Single and Multivariable. Calculus. Early Transcendentals .. A few figures in the pdf and print versions of the book are marked with “(AP)” at the end of the. MA Multivariable Calculus. Lecture Notes1. Wong Yan Loi. Department of Mathematics, National University of Singapore, Singapore This is the text for a two-semester multivariable calculus course. The setting The book's aim is to use multivariable calculus to teach mathematics as a blend of.

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Chapter One - Euclidean Three Space. Introduction. Coordinates in Three -Space. Some Geometry. Some More Geometry--Level Sets. Chapter. Multivariable calculus is just calculus which involves more than one variable. To do it properly, you have to use some linear algebra. Otherwise. 14, Non-independent variables, Week 6 summary (PDF). 15, Partial differential equations; review. III. Double integrals and line integrals in the plane. 16, Double .

In this research paper, we extend our previous study [6] and develop some important concepts; tan- gent planes and partial derivatives for multi-variable functions There are a few works on time scale analysis with on time scales with Mathematica. Ufuktepe first defined Keywords-partial delta derivative; time scale; mathematica; elements of time scales, and gave the delta derivative of tangent plane; a function[7] for Mathematica, B. Balllard and B. Ufuktepe and S. The codes of all Mathematica commands which have written by the authors can be seen in the Appendix. In this section and Section 2, we give some elements of the time scales calculus for one variabe, and in Section 3 To describe a time scale in our Mathematica package, we briefly give some basics of multi-variable calculus on we use a collection of three lists: a list of right-dense time scales. We present some foundational definitions and left-scattered points, a list of left-dense right-scattered results, for more details, one can see [1], [3], [4].

In general, the third coordinate z of a point x, y, z indicates the height of the point above or below the xy-plane. Notice how we use dotted lines in our picture to create a sense of perspective so that we can picture a three-dimensional concept on two-dimensional paper. The coordinates x, y, and z of the point P. An equation involving Cartesian coordinates is said to be a Cartesian equation. The terms rectangular coordinates and rectangular equation are sometimes used instead.

The set of all spatial points is denoted by JR3 to signify that three coordinates are needed to describe such points. Just as the coordinate axes in the plane divide the plane into four quadrants, so do the xy-plane, the xz-plane, and the yz-plane divide space into eight octants.

The octant that contains points with all three coordinates positive is called the first octant. It is sometimes helpful to think of the xy-plane as the floor of a room. Look at Figure 3, which depicts the first octant from this point of view. Let P x0, y0, zo be a point with all three coordinates positive. Then the first two coordinates of P tell which position P' x0, y0,0 on the floor the point Plies over.

The last coor dinate tells how high above the floor Pis. These points are graphed in Figure 4. Some polygonal paths connecting these points to the origin are also plotted for perspective. For example, to reach the point -1, -2,1 from 0 0, 0, 0 , we may go 2 units in the negative y-direction, arriving at the point 0, -2, 0 , then 1 unit in the negative x-direction, arriving at the point -1, -2, 0 , and finally 1 unit in the positive z-direction, finishing at -1, -2,1.

Two distinct points in space determine a box with sides parallel to the coordinate axes see Figure 5. The length of a main diagonal of this box coincides with the di. We have labeled length, width, and height of the box as a,. Translating this calculation into coordinates gives us a formula for the distance between two points in space: Notice that the left side of equation 9. The coefficients of I. There are no mixed products xy, xz, or yz. Solution We complete the squares for x and y because of the presence of the linear terms and 4y: We see that the given equation describes a sphere with center 3, -2,0 and radius v'7 see Figure 6.

The set. We call this set the closed. Solution This set can be identified by rearranging the given inequality as. But there is a geometric interpretation as well. Like a vector in the plane, a vector in space can be represented by a directed line segment that specifies magnitude and direction. We say that a directed line segment. What vectors are represented by the directed line segments PQ, PR, and?

PR 1s , 9- -1 , 7- 4 or -9, 10, and t vector represented by QS 1s , 1, or -9, 10, 3. Thus PQ and PR represent different vectors. However, PR and QS represent the same vector, -9, 10, 3. Solution The vector represented by -. As in the plane, the geometric interpretation of vector addition is obtained by using v. Scalar Multiplication. A is a real number i. A to be. Geometrically, we think of scalar multiplication as producing a vector that is parallel to.

They have opposite directions if one is a negative scalar multiple of the other. For now, these notions are intuitive gen eralizations of facts that we already know for vectors in the plane. They will be made more precise in the discussion that follows later in this section. Calculate 3v and -4v. A vector with length one is called a unit vector.

We sometimes refer to a unit vector as a direction vector or a direction. Therefore vector vcannot be a unit vector for any value of s.

We call dir v the direction of v. If we multiply each side of equation 9. The right side of equation 9. We say that vand ware parallel if either a vand whave the same direction or b v and w are opposite in direction.

In the case that ware nonparallel, we see from Figure 7 earlier that equation 9. In each of Exercises , explain how to recognize that the given Cartesian equation is not the equation of a sphere.

In each of Exercises , determine the center and radius of each sphere whose Cartesian equation is given. The closed ball with center 1,2,9 and radius 1. The set of points whose distance to the point 11', 7! The set of points whose distance to the point 1, -2, 0 is not greater than 3. The set of points outside the closed ball with center 2, 1, 0 and radius 2.

The set of points outside the open ball with center -3,-1, -5 and radius 5. Also determine if the two vectors are parallel. Calculate the specified vector. In each of Exercises , points P and Q are given. State a the vector v which is represented by PQ, b the vector that has the same length as v and direction opposite to that of v, c the length of v, d the direction dir v of v, and e the vector with length 12 that has the same direction as v.

See the instructions to Exercises for examples. What are the lengths of the diagonals of the parallelo gram determined by the vectors 1, 1,0 and 0, 1,2? Give a geometric description of the set of all vectors of the form.

In each of Exercises , the two given vectors deter mine a parallelogram P. Calculate the vectors with positive first entries that represent the diagonals of P. The yz-coordinate plane. The xz-coordinate plane. The plane that is parallel to the xy-plane and that passes through the point rr, -?

The open half space that contains all points that lie on the same side of the xz-coordinate plane as the point 1, 1, 1. The half space that comprises the yz-plane as well as all points that are on the same side of the yz-plane as the point 1, 1, 1. The set of points whose distance to the xy-coordinate plane is greater than 5. The set of points that lie in one or more of the three coordinate planes. In other words, the union of the three coordinate planes.

The set of points that are farther from the origin than from 2, -1, Which point in the xy-plane is nearest to P0? Which point in the yz-plane is nearest to P0? How about the xz-plane?

A student walking at the rate of 4 feet per second crosses a 16 foot high pedestrian bridge.

A car passes directly underneath traveling at constant speed 40 feet per second. How fast is the distance between the student and the car changing 2 seconds later? Suppose that P, Q, and R are three noncollinear points. Show that. On which sphere do the four points 5, 2,3 , 1,6, -1 , 3,-2,5 , and -1,2,-3 lie? Atomic particles may carry a positive charge like a proton or a negative charge like an electron.

A charged. Suppose that v1, v2, N are scalars. NvN is said to. Suppose that P is the location of a charge p and Q is the. Sketch the surface. It does not depend on the initial point A that is selected.

In this section, we will study an algebraic operation that can be used to. This terminology should not be confused with scalar multiplication, which we have discussed in the preceding two sections. The result of scalar multi plication is a vector. We never use a dot to signify this type of multiplication. By contrast, the scalar product is an operation between two vectors, and the result is a scalar. We always use a dot to signify this type of multiplication.

In the next theorem, we collect several simple algebraic rules for working with the dot product. Nevertheless,we will generally insert parentheses for clarity. Because we do not form the dot product of a vector v and a. Solution Using Theorem lb and then la, we have. We will use this formula to develop a method for calculating the angle between two vectors and for calculating projections.

To that end, recall that the Law of Cosines says that if the sides of a triangle measure first two sides, as in Figure b. This trigonometric formula tells us that llv-wll2. Substituting this expression for llv- wll2 into the left side of equation 9. We obtain equation llvll llwll, which is not 0.

The left side of. The left side of equation 9. Using Theorem le, we. This relationship, called the Cauchy-Schwarz Inequality, can also be deduced from the identity. Although the zero vector 0 has no direction, it is conventional to say that 0 is orthogonal to every vector. Our next theorem serves two important purposes. The definition of ortho gonality of two vectors is geometric, as it should be.

However, because the defi nition does not involve the entries of the vectors, it does not allow us to easily verify. By contrast, our treatment of parallel vectors has been, up until now, algebraic and easy to apply. What it has lacked has been a transparent geometric foundation that refers to the angle between the two vectors.

In this case, v and w have the same. Statement a is an immediate consequence of this observation. To prove statement b, recall from Theorem. Both sides are 0 in this case. Using this equation with the first equation of line 9. Similarly, using the last equation of line 9. Statement c Finally, from formula. Indeed, the equation Solution From the dot products. One of the most powerful constructions in geometry is the projection.

Figure 4a shows two nonzero vectors. The next theorem provides us with an analytic expression for the direction,. Substituting this expression for Pw v into equation 9. With this value for c, equation 9. By equating the lengths of the vectors on each side of formula 9. The result of these scalar multiplications, the projection Pw v of v onto w, is a vector.

Because equation 9. The scalar v dir w ,. The absolute value of the component of v in the direction of w is the length of the projection Pw v. Calculate the projection of v onto w, the projection of w onto v, and the lengths of these projections.

Also calculate the component of v in the direction of w and the component of w in the direction of v. To efficiently calculate Pw v and the other quantities associated with this projection, we determine two scalars: We first calculate. The absolute value of this quantity is the length of the projection.

Formula 9. As a practical matter, however, Example 6 shows that formula 9. Calculate the v onto u. What is the component of v in the direction of u? Because u is a unit vector, the component of v in the direction of u is v u, or 6, and llull.

In particular,. As a result, Suppose. The numbers a, 3, and "Y are called the direction angles for u, and the entries u1, u2, u3 are called the direction cosines of u. Notice that, by Theorem 2: Refer to Figure 6. The direction angles for v are the angles that v makes with the positive x-, y-, and z-axes.

The hitch is set up so that the force is exerted at an angle of 30 with the horizontal. How much work is performed? Solution We use a 30 triangle, as shown in Figure resolve the force vector direction. F as a sum of two orthogonal vectors, one of which is in the We find that the direction of F is given by see panel b of Figure 8. Because the force vector F has. Solution The component of P,.. We calculate. This is the resolution that we desire.

Calculate 1,2, -1 3,4,2. Use the arccosine to express the angle between 6,3,2 and 4,-7,4. For what value of a are 1,a,-1 and 3,4,a perpendicular? Calculate the projection of 3,1,-1 onto 8,4,1.

In each of Exercises , calculate the dot product of the given vectors and their lengths. Verify that the Cauchy Schwarz Inequality holds for the pair. Calculate how much work is performed if she exerts a force of pounds and succeeds in pulling the car feet. What is the angle between PQ and PR? Suppose that a and b are nonnegative. Let v1, v2 , and v3 be any numbers.

Find a suitable vector" such that the Cauchy-Schwarz Inequality applied. Suppose that a, b, and c are positive numbers. What are the intercepts A, B, and C with the x-axis, y-axis, and.

If v lies in the. Calculate the cosines of the three angles of 6. Verify the identity. Prove that for any two vectors vand w, the angle between u. Using vector methods, find an equation involving oP,j, the dot product, and norm to describe the cone with vertex at the origin, forming an angle of 60 at.

Using vector methods, find an equation involving oP, k, the dot product, and norm to describe. Show that the points 2,1, 4 , 5,3, 2 , and 7,4,6 are the vertices of a right triangle. Prove the Cauchy-Schwarz Inequality from this identity. Deduce that in any event, one of v and w is a scalar mul tiple of the other. Let T: V be defined by.

Of course such a vector is not unique: We now define a product of two spatial vectors that results in another vector. We will soon present an alternative approach to calculating cross products, but first we verify that v X w performs an important geometric job.

We calculate the dot product of v with v X w: Recall that "orthogonal" is a synonym for "perpendicular. There is a nice way to remember the formula for a cross product using the language of determinants. The determinant is a procedure for calculating a number based on the entries of a square array or square.

By "square," we mean that the number of rows in the array equals the number of columns. In the general theory of determinants, the number of rows and columns may be any positive integer. For our purposes, we need only consider.

The factor of each entry is the determinant of the obtained by crossing out the row and column of the entry.

This pattern of alternating sums is the reason for describing the determinant as an "alternating sum. An equivalent scheme for calculating the determinant of a sometimes used. As Figure. The products of the six indicated diagonals are then computed. The determinant is the sum of the products that arise from the diagonals that slope down to the right minus the sum of the products that arise from the diagonals that slope down to the left.

Here the determinant is expanded just as if i,j, k were scalars. Recall that the dot product v w of two vectors is a scalar; however, the cross product of two vectors is another vector. Texts that call v w the scalar product often call v X w the vector product. We will not use this alternative terminology. Vectors These algebraic properties for the cross product can be verified by routine calcula tion.

The properties of distribution are as expected. However, pay particular atten tion to the law. Next, suppose that u and v are parallel. The first step is to use v and w to determine a plane. If P0 is a point in space, then we may represent v and w by directed line segments that.

Next, we observe that, for any point P0 on a plane V in space, there is a unique line f, through. A nonzero vector n is normal to V if n can be represented by a directed line segment Figure. If n is also a unit vector, then it is said to be v and w. A pair of nonparallel vectors always has two unit one of them, then -n is the other see Figure 8. With the. The negative of this vector is the standard unit normal vector for the ordered pair w, v. The direction that the thumb points during this process is the direction of.

Thus the standard unit normal for the pair normal for. Example 4 provides particular cases of this general fact. Because v X w is perpendicular to both v and w, the vector dir v X w must be the standard unit normal for either the ordered pair v, w or the ordered pair w, v. Our next theorem tells us which. It also specifies the length of v X w and tells us precisely when the equation v X w 0 holds. Let v and w be vectors. If v and w are not parallel, then v x w points in the direction of the standard unit normal for the pair v, w.

The second equality in this chain is not obvious, but it may be verified by multiplying everything out. If neither v nor w is the zero vector, then the scalars ll v ll and ll w ll are nonzero. In this case, the identity of part a becomes. Because 0: Because 0 is the only zero length vector, assertion c follows.

For part d, we will limit our verification to a few special cases. Based on our observations in Example 4, we conclude that, for the basic vectors i,j, and k, the operation of cross product produces the standard unit normal.

Geometric reasoning can be used to show that the cross product v X w always points in the direction of the standard unit normal for the vectors v,w. What is the standard unit normal vector for v, w?

Now we learn a connection between cross products and areas of triangles and parallelograms. To be specific, if.

Notice that, when we declare OPQ to be the triangle deter mined by v and w, the third side QP represents v-w. Notice that the areas of triangles See Figure. Thus had we declared OQR as the triangle determined by the vectors v and w instead of 0PQ, we would have been speaking of a triangle with the same area.

The area of triangle OPQ, namely half the product of its base and height, is therefore llwll llvllsin O. This quantity is llvxwll by Theorem 3b. Vectors discussed, the second assertion follows from the first because the area of the parallelogram determined by v and. E X A M P L E 9 In the picture tube for an oscilloscope, a magnetic field is used to control the path of ions that transmit the image to the screen.

If s is the. Suppose that the velocity vector for the c, -c, 0 , where c is a physical constant. The magnetic field will have the form a, 1, 1 , where the value of a will be varied to force the ions to go in different. Calculating the cross product c, -c, 0 X a, 1, 1 , we find that. We have just seen that the cross product is useful for finding the areas of triangles and parallelograms.

We now develop this idea further by introducing a product that involves three vectors. Notice that the triple scalar product u Xv w of three vectors u, v, and w is a scalar because it is the dot product of the two vectors u Xv and w.

The parentheses in the triple scalar product are not really necessary: The association u X v w. The next theorem shows that we can compute a triple scalar product without first calculating a cross product. The solid region determined by u, v, and w is called a "parallelepiped.

Moreover, lluXvii is the area of the base. The height of at Figure. We calculate the. Because the volume of the parallelepiped is obviously independent of the order in which we take the vectors u, v, and w, we deduce that the absolute value of the triple.

We record the results of this investigation as a theorem: In case uxv and w point to the same side of the plane determined by u and v the vectors form a right-hand system , then equals the volume of the parallelepiped. Equivalently, they are coplanar if and. Proot According to Theorem 6, this determinant is parallelepiped determined by the three vectors has. Figure 14 to help you see this. Finally, by applying Theorem 3b, first to the perpendicular pair. Chapter 9 Vectors which is equation 9.

We obtain equation 9. Find the area of the parallelogram determ ordered pair 2,1, -2 , 1,1,0.

True or false: Verify v X w and are perpendicular to v X w by showing that v. In each exercise,. In each of Exercises , calculate u Xv,v X w, and the u Xv w.

In each of Exercises , use the triple scalar product to r. Namely , let F be a force vector applied at a point Q in space, and let P be another point in space.

Notice that r is a vector. The direction of r gives the axis of rotation. A bolt at point P will be driven by a wrench in the direction of r see Figure Exercises concern torque. Suppose that v and ware spatial vectors and that A and are scalars.

Prove that. A F is applied to the end of the handle point Q. P and Q determine a line i. Suppose i. Prove the converse to equation 9. Find the point P on the curve that is closest to the origin. Find a vector v that is tangent to the. OP do these calculations have? Plot significance for v and. The philosophy here is that each equation "removes a degree of freedom" or takes away a dimension.

The same philosophy works in three dimensions: The set of points in space satisfying one linear equation will be a plane, and the set of points in space satis fying two linear equations will usually be a line.

Notice that we had to add the word "usually" because sometimes two equations have no solution or have too many solutions. However, we specify the "direction" of a plane in a rather indirect way. Namely, we give a nonzero vector n that is perpendicular to V. By this, we mean that n is perpendicular to every vector that lies in V. Recall that such a vector n is said to be a normal vector for V. All normal vectors for V are then of the form. An for a nonzero scalar.

Notice that a normal vector determines the "tilt" of a plane in space, but not its position. Any two parallel planes will have the same normal vectors see Figure. That is the geometric reasoning that underlies the next theorem. Figure 1 Parallel planes have the same normal vectors. This equation may be written in the alternative form. P0P lies in the plane.

This is true precisely when the vector P0P is perpendicular to n. Equation 9. We are given one such point, P0. The chain of reasoning that we have used to determine a Cartesian equation for a given plane can be reversed.

We state this observation as a theorem. A,B,C as a normal vector and passes through the point x0,y0,z0. Solution By writing the given equation in the standard form may conclude that the vector. Thus the 7,5,1 is on V. Therefore the point 4,3,1 is also on V.

In general, each new equation imposed on. We therefore expect the solution set of one Cartesian equation among the variables x,y, z to be a two-dimensional surface.

Theorem 2 tells us that, if the p 2, -1, 4. It is intuitively clear that any three points that are not on the same straight line determine a plane; imagine that the three points are three fingertips and balance your notebook on the fingertips-that is the plane we seek.

To use equation or. The cross product of these two vectors will be perpendicular to both of them, hence perpendicular to V. Thus a normal vector will. According to formula 9.

Each point in the plane satisfies this equation, yielding be. We are given three such points. By choosing one, say R, we obtain D is a Thus. We started by choosing two pairs of points: P, Q and P,R. We used each pair of points to determine a vector in the plane, but we could just as well have chosen the opposite vector.

Cartesian equation for the plane V. We might have used equation 9. The coordinates of any one of the three given points could then have been used for x0,y0, z0.

To illustrate with just one of these combinations,. We define the angle between two planes to be the angle between their normals, as in Figure. Solution As we have noticed, 1, -1, -1 is a normal vector for the first plane, and -1, 1, -3 is a normal for the second plane. The angle between the two given planes satisfies. If we write out this vector equation in coordinates, then we get three scalar parametric equations for the plane: The parameters s and t may be thought of as coordinates in the plane V: Each point of V is determined by specifying the values of s and t.

We state these observations as a theorem. For example, we may take. The solution of Example 5 illustrates a simple procedure for finding two nonparallel vectors that are orthogonal to one given vector. A line f, in space can be described by a point that it passes through and a vector that is parallel to it.

Look at Figure 6. P0P is parallel to m. This happens if and only if P0P is a multiple of m. In coordinates, this last equation becomes. Notice that tis a scalar and will play the role of a parameter.

After we match up coordinates, the parametric equations for a line in space become. Does the point 1,2,3 lie on? How about the point ,23, - 3?

Give para metric equations for the line through the origin that is parallel to. To determine whether the point 1,2,3 lies on the line, we need to determine whether some value of. We need look no further: There is no single value of t which satisfies all three equations. Thus the point 1,2,3 does not lie on the line. We have better luck with the point ,23, - 3. We endeavor to find a single. As our next theorem shows, Cartesian equations of a line are obtained by elim inating the parameter from a parameterization.

We equate. For example, the point 1, -9, 5 lies on the line because each quotient in line 9. The next example shows how to find the symmetric form of a line that passes through two given points. Solution The symmetric form of a line requires a point on the line and a vector m that is parallel to the line. The piece of information we are initially missing is m, but the vector.

Solution We set each of these three equal quantities equal tot. We obtain. To pass to symmetric equations for f, we solve for equate the resulting expressions for. Each of these equations is, taken by itself, the equation of a plane. These three equations describe the very same line as in the last example, but they look quite different from the parametric equations that we found there.

What is the rela tionship between these two parameterizations? The answer is that the change of variable. Find the point R at which l intersects V. Solution First of all, let us note that we cannot expect that the point of intersection, if there is one, will correspond to the same value of.

Imagine two cars crossing each other's path at an intersection. When all goes well the two cars pass through the same point but. Now we equate the expressions for. There will be a point of intersection s and t also satisfy the third equation. That is the case s 1 corresponds to the point -4, 3,2 on the first line;.

The situation of intersecting planes and intersecting lines is a simple instance of this principle. S2 to be the minimum, if it exists, of all distances d P1, P2 where P1 belongs to P2 belongs to S2 We now use vector methods to find formulas for the.

Let n A, be a normal vector for V and let Q x1,y1,z1 be any point on V. Then the. That is, the distance between P and V is equal to distance between. This answer agrees with the distance that we calculated in the Insight following Example That computation was considerably more laborious because it required the calculation of the point R at which V and the line normal to V through P intersect. PQ in the direction of n. Let V be the plane containing both P and , let f' be the line in V that passes through P and that is perpendicular to f, and let R be the point of intersection of f and f,' see Figure 9.

Reasoning as in Theorem 6, we see that R is the point on f that is clos to P. To do this, we need to see that n is perpendicular to f and lies in V. Because n m X m XPQ , we know that n is perpendicular to the first operand of the cross product, m, which is to say n is perpendicular to f.

We also know that n is perpendicular to the second operand of the cross product, m XPQ. But m and PQ are nonparallel vectors in V. Therefore m XPQ is orthogonal to V. Because n is perpendicular to a normal for V, it follows that n lies in V. Thus the distance between P and f is 7. In Exercise 93, you are asked to verify this result by using the methods of Chapter 4. Suppose that m1 and m2 are nonparallel vectors, that 1 is a line through P1 x1,y1, z1 parallel to m1, and that 2 is a line through P2 x2,y2, z2 parallel to m2.

That is, the distance between f1 and 2 is equal to. Let R1 and R2 be the points on 1 and 2, respectively, that are closest to each other. Then, R iR2 is perpendicular to both 1 and 2 It follows that RiR2 is parallel to n, and the length llR1R;ll is precisely the length of the orthogonal projection of P1P; in the direction of n. Find the distance between 1 and 2. Thus the distance between 1 and 2 is 6. Vector projection allows us to avoid such calculations.

If v and w are nonparallel vectors represented by directed line segments with a common initial point, then they determine a plane V with v Xw as a normal. Because the triple vector product u X v Xw is perpendicular to the second operand v Xw, we see that u X v Xw lies in V.

Our next theorem provides specific formulas for these scalars. A shorter proof is discussed in Exercise 94; the argument uses tedious algebraic identities involving the nine entries of u, v, and w. The triple vector products u X v Xw and u Xv Xw satisfy. We prove the first equation and leave 9. For the left side of 9. For the right side of 9.

Thus equation 9. Now suppose that v and w are perpendicular. Let V be the plane they deter mine. Then 9. Starting with o: Next, working with 3w and noting that 3w v we have.

Now we make no assumptions about v.

As in equation 9. Calcu late uX vXw , and verify equation 9. Solution We have. The point 3, 15, -2 is on a plane that has 3, 1,6 as a normal and Cartesian. What is D? Problems for Practice In each of Exercises , a Cartesian equation that describes a plane V is given.

Using only the given equation,. In each of Exercises , describe all normals to the plane that is parameterized by the given equations. In each of Exercises , find a Cartesian equation for the plane with the given normal vector n and passing through the given point P. In each of Exercises , find a Cartesian equation for the plane determined by the three given points.

In each of Exercises , find symmetric equations for the line that passes through the two given points. In each of Exercises , find, in parametric form, the line of intersection of the two given planes. In each of Exercises , find symmetric equations for the line of intersection of the two given planes. In each of Exercises , find the cosine of the angle between the two given planes. How do you tell if lies in V? If does not lie in V, how do you tell. In each of parts a to d, apply your answer to each given line and plane.

If the line and plane do intersect, find the point of intersection. Find them. A vector in two dimensions is an ordered pair dimensions is an ordered triple points in the plane, and if. Geometrically, two vectors are added by following the displacement of the first by the displacement of the second.

In both cases,. The length of a vector is. By convention, 0 is both orthogonal. A unit vector is called a. The geometric interpretation of cross product is that. We may use determinant notation to express the cross product as. This is called the symmetric form for the line. If a, b, and c. Review Exercises for Chapter 9 In Exercises , calculate the distance between the given. Write v represented by PQ in the form ll vlldir v. For what value of a is a, 3, 1 perpendicular to 4, 5, 13? If a, b, is parallel to 27, 5,9 , then what is a?

In each of Exercises. In each of Exercises , find the area of the triangle determined by the two given vectors. In each of Exercises , vectors Calculate the triple scalar product Frustrated by the apathy with which his research was greeted, Grassmann did not waver in his belief of its importance. As he expressed it, I remain completely confident that the labor I have.

Hermann Giinther Grassmann was born and raised in Stettin, a city that is now known by its. Polish name, Szczecin. Grassmann acquired his educa tion in Berlin, where he studied philology and theology but received no mathematical training at the university level.

After returning to Stettin, he became a high school teacher and father. The central conception of all modem physics is the "'Hamiltonian. In the words of a prominent scientist of the time, this phenomenon was "unheard of and without analogy. Hamilton's introduction of vector calculus was a by-product of his creation of the quaternions. Addition and subtraction of quaternions was no problem. Finding a suitable multiplication proved dif ficult.

The stumbling block was Hamilton's initial reluctance to abandon the commutative law:. Every morning in the early part of [October. I walking Hamilton recognized that he would have to give up the commutative property of multiplication. To see how to multiply two quaternions, imagine the face of a clock. This terminology was derived from radius vector, a term that was already used in analytic. At the same time, he wrote that "the algeb raically real part [namely, the quantity tin the quaternion.

Hamilton also introduced the scalar product i. In other words, if u and v are vectors quaternions u and v with S. Here we evaluate a function rat a real number t, and we obtain a vector r t as the result. By writing out the vector r t in terms of its components x t ,y t , z t , we see that we can identify the value of r t with the point x t ,y t , z t in space. By developing the calculus of vector-valued functions, we will be able to analyze such space curves.

For example, if we differentiate each component of. That much is analogous to the scalar-valued theory that we already know. But the geometry of curves in three-dimensional space is very rich, and there will be much that is new in this chapter. Early in the 16th century, Johannes Kepler deduced the three laws of planetary motion that now bear his name.

They were among the first precise quantitative laws known to science. What was lacking, however, was any theory that provided a reason for those laws. Half a century later, Isaac Newton demonstrated the power of calculus, then in its infancy, by deducing Kepler's Laws from basic physical principles.

Chapter 10 concludes by applying the calculus of vector-valued functions to derive Kepler's Laws. A function whose range consists of vectors is called a vector-valued function.

Such functions will be the focus of our study in this chapter. One reason for our interest in this matter is that a vector-valued function r can be used to describe a curve C in space.

To do so, we draw r t as a directed line segment with initial point at the origin. Figure la shows such directed line segments for several values, t0, t1, The curve C is the collection of terminal points of the vectors r t as t runs through all values of the domain of r, as Figure lb illustrates. We say that the curve C is parameterized by r. We refer to the vector-valued function r as a parametric curve.

Solution From Section This straight line is therefore the curve described by r. Describe the curve C along which the particle moves.

If we think of the variable t as "time," then we can imagine a particle tracing the curve so that its position at time tis the terminal point of r t. With this interpretation of the curve, we say that r t is the position vector of the particle at time t. In this context, we often call C a trajectory. These five points are plotted in Figure 2; they do not, however, give us a very good idea of the particle's path!

Because curve Clies on this cylinder, we can picture how Cpasses through the five plotted points. See Figure 4 for the arc of Cthat joins the five plotted points and Figure 5 for the arc of C that is obtained by plotting r t for values oft between ' and '.

Some Geometry. Some More Geometry--Level Sets. Loomis and Shlomo Sternberg. This book is based on an honors course in advanced calculus that we gave in the.

Multivariable calculus is just calculus which involves more than one This book presents the necessary linear algebra and then uses it as a. This is the text for a two-semester multivariable calculus course. The setting The book's aim is to use multivariable calculus to teach mathematics as a blend of.

MA Multivariable Calculus. Lecture Notes1. Wong Yan Loi.