The Math Forum

Search All of the Math Forum:

Views expressed in these public forums are not endorsed by NCTM or The Math Forum.

Math Forum » Discussions » Courses » ap-calculus

Notice: We are no longer accepting new posts, but the forums will continue to be readable.

Topic: [ap-calculus] Is it possible to zoom in on a curve?
Replies: 0  

Advanced Search

Back to Topic List Back to Topic List  

Posts: 108
Registered: 12/8/04
[ap-calculus] Is it possible to zoom in on a curve?
Posted: Mar 10, 2011 10:55 AM
  Click to see the message monospaced in plain text Plain Text   Click to reply to this topic Reply

Yesterday I had a discussion withcolleagues over a topic that seems to be common practice in theteaching of the Calculus, but upon reflection seems to be mistaken.

The following quote comes from apopular calculus text. The argument it presents is quite common inboth calculus text books and calculus classrooms.

?We sometimes refer to the slope ofthe tangent line to a curve at a point as the slope of the curve atthe point. The idea is that if we zoom in far enough towards thepoint, the curve looks almost like a straight line. ? The more wezoom in the more the parabola looks like a line. In other words thecurve becomes almost indistinguishable from its tangent line.?(Steward, p. 150)

Let's consider a parabola given by theequation y = x^2 and the tangent line to this curve at the origin,namely the x-axis. We will need to consider two frames. The firstframe is defined by a rectangle with base from the origin, (0,0), tothe position (b,0). The height of the rectangle will start at theorigin and rise to the position (0,b^2). The second frame will scalethe first frame by a factor, f, where 0<f<1. So the secondrectangle has base from (0,0) to (fb,0) and height from (0,0) to(0,[fb]^2). The range of the scale factor f accounts for ?zoomingin?.

Next consider a point in the firstframe at x=a, and the height of the parabola at this point, y=a^2. Inthe scaled frame the point in the same relative position on the baseis given by x=fa. The height of the parabola in the scaled frame isthen y=(fa)^2. In the first frame the relative height of the parabolaat the relative position a/b is a^2/b^2. In the second frame therelative height of the parabola at the same relative position, fa/fb,is (fa)^2/(fb)^2. And consequently a^2/b^2=(fa)^2/(fb)^2. Therelative heights of the parabolas in the original and scaled framesat the same relative position are equal. The curve and the tangentline do not become indistinguishable. In fact the picture does noteven change! The original frame and the scaled frame look exactly thesame.

It seems that this misperception hasbecome the basis for an enduring misconception.

So it goes.

Bill Crombie

Course related websites:
To search the list archives for previous posts go to
To unsubscribe click here:
To change your subscription address or other settings click here:

Point your RSS reader here for a feed of the latest messages in this topic.

[Privacy Policy] [Terms of Use]

© The Math Forum at NCTM 1994-2018. All Rights Reserved.