Lecture 1 Introduction Rectangular Coordinate Systems Vectors Lecture 2 Length, Dot Product, Cross Product Length...


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1 CONTENTS i Contents Lecture Introduction. Rectangular Coordinate Sstems Vectors Lecture Length, Dot Product, Cross Product 5. Length Dot Product Orthogonal Projections Cross Product Lecture 3 Linear Functions, Lines and Planes 3 3. Linear Functions Linear mappings Lines Planes in 3space Lecture 4 Quadratic Surfaces 4. Quadratic Functions Quadratic Curves Quadratic Surfaces Sketching Surfaces Lecture 5 VectorValued Functions 3 5. VectorValued Functions and Their Graphs
2 ii CONTENTS 5. Limits, Continuit and Derivatives Lecture 6 Integration of VectorValued Functions, Arc Length Integration Arc Length Arc Length as a Parameter Lecture 7 Unit Tangent and Normal Vectors 4 7. Smooth curves Unit Tangent Vector Principal Normal Vector Lecture 8 Curvature 48 Lecture 9 Multivariable Functions Definition of multivariable functions and their natural domains Graphing multivariable functions Topological properties of domains Lecture Limits and Continuit 58 Lecture Differentiabilit of Two Variable Functions 66 Lecture Partial Derivatives 68 Lecture 3 The Chain Rules 74 Lecture 4 Tangent Planes and Total Differentials 78
3 CONTENTS iii Lecture 5 Directional Derivatives and Gradients 83 Lecture 6 Functions of Three and n Variables 86 Lecture 7 Multivariable Talor formula 9 Lecture 8 Parametric Problems (optional) 9 Lecture 9 Maima and Minima 96 Lecture Etrema Over a Given Region Lecture Lagrange Multipliers 3 Lecture Double Integrals 6 Lecture 3 Integration of Double Integrals Lecture 4 Double Integrals in Polar Coordinates, Surface Area. 3 Lecture 5 Triple Integrals 7 Lecture 6 Change of Variables Lecture 7 Triple Integrals in Clindrical and Spherical Coordinates7 Lecture 8 Line Integrals 3 Lecture 9 Line Integrals Independent of Path 36 Lecture 3 Green s Theorem 4 Lecture 3 Surface Integrals 46
4 iv CONTENTS Lecture 3 Surface Integrals of Vector Functions 49 Lecture 33 The Divergence Theorem 54 Lecture 34 Stokes Theorem 58 Lecture 35 Applications 6
5 CONTENTS v.
6 Lecture Introduction In onevariable calculus ou have studied functions of one real variable, in particular the concepts of continuit, differentiation and integration. Functions of one variable can capture the dependence of some quantit b onl one other quantit. In practice however, one often needs to investigate the dependence on one or more quantities on man variables, such as time, location, temperature, air pressure, or costs of different products etc. Therefore it in natural to consider functions that depend on man variables. We will write f(, ) for a function of two variables, f(,, z) for a function of three variables or, more generall, f(,,..., n ) for a function of n variables. Instead of f other letters can be used (such as g, h, F, G, H or f, f etc.). Here it is assumed that a domain is specified to which the argument variables belong. We will write R for the set of all pairs (, ) of real numbers, R 3 for all triples of real numbers and R n for all ntuples of real numbers. The domains of multivariable functions are subsets of those. It is convenient to plot functions of one variable as a graph in the twodimensional plane (and, vice versa, one can stud curves in the plane using functions). Similar to this, a function of two variables can be plotted as a surface in threedimensional space, and twodimensional curved surfaces can be studied b functions of two variables. Visualising functions of more than two variables is more difficult. In this unit we will cover the concepts of continuit, differentiation and integration of functions of man variables, as well as appling these concepts to stud the geometr of curves and surfaces.. Rectangular Coordinate Sstems First we will recall some concepts from linear algebra that were introduced in Math. Recall that a point P in space can be described b a pair of numbers, namel its Cartesian coordinates (, ). We use a coordinate sstem of two perpendicular aes, the ais and ais. Now, is the number on the ais that corresponds to the perpendicular projection of P to the ais (parallel to the ais) and is the number on the ais that corresponds to the perpendicular projection of P to the ais (parallel to the ais). Here we use the lower indices to indicate that these are the coordinates of the point P.
7 . Rectangular Coordinate Sstems (, ) In 3space, the situation is similar. Here we need an etra ais, the zais. Usuall we make the zais vertical and pointing upward, while we make the ais and ais to form a horizontal plane as follows z (,,z ) To find the coordinates a point P we need to project it perpendicularl to the corresponding ais (parallel to the plane spanned b the remaining aes). This can be done in two steps, we first find the point P that is the perpendicular projection of P to the plane. Then, are the coordinates of P in the plane and z is the distance between P and P with a positive sign if P is located above the plane and with a negative sign in the opposite case. To find a point with given coordinates (,, z ) first find P with coordinates (, ) in the plane and the go up b the distance of z z if z or down b z z if z < to find P. We will use the notation P (,, z ) to indicate that the point P has coordinates (,, z ). Note: The coordinate sstem as drawn above is called a righthanded sstem (when the fingers of the right hand are cupped so that the curve from the positive ais toward the positive ais, the thumb points (roughl) in the direction of the positive zais). If we interchange the positions of the ais and ais, then we obtain a lefthanded sstem. It can be shown that rectangular coordinate sstems in 3space fall into just these two categories: righthanded and lefthanded. In this unit, we will alwas use righthanded sstems, and we will alwas draw the coordinates with the zais vertical and pointing upward. We will call this the zcoordinate sstem. Eample Find the point with coordinates (,, ) in the zcoordinate sstem.
8 . Vectors 3 Solution: We first draw the coordinate aes to obtain the zcoordinate sstem. Then find (, ) on the plane, draw a vertical line through the point, and move down one unit (since z ), and we arrive at (,, ). z (,,). Vectors Shifts in the plane or space can be described b saing to what point P a given point P has been shifted. The line l connecting P and P gives the direction of the shift. An other point Q would be shifted along a line parallel to l b a distance equal to the distance between P and P to a point Q, so that P, P, Q, Q form a parallelogram. For this description it did not matter whether we started at P or Q, so instead of the pairs P (,, z )P (,, z ) or Q (X, Y, Z )Q (X, Y, Z ) we ma consider the object v,, z z X X, Y Y, Z Z called vector (in 3space). To indicate that a vector v is represented b an initial point P and terminal point P we write v P P. When the initial and terminal point are the same, no shift occurs and the corresponding vector is the zero vector,,. The same concept works in the dimensional plane and an ndimensional space. To keep things simple, but not too simple, we restrict here to 3dimensional space.
9 4. Vectors Points and vectors are described b in a similar wa b coordinates. In fact, a point P(,, z) can be identified with a vector v,, z that shifts the origin O(,, ) to P(,, z). Nevertheless, points and vectors are different objects. We cannot add points or multipl points with numbers, but we can add vectors: v + v,, z +,, z +, +, z + z. The geometric meaning of adding two vectors is to perform two shifts, first b v and then b v. The result is a single shift b v + v. Notice that the sum of two vectors does not depend on the order of the two shifts being performed. We can also multipl a vector v,, z b a number α (in this contet called a scalar as opposed to a vector): α v α,, z α, α, αz. Geometricall, the shift α v is in the same direction as v, if α > and in the opposite direction, if α <, b α times the original distance. It follows easil from the arithmetic of numbers that the following properties are satisfied: (a) u + v v + u (b) ( u + v) + w u + ( v + w) (c) u + + u u (d) u + ( u) (e) k(l u) (kl) u (f) k( u + v) k u + k u (g) (k + l) u k u + l u (h) u u. In our computations with vectors we will rel on these properties. One can define a more abstract notion of vector spaces b stipulating theses properties as aioms. This will be discussed in more detail in Linear Algebra Pmth3. You ma verif that the set of polnomials, or the set of polnomials of degree less or equal to 4 also constitute an abstract linear space.
10 5 Lecture Length, Dot Product, Cross Product. Length We know that in space, the distance between P (, ) and P (, ) is d ( ) + ( ). This is at the same time the length of the vector v P P, denoted b v. In 3space, there is a similar formula: the distance between P (,, z ) and P (,, z ) is d ( ) + ( ) + (z z ) which again is the length v of the vector v P P. Can ou prove these formulas using elementar geometr and the following diagrams? (Hint: Pthagoras theorem). z (,,z ) (, ) z z (, ) d   (,,z )   Thus, the length of a vector v v, v, v 3 (or v v, v, v 3,...v n ), also called the norm of v, is given b v v + v + v 3 ( ) or v v + v + v v n. The zero vector has length. A vector of length is called a unit vector. For eample, if v, then v v is a unit vector (wh? Please check). The following unit vectors are of special importance:
11 6. Dot Product i,, j, in space, i,,, j,,, k,, in 3space. This is because for an vector v v, v, we have v v, v v, +, v v, + v, v i + v j; and for an vector v v, v, v 3, we have v v, v, v 3 v i + v j + v 3 k. These vectors are sometimes called the unit coordinate vectors.. Dot Product The dot product assigns a scalar (number) to two vectors. It can be defined in an dimension. Let u u, u, v u, v. Then the dot product of u and v is given b u v u v + u v. Similarl, for u u, u, u 3, v v, v, v 3, u v u v + u v + u 3 v 3, or, generall, for u u, u,..., u n, v v, v,..., v n, u v u v + u v + + u n v n. Notice that the length of a vector v can be epressed as v v v. The relation between the dot product and the Euclidean geometric notion of angle is the subject of the Theorem below.
12 . Dot Product 7 Theorem Let u, v be nonzero vectors in space or 3space 3, and let θ be the angle between u and v. Then u v u v cosθ, i.e. cosθ u v u v. Proof: We prove the case that u v are space vectors. The 3space case can be proved similarl. As in the diagram, we can use u and v to form the triangle OPQ, where u u, u, v v, v, and OP u, OQ v. Note that the length of OP is u, that of OQ is v and the length of PQ is PQ v u since PQ v u, v u v u. The law of cosines applied to the triangle OPQ gives v u u + v u v cosθ which is equivalent to u O P v Q i.e. u v cosθ ( u + v v u ) { u + u + v + v [ (v u ) + (v u ) ]} u v + u v u v u v u v cosθ. Theorem shows that dot product can be used to calculate the angle between two vectors. In particular, two nonzero vectors u, v are perpendicular if and onl if u v, i.e. cosθ. Another consequence of Theorem is the important CauchSchwarz inequalit u v u v 3 In higher dimensional spaces the formula cosθ u v u v can be used to define the notion of angle.
13 8. Dot Product with equalit occuring onl if the two vectors are parallel or one of them is the zero vector. Eample. Let P (,, ), P (3,, ) and P 3 (,, 3). Find the angle between P P and P P 3. Solution: Let θ denote the angle. Then, b Theorem, cosθ P P P P 3 P P P P 3 We have P P 3,,,, P P 3,, 3,, P P P P 3 + ( ) + 3 P P + ( ) + 6 P P Hence cos θ 3 6 6, θ 6o (or π 3 ) Dot product has the usual arithmetic properties which we list below, the proof follows from the definition of dot product directl, and therefore will not be provided. You are encouraged to prove some of them, at least (d). If u, v, and w are vectors and k is a scalar, then (a) u v v u (b) u ( v + w) u v + u w (c) k( u v) (k u) v u (k v) (d) v v v (and equalit holds onl for v ). Propert (a) is called smmetr, (b) and (c) together bilinearit and (d) positivit. In a more abstract setting these properties are used to define dot products in arbitrar vector spaces. This concept will be developed in Pmth3.
14 .3 Orthogonal Projections 9.3 Orthogonal Projections Let u and b be two vectors in space or 3space. Then the vector w formed as in the diagram below is called the orthogonal projection of u on b, and is denoted b proj b u u w b If the angle θ between u and b is less than 9, i.e. direction of b. Therefore proj b u has the form π, then proj b u is in the proj b u k b for some scalar k >. Using the definition, we find that the length of proj b u is u b u cos θ u u b u b b. (b Theorem ) On the other hand, proj b u k b k b. Therefore k b u b u b, that is k b b. Substituting back, we obtain Theorem proj b u u b b b The above formula is also true if θ is greater than 9. Please check this b modifing the above proof. Let us now look at some of the uses of this formula. B definition, the distance between two sets P and Q in  or 3space is the infimum of all distances of a point P P and Q Q, i.e., roughl speaking, the
15 .3 Orthogonal Projections distance between the two points in P and Q that are closest to each other in the respective sets: dist(p, Q) inf PQ. P P,Q Q Eample. Find a formula for the distance D between the point P(, ) and the line A + B + C. Solution: Let us use the graph at the right to help with the argument. The distance from P to an arbitrar point Q on the line is the hpotenuse of a right triangle with one leg being the distance from P to the orthogonal projection (sa Q ) and the other being the distance between Q and Q. From PQ PQ we see that the infimum of distances PQ is attained for Q Q. Therefore we need to find the orthogonal projection of P on the line. Q n P a+b+c Let Q (, ) be an point on the line (i.e. A + B + C ) and position n so that Q is its initial point. Then QP n D proj n QP n (using Theorem ) n QP n n QP n n n (using k v k v ) But QP,, n A, B. Hence QP n A( ) + B( ) n A + B A B A + B + C (using A + B + C ), A + B and D QP n A + B + C n A + B Note: The point Q is introduced just for an intermediate step, it does not appear in the final formula.
16 .4 Cross Product Notice that the distance of a point P(, ) to the ais is just and the distance to the ais is just. This can be verified b the formula with A, B, C and A, B, C, respectivel..4 Cross Product For vectors in 3space onl, another kind of product, called the cross product, is defined. If u u, u, u 3 and v v, v, v 3, then the cross product u v is a vector given b u v u u 3 v v 3 i i j k u u u 3 v v v 3 u u 3 v v 3 j + u u u v k Note that u v is a number, but u v is a vector. The following theorem shows some of the properties of cross product are similar to dot product, but man are different. Theorem 3. Cross product has the following properties: () u v is orthogonal to both u and v, i.e. u ( u v), v ( u v). () u v u v sinθ (compare u v u v cosθ) Notice that this is the area of the parallelogram spanned b the vectors u and v. It follows u v u v where the equalit holds if and onl if u v. (3) (a) u v ( v u) (compare u v v u) (b) u ( v + w) ( u v) + ( u w) (c) ( u + v) w ( u w) + ( v w) (d) k( u v) (k u) v u (k v) (e) u u (f) u u (compare u u u ) similar to dot product
17 .4 Cross Product (4) If a a, a, a 3, b b, b, b 3 and c c, c, c 3, then a ( b c) a a a 3 b b b 3 c c c 3 (5) a ( b c) is the volume of the parallelepiped spanned on the vectors a, b and c. Proof: We onl prove (4) and (). We prove (4) first and then use (4) to deduce (). Proof of (4): b c i j k b b b 3 c c c 3 b b 3 c c 3 i b b 3 c c 3 j + b b c c k Hence a ( b b c) a b 3 c c 3 a b b 3 c c 3 a a a 3 b b b 3 c c c 3 + a 3 b b c c Proof of () (using (4)): u ( u v) u u u 3 u u u 3 v v v 3, because the first and second rows are the same.
18 3 Lecture 3 Linear Functions, Lines and Planes 3. Linear Functions A linear function of one variable is given b an equation of the form f() m + b, where m, b are real parameters. Its graph is a line in the plane with slope m and intercept b. Linear functions are eas to handle, et the can serve as good models for man processes in science, nature and econom. Even nonlinear processes can often be modelled approimatel using linear functions. The means to do this is differential calculus. A function f() that is differentiable at some point can be epressed as f() f( ) + f ( )( ) + e(, ), where e(, is a small error term. The function f( ) + f ( )( ) m + b is a linear function with m f ( ) and b f( ) f ( ). The error term is small in the sense that it tends faster to zero than the linear function when approaches. More precisel, even the ratio of the two small quantities tends to zero: e(, ) as. Indeed, e(, ) f() f( ) f ( )( ) lim lim holds if and onl if f() f( ) lim f ( ), which is the definition of the derivative f ( ). Linear functions of several variables have the form f(, ) a + b + d f(,, z) a + b + cz + d f(,,..., n ) a + a + + a n n + d lim f() f( ) f ( ) in the case, 3 or n variables, respectivel. Here a, b, c, d, a,...,a s are parameters. We will investigate their geometric meaning later. The role of linear functions of several variables is similar to linear functions of one variable.
19 4 3. Linear mappings 3. Linear mappings A linear mapping from ndimensional space C n to mdimensional space R m is given b m linear functions f (,..., n ) a + a + a n n + b f (,..., n ) a + a + a n n + b. m f m (,..., n ) a m + a m + a mn n + b n In matri notation this can be written as A + b, where is a column vector in R n, b and are column vectors in R m and A is an m n matri. The geometric meaning of b is a parallel displacement in direction of the vector b. The columns of A are the images of the standard vectors,,...,.... Important eamples of linear mappings are rotations. A rotation in the plane about the origin at an angle φ is described b cos φ sin φ sin φ + cosφ. Here b and Notice that det A. ( ) cos φ sin φ A. sin φ cosφ Rotations in 3dimensional space are a bit more complicated. An rotation in 3dimensional space would map the standard vectors i, j, k to a triple of mutuall perpendicular unit vectors i, j, k. Such rotation is completel determined b those vectors.
20 3. Linear mappings 5 Simple eamples are rotations about one of the coordinate aes. E.g. a rotation about the kais corresponds to cos φ sin φ A sin φ cosφ. Geometricall we can decompose an rotation into a sequences of 3 simple ones about the three aes. The determinant of an rotation matri in 3dimensional space is. We can now prove the fact stated in the previous lecture that u v u v sin θ. Assume u v. Let w be the unit vector w u v. u v Then w w w 3 u v w u v det u u u 3 v v v. 3 Now we appl a rotation in space that maps w to k and u to a multiple of i. Since v is perpendicular to w its image is in the i, j plane. The determinant of the matri above does not change since the determinant of the rotation matri is. But in the new coordinates we ma assume that u u,, v v, v, w,, and u v det u v v u v. This is clearl the area of the parallelogram spanned b u, v since u is the length of the base side and v is the hight. The following fact will be used later: The cube of volume spanned b the standard vectors i, j, j
21 6 3.3 Lines is mapped under a linear mapping with matri A to a parallelepiped spanned b the vectors a a, a 3 a a, a 3 a 3 a 3 a 33 of volume det A. Hence a linear mapping stretches the volume of a solid b a factor of det A. 3.3 Lines In space or 3space, a line is determined b a point P on it, and a direction parallel to it. Here the point P will be given b its coordinates and the direction b a nonzero vector v. First, consider the case of a line in space, which passes through P (, ) and parallel to v a, b. Let P (, ) be a general point on the line. Then P P, is parallel to v a, b. Now we use the fact that two vectors u and v are parallel if and onl if u t v for some scalar t. Hence, P P t v for some scalar t, i.e., t a, b or ta, tb. v P Now, when we let t run through (, ), the point (,, z) determined b the above formulas runs through the entire line. We sa + ta, + tb, t a parameter is the parametric equation for the line passing through P (, ), and parallel to v a, b. In the case of dimension, he two parametric equations can be reduced to one single equation b eliminating the parameter t. Multipling the first equation b b and subtracting the second equation multiplied b a we get b( ) a( ). Notice that the vector n b, a is perpendicular to v, since the dot product n v. Such vector is called normal vector for the line and therefore this equation is called the pointnormal equation. It can be rewritten in the form A + B + C ()
22 3.3 Lines 7 with A b, B a, C a b. If B we obtain the usual slopeintercept equation b solving for : A B C B. If B the line is vertical and has no slopeintercept equation. Remark. If we divide equation () b the length of the normal vector A + B we obtain the socalled Hesse normal form A A + B + B A + B + C A + B. There are angles φ, φ π φ A such that B A cosφ +B and A sin φ +B cosφ. The angles φ, φ are the ones formed b the normal and the and aes, C respectivel. The absolute value A gives the distance of the line to the origin. +B (Tr to prove this.) We rewrite now the parametric equation as vector equation. Let O be the origin and denote r OP, r OP. Then P P r r and the line can be represented b r r t v, or r r + t v, t a parameter. This equation can be used in 3 or, more generall, in an ndimensional space. In 3space, with P (,, z ), v a, b, c, r,, z and r,, z, the parametric equations become: + ta, + tb, z z + tc r r + t v vector equation. parametric equation. Notice that eliminating the parameter t from these equations leaves us still with equations. If a we find b a + b a z c a + z c a A dimensional line in 3space cannot be described b a single linear equation because one equation lowers the dimension onl b, so equations are needed 4. 4 This wouldn t be true if we permitted nonlinear equations. E.g. + z describes the ais {, z }.
23 8 3.4 Planes in 3space Eample. Find an equation of the line L which passes through P (,, z ) and P (,, z ) where P P. Solution: If we can find a vector to which the line is parallel, then we can use the formulas discussed above to find the equation. Clearl P P,, z z is such a vector. Hence the parametric equation is + t( ), + t( ), z z + t(z z ). In vector form r,, z + t P P. Eample. Show that the vector n A, B is perpendicular to the line A + B + C. Proof: Let P (, ) and P (, ) be two points on the line, i.e. A + B + C, A + B + C. Then it suffices to show n A, B is perpendicular to P P,, i.e. to show n P P. We calculate n P P A( ) + B( ) A + B (A + B ) C ( C) and hence prove what we wanted. 3.4 Planes in 3space A plane in 3space is uniquel determined b a point P (,, z ) on it and a vector n A, B, C perpendicular to it. Such a vector is called a normal vector of the plane.
24 3.4 Planes in 3space 9 A general point P (,, z) is on the plane if and onl if P P is perpendicular to n, i.e. P P n. Since P P,, z z and n A, B, C, we have P P n A( ) + B( ) + C(z z ), and the equation of the plane is P n P A( ) + B( ) + C(z z ) This is called the pointnormal form of the equation of the plane. Note that on simplifing the above equation for the plane, we see that the equation has the form A + B + Cz + D where D A B Cz. () Now, if C this can be rewritten as a graph equation z A C B C D C. (3) Notice that one linear equation in 3space lowers the dimension down b one to a dimensional plane. Remark. There is also a Hesse normal form for planes in 3space. Divide equation () b A + B + C A. Then there are angles φ, φ, φ 3 such that A +B +C B cosφ, C A cosφ +B +C, A cosφ +B +C 3. The angles φ, φ, φ 3 are the angles between the normal and the respective aes. The number of the plane to the origin O(,, ). Eample 3. If A, B, C, then the equation A + B + Cz + D D A +B +C is the distance describes a plane having normal vector n A, B, C. (Compare with eample above: A + B + C is a line having normal vector n A, B.) Proof: Since A, B, C, at least one of the three components A, B, C is not zero. Suppose C (the other cases can be proved similarl). Then for an given,, we can find a unique z such that the graph equation (3) and hence the plane equation is satisfied. Substituting this into A + B + Cz + D
25 3.4 Planes in 3space we see the equation is equivalent to A( ) + B( ) + C(z z ). This is a pointnormal form of the equation of a plane with normal vector n A, B, C passing through the point (,, z ).. Eample 4. Find an equation of the plane through three different points P (,, z ), P (,, z ) and P 3 ( 3, 3, z 3 ). Solution We need to find a normal vector for the plane. With the three given points, we can form two vectors P P and P P 3, both ling on the plane. According to the properties of cross product, P P P P 3 is perpendicular to both P P and P P 3, hence to the plane. Thus we can use n P P P P 3 as a normal vector. n P3 P P Since P P,, z z and P P 3 3, 3, z 3 z and the equation can be written as,, z z n we can use propert (4) for cross product to write the equation above in the following neat form: z z z z 3 3 z 3 z Eample 5. Find the distance d between P (,, z ) and the plane A + B + Cz + D. Solution We generalize the method used in Eample, Lecture.
26 3.4 Planes in 3space Let Q (,, z ) be an point on the plane, and hence it satisfies A + B + Cz + D. Then d proj n QP, where n A, B, C is a normal vector. We have proj n QP n QP n n a( ) + b( ) + c(z z ) n a + b + c proj n QP A( ) + B( ) + C(z z ) n A + B + C A + B + Cz (A + B + Cz ) A + B + C A + B + C A + C + Cz + D A + B + C Q n P (using A + B + Cz D) Thus d A + B + Cz + D A + B + C.
27 4. Quadratic Functions Lecture 4 Quadratic Surfaces 4. Quadratic Functions Quadratic functions of one variable q() a + b + c with parameters a, b, c are the net simplest after linear functions. The are also often used to model processes in nature, science and econom. The Talor formula tells us that a function f() that has continuous derivatives up to third order can be approimated b a quadratic function in the following wa: f() f( )+f ( )( )+ f ( )( ) +e(, ) a +b+c+e(, ), where a f ( ), b f ( ) f ( ), c f( ) f ( ) + f ( ) error term e(, ) 6 f (ξ)( ) 3 and the tends to even faster than the quadratic function ( ) as tends to, i.e. e(, ) lim ( ). This has been used to investigate critical points for local maima and minima. Since the graph of a quadratic function is a parabola with an absolute minimum at its verte if a > or a maimum if a < one concludes that a function that is two times differentiable has a local minimum or maimum at a critical point if a f ( ) > or a f ( ) <. Indeed, if f ( ) (for being critical) we have f() f ( )( ) + f( ). The situation of man variables is similar, but slightl more complicated because of the large amount of possible quadratic terms. Let us look first into the case of two variables. f(, ) a + b + c + d + e + f. We introduce a more sstematic notation that uses indices and turns out to be even more useful in higher dimensions. Denote the first variable b and the second variable b. Then denote a a, b a a, c a, d a, e a, f a. The indices of the new a s tell us immediatel how man and which factors or follow. Also we don t need to invent new letters for the huge amount
28 4. Quadratic Curves 3 of coefficients that occur in higher dimensions, and we ma use sigma notation. The equation becomes f(, ) a +a +a +a +a +a a ij i j + a i i +a. i j i In three dimensions we get f(,, 3 ) a + a + a a + a a a + a + a a a ij i j + a i i + a. i j i In higher dimension we onl need to replace 3 as upper bound of the summation b the dimension n n n n f(,,..., n ) a ij i j + a i i + a. i j i 4. Quadratic Curves In Math we have studied implicit equations of curves F(, ) A seconddegree equation in, has the general form A + B + C + D + E + F. (4) It gives a quadratic curve 5. B switching to an alternative coordinate sstem we can significantl simplif the equation (4). In a first step we will appl a rotation of the plane about the origin b a suitable angle φ in order to get rid of the mied term B. Recall that such rotation is performed b a mapping cos φ u + sin φ v (5) sin φ u + cosφv where, are the old coordinates and u, v are the new coordinates. You ma check that the unit vectors i, and j, are mapped to a pair of mutuall perpendicular unit vectors. 5 Quadratic curves are often called conic sections because the arise when a cone and a plane intersect in 3space.
29 4 4. Quadratic Curves Theorem. B a suitable rotation of the form (5) the equation (4) turns into A u + C v + D u + E v + F, where A, C, D, E, F are some new coefficients. Proof. B plugging the epressions (5) for, into the equation (4) we find the coefficient in front of uv to be B A cosφsin φ + B(cos φ sin φ) C sin φ cosφ (A C) sin φ + B cos φ. Here we used standard trigonometric formulae. To make this epression zero we need cot φ C A B. If B this determines a unique angle φ between and π. If B the mied term didn t occur in the first place. Without developing the relevant theor (this will be left to Linear Algebra Pmth3) we notice that the new coefficients written as a matri can be computed in the following wa: ( ) ( A B cosφ sin φ sin φ cos φ B C ) ( A B B C ) ( ) cosφ sin φ. sin φ cosφ (You ma check this b direct computation.) B our choice of φ we have B. In the classification of quadratic curves below ou will see that the tpe of curve depends on whether A and C are positive, negative or zero. Notice that A and C are the eigenvalues 6 Of the matri at the left hand side. It turns out that the old and new coefficient matrices have the same eigenvalues, determinant (which is the product of the eigenvalues) and trace (which is the sum of the entries at the main diagonal, hence the sum of the eigenvalues). It follows that the eigenvalues have the same sign if the determinant A C AC B is positive and opposite signs if the 4 determinant is negative. If both eigenvalues have the same sign then it is the sign of A. Assume now that the quadratic equation has no mied term. The second step depends on whether A, C are zero or not. If A and C we use a shift of the coordinate sstem to get rid of D and E. B completing the squares we find A + C + D + E + F A( + D A ) + C( + E C ) + F D 4A E 4C. 6 You ma recall the concept of eigenvalues from Math. A more thorough theor of eigenvalues will be developed in Pmth3.
30 4. Quadratic Curves 5 Hence the equation takes the form A( ) + C( ) F, where D A, E C, F D 4A + E 4C F. If A but C or A and C then we complete the square for the variable with the nonvanishing coefficient and leave the other unchanged to get C( ) + D F or A( ) + E F. The following are the representative eamples.. If A >, B > and F > (or A <, B < and F < ) we divide b F and get the equation of an ellipse with halfaes a F /A and b F /C centred at (, ). Ellipse: ( ) a + ( ) b b a. If A and E we ma divide b E and find a vertical parabola with verte (, ). Parabola: a( ) Here a A E and F E. (a>) 3. If C and D we ma divide b D and find a horizontal parabola with verte (, ). b( ) Here b C D and F D. (b>)
31 6 4. Quadratic Curves 4. If A >, C < and F > (or A <, C > and F < ) we divide b F and find a hperbola. Hperbola: ( ) ( ) a b Here a F /A, b F /C. 5. If A >, C < and F < (or A <, C > and F > ) we divide b F and find again a hperbola. ( ) ( ) a b Here a F /A, b F /C. 6. The remaining cases are in some sense degenerate and not reall quadratic curves. We list them for the sake of completeness. (a) If A >, C >, F (or A <, C <, F ), the equation becomes A( ) + C( ), which is a single point. (Which?) (b) If A >, C <, F (or A <, C >, F ), the equation A( ) + C( ) describes a pair of crossing lines, namel, ± A/C( ) (c) If A >, C > and F /A < (or A <, C < and F /A < ) we get A( ) + C( ) F, which is the empt set. (d) If A, C, E and F /A < (or A, C, D and F /C < ) we get ( ) F /A (or ( ) F /C), which is the empt set. (e) If A, C, E and F /A > (or A, C, D and F /C > ) we get ± F /A (or ± F /C), which is a pair of parallel lines. (f) If A, C, E and F (or A, C, D and F ) we get ( ) (or ( ) ), which is a single line.
32 4.3 Quadratic Surfaces Quadratic Surfaces A seconddegree equation in,, z has the general form A + B + Cz + D + Ez + Fz + G + H + Iz + J. It gives a quadric surface (a quadric for short). A classification of these surfaces would be similar to the classification of quadratic curves carried out above, but requires more effort and time. We ignore here the degenerate cases and list the 6 important tpes: z Ellipsoid: a + b + z c o z Hperboloid of one sheet: a + b z c o z Hperboloid of two sheets: a + b z c o z Elliptic cone: z a + b o z Elliptic paraboloid: z a + b o
33 8 4.4 Sketching Surfaces Hperbolic paraboloid: z b a z The most important quadratic surfaces for this unit are the paraboloids because the approimate the graphs of functions at its critical point and show whether the critical point is a maimum, a minimum or neither of them. Clearl, (z z ) A( ) + C( ) is a cuplike elliptic paraboloid with verte at (,, z ), i.e (,, z ) is a local minimum if both A, C are positive and an upside down cuplike elliptic paraboloid with verte at (,, z ), i.e (,, z ) is a local maimum if both A, C are negative. If A and C have opposite sign then the resulting surface is the saddlelike hperbolic paraboloid, i.e we have neither maimum nor minimum. The cases when A or C (or both) are zero are indecisive. If we started with a paraboloid of the form (z z ) A( ) + B( )( ) + C( ) we can perform a rotation in the, plane about (, ) eactl in the same wa as in the classification of quadratic curves to get rid of the mied term B( )( ). Without doing this we can decide whether the critical point (, ) is a maimum, minimum or saddle point b looking at the determinant det A B B C AC B and the coefficient A. IF AC B > we have a maimum or minimum depending on whether A < or A >. If AC B < we have a saddle. The case AC B is indecisive. 4.4 Sketching Surfaces Sketching the graph of a surface is usuall much more difficult than sketching a curve. One practical wa to sketch a surface is b using a method called mesh plot: one builds up the shape of the surface using curves obtained b cutting the surface with planes parallel to the coordinate planes.
34 4.4 Sketching Surfaces 9 The curve of intersection of a surface with a plane is called the trace of the surface in the plane. Let us now look at several eamples to see how the mesh plot method can be used in sketching surfaces. Eample. Sketch the graph of the surface 4 + z. Solution: For an fied value, the equation can be written as + z + 4 which gives a circle (for fied ) in the zplane. If we take k, and put the circle in the plane k (b moving the one in the zplane and keeping the motion parallel to the ais), then when we choose different k, we obtain man circles which are parallel but with different radii (the circle on the plane k has radius + k 4 ). z z k k k k k The above graphs show the circles (k,,,, ) and a sketch of the surface obtained b smoothl connecting these circles. Eample. Sketch the graph of z 4 9. Solution: Take k. Then the equation becomes z k 4 9, which is of the form (z z ) a and hence is a parabola. The following graphs show various parabolas obtained b varing k and a sketch of the surface b connecting these parabolas.
35 3 4.4 Sketching Surfaces z z Eample 3. Sketch the graph of z + 8 4z 4. Solution: We can change the equation into one of the standard forms to help us to know the shape of the surface. We use the completing squares method: i.e. 4 + ( ) + (z 4z + 4) , 4 + 4( + ) + (z ) 4, or + ( + ) (z ) +. 4 Thus we know it represents an ellipsoid with center (,, ). z z Write, + and z z. Then the equation becomes + + z 4. We can use the mesh plot method to sketch the surface in the z coordinate sstem and then move it to the zcoordinate sstem according to the relationship, + and z z. Note:,, z z z shows the,, z aes are parallel to the,, z aes, respectivel, and that the origin (,, z ) (,, ) in the z sstem is at the point (,, z) (,, z ) in the zsstem.
36 3 Lecture 5 VectorValued Functions 5. VectorValued Functions and Their Graphs. The function f() assigns to each from the domain D R a real value from the codomain B R, and hence it is a realvalued function. The equation of a line in space has the form + ta, + tb, which can also be written in vector form where r,, r,, v a, b. r r + t v, We can regard r r +t v as a function which assigns to each t a vector r r +t v. Thus we have a vectorvalued function here. In general, a vectorvalued function in space has the form r(t) (t), (t) (t) i + (t) j, in 3space, a vectorvalued function has the form r(t) (t), (t), z(t) (t) i + (t) j + z(t) k. The realvalued functions (t), (t) and z(t) are called the components of r(t). Clearl r(t) (t) i + (t) j + z(t) k is equivalent to (t), (t), z z(t). We call this later sstem of equations the parametric form of the vectorvalued function r(t) (t) i + (t) j + z(t) k. Let r(t) be a vectorvalued function in space or 3space defined on some closed interval [a, b]. Then the range of r describes a curve C, which we call the graph of r(t), or the graph of the equation r r(t). It might be helpful to visualise a vector function as the trajector of a point moving in space in dependence of time t. Clearl C has the parametric equation (t), (t) (in space) (t), (t), z z(t) (in 3space).
37 3 5. Limits, Continuit and Derivatives Eample. Describe the graph of the function r(t) cost i + sin t j, t π. Solution: The equation has the parametric form cost, sin t, t π which is just the parametric equation for the unit circle +. Thus the graph is the unit circle. Eample. Describe the graph of the vectorvalued function r ( + t) i + 3t j + (5 4t) k, t R. Solution: The equation is equivalent to + t, 3t, z 5 4t. We know this is the parametric equation for the line passing through (,, 5) and parallel to the vector, 3, 4. Or we can simpl rewrite the vector equation as r,, 5 + t, 3, 4. This is the equation of the line through (,, 5) parallel to, 3, Limits, Continuit and Derivatives The notions of limit, continuit and derivative for realvalued functions can all be passed to vectorvalued functions through the components. Namel, () Limit. For r(t) (t) i + (t) j, define ( ) lim r(t) lim (t) i + t a t a ( lim t a (t) ) j. For vectors in 3space, the definition is similar. If at least one of the limits of the component functions does not eist, then we sa lim t a r(t) does not eist. () Continuit. r(t) defined on some domain D R is said to be continuous at t D 7 if 7 Sometimes the condition that r(t) is defined at t is stated eplicitl, but we take the point of view that continuit or noncontinuit are notions that make sense onl for points that are a priori in the domain of the function.
38 5. Limits, Continuit and Derivatives 33 (a) lim t t r(t) eists and (b) lim t t r(t) r(t ). Clearl r(t) is continuous at t if and onl if all its component functions are continuous at t. (3) Derivative. The derivative of r(t ) is defined b This can be reformulated as r (t ) d r(t) lim dt h r(t + h) r(t ). h r(t) r(t ) + r (t )(t t ) + e(t, t ), where the error term e(t, t ) has the propert e(t, t ) lim, (6) t t t t i.e. e(t, t ) tends to zero faster than t t. In other words, is approimatel a linear function r(t) r + t v, where r r(t ) t r (t ) and v r (t ). The error is small, even compared to the small quantit t t. We introduce here a convenient notation that we will use later, namel to epress (6). More generall the o notation e(t, t ) o(t t ), has the meaning f(t) o(g(t)) as t t f(t) lim t t g(t). Notice that this notation is alwas related to a variable (here) t approaching a certain limit. This limiting process has to be stated or to be known b default. Eample. o() as ; cos o() as π ; m o( n ) as if m > n, we sa m tends to zero at a higher order that n. The convenience of the o is the simplicit of its arithmetic. Without knowing the particular functions we can state:. o( n ) ± o( n ) o( n ) (as.)
39 34 5. Limits, Continuit and Derivatives. o( n ) o( m ) o( n+m ). 3. o( n ) o( m ) if n m. 4. If f() is a bounded function defined in some neighbourhood of then f() o( n ) o( n ). Notice that o() is not the notation for a function, but onl epresses a propert of a function. In coordinates the derivative of a vector function is r (t) (t) i + (t) j if r(t) (t) i + (t) j, r (t) (t) i + (t) j + z (t) k if r(t) (t) i + (t) j + z(t) k. The following theorem collects some of the most important properties of derivatives for vectorvalued functions. Theorem () r (t ) is tangent to the curve r r(t) at r(t ) and points in the direction of increasing parameter. () (a) ( C), where C is a constant vector. (b) (k r(t)) k r (t), where k is a scalar. (c) [ r (t) ± r (t)] r (t) ± r (t) (d) [f(t) r(t)] f(t) r (t) + f (t) r(t) (product rule) (e) r(u(t)) r (u(t))u (t) (chain rule) (3) (a) [ r (t) r (t)] r (t) r (t)+ r (t) r (t) (product rule for dot product) (b) [ r (t) r (t)] r (t) r (t) + r (t) r (t) (product rule for cross product) (4) If r(t) is constant for all t, then r(t) r (t), i.e., r(t) and r (t) are alwas perpendicular. Proof: For () notice that the vector equation of a secant through r(t ) and r(t ) is r(t) r(t ) + vt where v r(t ) r(t ) t t. The tangent vector is the limiting position of secant vectors where t approaches t. This limit is, as in single variable calculus, b definition the derivative r (t ).
40 5. Limits, Continuit and Derivatives 35 The properties () (a)(e) can be proved in the same wa as in one variable calculus, or, even simpler, reduce to those properties. Although (3) (a) also reduces to properties of one variable functions we will give a proof that shows the advantage of the o notation. Notice that our proof below works for an dimension. We have r (t) r (t ) + r (t )(t t ) + o(t t ) r (t) r (t ) + r (t )(t t ) + o(t t ) B forming the dots product of both sides and using dot product rules we get r (t) r (t) r (t ) r (t ) + r (t ) r (t )(t t ) + r (t )(t t ) r (t )+ + ( r (t ) + r (t )(t t )) o(t t ) + ( r (t ) + r (t )(t t )) o(t t ) r (t ) r (t ) + ( r (t ) r (t ) + r (t ) r (t ))(t t ) + + o(t t ) Here we have used that r (t )+ r (t )(t t ) and r (t )+ r (t )(t t ) are bounded in some neighbourhood of t, i.e. r (t ) + r (t )(t t ) M r (t ) + r (t )(t t ) M and therefore, b the CauchSchwarz inequalities, ( r (t ) + r (t )(t t )) o(t t ) M o(t t ) o(t t ) ( r (t ) + r (t )(t t )) o(t t ) M o(t t ) o(t t ). This line proves the claim. The proof of (3) (b) is eactl the same with replaced b. To prove (4), let us first recall that r(t) r(t) r(t). Hence r(t) r(t) C for all t. It follows that (C) ( r(t) r(t)) r (t) r(t) + r(t) r (t) (using propert (3)(a)) r(t) r (t) This implies r(t) r (t) for all t. The proof for (4) is complete. Eample 3
41 36 5. Limits, Continuit and Derivatives (a) Find r (t) for r(t) ( + t) j + (3t) j + (4t ) k (b) Find r (t) and r () for r(t) sin t i + cost j. Solution: (a) (b) r (t) ( + t) i + (3t) j + (4t ) k i + 3 j + 4 k r (t) ( sin t) i + (cost) j cost i sin t j r () cos i sin j i Note: In part (a), the graph of r(t) is a straight line and r (t) is a constant vector which is parallel to the line. Recall that propert () in Theorem sas r (t) is tangent to the line and our eample confirms this. In part (b), r(t) ( sin t) + (cos t) and propert (4) in Theorem sas r(t) r (t). Here we can also check directl: r(t) r (t) ( sin t i + cost j) ( cost i sin t j) ( sin t)( cos t) + (cost)( sin t) sin t cost costsin t.
42 Lecture 6 Integration of VectorValued Functions, Arc Length Integration Recall that if r(t) (t) i + (t) j + z(t) k, then the derivative of r(t) is the vector function whose components are the derivatives of the components of r(t): r (t) (t) i + (t) j + z (t) k. The integral for r(t) is defined in the same fashion: b r(t)dt b (t)dt i + b (t)dt j + b a a a a z(t)dt k. The fundamental theorem of calculus is also true for vector functions, i.e. if R(t): [a, b] R 3 is a vector function such that then b a R (t) r(t) r(t)dt R(b) R(a). An such vector function is called antiderivative and the set of all antiderivatives is denoted b r(t)dt (t)dt i + (t)dt j + z(t)dt k. Notice that an two antiderivatives differ b a constant vector C c i + c j + c 3 k, i.e. an integration constant in each component. Using the definition and properties for integrals of realvalued functions, one can prove easil the following properties: () () (3) d dt C r(t)dt C r(t)dt [ r (t) ± r (t)]dt r(t)dt r(t) r (t)dt ± r (t)dt
43 38 6. Arc Length Eample. Find r(t)dt and r(t)dt, where Solution: or r(t)dt r(t) t i + (t + ) j t dt i + (t + )dt j ( ) t C i + ( ) t + t + C j t3 3 i + ( t + t ) j + C ( t 3 r(t)dt 3 i + ( t + t ) ) j 3 i + j. r(t)dt t dt i + t3 + (t 3 i + t) j 3 i + j (t + )dt j 6. Arc Length We know from first ear calculus that if (t) and (t) are continuous, then the curve given b (t), (t), a t b has arc length ((b),(b)) L b a (t) + (t) dt. ((a),(a)) This formula generalizes to 3space curves: The arc length of the curve (t), (t), z z(t), a t b z ((b),(b),z(b)) is given b ((a),(a),z(a)) L b a (t) + (t) + z (t) dt.
44 6.3 Arc Length as a Parameter 39 Eample. Find the arc length of the curve r a cost i + a sin t j, t π. Solution: L π π π (t) + (t) dt ( a sin t) + (a cost) dt a (sin t + cos (t)) dt π a dt πa. 6.3 Arc Length as a Parameter If we visualise a curve as the trajector of a moving object it is clear that the same trajector can be travelled at a different speed. This means that the same curve is represented in the parametric form with different parameters, and thus it has different parametric equations. For eample, (t) a cos t, (t) a sin t, t π (s) a cos(s ), (s) a sin(s ), s π (u) a cos(πe u ), (u) a sin(πe u ), u < all represent the same curve: a circle with center (, ) and radius a. a To avoid such ambiguit, it is desirable to have a universal parameter for the parametric equations. This can be done b stipulating that we travel the curve with speed, i.e. length unit i unit of time. In other words, the arc length is used as parameter. Let us now see how this can be done. Let C be a given smooth curve. We first introduce the arc length parameter using the following three steps:
45 4 6.3 Arc Length as a Parameter () choose a point P on the curve, called a reference point; () Starting from P, choose one direction along the curve to be the positive direction and the other to be the negative direction; P P (3) If P is a point on C, let s be the signed arc length along C from P to P, where s is positive if P is in the positive direction from P and s is negative if P is in the negative direction from P. Let us suppose that C is initiall given b the parametric equations (t), (t), z z(t), and P ((t ), (t ), z(t )), P ((t), (t), z(t)), and the positive direction of C is the direction of increasing t. Then we know from the last section that s t t (u) + (u) + z (u) du This gives s as a function of t. Differentiating we obtain ds dt (t) + (t) + z (t). Eample 3. Find parametric equations for a cost, a sin t, t π, using arc length s as a parameter, with reference point for s being (, a) in the plane. Solution: The point (, a) corresponds to t π on the curve. Therefore, s t π t π t π (u) + (u) du ( a sin u) + (a cosu) du ( a dt a t π ). Solving for t from s a ( t π ) we obtain t s a + π.
46 6.3 Arc Length as a Parameter 4 As t varies from to π, s varies from π a to 3 πa. Hence, the parametric equations in s are Or, in vector form ( s a cos a + π ( s, a sin ) a + π, ) π a s 3 πa. ( s r a cos a + π ) ( s i + a sin a + π ) j, π a s 3 πa.
47 4 7. Smooth curves Lecture 7 Unit Tangent and Normal Vectors 7. Smooth curves We call a curve smooth if it has a tangent at each point and the slope of the tangent changes in a continuous wa from point to point. In particular this means that the curve has no edges or cusps. If a curve in the plane is described as the graph of a function f(): (a, b) R that has continuous derivative then the curve is a smooth function. However, the following eample shows that if r(t) has continuous derivative, the curve r r(t) ma not be smooth. Eample. Let r(t) t i + t 3 j, then r (t) t i + 3t j is continuous. The parametric form for the graph of r(t) is t, t 3, which is equivalent to 3 and it represents a curve which is not smooth at (, ). Indeed, the direction of the tangent at r(t ) is given b (t ), (t ) t, 3t. For t, the cusp point, this is the zero vector which does not give an direction. Let us tr to look a this curve as a graph. Since the vertical line test gives two intersection points, the curve is not the graph of a function f() but it is the graph of a function g() 3. Notice that the derivative of g() is, according to the chain rule g () d d d dt t 3t and does not eist at t, i.e. (, ) (, ). The vanishing denominator 3t is clearl the problem here. If we have a parametric curve (t), (t) such that d (t dt ) then the function (t) has an inverse t h() on some (possibl ver small) interval containing (t ). In this case the derivative h ( ) (t. This is the statement of the ) d dt
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