Calculus: The Language of Change

Exercise Set 15.5
Differences and Displacements

solutions by Randall Childs and David Theobald

See the text for explicit instructions on using a Mathematica notebook to solve Exercise 2.

Exercise 1

Finish-Start

In Figure 15.52 (page 306), a displacement arrow (or oriented displacement) from the end of R to the end of S is shown.  Sketch the parallel position vector, and give an algebraic formula for it.

The vectors S and R and the green displacement arrow are given by the text.  Denote the parallel position vector (parallel to the displacement arrow) by T.  T = S - R.

T = (({{(Subscript[t, 1])}, {(Subscript[t, 2])}})) = (({{(Subscript[s, 1])}, {(Subscript[s, 2])}})) - (({{(Subscript[r, 1])}, {(Subscript[r, 2])}})) = (({{(Subscript[s, 1]) -  (Subscript[r, 1])}, {(Subscript[s, 2]) -  (Subscript[r, 2])}})) = S  -  R

The red arrow is what you should have sketched (the parallel position vector).

Exercise 3

Give a geometric condition so that the formula |T-S| =  |T| - |S|  is true.  Which of the vectors in the previous exercise (Exercise 2) satisfy your conditions?

In the following diagram, the red arrow is the copy of the vector T-S with the tail placed at the tip of S.

The blue, red, and black arrows trace the sides of a triangle.  There is a geometric condition known as the Triangle Inequality which states the length of any side of a triangle is less than the sum of the lengths of the other two sides.  This is just another way of saying that the shortest distance between two points is a straight line.

So the triangle inequality gives the following relationship between the lengths of T, S, and T-S:

|T-S| + |S| > |T|
or
|T-S| > |T| - |S|

So long as S, 'T-S', and T form a triangle, this inequality will hold.  The reason they form a triangle is because S and T point in different directions.  If they point in the same direction, then the following equation clearly holds:

|T-S| + |S| = |T|
or
|T-S| = |T| - |S|

What if X and Y pointed in diametrically opposite directions?

Convince yourself that if this were the case, then:

|T+S| = | |T| - |S| |
and
|T-S| = |T| + |S|