Mathematics

Finding the tangent

The derivative gives you the slope at one place. In three steps that becomes the equation of the line that hugs the curve there.

What you need first

Intro to derivatives Linear functions

The tells you how steep a curve is at a particular place. That is a number. But often you want more: the straight line that touches the curve exactly there and has the same slope. It is called the tangent, and it is the best straight approximation to the curve near that point. Anyone who can find a tangent can also extrapolate growth or join curves together without a kink at the seam.

Three steps, always the same

A straight line has the form $y = mx + b$ , so you need two numbers. You get the slope $m$ from the derivative: $m = f'(x_0)$ . You get the intercept $b$ by substituting the point of contact, because the tangent has to pass through it. And the point of contact itself is $(x_0 | f(x_0))$ .

An example: $f(x) = x^2$ at the place $x_0 = 3$ . First the point of contact: $f(3) = 9$ , so $(3|9)$ . Second the slope: $f'(x) = 2x$ , so $f'(3) = 6$ . Third substitute: $9 = 6 \cdot 3 + b$ , which gives $b = -9$ . The tangent is $y = 6x - 9$ .

Coordinate grid

y = 1x + 1Starts at 1, +1 per step right

Set the line to

m = 6

and

b = -9

, the tangent from the example. It runs through the point (3|9), which you can find on the grid, and meets the

x

axis at 1.5. Change

b

by a small amount: the line shifts in parallel and no longer touches the parabola. Exactly this one combination of slope and height fits, which is why the tangent is unique.

y = f'(x_0) \cdot (x - x_0) + f(x_0)

The same tangent in one line: slope times distance from the point of contact, plus the height there.

This short form saves you from solving for $b$ . Substitute the three numbers and expand, and the line is finished. For the example: $y = 6 \cdot (x - 3) + 9 = 6x - 18 + 9 = 6x - 9$ . Both routes give the same answer, take whichever comes more easily.

The normal

Sometimes the line you want is not the one that hugs the curve but the one perpendicular to it. It is called the normal and has slope $-\frac{1}{m}$ . So a tangent slope of 6 becomes $-\frac{1}{6}$ . The rest of the calculation is identical: the same point, the same three steps, only a different slope.

Exercises

0 of 6 solved

Time to try it yourself. You can't break anything, every attempt counts.

What is a tangent to the graph of a function?

For $f(x) = x^2$ the derivative is $f'(x) = 2x$ . What slope does the tangent have at $x_0 = 5$ ?

For $f(x) = x^2$ the tangent at $x_0 = 2$ is wanted. What is it?

Put the steps for finding a tangent into the right order.

1Write down the finished line equation and do the point check.
2Put the point and the slope into $y = mx + b$ and solve for $b$ .
3Substitute $x_0$ into the derivative, that is the slope $m$ .
4Find the derivative $f'(x)$ .
5Work out the function value $f(x_0)$ , the height of the point of contact.

For $f(x) = x^3$ the derivative is $f'(x) = 3x^2$ . What slope does the tangent have at $x_0 = -2$ ?

The normal is … to the tangent.

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