Pearson Edexcel International Advanced Level Mathematics: Parametric and implicit differentiation (YMA01)

This guide follows the Pearson Edexcel International Advanced Subsidiary/Advanced Level Mathematics specification (Issue 3, April 2019) and covers Pure Mathematics 4 (P4) sections 3.1, 5.1 and 5.2. P4 is a compulsory unit for IAL Mathematics and Pure Mathematics; the specification gives its IAS weighting as N/A, so it is not part of IAS. The specification assumes P1, P2 and P3 knowledge, so the chain, product and quotient rules are taken as known.

Use it with the revision notes and the practice questions. See the Edexcel A-level Mathematics hub, the printable checklist and the free 10-minute diagnostics. Gradients, tangents and normals from P1 are in differentiating powers of x.

What this unit covers

Spec	You must be able to	Where
3.1	Use parametric equations of curves; convert between cartesian and parametric forms	Section 1
5.1	Differentiate simple functions defined implicitly; find tangents and normals	Section 2
5.1	Differentiate simple functions defined parametrically; find tangents and normals	Section 3
5.2	Form simple differential equations	Section 4
5.2	Solve connected rates of change problems	Section 5

Solving a differential equation is spec 6.4, and volume of revolution (6.1) and area under a curve (6.5) from parametric equations are integration topics. None of these is covered here.

1. Parametric equations (3.1)

A curve is given parametrically when x and y are both functions of a third variable, the parameter, usually t or θ. Every parameter value picks out a single point (x, y). The parameter often has a range, such as 0 ≤ t ≤ 2π, and that range limits which part of the curve you get.

To convert to a cartesian equation, eliminate the parameter. Three routes cover most questions:

• Make t the subject of one equation and substitute into the other.
• Use sin²t + cos²t = 1 (or 1 + tan²t = sec²t) when the equations contain trig functions.
• Use a double-angle formula first if one equation has 2t and the other t.

Worked example 1: algebraic elimination

A curve has x = 3t − 1 and y = 2t² − 5t. Find y in terms of x.

t = (x + 1)/3
y = 2(x + 1)²/9 − 5(x + 1)/3
  = [2(x² + 2x + 1) − 15(x + 1)]/9
  = (2x² − 11x − 13)/9

Worked example 2: trig elimination

A curve has x = 2 sin t + 1 and y = 3 cos t − 2. Find its cartesian equation.

sin t = (x − 1)/2     cos t = (y + 2)/3
sin²t + cos²t = 1
((x − 1)/2)² + ((y + 2)/3)² = 1

This is an ellipse with centre (1, −2).

Worked example 3: double angle

A curve has x = 3 cos 2t and y = 4 sin t. Find x in terms of y.

cos 2t = 1 − 2 sin²t
x = 3(1 − 2 sin²t) = 3 − 6(y/4)² = 3 − 3y²/8

So 8x = 24 − 3y². Because −1 ≤ sin t ≤ 1, y lies between −4 and 4, so the cartesian curve is limited to that part.

Going from cartesian to parametric

Choose a parameter that makes both coordinates simple. For y = x³ − 2x you can write x = t, y = t³ − 2t. For x² + y² = 25 write x = 5 cos θ, y = 5 sin θ. Always check that your equations reproduce the original.

Domain matters

Eliminating t can lose a restriction. If x = 1/(t + 2) and y = t/(t + 2), then t = 1/x − 2 and y = 1 − 2x. But x can never be 0, so the point (0, 1) is not on the curve.

2. Implicit differentiation (5.1)

A curve is defined implicitly when x and y are mixed in one equation, such as x² + 2xy + 3y² = 18. You differentiate every term with respect to x. The key rule, from the chain rule, is:

d/dx [f(y)] = f ′(y) × dy/dx

Term	Derivative with respect to x
y²	2y dy/dx
sin y	cos y dy/dx
e^y	e^y dy/dx
ln y	(1/y) dy/dx
xy	y + x dy/dx (product rule)
x²y³	2xy³ + 3x²y² dy/dx

After differentiating, collect every dy/dx term on one side, factorise, then divide.

Worked example 4: tangent and normal

The curve C has equation x² + 2xy + 3y² = 18. Find the tangent and normal at (3, 1).

Check: 9 + 6 + 3 = 18, so (3, 1) is on C.
2x + 2y + 2x dy/dx + 6y dy/dx = 0   (the 2xy term gives 2y + 2x dy/dx)
dy/dx = −(x + y)/(x + 3y)
At (3, 1): dy/dx = −4/6 = −2/3
Tangent: y − 1 = −(2/3)(x − 3)   →   2x + 3y = 9
Normal gradient 3/2: y − 1 = (3/2)(x − 3)   →   3x − 2y = 7

Worked example 5: stationary points

For the same curve, stationary points need dy/dx = 0, so x + y = 0 and y = −x. Substitute into the equation:

x² + 2x(−x) + 3x² = 18
2x² = 18
x = ±3,   y = ∓3

The stationary points are (3, −3) and (−3, 3). The denominator x + 3y is not zero at either.

Worked example 6: exponentials

Find dy/dx at (0, 1) on e^(2x) y + y³ = 2.

Check: e⁰ × 1 + 1 = 2.
2e^(2x) y + e^(2x) dy/dx + 3y² dy/dx = 0
At (0, 1): 2 + dy/dx + 3 dy/dx = 0
dy/dx = −1/2

You can also use dy/dx = 1/(dx/dy) when x is easier to write in terms of y.

3. Parametric differentiation (5.1)

For a curve with x = f(t) and y = g(t), apply the chain rule to get:

dy/dx = (dy/dt) ÷ (dx/dt)

The result is usually a function of t. The tangent is horizontal where dy/dt = 0 (and dx/dt ≠ 0), and vertical where dx/dt = 0 (and dy/dt ≠ 0).

Worked example 7: tangent and normal

A curve has x = t² + t and y = t³ − 9t. Find the tangent and normal at t = 2.

dx/dt = 2t + 1     dy/dt = 3t² − 9
At t = 2: dx/dt = 5, dy/dt = 3, so dy/dx = 3/5
Point: x = 4 + 2 = 6,  y = 8 − 18 = −10
Tangent: y + 10 = (3/5)(x − 6)    →   3x − 5y = 68
Normal gradient −5/3: y + 10 = −(5/3)(x − 6)   →   5x + 3y = 0

Worked example 8: stationary point

A curve has x = e^(2t) and y = t e^t. Find its stationary point.

dx/dt = 2e^(2t)     dy/dt = e^t (1 + t)
dy/dx = e^t (1 + t) / 2e^(2t) = (1 + t)e^(−t)/2

dy/dx = 0 when t = −1. Then x = e⁻² and y = −e⁻¹, so the stationary point is (e⁻², −e⁻¹). Since dx/dt is never 0, the tangent is never vertical.

4. Forming differential equations (5.2)

A differential equation (DE) links a rate of change to the quantities involved. Translate the wording like this:

Statement	Equation
Rate of increase of N is proportional to N	dN/dt = kN, k > 0
Mass M decreases at a rate proportional to M	dM/dt = −kM, k > 0
Rate of change of V is inversely proportional to √V	dV/dt = k/√V
Temperature θ falls at a rate proportional to the excess over 20	dθ/dt = −k(θ − 20)
Inflow 20, outflow proportional to √h	dV/dt = 20 − k√h

The word “decreases” or “falls” puts a minus sign in the equation when k is positive. If the question gives the rate at one instant, substitute to find k.

Worked example 9: a dissolving cube

A sugar cube of side x cm dissolves so that its volume V decreases at a rate proportional to its surface area. Show that dx/dt is constant.

V = x³,  surface area S = 6x²
dV/dt = −λ(6x²)       (λ > 0)
dV/dt = 3x² dx/dt     (chain rule)
3x² dx/dt = −6λx²   →   dx/dt = −2λ

5. Connected rates of change (5.2)

Connect the changing quantities through a chain of derivatives:

dy/dt = dy/dx × dx/dt

Method: write the quantity whose rate you want, find a formula linking it to the quantity whose rate you know, differentiate that formula, then substitute.

Worked example 10: a balloon

A spherical balloon is inflated at 30 cm³ s⁻¹. Find the rates of change of radius and surface area when r = 6.

V = (4/3)πr³    dV/dr = 4πr²
dV/dt = dV/dr × dr/dt   →   30 = 4π(36) dr/dt
dr/dt = 5/(24π) cm s⁻¹
S = 4πr²    dS/dt = 8πr × dr/dt = 8π(6) × 5/(24π) = 10 cm² s⁻¹

Worked example 11: a conical tank with a leak

A tank is an inverted cone with radius equal to h/3 when the water depth is h. Water enters at 20 cm³ s⁻¹.

(a) Find dh/dt when h = 12 with no leak.

V = (1/3)π(h/3)²h = πh³/27     dV/dh = πh²/9
20 = (πh²/9) dh/dt   →   dh/dt = 180/(πh²)
At h = 12: dh/dt = 180/(144π) = 5/(4π) cm s⁻¹

(b) Water also leaks out at a rate k√h. Form a DE for h.

dV/dt = 20 − k√h
(πh²/9) dh/dt = 20 − k√h
dh/dt = 9(20 − k√h)/(πh²)

Common errors

• Forgetting dy/dx on a y term, or writing 2y without it.
• Missing the product rule on xy.
• Putting the fraction upside down in parametric differentiation (it is dy/dt over dx/dt).
• Using the gradient of the tangent as the gradient of the normal.
• Substituting a value of t into dy/dx but using the wrong point for the line.
• Giving a decreasing rate without a minus sign.
• Differentiating a volume formula that still has two variables in it.
• Ignoring a restriction on t after eliminating the parameter.

Official syllabus

Pearson Edexcel International Advanced Subsidiary/Advanced Level in Mathematics, Further Mathematics and Pure Mathematics, Specification, Issue 3, April 2019, unit P4 sections 3.1, 5.1 and 5.2.