The derivative gives a gradient at a point. From there you can write the tangent, write the normal, or read the angle the line makes with the horizontal.
One number, $f'(a)$, unlocks three answers: the equation of the tangent, the equation of the line perpendicular to it, and the angle the curve is climbing at.
Meet the destination, bring back what you already know, and gather the terms and formulas this lesson leans on.
A line has gradient $1$. What angle does it make with the $x$-axis? Now try gradient $\sqrt{3}$. What operation turns a gradient into an angle?
Before you work it out, what is your instinct? Write it down, then check it against the lesson.
Work through the core explanation before applying it.
At $x = a$ the curve has gradient $m = f'(a)$. The tangent is $y - f(a) = m(x - a)$. The normal is perpendicular, with gradient $-\dfrac{1}{m}$. The angle of inclination $\theta$ satisfies $m = \tan\theta$.
Meet the destination, bring back what you already know, and gather the terms and formulas this lesson leans on.
Work through the core explanation before applying it.
At $x = a$ you need two things: the point $(a, f(a))$ and the gradient $m = f'(a)$. Then use point-gradient form, $y - f(a) = m(x - a)$.
For $y = x^2$ at $x = 2$: the point is $(2, 4)$ and $f'(x) = 2x$ gives $m = 4$. So $y - 4 = 4(x - 2)$, which tidies to $y = 4x - 4$.
A common slip is to substitute $x = a$ into $f'(x)$ and then use $(a, f'(a))$ as the point. The gradient comes from $f'$, the point comes from $f$.
A tangent needs the point $(a, f(a))$ from $f$ and the gradient $f'(a)$ from $f'$, then $y - f(a) = f'(a)(x - a)$. For $y = x^2$ at $x = 2$ that gives $y = 4x - 4$.
Pause, copy the two-substitutions rule and the worked $y = 4x - 4$, into your book.
We just saw how to write the tangent. That raises a question: what about the line perpendicular to it? This card answers it → same point, negative reciprocal gradient.
Perpendicular lines have gradients multiplying to $-1$. So if the tangent has gradient $m$, the normal has gradient $-\dfrac{1}{m}$.
Continuing the example: the tangent to $y = x^2$ at $(2, 4)$ has gradient $4$, so the normal has gradient $-\dfrac{1}{4}$ and its equation is $y - 4 = -\dfrac{1}{4}(x - 2)$.
The normal passes through the same point of contact. Only the gradient changes.
The normal shares the point of contact and has gradient $-\dfrac{1}{m}$, the negative reciprocal of the tangent gradient. When $m = 0$ the normal is vertical, $x = a$.
Pause, copy the negative-reciprocal rule, the worked normal $y - 4 = -\dfrac{1}{4}(x - 2)$, and the vertical-normal case, into your book.
We just saw two lines through one point. That raises a question: how steep is that, in degrees? This card answers it → the gradient is the tangent of the angle.
The angle of inclination $\theta$ is the angle the line makes with the positive $x$-axis. Gradient and angle are linked by $m = \tan\theta$.
So $\theta = \tan^{-1}(m)$. A gradient of $1$ gives $45^\circ$; a gradient of $\sqrt{3}$ gives $60^\circ$.
If $m$ is negative the calculator returns a negative angle. Add $180^\circ$ to express the inclination in the usual range $0^\circ \leq \theta < 180^\circ$. For $m = -1$, $\tan^{-1}(-1) = -45^\circ$, so $\theta = 135^\circ$.
The angle of inclination satisfies $m = \tan\theta$, so $\theta = \tan^{-1}(m)$. A negative gradient gives a negative calculator value, and adding $180^\circ$ puts it in the range $0^\circ$ to $180^\circ$.
Pause, copy $m = \tan\theta$, the $45^\circ$ and $60^\circ$ cases, and the add-$180^\circ$ rule for negative gradients, into your book.
Worked examples · 3 in a row, reveal as you go
Follow the reasoning through complete worked solutions.
Find the equations of the tangent and the normal to $y = x^2$ at the point where $x = 2$.
Find the angle of inclination of the tangent to $y = x^2$ at $x = -1$, to the nearest degree.
Find the point on $y = x^2 - 4x$ where the tangent makes an angle of $45^\circ$ with the $x$-axis.
Work through the core explanation before applying it.
Run the quick drill and copy the summary into your book.
At the start you were asked what angle a line of gradient $1$ makes with the $x$-axis, and then gradient $\sqrt{3}$. State both angles, name the operation that converts a gradient to an angle, and say what extra step a negative gradient needs.
Answer the drill bank and rate your confidence.
Pick your answer, then rate your confidence, that tells the system what to drill next. Each retry pulls a fresh mix from the bank.
Write full responses, then check them against the model answers.
Q1. Find the equations of the tangent and the normal to $y = x^3 - 2x$ at the point where $x = 2$. (4 marks)
Q2. Find the angle of inclination, to the nearest degree, of the tangent to $y = x^2 - 6x$ at $x = 1$. (3 marks)
Q3. Explain why the normal to a curve at a point where the tangent is horizontal cannot be written in the form $y = mx + b$, and state its equation instead. (3 marks)
Practice 1: $f'(x) = 3x^2$, so $f'(2) = 12$. Practice 2: $-\dfrac{4}{3}$. Practice 3: $45^\circ$. Practice 4: $2x = 10$, so $x = 5$.
Q1 (4 marks): $f(2) = 8 - 4 = 4$, so the point is $(2, 4)$ [1]. $f'(x) = 3x^2 - 2$, so $m = 12 - 2 = 10$ [1]. Tangent: $y - 4 = 10(x - 2)$, so $y = 10x - 16$ [1]. Normal: gradient $-\dfrac{1}{10}$, so $y - 4 = -\dfrac{1}{10}(x - 2)$ [1].
Q2 (3 marks): $f'(x) = 2x - 6$, so $m = 2 - 6 = -4$ [1]. $\tan^{-1}(-4) = -75.96^\circ$ [1]. Adding $180^\circ$ gives $\theta \approx 104^\circ$ [1].
Q3 (3 marks): A horizontal tangent has gradient $0$, so the normal gradient would be $-\dfrac{1}{0}$, which is undefined [1]. The normal is therefore vertical, and a vertical line has no gradient so cannot be written as $y = mx + b$ [1]. Its equation is $x = a$, where $a$ is the $x$-coordinate of the point of contact [1].
Take the module quiz if you are ready, then mark the lesson complete.
Write tangents and normals, convert between gradient and angle, and locate points with a required gradient. Beat the boss to bank a tier, gold (90% + speed), silver (75%), or bronze (50%). Replays welcome.
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