Linear Approximations

We can approximate a differentiable function near a point by using a tangent line. Let f (x) be a differentiable function and let (a, f (a)) be a point on the curve representing f. Then the slope at this point is f ' (a). Using the point-slope form of the equation for a line, the equation of the tangent line is then y - f (a) = f ' (a)(x - a). If we solve this for y and rename it as a new function, L(x), we have L(x) = f ' (a)(x - a) + f (a) . This is just the equation for the line tangent to f at x = a. For values of x near a, L(x) is a pretty good approximation to f (x).

Try the following:

1. The applet initially shows the graph of a parabola. The magenta point is the point of tangency and its position is the point (a, f (a)). The blue point is on the curve and is the point (x, f (x)). The black point is on the tangent line and is an approximation to the blue point. The coordinates of the black point are (x, L(x)), where L is the function represented by the tangent line. The vertical black line segment represents the error between f (x) and L(x). Values of f (x), L(x), and the error are shown in the upper right corner of the graph. Click on "Restore Limits" then click on "Zoom In." Note that as you zoom in, the curve looks more and more like a straight line. Move the x slider so that x is close to a, and notice that the error is quite small.
   
   
2. Select the second example, which shows a sine curve and a tangent line at 0. Move the x slider and notice that, if you are close to 0 the error is fairly small, but as you move away from 0 the error can become huge. If you zoom in a couple of times, you will notice that the sine curve in this region becomes like a straight line very quickly, because it isn't curved very much.
   
   
3. Select the third example, which shows the same sine curve but with a tangent line at π / 2. Move the x slider and notice that you need to be quite close to a for the error to be small. Zoom in a few times and you should notice that you need to zoom many more times before the curve looks like a line, because in this region the sine curve is very curvy (or, in other words, the second derivative is farther from 0).
   
   
4. You might have wondered why, in the displayed equation for L(x), the slope used isn't 0 but some very small number. This is caused by rounding errors in two places. The first is due to a = 1.5708 which isn't quite the same as π / 2. But, type "pi / 2" into the a input box.You will notice that the slope is now an even smaller number, but still not 0. That's because π is rounded off by the computer and software to a finite number of digits, so evaluating the derivative happens to give a number really close to, but not quite equal to, zero.
   
   
5. You can enter your own function, pick a point of tangency using the a input box or slider, then pick a point near a using the x input box or slider. Try exponential functions, power functions, etc.

For more information on rights and downloading, refer to http://www.calculusapplets.com/download.html.

© Copyright 2001 David J. Eck

© Copyright 2007 Thomas S. Downey

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