The Tangent Line Approximation

Suppose we want to find the tangent to the curve shown at point P.
How can we go about finding the tangent line?

The curve with which we'll work is the parabola

f(x) = x2

Let's find the equation of the line tangent to the parabola at (1,1).

The slope of the tangent is just f′(x) evaluated at x.

f′(x) = 2x
f′(1) = 2

Now, the equation of the line can be written in point-slope form like this:

y-y0 = m(x-x0)
y-y0 = f′(x0)(x-x0)
y-1 = 2(x-1)
since the line passes through the point (1,1) and has slope 4.

In slope-intercept form, the equation of the tangent line becomes

y = 2x-1

  • Notice how the equation of the tangent line changes as you move point P.

What happens when x = 0 for this function? What about as |x| gets large?

Now that we can find the tangent to a curve at a point, of what use is this?

  • "Magnify" the parabola by zooming in on point P. Here a zoom of factor 4 of the prevoius graph:

Do you notice that as you zoom in on P the curve looks more and more linear and is approximated better and better by the tangent line?

Let's get more specific:

Near x0, we saw that y = f(x) can be approximated by the tangent line y-y0 = f′(x0)(x-x0). Writing this as y = y0 + f′(x0)(x-x0) and noting that y = f(x0), we find that

f(x) ≈ f(x0) + f′(x0)(x-x0)

(we shall see that the right-hand side is just the 2-term Taylor Expansion of f(x); f′(x0)(x-x0) è il differenziale di f in x0)

If we know the value of f at x0, this gives us a way to approximate the value of f at x near x0. We do this by starting at (x0,f(x0)) and moving along the tangent line to approximate the value of the function at x.

Look at f(x) = arctan(x).

We know that f(1) = π/4 since tan(π/4) = 1.

Let's use the tangent approximation f(x) ≈ f(x0) +f′(x0)(x-x0) to approximate f(1.04):

  • Now f′(x) = [1/(1+x2)] so f′(1) = [1/(1+12)] = 1/2.

  • Let x0 = 1 and x = 1.04.

  • Then f(1.04) ≈ f(1) + f′(1)(1.04 - 1) ≈ π/ 4 + 1/2 (0.04) ≈ 0.81

How well does this approximate arctan(1.04)?

  • Display the tangent through (1, π/4).

  • Zoom in on the point to see geometrically how close together the curve and the tangent line are at x = 1.04.

Key Concepts [index]

  • For the curve y = f(x), the slope of the tangent line at a point (x0,y0) on the curve is f′(x0). The equation of the tangent line is given by

    y-y0 = f′(x0)(x-x0)

  • For x close to x0, the value of f(x) may be approximated by

    f(x) ≈ f(x0) + f′(x0)(x-x0)