Maths with Lemon

Implicit Differentiation & Related Rates

IB Mathematics: Analysis and Approaches Higher Level · AHL 5.14 · Lesson 10.1. Differentiate equations involving both variables, connect changing quantities through time, and solve geometric rate problems with correct units and signs.

Lesson 10.1 worksheet (PDF)AA Calculus overviewDifferentiation rules

1. The Idea Behind Implicit Differentiation

What you need to know

  • Explicit functions are written as \(y=f(x)\).
  • An implicit relation may mix \(x\) and \(y\), such as \(x^2+y^2=4\).
  • When differentiating terms involving \(y\), use the chain rule.

Key points

Treat \(y\) as a function of \(x\). Differentiate both sides with respect to \(x\).

\[\frac{d}{dx}(y^n)=ny^{n-1}\frac{dy}{dx}\]
Worked example: a circle

Differentiate the equation \(x^2+y^2=4\).

\[2x+2y\frac{dy}{dx}=0\]
\[\boxed{\frac{dy}{dx}=-\frac{x}{y}}\]

The gradient is undefined where \(y=0\), corresponding to vertical tangents on the circle.

A common error is differentiating \(y^2\) as \(2y\). The missing factor is \(\frac{dy}{dx}\).

2. Differentiating Implicit Equations

From your worksheet · Page 2

  • Differentiate each side term by term.
  • Collect terms containing \(\frac{dy}{dx}\).
  • Factor and isolate the derivative.
  • Keep your answer in terms of \(x\) and \(y\), unless asked otherwise.

Step-by-step example

Find the derivative of \(x^3=5y^3+3\)

Step 1: Differentiate with respect to \(x\).

\[3x^2=15y^2\frac{dy}{dx}\]

Step 2: Make the derivative the subject.

\[\boxed{\frac{dy}{dx}=\frac{x^2}{5y^2}}\]
You try: \(4x^3=-2y^2+2\)
\[12x^2=-4y\frac{dy}{dx}\]
\[\frac{dy}{dx}=-\frac{3x^2}{y}\]
You try: \(5x^3=y^2+5\)
\[\frac{dy}{dx}=\frac{15x^2}{2y}\]

3. Finding the Gradient at a Point

From your worksheet · Page 2

  • Differentiate the entire relation first.
  • Substitute the given coordinates into the derivative.
  • Check that the point satisfies the original relation.
  • Use the gradient to write a tangent or normal equation if required.

Worked example

Gradient of \(3x^3+2y^2=5\) at \((-1,2)\)

Differentiate implicitly:

\[9x^2+4y\frac{dy}{dx}=0\]
\[\frac{dy}{dx}=-\frac{9x^2}{4y}\]

Substitute \(x=-1\), \(y=2\):

\[\boxed{m=-\frac98}\]

Since \(3(-1)^3+2(2)^2=5\), the point lies on the curve.

You try: \(x=y^3+1\) at \((2,1)\)
\[1=3y^2\frac{dy}{dx}\quad\Rightarrow\quad\boxed{\frac{dy}{dx}=\frac13}\]

4. Related Rates: The Four-Step Method

What you need to know

  • Rates describe change with respect to time \(t\).
  • Increasing quantities have positive rates; decreasing quantities have negative rates.
  • Build an equation connecting the quantities before differentiating.
  • Keep units consistent: length/time, area/time or volume/time.

Method

Four steps for success
  1. Sketch and label the quantities that change.
  2. Write a geometric or algebraic relationship.
  3. Differentiate both sides with respect to \(t\).
  4. Substitute the instantaneous values and solve, including units.
\[\frac{d}{dt}(x^2+y^2)=2x\frac{dx}{dt}+2y\frac{dy}{dt}\]
Substitute the numerical dimensions after differentiating. A changing dimension is not a constant.

5. Rectangles and Right Triangles

Worksheet · Pages 3–4

  • Rectangle area and perimeter.
  • Right-triangle area.
  • Changing hypotenuse via Pythagoras.
  • Negative rates represent shrinking lengths.

Essential formulas

\[A=lw\quad\Rightarrow\quad\frac{dA}{dt}=l\frac{dw}{dt}+w\frac{dl}{dt}\]
\[c^2=a^2+b^2\quad\Rightarrow\quad c\frac{dc}{dt}=a\frac{da}{dt}+b\frac{db}{dt}\]
Worksheet example: rectangle area

At an instant, \(l=7\) cm and \(w=1\) cm. The length increases at 3 cm/s and the width decreases at 5 cm/s.

\[\frac{dA}{dt}=7(-5)+1(3)\]
\[\boxed{\frac{dA}{dt}=-32\ \mathrm{cm^2/s}}\]

The area is decreasing at that instant.

You try: rectangle \(l=9\), \(w=5\), rates \(4\) and \(2\) in/s
\[\frac{dA}{dt}=9(2)+5(4)=\boxed{38\ \mathrm{in^2/s}}\]

6. Circles, Spheres, Cylinders and Cones

Worksheet · Pages 5–6

  • Differentiate area and volume formulas with respect to time.
  • For cylinders and cones, both radius and height may change.
  • If volume is constant, set \(\frac{dV}{dt}=0\).
  • Express answers in the correct squared or cubed units per time.

Formula collection

\[A_{\mathrm{circle}}=\pi r^2\quad\Rightarrow\quad\frac{dA}{dt}=2\pi r\frac{dr}{dt}\]
\[V_{\mathrm{sphere}}=\frac43\pi r^3\quad\Rightarrow\quad\frac{dV}{dt}=4\pi r^2\frac{dr}{dt}\]
\[V_{\mathrm{cylinder}}=\pi r^2h\]
\[\frac{dV}{dt}=2\pi rh\frac{dr}{dt}+\pi r^2\frac{dh}{dt}\]
\[V_{\mathrm{cone}}=\frac13\pi r^2h\]
Worked example: shrinking sphere

A sphere has radius 8 cm and the radius decreases at 7 cm/min.

\[\frac{dV}{dt}=4\pi(8)^2(-7)\]
\[\boxed{\frac{dV}{dt}=-1792\pi\ \mathrm{cm^3/min}}\]

7. Interactive Related-Rates Calculator

Explore how rates interact

  • Choose a rectangle or circle.
  • Enter the dimensions at a particular instant.
  • Enter positive or negative rates of change.
  • Compare the calculated rate with your own differentiation.

Instantaneous rate explorer

Use a consistent length unit and time unit (for example cm and seconds).

8. Multi-Equation Related Rates and HL Practice

Worksheet · Pages 7–8

  • Connect area, perimeter and side length of squares.
  • Connect volume, surface area and side length of cubes.
  • Relate sphere volume to its surface area.
  • Use the chain rule across multiple equations.

Worked example: changing cube

A cube grows in volume at 100 ft³/min when its side length is 2 ft

Start with \(V=s^3\) and \(S=6s^2\).

\[\frac{dV}{dt}=3s^2\frac{ds}{dt}\quad\Rightarrow\quad\frac{ds}{dt}=\frac{100}{3(2)^2}=\frac{25}{3}\]
\[\frac{dS}{dt}=12s\frac{ds}{dt}=12(2)\left(\frac{25}{3}\right)\]
\[\boxed{\frac{dS}{dt}=200\ \mathrm{ft^2/min}}\]
Quick check: if a circle’s radius decreases at 3 in/min when \(r=8\), how quickly is its area changing?
\[\frac{dA}{dt}=2\pi(8)(-3)=\boxed{-48\pi\ \mathrm{in^2/min}}\]

Open the full Lesson 10.1 worksheet

9. Revision Checklist

Before your assessment

  • Apply the chain rule to expressions involving \(y\).
  • Find derivatives and gradients at specified points.
  • Differentiate with respect to time in related-rates problems.
  • Use product rule when two changing dimensions are multiplied.
  • Interpret signs, units and physical constraints.

Common mistakes to avoid

Implicit differentiation

Do not forget \(\frac{dy}{dx}\) when differentiating terms in \(y\).

Related rates

Do not replace changing variables by their instantaneous values before differentiation.

For optimisation problems, also check endpoints when they are included in the feasible domain, as noted in AHL 5.14.

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