Multivariable Calculus: Surfaces, Coordinates, and Tangents, Lecture notes of Mathematics

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MAT3100: MULTIVARIABLE CALCULUS

MWALE David

Contents

• 1 FUNCTIONS OF SEVERAL VARIABLES
• 2 Further Differential Calculus
• 3 Surfaces - 3.0.1 Surfaces of Revolution
  • 3.1 Cylindrical and Spherical Coordinates
    • 3.1.5 Tangent and Normal lines
• Bibliography

Chapter 2

Further Differential Calculus

Chapter 3

Surfaces

The aim of this chapter is to present the study of surfaces. However, in these notes we will look at some basic concepts regarding surfaces.

3.0.1 Surfaces of Revolution

In this section, we will look at how to find the equation of a surface obtained after rotating a given curve about one of the axes. The following examples will help us understand the notion here.

Example 3.0.2. Write the equation for the surface obtained by revolving the curve z = y^2 (in the yz -plane) about the z− axis. Solution We begin first by sketching the graph of z = y^2 : We then take any point (0 ; y?; z? ) ; different from the origin, on z = y^2 and consider the resulting point ( x; y; z ) : It is easy to see that z = z?^ and √ x^2 + y^2 = y?: Consequently,

x^2 + y^2 = ( y? )^2 = z?^ = z:

Thus the resulting points satisfy the equation z = x^2 + y^2 :

Example 3.0.3. Write an equation for the surface obtained by rotating the

Figure 3.1: Graph Graph showing relationship between Cylindrical and Rectangular coordi-nates

gular coordinates (2 ; 2 √ 3 ; 8) :

2. Find rectangular coordinates for the point with cylindrical coordinates (5 ; π 6 ; 2) : Solution
3. r =

22 + (2 √ 3)^2 = 4 ; tan θ = √ 3 ; so that θ = π 3 : Thus the set of cylindrical coordinates is (4 ; π 3 ; 8) :

2. From x = r cos θ and y = r sin θ; it is easy to show that (5 ; π 6 ; 2) becomes ( 5 √ 3 2 ;^

2 ;^^2

Example 3.1.2. 1. Find a cylindrical equation for the ellipsoid x^2 + y^2 +4 z^2 = 5 :

2. Find a rectangular equation for the surface whose cylindrical equation is

z = r^2 cos 2 θ: Solution

1. Since r^2 = x^2 + y^2 ; x^2 + y^2 + 4 z^2 = 5 becomes

r^2 + 4 z^2 = 5 :

z = r^2 cos 2 θ = r^2 (cos^2 θ − sin^2 θ ) = r^2 cos^2 θ − r^2 sin^2 θ = x^2 − y^2 ;

and so the surface is the hyperbolic paraboloid.

Example 3.1.3. 1. Find a set of spherical coordinates for the point whose rectangular coordinates are (1 ; 1 ; √ 6) :

2. Find the rectangular coordinates of the point with spherical coordinates (3 ; π 2 ; π 2 ) : Solution
3. Using the above equations ρ^2 = 1^2 + 1^2 + ( √ 6)^2 = 8 : tan φ = √^13 ; so that φ = π 6 : tan θ = 11 = 1 ; so that θ = π 4 : Thus the spherical coordinates are (2 √ 2 ; π 6 ; π 4 ) :

Example 3.1.4. 1. Find a spherical equation for the surface whose rectan- gular equation is x^2 + y^2 + z^2 + 6 z = 0 :

2. Find an equation in spherical coordinates of the ellipsoid

x^2 + y^2 + 9 z^2 = 9 :

3.1.5 Tangent and Normal lines

Let S be a surface F ( x; y; z ) = 0 and let P ( x 0 ; y 0 ; z 0 ) be a point on S: Let C be a curve on S which passes through P given parametrically by g ( t ) = x ( t ) i + y ( t ) j + z ( t ) k. Then, for all t; F ( x ( t ) ; y ( t ) ; z ( t )) = 0 if F is differentiable and x′ ( t ) ; y′ ( t ) and z′ ( t ) all exists, then differentiating using the chain rule gives

∂F

∂t =^

∂F

∂x

dx dt +^

∂F

∂y

dy dt +^

∂F

∂z

dz dt = 0^ or ∂F ∂t =

i ∂F∂x + j ∂F∂y + k∂F∂z

( dx dt i^ +^

dy dt j^ +^

dz dt k

and at P ( x 0 ; y 0 ; z 0 ) ∇F ( x 0 ; y 0 ; z 0 ) · g′ ( t 0 ) = 0 : But g′ ( t 0 ) is a tangent vector to the curve through P: This result then says that the gradient at P; ∇F; is perpendicular to the tangent vector of every curve on S passing through P: Thus, all tangent lines at P lie in a plane that is normal to ∇F ( x 0 ; y 0 ; z 0 ) and contains P: Therefore, if F is differentiable at the point P ( x 0 ; y 0 ; z 0 ) on the surface S given by the equation F ( x; y; z ) = 0 such that ∇F ( x 0 ; y 0 ; z 0 ) 6 = 0 ; then

1. The plane through P which is normal to ∇F ( x 0 ; y 0 ; z 0 ) is called the tangent plane to the surface S at P:
2. The line through P having the direction of ∇F ( x 0 ; y 0 ; z 0 ) is called the normal line to the surface S at P:

Example 3.1.6. Find an equation of the tangent plane to the hyperboloid z^2 − 2 x^2 − 2 y^2 = 12 at the point (1 ; − 1 ; 4) : Find also the unit normal vector to the surface at (1 ; − 1 ; 4) : Solution Let F ( x; y; z ) = z^2 − 2 x^2 − 2 y^2 − 12 then ∇F ( x; y; z ) = − 4 x i − 4 y j + 2 z k: It follows that ∇F (1 ; − 1 ; 4) = − 4 i + 4 j + 8 k is a vector perpendicular to any vector lying in the plane. Let ( x; y; z ) be a point on the plane, then v = ( x −

1. i + ( y + 1) j + ( z − 4) k is a vector lying in the plane. Since the two vectors ∇F and v are perpendicular, we get the equation of the tangent plane by

( − 4 i +4 j +8 k ) · [( x − 1) i + ( y + 1) j + ( z − 4) k ] = − 4( x− 1)+4( y +1)+8( z− 4) = 0 :

equation of the normal line to the surface at the point ( x 0 ; y 0 ; z 0 ) ; we use the gradient ∇F ( x 0 ; y 0 ; z 0 ) for direction numbers, and is given by

x − x 0 Fx ( x 0 ; y 0 ; z 0 ) =^

y − y 0 Fy ( x 0 ; y 0 ; z 0 ) =^

z − z 0 Fz ( x 0 ; y 0 ; z 0 )

Example 3.1.8. Find a set of symmetric equations for the normal line to the surface given by xyz = 12 at the point (2 ; − 2 ; − 3) : Solution From F ( x; y; z ) = xyz − 12 ; it is to show that

∇F (2 ; − 2 ; − 3) = 6 i − 6 j − 4 k:

Consequently, x − 2 6 =^

x + 2 − 6 =^

z + 3 − 4 : Most of the examples done here maybe be found in one or all of the recommended and prescribed readings.

Bibliography

Prescribed Readings

1. Grossman S.T., (1996) Multivariable Calculus, Linear Algebra and Differential Equations, 3rd ed. HBL Publishers. ISBN: 0 155 64751 2 Recommended Readings
2. Stewart J., (2009) Calculus 6th ed. Brooks/Cole Cengage Learning, ISBN: 0 495 38362 8
3. Wylie R.C. and Barret L.C. (1995). Advanced Engineering Mathematics. McGraw-Hill. Company. ISBN: 0 070 72206 4
4. Kreyzig E. (2011). Advanced Engineering Mathematics. 10th Ed. John Willey and Sons. ISBN: 0 470 45836 5