

Full worked example — x = t², y = 2t

Step 1: Differentiate x with respect to t:

dx/dt = 2t

Step 2, Differentiate y with respect to t:

dy/dt = 2

Step 3: Divide to find dy/dx:

dy/dx = 2 / (2t) = 1/t The gradient of this parabola at parameter value t is simply 1/t. When t is large (far from the vertex), the curve is nearly flat. When t is small, the curve is steep — which makes geometric sense.

Step 4, At t = 3, the gradient is:

dy/dx = 1/3 The Cartesian point is x = 9, y = 6. You can verify: eliminating t gives y² = 4x. At (9, 6): implicit differentiation gives dy/dx = 2/y = 2/6 = 1/3 ✓

§ 04The Second Derivative d²y/dx²

The second derivative tells you about concavity, whether the curve bends upward or downward. Getting it right for parametric equations requires one extra insight.

The second derivative d²y/dx² means: differentiate dy/dx with respect to x. But dy/dx is a function of t, not x. So we need to use the chain rule again, we differentiate dy/dx with respect to t first, then divide by dx/dt.

Second derivative — parametric form

d²y/dx² = d/dx[dy/dx]

= [d/dt (dy/dx)] ÷ (dx/dt)

In plain English: differentiate dy/dx with respect to t, then divide by dx/dt. You do not differentiate dy/dt a second time, that is a very common mistake.

The most important thing about the second derivative You must differentiate the expression for dy/dx (which is a function of t) with respect to t. Do not try to differentiate dy/dt and dx/dt again separately and divide them — that would give d(dy/dt)/d(dx/dt), which is something completely different and wrong.

Example — finding d²y/dx² for x = t², y = 2t

We found dy/dx = 1/t above. Now find d²y/dx².

Second derivative for x = t², y = 2t

Start from dy/dx = 1/t. Differentiate this with respect to t:

d/dt [dy/dx] = d/dt [1/t] = −1/t²

We already know dx/dt = 2t. Now divide:

d²y/dx² = (−1/t²) / (2t) = −1 / (2t³)

The negative sign means the curve is concave downward for t > 0. The original curve y² = 4x opens to the right, and the upper branch (t > 0) is indeed concave down ✓

d²y/dx² = −1/(2t³)

§ 05Tangents and Normals

Once you have dy/dx in terms of t, finding the tangent or normal at a specific point is exactly the same process as for any other curve: just evaluate at the right value of t.

Tangent at parameter value t = t₀

The tangent line at a point passes through that point with the same slope as the curve at that point. For a parametric curve:

The point is (x₀, y₀) = (f(t₀), g(t₀)), substitute t₀ into both parametric equations.
The slope is m = dy/dx evaluated at t = t₀.
The tangent equation is y − y₀ = m(x − x₀).

Normal at parameter value t = t₀

The normal is perpendicular to the tangent. Its slope is −1/m (the negative reciprocal). So:

Normal slope = −(dx/dt) / (dy/dt) , flip the fraction and change the sign.
Normal equation: y − y₀ = (−1/m)(x − x₀).

Special cases

When dx/dt = 0 at a point, the tangent is vertical (dy/dx is undefined). When dy/dt = 0 at a point, the tangent is horizontal, this is a stationary point of the curve.

Identifying special points on a parametric curve

dy/dt = 0 → horizontal tangent (stationary point)
dx/dt = 0 → vertical tangent
dy/dt = 0 and dx/dt = 0 → singular point (investigate separately)

§ 06Interactive Graph — Parametric Curves and Tangent Lines

Select a curve and drag the slider to move a point along it. The tangent line at that point is drawn in teal, and the gradient dy/dx is shown. Try to predict where the tangent will be horizontal or vertical before you drag there.

← Move the slider to explore the curve

Curve:

t = 0.00

§ 07Twelve Worked Exercises

Work through each problem before revealing the answer. Solutions are explained in plain English at every step.

Exercises opened: 0 / 12

Group A — Finding dy/dx (Exercises 1–4)

Exercise 01 Basic

Find dy/dx for x = t³, y = t²

Differentiate x with respect to t using the power rule: dx/dt = 3t²
Differentiate y with respect to t: dy/dt = 2t
Divide dy/dt by dx/dt: dy/dx = 2t / (3t²) = 2 / (3t)
The t in the numerator and one power of t in the denominator cancel: dy/dx = 2/(3t) (valid for t ≠ 0)

Exercise 02 Basic

Find dy/dx for x = 4cos(t), y = 3sin(t)

Differentiate x with respect to t. Remember d/dt[cos t] = −sin t: dx/dt = −4 sin(t)
Differentiate y with respect to t. d/dt[sin t] = cos t: dy/dt = 3 cos(t)
Form dy/dx: dy/dx = 3 cos(t) / (−4 sin(t))
Simplify. The ratio cos t / sin t = cot t, so: dy/dx = −(3/4) cot(t) This is an ellipse with semi-axes 4 (horizontal) and 3 (vertical). The gradient formula confirms the tangent is horizontal when sin t = 0 (at the top and bottom of the ellipse) and vertical when cos t = 0 (at the left and right endpoints). ✓

Exercise 03 Basic

Find dy/dx for x = e^t, y = e^(2t)

Differentiate both x and y with respect to t using d/dt[eᵏᵗ] = k·eᵏᵗ. Then divide. You may also recognise that y = (eᵗ)² = x², so there is a nice way to check your answer.

Differentiate x with respect to t: dx/dt = eᵗ
Differentiate y. Using the chain rule on e^(2t): dy/dt = 2e^(2t)
Divide: dy/dx = 2e^(2t) / eᵗ = 2eᵗ
Since x = eᵗ, we can also write this in terms of x: dy/dx = 2eᵗ = 2x Check: eliminating t gives y = e^(2t) = (eᵗ)² = x². So dy/dx = 2x ✓

Exercise 04 Basic

Find dy/dx for x = t + 1/t, y = t − 1/t

Write 1/t as t⁻¹ so d/dt[1/t] = −t⁻² = −1/t². Be careful with signs for both dx/dt and dy/dt.

Differentiate x with respect to t. Write 1/t = t⁻¹: dx/dt = 1 + (−1/t²) = 1 − 1/t²
Differentiate y. The 1/t term now has a minus sign in front in y, so when differentiated it becomes +1/t²: dy/dt = 1 − (−1/t²) = 1 + 1/t²
Divide: dy/dx = (1 + 1/t²) / (1 − 1/t²)
Multiply numerator and denominator by t² to clear the fractions: = (t² + 1) / (t² − 1) dy/dx = (t² + 1) / (t² − 1)

Group B — Second Derivatives and Special Points (Exercises 5–8)

Exercise 05 Medium

Find d²y/dx² for x = t², y = t³

First find dy/dx. dx/dt = 2t, dy/dt = 3t²: dy/dx = 3t² / (2t) = 3t/2
Now differentiate dy/dx with respect to t: d/dt [3t/2] = 3/2
Divide by dx/dt = 2t: d²y/dx² = (3/2) / (2t) = 3/(4t)
For t > 0 the second derivative is positive (concave up). For t < 0 it is negative (concave down). This matches the shape of the curve y = x^(3/2): d²y/dx² = 3/(4t)

Exercise 06 Medium

For x = t − sin(t), y = 1 − cos(t), find where the tangent is horizontal

Horizontal tangent means dy/dx = 0, which means dy/dt = 0 (and dx/dt ≠ 0 at the same point). Find dy/dt, set it to zero, and solve for t.

This is the cycloid, the curve traced by a point on the rim of a rolling circle. Find dy/dt: dy/dt = sin(t)
Find dx/dt: dx/dt = 1 − cos(t)
Horizontal tangent when dy/dt = 0, so sin(t) = 0: t = 0, ±π, ±2π, ...
Check dx/dt at these values. At t = 0, 2π, ...: cos(t) = 1, so dx/dt = 1 − 1 = 0. Both derivatives are zero, these are the cusp points where the cycloid touches the ground. At t = π, 3π, ...: cos(t) = −1, so dx/dt = 1 − (−1) = 2 ≠ 0. These are genuine horizontal tangents, the highest points of the cycloid. Horizontal tangents at t = π + 2nπ (the peaks). The cusps at t = 2nπ are singular points, not regular horizontal tangents.

Exercise 07 Medium

Find d²y/dx² for the ellipse x = 3cos(t), y = 2sin(t)

dx/dt = −3sin(t), dy/dt = 2cos(t). So: dy/dx = 2cos(t) / (−3sin(t)) = −(2/3) cot(t)
Differentiate dy/dx with respect to t. Recall d/dt[cot t] = −cosec²(t): d/dt [−(2/3) cot t] = −(2/3) · (−cosec²t) = (2/3) cosec²(t)
Divide by dx/dt = −3sin(t): d²y/dx² = [(2/3) cosec²t] / (−3 sin t) = (2/3) · (1/sin²t) / (−3 sin t) = −2 / (9 sin³t)
For 0 < t < π (upper half of ellipse), sin t > 0, so d²y/dx² < 0: the upper half is concave downward, as expected for the top of an ellipse: d²y/dx² = −2 / (9 sin³t)

Exercise 08 Medium

For x = t², y = t³ − 3t, find the values of t where the curve has a stationary point

Stationary points occur where dy/dx = 0. Since dy/dx = (dy/dt)/(dx/dt), this happens when dy/dt = 0 (and dx/dt ≠ 0). Find dy/dt, set it to zero, and solve for t.

dx/dt = 2t. dy/dt = 3t² − 3: dy/dx = (3t² − 3) / (2t)
Set dy/dt = 0 for stationary points: 3t² − 3 = 0, so t² = 1: t = 1 or t = −1
Check dx/dt ≠ 0 at these values. dx/dt = 2t: at t = 1, dx/dt = 2 ≠ 0 ✓. At t = −1, dx/dt = −2 ≠ 0 ✓
Find the Cartesian coordinates: at t = 1: (1, −2). At t = −1: (1, 2). These are both stationary points: Stationary points at t = 1 → (1, −2) and t = −1 → (1, 2) Both points have the same x-coordinate (x = 1), which means the curve loops, it passes through the vertical line x = 1 twice, once going up and once going down.

Group C — Tangents, Normals, and Combined Problems (Exercises 9–12)

Exercise 09 Hard

Find the equation of the tangent to x = t², y = 2t at the point where t = 2

Find the Cartesian point at t = 2: x = 4, y = 4 → Point: (4, 4)
Find dy/dx at t = 2. We showed dy/dx = 1/t for this curve: dy/dx = 1/2 at t = 2
Tangent equation: y − y₁ = m(x − x₁) with (x₁, y₁) = (4, 4) and m = 1/2: y − 4 = (1/2)(x − 4) y = (1/2)x − 2 + 4 = (1/2)x + 2
Tangent: y = x/2 + 2 (or equivalently x − 2y + 4 = 0)

Exercise 10 Hard

Find the equation of the normal to the ellipse x = 5cos(t), y = 3sin(t) at t = π/4

Find the tangent slope first (it will involve cot(π/4)), then use the negative reciprocal for the normal slope. Remember cot(π/4) = 1 and the normal slope = −1/m_tangent.

Point at t = π/4: x = 5cos(π/4) = 5/√2, y = 3sin(π/4) = 3/√2
dx/dt = −5sin(t), dy/dt = 3cos(t). Tangent slope: dy/dx = 3cos(t) / (−5sin(t)) = −(3/5)cot(t) At t = π/4: cot(π/4) = 1, so m_tan = −3/5
Normal slope = −1/m_tan = −1/(−3/5) = 5/3
Normal equation through (5/√2, 3/√2) with slope 5/3: y − 3/√2 = (5/3)(x − 5/√2) y = (5/3)x − 25/(3√2) + 3/√2 y = (5/3)x − 25/(3√2) + 9/(3√2) y = (5/3)x − 16/(3√2) or equivalently y = (5/3)x − 8√2/3

Exercise 11 Hard

For x = t², y = t³ − 3t: find d²y/dx² and determine the nature of each stationary point found in Exercise 8

From Exercise 8: dy/dx = (3t² − 3)/(2t). Now differentiate this with respect to t using the quotient rule: Numerator: (3t² − 3), Denominator: 2t d/dt [(3t²−3)/(2t)] = [6t·2t − (3t²−3)·2] / (2t)² = [12t² − 6t² + 6] / (4t²) = (6t² + 6) / (4t²) = (3t² + 3)/(2t²)
Divide by dx/dt = 2t: d²y/dx² = [(3t²+3)/(2t²)] / (2t) = (3t²+3)/(4t³)
At t = 1: d²y/dx² = (3+3)/(4) = 6/4 = 3/2 > 0 → minimum at (1, −2)
At t = −1: d²y/dx² = (3+3)/(4·(−1)) = 6/(−4) = −3/2 < 0 → maximum at (1, 2): (1, −2) is a local minimum. (1, 2) is a local maximum.

Exercise 12 Hard

Show that the tangent to the astroid x = cos³(t), y = sin³(t) at parameter t has constant length 1 between the two coordinate axes

Find dy/dx using the chain rule on cos³t and sin³t. Then find the tangent line at general t, find where it crosses the x-axis (y=0) and y-axis (x=0), and compute the distance between those two intercepts.

dx/dt = −3cos²(t)·sin(t) (chain rule on cos³t). dy/dt = 3sin²(t)·cos(t) (chain rule on sin³t).
Form dy/dx: dy/dx = 3sin²(t)cos(t) / (−3cos²(t)sin(t)) = −sin(t)/cos(t) = −tan(t)
Tangent at point (cos³t, sin³t) with slope −tan t: y − sin³t = −tan(t) · (x − cos³t) y = −tan(t)·x + cos³t·tan(t) + sin³t y = −tan(t)·x + cos²t·sin t + sin³t y = −tan(t)·x + sin t(cos²t + sin²t) = −tan(t)·x + sin t
x-intercept (y=0): 0 = −tan(t)·x + sin t → x = sin t / tan t = cos t. So the tangent meets the x-axis at (cos t, 0).
y-intercept (x=0): y = sin(t). So the tangent meets the y-axis at (0, sin t).
Length of the tangent segment between these two axis points: L = √((cos t)² + (sin t)²) = √(cos²t + sin²t) = √1 = 1 The length of the tangent segment between the two coordinate axes is always 1, regardless of t. ✓

§ 08Common Mistakes

Parametric differentiation has a short list of very specific errors that catch students out. Know them before the exam.

✗ Dividing the wrong way — writing dx/dy instead of dy/dx

Wrong: dy/dx = (dx/dt) / (dy/dt)

Correct: dy/dx = (dy/dt) / (dx/dt), the y derivative on top, the x derivative on the bottom. Think: "I want dy/dx, so dy goes on top."

✗ Finding d²y/dx² by differentiating dy/dt and dx/dt again separately

Wrong: d²y/dx² = (d²y/dt²) / (d²x/dt²). This is not the second derivative of y with respect to x.

Correct: Find dy/dx as a function of t first. Then differentiate that expression with respect to t, and divide by dx/dt. The correct formula is d²y/dx² = [d/dt(dy/dx)] / (dx/dt).

✗ Not substituting t to find the actual point

When a question asks for the tangent "at t = 2", many students find the gradient dy/dx but forget to also find the coordinates (x, y) by substituting t = 2 into both parametric equations. You need the point as well as the slope to write the line equation.

✗ Ignoring the condition dx/dt ≠ 0

The formula dy/dx = (dy/dt)/(dx/dt) requires dx/dt ≠ 0. Where dx/dt = 0, the curve has a vertical tangent and dy/dx is undefined. Always check this, especially when asked about special points on a curve.

✗ Confusing the parameter with the Cartesian variable

When you find that dy/dx = 1/t for the parabola x = t², y = 2t, this is the gradient in terms of t: not in terms of x. If you want the gradient at the point (9, 6) on the curve, you need to find what t gives that point (t = 3), then substitute. You cannot just put x = 9 into 1/t.

§ 09Eliminating the Parameter — and Why It Often Makes Things Worse

Some parametric curves can be converted to a single Cartesian equation by eliminating t. This is sometimes useful for sketching, but it is rarely the best approach for differentiation.

For x = t², y = 2t: from the second equation t = y/2. Substitute into the first: x = (y/2)² = y²/4, so y² = 4x. This is the Cartesian form of the parabola. We could now differentiate implicitly to find dy/dx.

But notice the parametric approach was faster: two easy differentiations and a division, versus implicit differentiation of y² = 4x. For trigonometric parametric curves like x = cos t, y = sin t the Cartesian form x² + y² = 1 requires implicit differentiation anyway, and for complex curves like the cycloid there is no neat Cartesian form at all.

When to eliminate and when not to Eliminate the parameter only when it simplifies things significantly — for example when one equation gives t directly in terms of x and substitution is clean. For trigonometric parametric equations, stick with the parametric differentiation formula. It is almost always faster and less error-prone.

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