Recently, I had to write a brief introduction to the precalculus topic "Vertical Translation of Graphs." Figure 1 ( in black, in red) says just about everything. 


Figure 1   The red curve () is the black curve () vertically translated upward by one unit. 


But is the issue all that trivial? Although the curves are vertically separated by one unit, they don't look uniformly spaced. The animation in Figure 2 helps overcome the optical illusion that makes it seem like the black curve bends towards the red curve, even though the curves are congruent. 



Figure 2   The length of the vertical green segment remains 1 


For sure, the two curves get "closer" as increases.  For this, the distance between the curves has to be measured along the normal to one of them.  To measure along the normal to , we need the intersection of the line normal to the graph of , namely, 



and the curve . Maple computes this intersection in Table 1, where we return the endpoints of the segment whose length is the measurement we want. 


[a, `*`(`^`(a, 2))]

[`+`(`/`(`*`(`/`(1, 2), `*`(`+`(`-`(1), `*`(2, `*`(`^`(a,
                        `*`(a))), `+`(`/`(`*`(`/`(1, 4),
                        `*`(`+`(1, `*`(4, `*`(`^`(a, 4)))))),
                        `*`(`^`(a, 2))))]


Table 1   Endpoints of the line segment between and , the segment being normal to the graph of  


The length of this line segment, given in Table 2, clearly isn't constant, and tends to zero as increases 

`+`(`/`(`*`(`/`(1, 4), `*`(`^`(`+`(`*`(4, `*`(`^`(a, 2))), 1), `/`(1, 2)))), `*`(`^`(a, 2))))

Table 2   Distance between graphs of and , measured along normal to graph of  

 Given the graph of , what curve would be equidistant, measured along the normal to the graph of ? This is the parallel curve, whose derivation appears in Maple via Calculus, a Birkhäuser text I wrote in 1994.  There, the derivation is based on a prior calculation of the coordinates of the center of curvature. Not recalling that detail, I re-derived the result I wanted by evaluating the arc length integral for the segment.  This calculation is given in Table 3. 


`/`(`*`(`+`(`*`(`^`(`+`(`*`(4, `*`(`^`(a, 2))), 1), `/`(1, 2))), 2), `*`(a)), `*`(`^`(`+`(`*`(4, `*`(`^`(a, 2))), 1), `/`(1, 2))))

`/`(`*`(`+`(`*`(`^`(a, 2), `*`(`^`(`+`(`*`(4, `*`(`^`(a, 2))), 1),
                        `/`(1, 2)))), `-`(1))), `*`(`^`(`+`(`*`(4, `*`(`^`(a, 2))), 1), `/`(1,


Table 3   Parametric representation of the curve parallel to and one unit away 


Figure 3 contains a graph of (in black) and the parallel curve in red.  The blue segments on the right are all of length one, and are orthogonal to the black curve.  The increasing lengths of the vertical green segments on the left show that the parallel curve is not a vertical translation. 



Figure 3   In red, the curve parallel to the graph of . The blue segments are all of length 1; the vertical green segments are of nonconstant length. 

So, what started out as a writing assignment I would have been happier to skip, turned out to be useful in two ways.  First, I found something interesting to investigate; second, I again noted the value of Maple in mathematical experimentation and investigation.  Although the calculations are not beyond technology-free manipulations, they would have been tedious. And certainly, the graphs would not have been easily drawn by hand. 

I'm particularly struck by the use of the arc length integral to move along a line a fixed distance. It was a sudden flash of an idea, and with Maple its implementation was quick and effective. That was the notion that infused the fifteen years of curriculum development with Maple while teaching at RHIT - the conviction that "it's better with Maple than without." 

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