16
This presentation is the property of its rightful owner.
Sponsored Links
1 / 107

VECTOR CALCULUS PowerPoint PPT Presentation


  • 181 Views
  • Uploaded on
  • Presentation posted in: General

16. VECTOR CALCULUS. VECTOR CALCULUS. So far, we have considered special types of surfaces: Cylinders Quadric surfaces Graphs of functions of two variables Level surfaces of functions of three variables. VECTOR CALCULUS.

Download Presentation

VECTOR CALCULUS

An Image/Link below is provided (as is) to download presentation

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.


- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -

Presentation Transcript


Vector calculus

16

VECTOR CALCULUS


Vector calculus

VECTOR CALCULUS

  • So far, we have considered special types of surfaces:

    • Cylinders

    • Quadric surfaces

    • Graphs of functions of two variables

    • Level surfaces of functions of three variables


Vector calculus1

VECTOR CALCULUS

  • Here, we use vector functions to describe more general surfaces, called parametric surfaces, and compute their areas.


Vector calculus2

VECTOR CALCULUS

  • Then, we take the general surface area formula and see how it applies to special surfaces.


Vector calculus

VECTOR CALCULUS

16.6

Parametric Surfaces and their Areas

  • In this section, we will learn about:

  • Various types of parametric surfaces

  • and computing their areas using vector functions.


Introduction

INTRODUCTION

  • We describe a space curve by a vector function r(t) of a single parameter t.

  • Similarly,we can describe a surface by a vector function r(u, v) of two parameters u and v.


Introduction1

INTRODUCTION

Equation 1

  • We suppose that r(u, v) = x(u, v) i + y(u, v) j + z (u, v) kis a vector-valued function defined on a region D in the uv-plane.


Introduction2

INTRODUCTION

  • So x, y, and z—the component functions of r—are functions of the two variables u and v with domain D.


Parametric surface

PARAMETRIC SURFACE

Equations 2

  • The set of all points (x, y, z) in such that x = x(u, v) y = y(u, v) z = z(u, v)and (u, v) varies throughout D, is called a parametric surface S.

    • Equations 2 are called parametric equationsof S.


Parametric surfaces

PARAMETRIC SURFACES

  • Each choice of u and v gives a point on S.

  • By making all choices, we get all of S.


Parametric surfaces1

PARAMETRIC SURFACES

  • In other words, the surface S is traced out by the tip of the position vector r(u, v) as (u, v) moves throughout the region D.


Parametric surfaces2

PARAMETRIC SURFACES

Example 1

  • Identify and sketch the surface with vector equation r(u, v) = 2 cos ui + vj + 2 sin uk

    • The parametric equations for this surface are: x = 2 cos uy = vz = 2 sin u


Parametric surfaces3

PARAMETRIC SURFACES

Example 1

  • So, for any point (x, y, z) on the surface, we have:x2 + z2 = 4 cos2u + 4 sin2u = 4

    • This means that vertical cross-sections parallel to the xz-plane (that is, with y constant) are all circles with radius 2.


Parametric surfaces4

PARAMETRIC SURFACES

Example 1

  • Since y = v and no restriction is placed on v, the surface is a circular cylinder with radius 2 whose axis is the y-axis.


Parametric surfaces5

PARAMETRIC SURFACES

  • In Example 1, we placed no restrictions on the parameters u and v.

  • So,we obtained the entire cylinder.


Parametric surfaces6

PARAMETRIC SURFACES

  • If, for instance, we restrict u and v by writing the parameter domain as 0 ≤ u ≤ π/2 0 ≤ v ≤ 3then x≥ 0 z ≥ 0 0 ≤ y ≤ 3


Parametric surfaces7

PARAMETRIC SURFACES

  • In that case, we get the quarter-cylinder with length 3.


Parametric surfaces8

PARAMETRIC SURFACES

  • If a parametric surface S is given by a vector function r(u, v), then there are two useful families of curves that lie on S—one with u constant and the other with v constant.

    • These correspond to vertical and horizontal lines in the uv-plane.


Parametric surfaces9

PARAMETRIC SURFACES

  • Keeping u constant by putting u = u0, r(u0, v) becomes a vector function of the single parameter v and defines a curve C1 lying on S.


Grid curves

GRID CURVES

  • Similarly, keeping v constant by putting v = v0, we get a curve C2 given by r(u, v0) that lies on S.

    • We call these curves grid curves.


Grid curves1

GRID CURVES

  • In Example 1, for instance, the grid curves obtained by:

    • Letting u be constant are horizontal lines.

    • Letting v be constant are circles.


Grid curves2

GRID CURVES

  • In fact, when a computer graphs a parametric surface, it usually depicts the surface by plotting these grid curves—as we see in the following example.


Grid curves3

GRID CURVES

Example 2

  • Use a computer algebra system to graph the surface

  • r(u, v) = <(2 + sin v) cos u, (2 + sin v) sin u, u + cos v>

    • Which grid curves have u constant?

    • Which have v constant?


Grid curves4

GRID CURVES

Example 2

  • We graph the portion of the surface with parameter domain 0 ≤ u ≤ 4π, 0 ≤v≤ 2π

    • It has the appearance of a spiral tube.


Grid curves5

GRID CURVES

Example 2

  • To identify the grid curves, we write the corresponding parametric equations:x = (2 + sin v) cos u y = (2 + sin v) sin uz = u + cos v


Grid curves6

GRID CURVES

Example 2

  • If v is constant, then sin v and cos v are constant.

    • So, the parametric equations resemble those of the helix in Example 4 in Section 13.1


Grid curves7

GRID CURVES

Example 2

  • So, the grid curves with v constant are the spiral curves.

    • We deduce that the grid curves with u constant must be the curves that look like circles.


Grid curves8

GRID CURVES

Example 2

  • Further evidence for this assertion is that, if u is kept constant, u = u0, then the equation z = u0 + cos vshows that the z-values vary from u0 – 1 to u0 + 1.


Parametric representation

PARAMETRIC REPRESENTATION

  • In Examples 1 and 2 we were given a vector equation and asked to graph the corresponding parametric surface.

    • In the following examples, however, we are given the more challenging problem of finding a vector function to represent a given surface.

    • In the rest of the chapter, we will often need to do exactly that.


Parametric representations

PARAMETRIC REPRESENTATIONS

Example 3

  • Find a vector function that represents the plane that:

    • Passes through the point P0 with position vector r0.

    • Contains two nonparallel vectors a and b.


Parametric representations1

PARAMETRIC REPRESENTATIONS

Example 3

  • If P is any point in the plane, we can get from P0 to P by moving a certain distance in the direction of a and another distance in the direction of b.

    • So, there are scalars u and v such that: = ua + vb


Parametric representations2

PARAMETRIC REPRESENTATIONS

Example 3

  • The figure illustrates how this works, by means of the Parallelogram Law, for the case where u and v are positive.

    • See also Exercise 40 in Section 12.2


Parametric representations3

PARAMETRIC REPRESENTATIONS

Example 3

  • If r is the position vector of P, then

    • So, the vector equation of the plane can be written as: r(u, v) = r0 + ua + vbwhere u and v are real numbers.


Parametric representations4

PARAMETRIC REPRESENTATIONS

Example 3

  • If we write r = <x, y, z>r0 = <x0, y0, z0>a = <a1, a2, a3>b = <b1, b2, b3>we can write the parametric equations of the plane through the point (x0, y0, z0) as:

  • x = x0 + ua1 + vb1y = y0 + ua2 + vb2z = z0 + ua3 + vb3


Parametric representations5

PARAMETRIC REPRESENTATIONS

Example 4

  • Find a parametric representation of the spherex2 + y2 + z2 = a2

    • The sphere has a simple representation ρ = ain spherical coordinates.

    • So, let’s choose the angles Φand θin spherical coordinates as the parameters (Section 15.8).


Parametric representations6

PARAMETRIC REPRESENTATIONS

Example 4

  • Then, putting ρ = ain the equations for conversion from spherical to rectangular coordinates (Equations 1 in Section 15.8), we obtain:

  • x = a sin Φ cos θy = asin Φ sin θ

  • z = a cos Φ

  • as the parametric equations of the sphere.


Parametric representations7

PARAMETRIC REPRESENTATIONS

Example 4

  • The corresponding vector equation is: r(Φ, θ) = a sin Φ cos θi + a sin Φ sin θj + a cos Φk

    • We have 0 ≤ Φ ≤ πand 0 ≤ θ≤ 2π.

    • So,the parameter domain is the rectangle D = [0, π] x [0, 2π]


Parametric representations8

PARAMETRIC REPRESENTATIONS

Example 4

  • The grid curves with:

    • Φ constant are the circles of constant latitude (including the equator).

    • θ constant are the meridians (semicircles), which connect the north and south poles.


Applications computer graphics

APPLICATIONS—COMPUTER GRAPHICS

  • One of the uses of parametric surfaces is in computer graphics.


Computer graphics

COMPUTER GRAPHICS

  • The figure shows the result of trying to graph the sphere x2 + y2 + z2 = 1 by:

    • Solving the equation for z.

    • Graphing the top and bottom hemispheres separately.


Computer graphics1

COMPUTER GRAPHICS

  • Part of the sphere appears to be missing because of the rectangular grid system used by the computer.


Computer graphics2

COMPUTER GRAPHICS

  • The much better picture here was produced by a computer using the parametric equations found in Example 4.


Parametric representations9

PARAMETRIC REPRESENTATIONS

Example 5

  • Find a parametric representation for the cylinderx2 + y2 = 4 0 ≤ z≤ 1

    • The cylinder has a simple representation r = 2 in cylindrical coordinates.

    • So, we choose as parameters θand zin cylindrical coordinates.


Parametric representations10

PARAMETRIC REPRESENTATIONS

Example 5

  • Then the parametric equations of the cylinder are x = 2 cos θy = 2 sin θz = z

  • where:

    • 0 ≤ θ≤ 2π

    • 0 ≤z≤ 1


Parametric representations11

PARAMETRIC REPRESENTATIONS

Example 6

  • Find a vector function that represents the elliptic paraboloid z = x2 + 2y2

    • If we regard x and y as parameters, then the parametric equations are simply x = xy = yz = x2 + 2y2and the vector equation isr(x, y) = xi + yj + (x2 + 2y2) k


Parametric representations12

PARAMETRIC REPRESENTATIONS

  • In general, a surface given as the graph of a function of x and y—an equation of the form z = f(x, y)—can always be regarded as a parametric surface by:

    • Taking x and y as parameters.

    • Writing the parametric equations as x = xy = yz = f(x, y)


Parametrizations

PARAMETRIZATIONS

  • Parametric representations (also called parametrizations) of surfaces are not unique.

    • The next example shows two ways to parametrize a cone.


Parametrizations1

PARAMETRIZATIONS

Example 7

  • Find a parametric representation for the surface that is, the top half of the cone z2 = 4x2 + 4y2


Parametrizations2

PARAMETRIZATIONS

E. g. 7—Solution 1

  • One possible representation is obtained by choosing x and y as parameters:x = xy = y

    • So, the vector equation is:


Parametrizations3

PARAMETRIZATIONS

E. g. 7—Solution 2

  • Another representation results from choosing as parameters the polar coordinates r and θ.

    • A point (x, y, z) on the cone satisfies: x = r cos θy = r sin θ


Parametrizations4

PARAMETRIZATIONS

E. g. 7—Solution 2

  • So, a vector equation for the cone is r(r, θ) = r cos θi + r sin θj + 2rkwhere:

    • r≥ 0

    • 0 ≤ θ≤ 2π


Parametrizations5

PARAMETRIZATIONS

  • For some purposes, the parametric representations in Solutions 1 and 2 are equally good.

  • In certain situations, though, Solution 2 might be preferable.


Parametrizations6

PARAMETRIZATIONS

  • For instance, if we are interested only in the part of the cone that lies below the plane z = 1, all we have to do in Solution 2 is change the parameter domain to: 0 ≤ r ≤ ½ 0 ≤θ≤ 2π


Surfaces of revolution

SURFACES OF REVOLUTION

  • Surfaces of revolution can be represented parametrically and thus graphed using a computer.


Surfaces of revolution1

SURFACES OF REVOLUTION

  • For instance, let’s consider the surface S obtained by rotating the curve y = f(x) a≤ x ≤ babout the x-axis, where f(x) ≥ 0.


Surfaces of revolution2

SURFACES OF REVOLUTION

  • Let θbe the angle of rotation as shown.


Surfaces of revolution3

SURFACES OF REVOLUTION

Equations 3

  • If (x, y, z) is a point on S, then x = xy = f(x) cos θz = f(x) sin θ


Surfaces of revolution4

SURFACES OF REVOLUTION

  • Thus, we take x and θas parameters and regard Equations 3 as parametric equations of S.

    • The parameter domain is given by: a≤ x ≤ b 0 ≤ θ≤ 2π


Surfaces of revolution5

SURFACES OF REVOLUTION

Example 8

  • Find parametric equations for the surface generated by rotating the curve y = sin x, 0 ≤ x ≤ 2π, about the x-axis.

  • Use these equations to graph the surface of revolution.


Surfaces of revolution6

SURFACES OF REVOLUTION

Example 8

  • From Equations 3,

    • The parametric equations are: x = xy = sin x cos θz = sin x sin θ

    • The parameter domain is: 0 ≤ x ≤ 2π 0 ≤θ≤ 2π


Surfaces of revolution7

SURFACES OF REVOLUTION

Example 8

  • Using a computer to plot these equations and rotate the image, we obtain this graph.


Surfaces of revolution8

SURFACES OF REVOLUTION

  • We can adapt Equations 3 to represent a surface obtained through revolution about the y- or z-axis.

    • See Exercise 30.


Tangent planes

TANGENT PLANES

  • We now find the tangent plane to a parametric surface S traced out by a vector function r(u, v) = x(u, v) i + y(u, v) j + z(u, v) kat a point P0 with position vector r(u0, v0).


Tangent planes1

TANGENT PLANES

  • Keeping u constant by putting u = u0, r(u0, v) becomes a vector function of the single parameter v and defines a grid curve C1lying on S.


Tangent planes2

TANGENT PLANES

Equation 4

  • The tangent vector to C1 at P0 is obtained by taking the partial derivative of r with respect to v:


Tangent planes3

TANGENT PLANES

  • Similarly, keeping v constant by putting v = v0, we get a grid curve C2 given by r(u, v0) that lies on S.


Tangent planes4

TANGENT PLANES

Equation 5

  • Its tangent vector at P0 is:


Smooth surface

SMOOTH SURFACE

  • If ru x rv is not 0, then the surface is called smooth(it has no “corners”).

    • For a smooth surface, the tangent plane is the plane that contains the tangent vectors ru and rv , and the vector ru x rv is a normal vector to the tangent plane.


Tangent planes5

TANGENT PLANES

Example 9

  • Find the tangent plane to the surface with parametric equations x = u2y = v2z = u + 2vat the point (1, 1, 3).


Tangent planes6

TANGENT PLANES

Example 9

  • We first compute the tangent vectors:


Tangent planes7

TANGENT PLANES

Example 9

  • Thus, a normal vector to the tangent plane is:


Tangent planes8

TANGENT PLANES

Example 9

  • Notice that the point (1, 1, 3) corresponds to the parameter values u = 1 and v = 1.

    • So, the normal vector there is: –2 i + 4 j + 4 k


Tangent planes9

TANGENT PLANES

Example 9

  • Therefore, an equation of the tangent plane at (1, 1, 3) is: –2(x – 1) – 4(y – 1) + 4(z – 3) = 0orx + 2y – 2z + 3 = 0


Tangent planes10

TANGENT PLANES

  • The figure shows the self-intersecting surface in Example 9 and its tangent plane at (1, 1, 3).


Surface area

SURFACE AREA

  • Now, we define the surface area of a general parametric surface given by Equation 1.


Surface areas

SURFACE AREAS

  • For simplicity, we start by considering a surface whose parameter domain Dis a rectangle, and we divide it into subrectangles Rij.


Surface areas1

SURFACE AREAS

  • Let’s choose (ui*, vj*) to be the lower left corner of Rij.


Patch

PATCH

  • The part Sij of the surface S that corresponds to Rij is called a patchand has the point Pij with position vector r(ui*, vj*) as one of its corners.


Surface areas2

SURFACE AREAS

  • Let ru* = ru(ui*, vj*) and rv* = rv(ui*, vj*) be the tangent vectors at Pij as given by Equations 5 and 4.


Surface areas3

SURFACE AREAS

  • The figure shows how the two edges of the patch that meet at Pij can be approximated by vectors.


Surface areas4

SURFACE AREAS

  • These vectors, in turn, can be approximated by the vectors Δuru* and Δvrv* because partial derivatives can be approximated by difference quotients.

    • So, we approximate Sij by the parallelogram determined by the vectors Δuru* and Δvrv*.


Surface areas5

SURFACE AREAS

  • This parallelogram is shown here.

    • It lies in the tangent plane to S at Pij.


Surface areas6

SURFACE AREAS

  • The area of this parallelogram is: So, an approximation to the area of S is:


Surface areas7

SURFACE AREAS

  • Our intuition tells us that this approximation gets better as we increase the number of subrectangles.

  • Also, we recognize the double sum as a Riemann sum for the double integral

    • This motivates the following definition.


Surface areas8

SURFACE AREAS

Definition 6

  • Suppose a smooth parametric surface S is:

    • Given byr(u, v) = x(u, v) i + y(u, v) j + z(u, v) k (u, v) D

    • Covered just once as (u, v) ranges throughout the parameter domain D.


Surface areas9

SURFACE AREAS

Definition 6

  • Then, the surface areaof S iswhere:


Surface areas10

SURFACE AREAS

Example 10

  • Find the surface area of a sphere of radius a.

    • In Example 4, we found x = a sin Φ cos θ, y = a sin Φ sin θ,z = a cos Φwhere the parameter domain is:D = {(Φ, θ) | 0 ≤ Φ ≤ π, 0 ≤ θ ≤ 2π)


Surface areas11

SURFACE AREAS

Example 10

  • We first compute the cross product of the tangent vectors:


Surface areas12

SURFACE AREAS

Example 10


Surface areas13

SURFACE AREAS

Example 10

  • Thus,

  • since sin Φ ≥ 0 for 0 ≤ Φ≤π.


Surface areas14

SURFACE AREAS

Example 10

  • Hence, by Definition 6, the area of the sphere is:


Surface area of the graph of a function

SURFACE AREA OF THE GRAPH OF A FUNCTION

  • Now, consider the special case of a surface S with equation z = f(x, y), where (x, y) lies in Dand f has continuous partial derivatives.

    • Here, we take x and y as parameters.

    • The parametric equations are:x = xy = yz = f(x, y)


Graph of a function

GRAPH OF A FUNCTION

Equation 7

  • Thus,and


Graph of a function1

GRAPH OF A FUNCTION

Equation 8

  • Thus, we have:


Graph of a function2

GRAPH OF A FUNCTION

Formula 9

  • Then, the surface area formula in Definition 6 becomes:


Graph of a function3

GRAPH OF A FUNCTION

Example 11

  • Find the area of the part of the paraboloid z = x2 + y2 that lies under the plane z = 9.

    • The plane intersects the paraboloid in the circle x2 + y2 = 9, z = 9


Graph of a function4

GRAPH OF A FUNCTION

Example 11

  • Therefore, the given surface lies above the disk Dwith center the origin and radius 3.


Graph of a function5

GRAPH OF A FUNCTION

Example 11

  • Using Formula 9, we have:


Graph of a function6

GRAPH OF A FUNCTION

Example 11

  • Converting to polar coordinates, we obtain:


Surface area1

SURFACE AREA

  • The question remains:

    • Is our definition of surface area (Definition 6) consistent with the surface area formula from single-variable calculus (Formula 4 in Section 8.2)?


Surface area2

SURFACE AREA

  • We consider the surface S obtained by rotating the curve y = f(x), a≤ x ≤ b about the x-axis, where:

    • f(x) ≥ 0.

    • f’ is continuous.


Surface area3

SURFACE AREA

  • From Equations 3, we know that parametric equations of S are:

  • x = xy = f(x) cos θz = f(x) sin θa≤ x ≤ b 0 ≤ θ≤ 2π


Surface area4

SURFACE AREA

  • To compute the surface area of S, we need the tangent vectors


Surface area5

SURFACE AREA

  • Thus,


Surface area6

SURFACE AREA

  • Hence,because f(x) ≥ 0.


Surface area7

SURFACE AREA

  • Thus, the area of S is:


Surface area8

SURFACE AREA

  • This is precisely the formula that was used to define the area of a surface of revolution in single-variable calculus (Formula 4 in Section 8.2).


  • Login