Tangents and normals AI HL

IB Mathematics: Applications and Interpretation HL · Unit 5: Calculus

Lesson 5.4 · Tangents and Normals

Find the gradients and equations of lines that touch a curve or meet it at right angles.

How will they hear?

Paul asks, “how shall they hear without a preacher?” (Romans 10:14, KJV). People need opportunities to hear the good news of Jesus Christ: that he died for our sins, rose again and calls us to turn to him in faith. Jesus sends his followers to share this message with people near and far.

When Paul spoke in Athens, he began with what his listeners knew and led them toward the truth about God, repentance and the resurrection (Acts 17:16–34). We can also listen to people, speak clearly and point them to Jesus. As we learn how a tangent meets a curve at a particular point, let us notice the people we meet and be ready to share the hope we have in Christ.

Learning objectives

By the end of this lesson, you should be able to:

find the gradient of a tangent to $y=f(x)$ at a given point;
find the equation of a tangent using its gradient and point of contact;
find the gradient and equation of the normal at that point;
recognise the special case of a horizontal tangent and vertical normal;
find points where a tangent is parallel or perpendicular to a given line;
check tangent and normal equations using graphing technology.

1. Tangent and normal gradients

A tangent has the same gradient as a differentiable curve at its point of contact. At $x=a$ :

$m_{\mathrm{tangent}}=f'(a)$

A normal passes through the same point and is perpendicular to the tangent. If $f'(a)\neq0$ , its gradient is the negative reciprocal:

$m_{\mathrm{normal}}=-\frac{1}{f'(a)}$

The two gradients satisfy $m_{\mathrm{tangent}}m_{\mathrm{normal}}=-1$ when both gradients are defined and nonzero.

Tangent gradient	Normal gradient	Relationship
$2$	$-\dfrac12$	Perpendicular
$-4$	$\dfrac14$	Perpendicular
$0$	Undefined	Horizontal tangent; vertical normal

Special case: If $f'(a)=0$ , the tangent is horizontal and has equation $y=f(a)$ . The normal is vertical and has equation $x=a$ . Do not attempt to divide by zero.

2. Equations at a given point

Both lines pass through the point $(a,f(a))$ . Use the point-gradient form of a straight line:

$y-y_1=m(x-x_1)$

For the tangent:

$\boxed{y-f(a)=f'(a)(x-a)}$

For the normal, provided $f'(a)\neq0$ :

$\boxed{y-f(a)=-\frac{1}{f'(a)}(x-a)}$

Worked example · Tangent and normal to a parabola.

Find both lines to $f(x)=x^2$ at $x=-2$ .

First find the point and tangent gradient:

$f(-2)=4,\qquad f'(x)=2x,\qquad f'(-2)=-4$

The tangent passes through $(-2,4)$ with gradient $-4$ :

$y-4=-4(x+2) \quad\Longrightarrow\quad \boxed{y=-4x-4}$

The normal gradient is $-1/(-4)=1/4$ :

$y-4=\frac14(x+2) \quad\Longrightarrow\quad \boxed{y=\frac14x+\frac92}$

Check that both equations give $y=4$ when $x=-2$ . Their gradients multiply to $-1$ .

A reliable four-step method

Point: Calculate $f(a)$ to find $(a,f(a))$ .
Gradient: Differentiate and evaluate $f'(a)$ .
Tangent: Substitute the point and gradient into $y-y_1=m(x-x_1)$ .
Normal: Use the negative reciprocal, or write $x=a$ if the tangent is horizontal.

3. A horizontal tangent and a vertical normal

Worked example · At the top of a curve.

Find the tangent and normal to $f(x)=4-x^2$ at $x=0$ .

$f(0)=4,\qquad f'(x)=-2x,\qquad f'(0)=0$

The tangent through $(0,4)$ is horizontal:

$\boxed{y=4}$

The normal through the same point is vertical:

$\boxed{x=0}$

A vertical line does not have a finite gradient, so it cannot be written in the form $y=mx+c$ .

4. Finding where a specified gradient occurs

If a tangent must have gradient $k$ , solve:

$f'(x)=k$

There may be more than one solution. Find the corresponding point on the curve for every valid solution.

Worked example · Two tangents with the same gradient.

For $f(x)=1/x$ , where $x\neq0$ , find the points where the tangent gradient is $-1/4$ .

$f'(x)=-\frac{1}{x^2}$

$-\frac{1}{x^2}=-\frac14 \quad\Longrightarrow\quad x^2=4 \quad\Longrightarrow\quad x=-2\ \text{or}\ x=2$

The points are $(-2,-1/2)$ and $(2,1/2)$ . Their tangent equations are:

$\boxed{y=-\frac14x-1} \qquad\text{and}\qquad \boxed{y=-\frac14x+1}$

The same tangent gradient occurs at two different points. At both points, the normal gradient is $4$ .

5. Parallel and perpendicular to a given line

First find the gradient $m$ of the given line.

For a parallel tangent, solve $f'(x)=m$ .
For a perpendicular tangent, solve $f'(x)=-1/m$ , provided $m\neq0$ .

Then substitute each $x$ -value into $f(x)$ to find the points of contact.

Worked example · A tangent parallel to a line.

Find the point on $f(x)=x^2-4x+1$ where its tangent is parallel to $y=2x+5$ .

The given line has gradient $2$ . Since $f'(x)=2x-4$ , solve:

$2x-4=2 \quad\Longrightarrow\quad x=3$

As $f(3)=9-12+1=-2$ , the point is $(3,-2)$ . The tangent is:

$y+2=2(x-3) \quad\Longrightarrow\quad \boxed{y=2x-8}$

Checking with a TI-84 Plus CE

Enter the curve as Y1. Use MATH → 8:nDeriv( to check its gradient at the required $x$ -value.
Enter the tangent and normal as Y2 and Y3, if neither line is vertical.
Graph the equations in the same viewing window. Both lines should pass through the stated point; the tangent should follow the curve’s direction there.
For a vertical normal $x=a$ , use the calculator’s vertical-line drawing feature or check its equation separately. It cannot be entered as $Y3$ .

A graph is useful for checking your answer, but your written solution should show the derivative, point and line equations.

IB-style Paper 1 · Skill practice

Show your method. Mark allocations are shown in square brackets.

Question 1 · Tangent and normal to a parabola [7 marks]

Let $f(x)=x^2$ . At $x=-2$ :

Find the coordinates of the point on the curve and its tangent gradient. [2]
Find the equation of the tangent. [2]
Find the equation of the normal. [3]

Show solution

(a) The point is $(-2,4)$ and $f'(-2)=2(-2)=-4$ . (b) $y-4=-4(x+2)$ , so $y=-4x-4$ . (c) The normal gradient is $1/4$ ; therefore $y-4=\frac14(x+2)$ , or $y=\frac14x+\frac92$ .

Question 2 · Normal at a given point [5 marks]

Find the equation of the normal to $y=4-x^2$ at $x=2$ .

Show solution

At $x=2$ , $y=0$ . The derivative is $\mathrm{d}y/\mathrm{d}x=-2x$ , so the tangent gradient is $-4$ and the normal gradient is $1/4$ . Through $(2,0)$ , the normal is $y=\frac14(x-2)$ .

Question 3 · A horizontal tangent [5 marks]

Let $g(x)=x^3-3x$ .

Find all points at which the tangent is horizontal. [3]
State the equations of the normal lines at those points. [2]

Show solution

(a) $g'(x)=3x^2-3=0$ gives $x=-1$ or $x=1$ . The points are $(-1,2)$ and $(1,-2)$ . (b) Since the tangents are horizontal, the normals are vertical: $x=-1$ and $x=1$ .

Question 4 · Parallel and perpendicular tangents [8 marks]

The curve is $y=x^2-4x+1$ . A line $l$ has equation $y=2x+5$ .

Find the point where the tangent to the curve is parallel to $l$ . [4]
Find the point where the tangent to the curve is perpendicular to $l$ . [4]

Show solution

The derivative is $2x-4$ . (a) For a parallel tangent, $2x-4=2$ , giving $x=3$ and $y=-2$ . The point is $(3,-2)$ . (b) A line perpendicular to $l$ has gradient $-1/2$ . Solve $2x-4=-1/2$ to obtain $x=7/4$ . Substitution into the curve gives $y=-47/16$ . The point is $(7/4,-47/16)$ .

IB-style Paper 2 · Applications

Question 5 · A hillside path [9 marks]

The height of a hillside path above a reference level is modelled by $H(x)=-0.02x^2+1.2x+5$ , where $H$ and the horizontal distance $x$ are measured in metres, $0\leq x\leq40$ . Engineers consider a short straight section touching the path at $x=20$ .

Find $H(20)$ and $H'(x)$ . [3]
Find the gradient and equation of the tangent at $x=20$ . [3]
Find the equation of the normal at that point. [2]
Interpret the tangent gradient in this context. [1]

Show solution

(a) $H(20)=-0.02(400)+1.2(20)+5=21$ m, and $H'(x)=-0.04x+1.2$ . (b) $H'(20)=0.4$ . The tangent through $(20,21)$ is $H-21=0.4(x-20)$ , or $H=0.4x+13$ . (c) The normal gradient is $-1/0.4=-2.5$ , so its equation is $H-21=-2.5(x-20)$ . (d) Near $x=20$ , the path rises approximately $0.4$ m for each additional horizontal metre.

Question 6 · An unknown coefficient [8 marks]

The curve $y=kx^3-4x+5$ has a normal with gradient $-1/8$ at the point $(2,b)$ .

Find the tangent gradient at this point. [2]
Find $k$ . [3]
Find $b$ and hence write the equation of the normal. [3]

Show solution

(a) The tangent is perpendicular to the normal, so its gradient is $8$ . (b) The derivative is $3kx^2-4$ . At $x=2$ , $12k-4=8$ , giving $k=1$ . (c) $b=1(2)^3-4(2)+5=5$ . The normal passes through $(2,5)$ with gradient $-1/8$ , so $y-5=-\frac18(x-2)$ .

Question 7 · Tangents at two points [8 marks]

For $x\neq0$ , the curve has equation $y=1/x$ .

Find its derivative and the $x$ -coordinates where the tangent gradient is $-1/4$ . [3]
Find the coordinates of both points of contact. [2]
Find the equation of each tangent. [3]

Show solution

(a) $\mathrm{d}y/\mathrm{d}x=-1/x^2$ . Solving $-1/x^2=-1/4$ gives $x=-2$ or $x=2$ . (b) The points are $(-2,-1/2)$ and $(2,1/2)$ . (c) At $(-2,-1/2)$ , $y+\frac12=-\frac14(x+2)$ , giving $y=-\frac14x-1$ . At $(2,1/2)$ , $y-\frac12=-\frac14(x-2)$ , giving $y=-\frac14x+1$ .

Exam checklist

Calculate the point of contact from the original function and the tangent gradient from the derivative. The normal uses the negative reciprocal when the tangent gradient is nonzero. For a horizontal tangent, write a vertical normal as $x=a$ . If a gradient condition gives several $x$ -values, check every valid point on the curve.

Lesson 5.4 · Tangents and Normals

How will they hear?

Learning objectives

1. Tangent and normal gradients

2. Equations at a given point

A reliable four-step method

3. A horizontal tangent and a vertical normal

4. Finding where a specified gradient occurs

5. Parallel and perpendicular to a given line

Checking with a TI-84 Plus CE

IB-style Paper 1 · Skill practice

Question 1 · Tangent and normal to a parabola [7 marks]

Question 2 · Normal at a given point [5 marks]

Question 3 · A horizontal tangent [5 marks]

Question 4 · Parallel and perpendicular tangents [8 marks]

IB-style Paper 2 · Applications

Question 5 · A hillside path [9 marks]

Question 6 · An unknown coefficient [8 marks]

Question 7 · Tangents at two points [8 marks]

Exam checklist

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IB Mathematics: Applications and Interpretation HL 6 Sections80 LessonsLifetime

Curriculum

Tangents and normals AI HL

How will they hear?

Learning objectives

1. Tangent and normal gradients

2. Equations at a given point

A reliable four-step method

3. A horizontal tangent and a vertical normal

4. Finding where a specified gradient occurs

5. Parallel and perpendicular to a given line

Checking with a TI-84 Plus CE

IB-style Paper 1 · Skill practice

Question 1 · Tangent and normal to a parabola [7 marks]

Question 2 · Normal at a given point [5 marks]

Question 3 · A horizontal tangent [5 marks]

Question 4 · Parallel and perpendicular tangents [8 marks]

IB-style Paper 2 · Applications

Question 5 · A hillside path [9 marks]

Question 6 · An unknown coefficient [8 marks]

Question 7 · Tangents at two points [8 marks]

Exam checklist

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