Chapter 3: Many Variables; Section 1: Differentiation; page 1
Function of Many Variables, Exercise 4
Question
Sketch the surfaces obtained by the revolution of the hyperbola
about x and about y!
Both surfaces are called hyperboloid, but one of them is "of one sheet" and the other - "of two sheets". Match them according to their names!
In three dimensions the surfaces are written as
Prove that this is correct ! Which of them is of one sheet ?
The hyperboloid of one sheet, expressed as ,
has a saddle point at x=y=0. Draw the level lines for the z values of 0, 1, 2, and check the consistency with this saddle point.
Reminder
In this three dimensional coordinate system, points corresponding to a constant height (z) form a curve, called a level line that can be plotted on the (x, y) plane, as we find on topographic maps. The level line is obtained by substituting the z of the function by a constant. Mathematically it represents the intersection of the function with the plane of constant z.
Part 1
Solution of question 1
The hyperbola and the two surfaces of revolution are shown at Fig. Sketch for exercise 4.
Part 2
Solution of question 2
As seen from this sketch, the revolution about the x axis yields a hyperboloid of two sheets, and the revolution about y - of one sheet.
Parts 3-6
Solution of question 3
If (a) and (b) are obtained by revolution of the hyperbola about x and the y axes, both should yield back the hyperbola by substituting z=0. Indeed both yield:
The surface of revolution, corresponding to the hyperboloid of one sheet, should yield concentric circles for constant values of y (y=yc). This is obtained for (a):
while (b) yields
.
The surface of revolution, corresponding to the hyperboloid of two sheets, should yield concentric circles for constant values of x (x=xc). This is obtained for (b):
for values of |xc|>1 , while (a) yields
As a consequence of the previous parts, (a) represents the hyperboloid of one sheet, while (b) represents the hyperboloid of two sheets.
Parts 7-9
Solution of question 4
As already found at part 3, for z=0, (a) yields
For z=1 one obtains
The substitution of z=2 in (a) yields a hyperbola
The required level lines are also drawn at the Fig. Sketch for exercise 4. They are consistent with a maximum of z along the x axis at point x=0, and also as a minimum of z along the y axis at point y=0, in agreement with a saddle point at the origin.
Score
By parts.
Parts 3, 4, 5, 6, 7, 8 are worth ½ point each.
Part 2 is worth one point.
Parts 1 and 9 are worth 3 points each.
If the addition is not an integer, add half a point.
By questions.
Questions 1 is worth 3 points.
Questions 2 is worth one point.
Questions 3 is worth 2 points.
Questions 4 is worth 4 points.