IB Mathematics: Analysis and Approaches SL/HL — Topic 5 Calculus

Tangents and Normals

Use derivatives to turn a curve problem into a straight-line problem: first find the point and gradient, then write the tangent or perpendicular normal.

AA SL/HL · SL 5.4

Learning goal

Find equations of tangents and normals to curves using derivatives, straight-line equations and perpendicular-gradient relationships.

Syllabus link

IB Mathematics AA SL/HL: SL 5.4. Tangents and normals at a given point; application of derivatives to gradients.

Big idea

The derivative supplies the tangent gradient. The normal passes through the same point and is perpendicular to the tangent.

Key relationships

\(m_t=f'(a)\), \(m_n=-1/m_t\) when \(m_t\ne0\), and \(y-y_1=m(x-x_1)\).

What are a tangent and a normal?

At a chosen point on a smooth curve, the tangent captures the curve's instantaneous direction. The normal is the straight line through the same point that is perpendicular to that tangent.

Tangent

The tangent touches the curve at the point and has the same instantaneous gradient as the curve there.

\[m_t=f'(a).\]

Normal

The normal passes through the same point and is perpendicular to the tangent. When the tangent gradient is non-zero,

\[\boxed{m_n=-\frac1{m_t}}.\]

Curve with tangent and perpendicular normal at a common point — Tangent and normal share the point of contact; their directions are perpendicular.

Straight-line facts we will use

Point-gradient form

\[\boxed{y-y_1=m(x-x_1)}\]

This is usually the quickest form once you know a point and gradient.

Perpendicular gradients

\[m_1m_2=-1\]

Equivalent to \(m_n=-1/m_t\) when neither line is vertical.

Do not forget the point

A gradient alone does not determine a line. A tangent or normal question therefore normally needs both the gradient and the point on the curve.

A reliable tangent method

Tangent checklist

Find the point on the curve by evaluating \(f(a)\).
Differentiate to find \(f'(x)\).
Evaluate \(f'(a)\) to get the tangent gradient.
Use \(y-y_1=m(x-x_1)\).
Simplify the equation if required.

Worked example 1: tangent to a cubic

Find the tangent to \(f(x)=x^3-4x+1\) at \(x=2\).

Step 1: point.

\[f(2)=8-8+1=1,\]

so the point is \((2,1)\).

Step 2: gradient.

\[f'(x)=3x^2-4\quad\Rightarrow\quad f'(2)=12-4=8.\]

Step 3: line equation.

\[y-1=8(x-2)\quad\Rightarrow\quad\boxed{y=8x-15}.\]

Cubic curve with tangent line at x equals 2 — The tangent passes through \((2,1)\) and has gradient 8.

Video: Equation of a tangent

Follow the full derivative → gradient → point-gradient sequence in a worked example.

A reliable normal method

Normal checklist

Find the point on the curve.
Find the tangent gradient from the derivative.
Convert it to the perpendicular normal gradient.
Use point-gradient form through the same point.

Worked example 2: normal to a quadratic

Find the normal to \(y=x^2-4x+2\) at \(x=1\).

The point is

\[y(1)=1-4+2=-1,\]

so \(P=(1,-1)\).

Differentiate:

\[\frac{dy}{dx}=2x-4\quad\Rightarrow\quad m_t=2(1)-4=-2.\]

The normal gradient is

\[m_n=-\frac1{-2}=\frac12.\]

Therefore

\[y+1=\frac12(x-1).\]

Multiplying by 2 and rearranging:

\[\boxed{x-2y-3=0}.\]

Quadratic curve and its normal at x equals 1 — The normal has gradient \(1/2\), the negative reciprocal of the tangent gradient \(-2\).

Video: Equation of a normal

See how the tangent gradient is converted into the perpendicular normal gradient before writing the line equation.

Tangent and normal at the same point

Worked example 3: find both lines

Find the tangent and normal to

\[y=\frac4x+x^2\]

at \(x=2\).

Point:

\[y(2)=\frac42+4=6,\]

so \(P=(2,6)\).

Tangent gradient:

\[\frac{dy}{dx}=-\frac4{x^2}+2x\quad\Rightarrow\quad m_t=-1+4=3.\]

Tangent:

\[y-6=3(x-2)\quad\Rightarrow\quad\boxed{y=3x}.\]

Normal: \(m_n=-1/3\), so

\[y-6=-\frac13(x-2)\quad\Rightarrow\quad\boxed{x+3y-20=0}.\]

Curve with both tangent and normal drawn at x equals 2 — Both straight lines pass through \((2,6)\); their gradients multiply to \(-1\).

Special case: horizontal tangent and vertical normal

If the tangent gradient is zero, the negative-reciprocal formula would involve division by zero. Geometrically, the answer is straightforward: a horizontal tangent has a vertical normal.

Worked example 4: horizontal tangent

For \(f(x)=x^3-3x+2\), find the tangent and normal at \(x=1\).

First find the point:

\[f(1)=1-3+2=0,\]

so the point is \((1,0)\).

Differentiate:

\[f'(x)=3x^2-3\quad\Rightarrow\quad f'(1)=0.\]

Therefore the tangent is horizontal:

\[\boxed{y=0}.\]

The normal is vertical through \(x=1\):

\[\boxed{x=1}.\]

Curve with horizontal tangent and vertical normal at x equals 1 — When \(m_t=0\), write the vertical normal directly as \(x=x_1\).

Exam-style unknown constants

Worked example 5: use a point and a normal gradient to determine parameters

The curve

\[y=ax^2+bx+1\]

passes through \((2,7)\). At that point, the normal has gradient \(-1/5\). Find \(a\), \(b\), and the equation of the normal.

Equation 1: use the point.

\[7=4a+2b+1\quad\Rightarrow\quad 2a+b=3.\]

Equation 2: use the normal gradient. If \(m_n=-1/5\), then the tangent gradient is 5.

\[\frac{dy}{dx}=2ax+b.\]

At \(x=2\):

\[4a+b=5.\]

Subtracting the equations gives \(2a=2\), so \(a=1\), then \(b=1\).

The normal through \((2,7)\) with gradient \(-1/5\) is

\[y-7=-\frac15(x-2)\quad\Rightarrow\quad\boxed{x+5y-37=0}.\]

Quadratic curve and normal used in an unknown constants problem — Derivative information can generate an extra equation for unknown coefficients.

Practice

A. Tangents

Find the tangent to \(f(x)=x^2+3x-2\) at \(x=1\).
Find the tangent to \(f(x)=2x^3-x^2+4\) at \(x=-1\).
Find the tangent to \(y=\frac3x+x^2\) at \(x=1\).
The tangent to \(y=x^2+kx\) at \(x=2\) is parallel to \(y=5x-1\). Find \(k\), then find the tangent equation.

B. Normals

Find the normal to \(y=x^2-4x+2\) at \(x=1\).
Find the normal to \(f(x)=x^3-6x\) at \(x=2\).
For \(y=\frac4x+x\), find the tangent and normal at \(x=2\).
The normal to \(y=x^2+kx+3\) at \(x=1\) is parallel to \(y=-\frac14x+10\). Find \(k\), then find the normal equation.

C. Extended

For \(f(x)=x^3-3x^2+4\), find the tangent and normal at \(x=3\). Hence find where the normal crosses the y-axis.
The curve \(y=ax^2+bx+1\) passes through \((2,7)\), and the normal there has gradient \(-1/5\). Find \(a\), \(b\) and the normal equation.

Concise answer key

1. \(y=5x-3\)

2. \(y=8x+9\)

3. \(y=-x+5\)

4. \(k=1\), tangent \(y=5x-4\)

5. \(x-2y-3=0\)

6. \(x+6y+22=0\)

7. Point \((2,4)\), tangent \(y=4\), normal \(x=2\)

8. \(k=2\), normal \(x+4y-25=0\)

9. Point \((3,4)\), tangent \(y=9x-23\), normal \(x+9y-39=0\); y-intercept \((0,13/3)\).

10. \(a=1\), \(b=1\), normal \(x+5y-37=0\).

Summary

The complete method

\[ \text{curve}\ \xrightarrow{\text{evaluate}}\ \text{point}\qquad \text{curve}\ \xrightarrow{\text{differentiate}}\ \text{tangent gradient}\qquad m_n=-\frac1{m_t}. \]

Finish by using the relevant straight-line gradient with the point-gradient equation \(y-y_1=m(x-x_1)\).

Final checks

Tangent and normal must pass through the same point on the curve.
If both gradients are finite and non-zero, their product should be \(-1\).
If the tangent is horizontal, write the normal as a vertical line rather than trying to compute \(-1/0\).