Hints for Lesson VC.02

• G.1.c.i This problem on finding the line that is the intersection of two planes has a pretty cryptic hint. What you are trying to do is find the equation of the line in which the two planes intersect. There are a lot of ways to do this, including:
  • find two points (too easy),
  • find the slope (this vector has to be perpendicular to both planes' normal vectors since it is in both planes),
• G.4 "Experiments with linearizations". In section G.4.b.ii, many astute observers note that some of the plots involving linearf are not actually straight lines. The question is, "Why doesn't `linearf' give straight lines?"

  What is really going on (in my opinion), which the lesson does not make at all clear, is that you are projecting a curve in the xy-plane up onto two different surfaces. I'll try to explain that here.

  You start with a curve in the xy-plane given by {x[t],y[t]}. In this problem t ranges from -0.1 to 0.1, so take a look at this curve:

       ParametricPlot[{x[t],y[t]},{t,-0.1,0.1}];
  

  When you use the first path in the lesson,

       {x[t_],y[t_]} = {0.7 + 4 Sin[t], 0.4 E^t};
  

  you can probably not see it curving for this t range. (If you want to see it curve, increase range.) But for the second path,

       {x[t_],y[t_]} = {0.7 Cos[3 t], 0.4 - t};
  

  the curving is quite visible even in this small range.

  Try these plots and see for yourself how straight the lines look.

  Now let's figure out what these functions f and linearf are doing. If you just plot {x,y,f[x,y]}, you are plotting the part of the surface that is above the xy range you plot. When you plot {x[t],y[t],f[x[t],y[t]}, you are plotting the part of the surface that is directly over the curve you drew in the xy-plane.

  Imagine that you have a sheet of glass underneath your parabola, and a big light that shines directly up through it. When your draw a path on the sheet of glass, you get a shadow on the parabola. The path on the sheet of glass is the {x[t],y[t]} path in the xy-plane. The shadow on the parabola is the {x[t],y[t],f[x[t],y[t]]} path.

  The same story goes for the linearf[x,y] function. When you plot linearf[x,y], you get a plane that touches your parabola at the point you linearized at. You can shine a light up through the sheet of glass (xy-plane) and see the shadow of your path on that plane as well, and that's the {x[t],y[t],linearf[x[t],y[t]]} path.

  • Do you think that the shadow on the linearf plane should be look straight even though the path in the xy-plane is curved? (Hint: no way!)
  • When you draw the plane from linearf and the parabola from f, where do they touch?

  In the lesson, you see only the z-coordinates of these shadow points plotted, so it is sort of hard to get this picture. But now that you have the big picture of what is going on, it should be easier to understand the lesson.

  The question asked by the problem is: when you look at the z-coordinates of the shadows on the plane (from linearf) and the parabola (from f) of any curve that goes through that point you linearized at, ({0.7,0.4} in the lesson) why are they near each other?