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- 1. Midterm I Review Math 21a March 5, 2008 . . Image: Flickr user Mr. Theklan . . . . . .
- 2. Announcements ◮ Midterm covers up to Section 11.4 in text ◮ Any topics not covered in class or on HW will not be on test (i.e., curvature) ◮ Complete list of learning objectives on the Midterm I Review sheet ◮ Odd problems, chapter reviews, old exams, reviews . . . . . .
- 3. Outline Vectors and the Geometry of Derivatives and Integrals of Space Vector Functions Three-Dimensional Arc Length (not Curvature) Coordinate Systems Motion in Space: Velocity and Vectors Acceleration The Dot Product Parametric Surfaces The Cross Product Partial Derivatives Equations of Lines and Planes Functions of Several Variables Functions and surfaces Utility Functions and Cylindrical and Spherical indifference curves Coordinates Limits and Continuity Vector Functions Partial Derivatives Vector Functions and Space Tangent Planes and Linear Curves Approximations . . . . . .
- 4. Three-Dimensional Coordinate Systems Section 9.1 Learning Objectives ◮ To understand and be able to apply rectangular coordinate systems in R3 and the right hand rule. ◮ To understand and be able to ﬁnd the distance d between two points (x1 , y1 , z1 ) and (x2 , y2 , z2 ) in R3 and to be able to use the distance formula √ d = (x1 − x2 )2 + (y1 − y2 )2 + (z1 − z2 )2 . ◮ To understand and be able to apply the equation of a sphere of radius r and center (x0 , y0 , z0 ), (x − x0 )2 + (y − y0 )2 + (z − z0 )2 = r2 . . . . . . .
- 5. Axes in three-dimensional space z . . z y . . y . x . x . x . y . . y x . z . . z . . . . . .
- 6. The right-hand rule z . x . . . y . . . . . . .
- 7. Distance in space Example Find the distance between the points P1 (3, 2, 1) and P2 (4, 4, 4). . . . . . .
- 8. Distance in space Example Find the distance between the points P1 (3, 2, 1) and P2 (4, 4, 4). z . Solution ..2 P . d . 3 . 2 . . y 1 .. √ P .1 . 5 x . √ √ d= 5 + 32 = 1+4+9 . . . . . .
- 9. Distance in space—General Theorem The distance between (x1 , y1 , z1 ) and (x2 , y2 , z2 ) is √ (x2 − x1 )2 + (y2 − y1 )2 + (z2 − z1 )2 In space, the locus of all points which are a ﬁxed distance from a ﬁxed point is a sphere. . . . . . .
- 10. Munging an equation to see its surface Example Find the surface is represented by the equation x2 + y2 + z2 − 4x + 8y − 10z + 36 = 0 . . . . . .
- 11. Munging an equation to see its surface Example Find the surface is represented by the equation x2 + y2 + z2 − 4x + 8y − 10z + 36 = 0 Solution We can complete the square: 0 = x2 − 4x + 4 + y2 + 8y + 16 + z2 − 10z + 25 + 36 − 4 − 16 − 25 = (x − 2)2 + (y + 4)2 + (z − 5)2 − 9 So (x − 2)2 + (y + 4) + (z − 5)2 = 9 =⇒ |(x, y, z)(2, −4, 5)| = 3 This is a sphere of radius 3, centered at (2, −4, 5). . . . . . .
- 12. Vectors Learning objectives for Section 9.2 ◮ To understand and be able to apply the deﬁnition of a scalar and a vector. ◮ To be able to represent vectors geometrically as directed line segments or algebraically as ordered pairs or triples of numbers. ◮ To understand and be able to apply the axioms or vector addition and scalar multiplication. ◮ To be understand and be able to determine the length of a vector. ◮ To understand the deﬁnitions of unit vectors, standard basis vectors, and position vectors and to be able to apply these deﬁnitions. . . . . . .
- 13. What is a vector? Deﬁnition ◮ A vector is something that has magnitude and direction ◮ We denote vectors by boldface (v) or little arrows (⃗ One is v). good for print, one for script ◮ Given two points A and B in ﬂatland or spaceland, the vector which starts at A and ends at B is called the displacement −→ vector AB. ◮ Two vectors are equal if they have the same magnitude and direction (they need not overlap) B . D . . v . u .. A C . . . . . . .
- 14. Vector or scalar? Deﬁnition A scalar is another name for a real number. Example Which of these are vectors or scalars? (i) Cost of a theater ticket (ii) The current in a river (iii) The initial ﬂight path from Boston to New York (iv) The population of the world . . . . . .
- 15. Vector or scalar? Deﬁnition A scalar is another name for a real number. Example Which of these are vectors or scalars? (i) Cost of a theater ticket scalar (ii) The current in a river vector (iii) The initial ﬂight path from Boston to New York vector (iv) The population of the world scalar . . . . . .
- 16. Vector addition Deﬁnition If u and v are vectors positioned so the initial point of v is the terminal point of u, the sum u + v is the vector whose initial point is the initial point of u and whose terminal point is the terminal point of v. . u . v . v . +v u . +v u . v . u . u . . The triangle law The parallelogram law . . . . . .
- 17. Opposite and difference Deﬁnition Given vectors u and v, ◮ the opposite of v is the vector −v that has the same length as v but points in the opposite direction ◮ the difference u − v is the sum u + (−v) . v . u . − . v . −v u . . . . . .
- 18. Scaling vectors Deﬁnition If c is a nonzero scalar and v is a vector, the scalar multiple cv is the vector whose ◮ length is |c| times the length of v ◮ direction is the same as v if c > 0 and opposite v if c < 0 If c = 0, cv = 0. 2 .v . v −2 . 1v . . . . . . .
- 19. Components deﬁned Deﬁnition ◮ Given a vector a, it’s often useful to move the tail to O and measure the coordinates of the head. These are called the components of a, and we write them like this: a = ⟨a1 , a2 , a3 ⟩ or just two components if the vector is the plane. Note the angle brackets! ◮ Given a point P in the plane or space, the position vector of − → P is the vector OP. Fact − → Given points A(x1 , y1 , z1 ) and B(x2 , y2 , z2 ) in space, the vector AB has components − → AB = ⟨x2 − x1 , y2 − y1 , z2 − z1 ⟩ . . . . . .
- 20. Vector algebra in components Theorem If a = ⟨a1 , a2 , a3 ⟩ and b = ⟨b1 , b2 , b3 ⟩, and c is a scalar, then ◮ a + b = ⟨a1 + b1 , a2 + b2 , a3 + b3 ⟩ ◮ a − b = ⟨a1 − b1 , a2 − b2 , a3 − b3 ⟩ ◮ ca = ⟨ca1 , ca2 , ca3 ⟩ . . . . . .
- 21. Properties Theorem Given vectors a, b, and c and scalars c and d, we have 1. a+b=b+a 6. (c + d)a = ca + da 2. a + (b + c ) = ( a + b ) + c 7. (cd)a = c(da) 3. a+0=a 4. a + (−a) = 0 8. 1a = a 5. c(a + b) = ca + cb These can be veriﬁed geometrically. . . . . . .
- 22. Length Deﬁnition Given a vector v, its length is the distance between its initial and terminal points. Fact The length of a vector is the square root of the sum of the squares of its components: √ |⟨a1 , a2 , a3 ⟩| = a2 + a2 + a2 1 2 3 . . . . . .
- 23. Useful vectors Deﬁnition ◮ A unit vector is a vector whose length is one ◮ We deﬁne the standard basis vectors i = ⟨1, 0, 0⟩, j = ⟨0, 1, 0⟩, k = ⟨0, 0, 1⟩. In script, they’re often written as ˆ, ȷ, ı ˆ ˆ k. Fact Any vector a can be written as a linear combination of the standard basis vectors ⟨a1 , a2 , a3 ⟩ = a1 i + a2 j + a3 k. . . . . . .
- 24. The Dot Product I Learning objectives for Section 9.3 ◮ To understand and be able to use the deﬁnitions of the dot product to measure the length of a vector and the angle θ between two vectors. Given two vectors a = ⟨a1 , a2 , a3 ⟩ and b = ⟨b1 , b2 , b3 ⟩, a · b = |a||b| cos θ = a1 b1 + a2 b2 + a3 b3 . ◮ To understand and be able to apply properties of the dot product. . . . . . .
- 25. The Dot Product II Learning objectives for Section 9.3 ◮ To understand that two vectors are orthogonal if their dot product is zero. ◮ To understand and be able to determine the projection of a vector a onto a vector b, a·b proja b = a |a|2 ◮ To be able to use vectors and the dot product in applications. . . . . . .
- 26. Deﬁnition If a and b are any two vectors in the plane or in space, the dot product (or scalar product) between them is the quantity a · b = |a| |b| cos θ, where θ is the angle between them. Another way to say this is that a · b is |b| times the length of the projection of a onto b. . a . · b is |b| times this length. a . b In components, if a = ⟨a1 , a2 , a3 ⟩ and b = ⟨b1 , b2 , b3 ⟩, then a · b = a1 b1 + a2 b2 + a3 b3. . . . . .
- 27. Algebraic properties of the dot product Fact If a, b and c are vectors are c is a scalar, then 1. a · a = |a|2 4. (ca) · b = c(a · b) = a · (cb) 2. a · b = b · a 5. 0 · a = 0 3. a · (b + c) = a · b + a · c . . . . . .
- 28. Geometric properties of the dot product Fact The projection of b onto a is given by a·b proja b = a |a|2 Fact The angle between two nonzero vectors a and b is given by a·b cos θ = , |a| |b| where θ is taken to be between 0 and π . . . . . . .
- 29. More geometric properties of the dot product Fact The angle between two nonzero vectors a and b is acute if a · b > 0. obtuse if a · b < 0, right if a · b = 0. The vectors are parallel if a · b = ± |a| |b|. ◮ b is a positive multiple of a if a · b = |a| |b| ◮ b is a negative multiple of a if a · b = − |a| |b| . . . . . .
- 30. Other uses of the dot product ◮ similarity ◮ Work W=F·d . . . . . .
- 31. The Cross Product Learning objectives for Section 9.4 ◮ To understand and be able to use the deﬁnitions of the cross product to ﬁnd a vector that is orthogoal two vectors. Given two vectors a = ⟨a1 , a2 , a3 ⟩ and b = ⟨b1 , b2 , b3 ⟩, a × b = (|a||b| sin θ)n = ⟨a2 b3 − a3 b2 , a3 b1 − a1 b3 , a1 b2 − a2 b1 ⟩, where n is a unit vector perpendicular to both a and b. ◮ To understand and be able to apply properties of the dot product, especially a × b = −b × a. ◮ To understand that two vectors are parallel if and only if their cross product is zero. ◮ To be able to use vectors and the cross product in applications. . . . . . .
- 32. Deﬁnition of the cross product Deﬁnition Given vectors a and b in space, the cross product of a and b is the vector a × b = |a| |b| (sin θ) n, where n is a vector perpendicular to a and b such that (a, b, n) is a right-handed set of three vectors. In components, if a = ⟨a1 , a2 , a3 ⟩= a1 i + a2 j + a3 k b = ⟨b1 , b2 , b3 ⟩= b1 i + b2 j + b3 k Then a × b = (a2 b3 − b2 a3 )i + (a3 b1 − b3 a1 )j + (a1 b2 − b1 a2 )k = ⟨a 2 b 3 − b 2 a 3 , a 3 b 1 − b 3 a 1 , a 1 b 2 − b 1 a 2 ⟩ . . . . . .
- 33. Determinant formula This is only to help you remember, in case you’ve seen determinants of 3 × 3 matrices: i j k a a a a a a a1 a2 a3 = i 2 3 − j 1 3 + k 1 2 b2 b3 b1 b 3 b1 b2 b1 b2 b3 = (a2 b3 − b2 a3 )i − (b3 a1 − a3 b1 )j + (a1 b2 − b1 a2 )k =a×b . . . . . .
- 34. Algebraic Properties of the Cross Product If a, b, and c are vectors and c is a scalar, then 1. a × b = −b × a 2. (ca) × b = c(a × b) = a × (cb) 3. a × (b + c) = a × b + a × c 4. (a + b) × c = a × c + b × c . . . . . .
- 35. Geometric Properties of the cross product ◮ a × b = 0 exactly when a and b are parallel. ◮ The magnitude of the cross product a × b is the area of the parallelogram with sides a and b. . b | . b| sin θ θ . . . a . . . . . .
- 36. Applications of the cross product ◮ Area of a parallelogram ◮ Torque τ = r × F ◮ Volume of a parallelipiped (scalar triple product) V = |a · (b × c)| . . . . . .
- 37. The scalar triple product in action Example Find the volume of the parallelipiped spanned by a = ⟨2, 0, −3⟩, b = ⟨1, 1, 1⟩, and c = ⟨0, 4, −1⟩ . . . . . .
- 38. The scalar triple product in action Example Find the volume of the parallelipiped spanned by a = ⟨2, 0, −3⟩, b = ⟨1, 1, 1⟩, and c = ⟨0, 4, −1⟩ Solution The scalar triple product can be calculated by . b 2 0 −3 . c 1 1 1 = −22 . 0 4 −1 It’s negative because the triple . a (a, b, c) is left-handed. So the volume is 22. . . . . . .
- 39. Equations of Lines and Planes I ◮ To understand and be able to represent a line ℓ in R3 using ◮ the vector equation r = r0 + tv ◮ the parametric equations x = x0 + at y = y0 + bt z = z0 + ct ◮ the symmetric equations x − x0 y − y0 z − z0 = = a b c . . . . . .
- 40. Equations of Lines and Planes II ◮ To be able to represent line segment in R3 from r0 to r1 , r(t) = (1 − t)r0 + tr1 , where 0 ≤ t ≤ 1. ◮ To understand and be able to represent a plane in R3 given a normal vector n = ⟨a, b, c⟩ and a point P0 = (x0 , y0 , z0 ) in the plane, n · ⟨x − x 0 , y − y 0 , z − z 0 ⟩ = 0 or a(x − x0 ) + b(y − y0 ) + c(z − z0 ) = 0. . . . . . .
- 41. Equations of Lines and Planes III ◮ To understand and be able to calculate the distance D from a point P1 = (x1 , y1 , z1 ) to the plane ax + by + cz + d = 0, |ax1 + by1 + cz1 + d| D= √ . a 2 + b2 + c2 . . . . . .
- 42. Applying the deﬁnition Example Find the vector, parametric, and symmetric equations for the line that passes through (1, 2, 3) and (2, 3, 4). . . . . . .
- 43. Applying the deﬁnition Example Find the vector, parametric, and symmetric equations for the line that passes through (1, 2, 3) and (2, 3, 4). Solution ◮ Use the initial vector ⟨1, 2, 3⟩ and direction vector ⟨2, 3, 4⟩ − ⟨1, 2, 3⟩ = ⟨1, 1, 1⟩. Hence r(t) = ⟨1, 2, 3⟩ + t ⟨1, 1, 1⟩ ◮ The parametric equations are x=1+t y=2+t z=3+t ◮ The symmetric equations are x−1=y−2=z−3 . . . . . .
- 44. Planes The set of all points whose displacement vector from a ﬁxed point is perpendicular to a ﬁxed vector is a plane. z . . n .0 r . y . . x . . . . . .
- 45. Equations for planes The plane passing through the point with position vector r0 = ⟨x0 , y0 , z0 ⟩ perpendicular to ⟨a, b, c⟩ has equations: ◮ The vector equation n · (r − r0 ) = 0 ⇐⇒ n · r = n · r0 ◮ Rewriting the dot product in component terms gives the scalar equation a(x − x0 ) + b(y − y0 ) + c(z − z0 ) = 0 The vector n is called a normal vector to the plane. ◮ Rearranging this gives the linear equation ax + by + cz + d = 0, where d = −ax0 − by0 − cz0 . . . . . . .
- 46. Example Find an equation of the plane that passes through the points P(1, 2, 3), Q(3, 5, 7), and R(4, 3, 1). . . . . . .
- 47. Example Find an equation of the plane that passes through the points P(1, 2, 3), Q(3, 5, 7), and R(4, 3, 1). Solution− → − → − → Let r0 = OP = ⟨1, 2, 3⟩. To get n, take PQ × PR: i j k − → − → PQ × PR = 2 3 4 = ⟨−10, 16, −7⟩ 3 1 −2 So the scalar equation is −10(x − 1) + 16(y − 2) − 7(z − 3) = 0. . . . . . .
- 48. Distance from point to line Deﬁnition The distance between a point and a line is the smallest distance from that point to all points on the line. You can ﬁnd it by projection. Q . . b·v .b − v v·v . b b·v . rojv b = p v . v v·v θ . . P .0 . . . . . .
- 49. Distance from a point to a plane Deﬁnition The distance between a point and a plane is the smallest distance from that point to all points on the line. . . Q . b |n · b| . |n| . n . P .0 To ﬁnd the distance from the a point to a plane, project the displacement vector from any point on the plane to the given point onto the normal vector. . . . . . .
- 50. Distance from a point to a plane II We have |n · b| D= |n| If Q = (x1 , y1 , z1 ), and the plane is given by ax + by + cz + d = 0, then n = ⟨a, b, c⟩, and n · b = ⟨a, b, c⟩ · ⟨x1 − x0 , y1 − y0 , z1 − z0 ⟩ = ax1 + by1 + cz1 − ax0 − by0 − cz0 = ax1 + by1 + cz1 + d So the distance between the plane ax + by + cz + d = 0 and the point (x1 , y1 , z1 ) is |ax1 + by1 + cz1 + d| D= √ a 2 + b 2 + c2 . . . . . .
- 51. Functions and surfaces Learning objectives for Section 9.6 ◮ To understand and be able to represent a function of two variables, z = f(x, y), graphically. Section 9.6) ◮ To understand and be able to use the trace (cross-section) of a function. Section 9.6) ◮ To understand and be able to determine the domain of a function z = f(x, y). Section 9.6) ◮ To understand and be able to represent a selection of quadric surfaces graphically. . . . . . .
- 52. function, domain, range Deﬁnition A function f of two variables is a rule that assigns to each ordered pair of real numbers (x, y) in a set D a unique real number denoted by f(x, y). The set D is the domain of f and its range is the set of values that f takes on. That is range f = { f(x, y) | (x, y) ∈ D } . . . . . .
- 53. Example Example √ Find the domain and range of f(x, y) = xy. . . . . . .
- 54. Example Example √ Find the domain and range of f(x, y) = xy. Solution ◮ Working from the outside in, we see that xy must be nonnegative, which means x ≥ 0 and y ≥ 0 or x ≤ 0 and y ≤ 0. Thus the domain is the union of the coordinate axes, and the ﬁrst and third quadrants. ◮ The range of f is the set of all “outputs” of f. Clearly the range of f is restricted to the set of nonnegative numbers. To make sure that we can get all nonnegative numbers x, notice x = f(x, x). . . . . . .
- 55. Solution continued Here is a sketch of the domain: . . . . . . .
- 56. Traces and surfaces Example Describe and sketch the surfaces. (i) z = 4x2 + y2 (vi) y2 − 9x2 − 4z2 = 36 (ii) z = 4 − x 2 − y2 (vii) x2 + 4y2 + 2z2 = 4 (iii) z 2 = 4 (x 2 + y 2 ) (iv) x = 2y2 − z2 (viii) y2 + z2 = 1 (v) x2 + y2 − 9z2 = 9 . . . . . .
- 57. Sketch of z = 4x2 + y2 The trace in the xz-plane (y = 0) is the parabola z = 4x2 . The trace in the yz-plane is the parabola z = y2 . The trace in the xy-plane is the curve 4x2 + y2 = 0, which is just a point (0, 0). But the trace in the plane z = k (k > 0) is the ellipse 4x2 + y2 = k. So we get ellipses crossing the parabolas, an elliptic paraboloid. 2 1 0 1 2 2.0 1.5 1.0 0.5 0.0 1.0 0.5 0.0 0.5 1.0 . . . . . .
- 58. Sketch of z2 = 4(x2 + y2 ) The trace in the xz-plane is the equation z2 = 4x2 , or z = ±2x. The trace in the yz-plane is the equation z = ±2y. Traces in the plane z = k are circles k2 = 4(x2 + y2 ). So we have a cone. 1.0 0.5 0.0 0.5 1.0 1.0 0.5 0.0 0.5 1.0 1.0 0.5 0.0 0.5 1.0 . . . . . .
- 59. Sketch of x = 2y2 − z2 The trace in the xy-plane is a parabola x = 2y2 . The trace in the xz-plane is a parabola opening the other direction: x = −z2 . Traces x = k give hyperbolas. So we have a hyperbolic paraboloid. 1.0 0.5 0.0 0.5 1.0 1.0 0.5 0.0 0.5 1.0 1.0 0.5 0.0 0.5 1.0 . . . . . .
- 60. Sketch of x2 + y2 − 9z2 = 9 The x2 + y2 piece tells us that this is a surface of revolution, where the z-axis is the axis of revolution. If we ﬁnd the xz-trace, we get x2 − 9z2 = 9, a hyperbola. We get a hyperboloid of one sheet. 2 1 0 4 1 2 2 4 0 2 0 2 2 4 4 . . . . . .
- 61. Sketch of y2 − 9x2 − 4z2 = 36 Traces in the xy and yz planes are hyperbolas. Traces in the plane y = k are ellipses 9x2 + 4z2 = k2 − 36. We have a hyperboloid of two sheets. 10 5 0 5 10 4 2 0 2 4 2 0 2 . . . . . .
- 62. Sketch of x2 + 4y2 + 2z2 = 4 The traces in the planes x = 0, y = 0, and z = 0 are all ellipses. We get an ellipsoid. 1.0 2 0.5 1 0.0 0 0.5 1 1.0 2 2 1 0 1 2 . . . . . .
- 63. Traces and surfaces Example Describe and sketch the surfaces. (i) z = 4x2 + y2 elliptic (v) x2 + y2 − 9z2 = 9 paraboloid hyperboloid of one sheet (ii) z = 4 − x2 − y2 elliptic (vi) y2 − 9x2 − 4z2 = 36 paraboloid hyperboloid of two sheets (iii) z2 = 4(x2 + y2 ) cone (vii) x2 + 4y2 + 2z2 = 4 ellipsoid (iv) x = 2y2 − z2 hyperbolic (viii) y2 + z2 = 1 cylinder paraboloid . . . . . .
- 64. Cylindrical and Spherical Coordinates I Learning objectives for Section 9.7 ◮ To understand and be able to apply the polar coordinate system in R2 , and to understand the relationship to rectangular coordinates: x = r cos θ y = r sin θ 2 2 2 r =x +y tan θ = y/x, where r ≥ 0 and 0 ≤ θ ≤ 2π . . . . . . .
- 65. Cylindrical and Spherical Coordinates II Learning objectives for Section 9.7 ◮ To understand and be able to apply the cylindrical coordinate system in R3 , and to understand the relationship to rectangular coordinates: x = r cos θ y = r sin θ z=z r 2 = x2 + y 2 tan θ = y/x z = z, where r ≥ 0 and 0 ≤ θ ≤ 2π . . . . . . .
- 66. Cylindrical and Spherical Coordinates III Learning objectives for Section 9.7 ◮ To understand and be able to apply the spherical coordinate system in R3 , and to understand the relationship to rectangular coordinates: x = ρ sin φ cos θ y = ρ sin φ sin θ z = ρ cos φ 2 2 2 2 ρ =x +y +z , where r ≥ 0, 0 ≤ θ ≤ 2π , and 0 ≤ φ ≤ π . . . . . . .
- 67. Why different coordinate systems? ◮ The dimension of space comes from nature ◮ The measurement of space comes from us ◮ Different coordinate systems are different ways to measure space . . . . . .
- 68. Vectors and the Geometry of Space Conversion from polar to cartesian (rectangular) x = r cos θ y = r sin θ Conversion from cartesian to . r . y θ polar: . . . x √ r = x2 + y 2 x y y cos θ = sin θ = tan θ = r r x . . . . . .
- 69. Cylindrical Coordinates Just add the vertical dimension Conversion from cylindrical to cartesian (rectangular): z . x = r cos θ y = r sin θ ( . r, θ, z) z=z Conversion from cartesian to z z . . cylindrical: . r . . y √ θ . y . . r = x2 + y 2 x . . x y y x . cos θ = sin θ = tan θ = r r x z=z . . . . . .
- 70. Spherical Coordinates like the earth, but not exactly Conversion from spherical to cartesian (rectangular): z . x = ρ sin φ cos θ . y = ρ sin φ sin θ ( . r, θ, φ) z = ρ cos φ . Conversion from cartesian to φ ρ . . z . spherical: . . y √ √ x . θ . r = x2 + y2 ρ = x2 + y2 + z2 y . x y y x . cos θ = sin θ = tan θ = r r x z cos φ = ρ . . . . . .
- 71. Outline Vectors and the Geometry of Derivatives and Integrals of Space Vector Functions Three-Dimensional Arc Length (not Curvature) Coordinate Systems Motion in Space: Velocity and Vectors Acceleration The Dot Product Parametric Surfaces The Cross Product Partial Derivatives Equations of Lines and Planes Functions of Several Variables Functions and surfaces Utility Functions and Cylindrical and Spherical indifference curves Coordinates Limits and Continuity Vector Functions Partial Derivatives Vector Functions and Space Tangent Planes and Linear Curves Approximations . . . . . .
- 72. Vector Functions and Space Curves Learning objectives for Section 10.1 ◮ To understand and be able to apply the concept of a vector-valued function, r(t) = ⟨f(t), g(t), h(t)⟩ = f(t)i + g(t)j + h(t)k. ◮ To be able to apply the concepts of limits and continuity to vector-valued functions. ◮ To understand that a curve in R3 can be represented parametrically x = f(t) y = g(t) z = h (t) or by a vector-valued function r(t) = ⟨f(t), g(t), h(t)⟩ . . . . . .
- 73. Example Given the plane curve described by the vector equation r(t) = sin(t)i + 2 cos(t)j (a) Sketch the plane curve. . . . . . .
- 74. Solution r(t) = r(t) = sin(t)i + 2 cos(t)j y . t r (t) 0 2j π/2 i . (π/4) r π −2j . x . 3π/2 −i 2π 2j . . . . . .
- 75. Derivatives and Integrals of Vector Functions Learning objectives for Section 10.2 ◮ To understand and be able to calculate the derivative of a vector-valued function. ◮ To understand and be able to apply the basic rules of differentiation for vector-valued functions. ◮ To understand and be able to ﬁnd the deﬁnite integral of a vector-valued function. . . . . . .
- 76. Derivatives of vector-valued functions Deﬁnition Let r be a vector function. ◮ The limit of r at a point a is deﬁned componentwise: ⟨ ⟩ lim r(t) = lim f(t), lim g(t), lim h(t) t→a t→a t→a t→a ◮ The derivative of r is deﬁned in much the same way as it is for real-valued functions: dr r(t + h) − r(t) = r′ (t) = lim dt h→0 h . . . . . .
- 77. Rules for differentiation Theorem Let u and v be differentiable vector functions, c a scalar, and f a real-valued function. Then: d 1. [u(t) + v(t)] = u′ (t) + v′ (t) dt d 2. [cu(t)] = cu′ (t) dt d 3. [f(t)u(t)] = f′ (t)u(t) + f(t)u′ (t) dt d 4. [u(t) · v(t)] = u′ (t) · v(t) + u(t) · v′ (t) dt d 5. [u(t) × v(t)] = u′ (t) × v(t) + u(t) × v′ (t) dt d 6. [u(f(t))] = f′ (t)u′ (f(t)) dt . . . . . .
- 78. Example Given the plane curve described by the vector equation r(t) = sin(t)i + 2 cos(t)j (a) Sketch the plane curve. (b) Find r′ (t) . . . . . .
- 79. Solution r(t) = r(t) = sin(t)i + 2 cos(t)j r′ (t) = cos(t)i − 2 sin(t)j y . t r (t) 0 2j π/2 i . (π/4) r π −2j . x . 3π/2 −i 2π 2j . . . . . .
- 80. Example Given the plane curve described by the vector equation r(t) = sin(t)i + 2 cos(t)j (a) Sketch the plane curve. (b) Find r′ (t) (c) Sketch the position vector r(π/4) and the tangent vector r′ (π/4). . . . . . .
- 81. Solution r(t) = r(t) = sin(t)i + 2 cos(t)j r′ (t) = cos(t)i − 2 sin(t)j y . t r (t) 0 2j π/2 i . (π/4) r . ′ (π/4) r π −2j . . x 3π/2 −i 2π 2j . . . . . .
- 82. Integrals of vector-valued functions Deﬁnition Let r be a vector function deﬁned on [a, b]. For each whole number n, divide the interval [a, b] into n pieces of equal width ∆t. Choose a point t∗ on each subinterval and form the Riemann sum i ∑ n Sn = r(t∗ ) ∆t i i=1 Then deﬁne ∫ b ∑ n r(t) dt = lim Sn = lim r(t∗ ) ∆t i a n→∞ n→∞ i=1 [ ] ∑ n ∑ n ∑ n = lim f(t∗ ) ∆ti i + g(t∗ ) ∆tj i + h(t∗ ) ∆tk i n→∞ i=1 i=1 i=1 (∫ b ) (∫ b ) (∫ b ) = f(t) dt i + g(t) dt j + h(t) dt k a a a . . . . . .
- 83. FTC for vector functions Theorem (Second Fundamental Theorem of Calculus) If r(t) = R′ (t), then ∫ b r(t) dt = R(b) − R(a) a Example ∫ π Given r(t) = ⟨t, cos 2t, sin 2t⟩, ﬁnd r(t) dt. 0 Answer ⟨∫ π ∫ π ∫ π ⟩ ⟨ ⟩ π2 t dt, cos 2t dt, sin 2t dt = , 0, 0 0 0 0 2 . . . . . .
- 84. Arc Length (not Curvature) Learning objectives for Section 10.3 ◮ To understand and be able to apply the arc length of a curve, r(t) = ⟨f(t), g(t), h(t)⟩, over an interval a ≤ t ≤ b, ∫ b ∫ b√ ′ L= |r (t)| dt = [f′ (t)]2 + [g′ (t)]2 + [h′ (t)]2 dt. a a ◮ To understand and be able to apply the arc length function of a curve, r(t) = ⟨f(t), g(t), h(t)⟩, over an interval a ≤ t ≤ b, ∫ t ∫ t√ ′ s (t) = |r (u)| du = [f′ (u)]2 + [g′ (u)]2 + [h′ (u)]2 du. a a . . . . . .
- 85. Length of a curve Break up the curve into pieces, and approximate the arc length with the sum of the lengths of the pieces: y . . . x . . . . . .
- 86. Length of a curve Break up the curve into pieces, and approximate the arc length with the sum of the lengths of the pieces: y . . . x . . . . . .
- 87. Length of a curve Break up the curve into pieces, and approximate the arc length with the sum of the lengths of the pieces: y . . . x . . . . . .
- 88. Length of a curve Break up the curve into pieces, and approximate the arc length with the sum of the lengths of the pieces: y . . . x . . . . . .
- 89. Length of a curve Break up the curve into pieces, and approximate the arc length with the sum of the lengths of the pieces: y . . . x . . . . . .
- 90. Length of a curve Break up the curve into pieces, and approximate the arc length with the sum of the lengths of the pieces: y . ∑√ n . ≈ L (∆xi )2 + (∆yi )2 i=1 . x . . . . . . .
- 91. Sum goes to integral If ⟨x, y⟩ is given by a vector-valued function r(t) = ⟨f(t), g(t), ⟩ with domain [a, b], we can approximate: ∆xi ≈ f′ (ti )∆ti ∆xi ≈ g′ (ti )∆ti So ∑√ n ∑ √[ n ]2 L≈ (∆xi )2 + (∆yi )2 ≈ f′ (ti )∆ti + [g′ (ti )∆ti ]2 i=1 i=1 ∑ √[ n ]2 = f′ (ti ) + [g′ (ti )]2 ∆ti i=1 As n → ∞, this converges to ∫ b√ L= [f′ (t)]2 + [g′ (t)]2 dt a In 3D, r(t) = ⟨f(t), g(t), h(t)⟩, and ∫ b√ L= [f′ (t)]2 + [g′ (t)]2 + [h′ (t)]2 dt a . . . . . .
- 92. Example Example Find the length of the parabola y = x2 from x = 0 to x = 1. . . . . . .
- 93. Example Example Find the length of the parabola y = x2 from x = 0 to x = 1. Solution ⟨ ⟩ Let r(t) = t, t2 . Then ∫ 1√ √ 5 1 √ L= 1 + (2t)2 = + ln 2 + 5 0 2 4 . . . . . .
- 94. Motion in Space: Velocity and Acceleration Learning objectives for Section 10.4 ◮ To understand the ﬁrst derivative as the velocity of a curve and the second derivative as the acceleration of a curve and be able to apply these deﬁnitions. . . . . . .
- 95. Velocity and Acceleration Deﬁnition Let r(t) be a vector-valued function. ◮ The velocity v(t) is the derivative r′ (t) ◮ The speed is the length of the derivative |r′ (t)| ◮ The acceleration is the second derivative r′′ (t). . . . . . .
- 96. Example Find the velocity, acceleration, and speed of a particle with position function r(t) = ⟨2 sin t, 5t, 2 cos t⟩ . . . . . .
- 97. Example Find the velocity, acceleration, and speed of a particle with position function r(t) = ⟨2 sin t, 5t, 2 cos t⟩ Answer ◮ r′ (t) = ⟨2 cos(t), 5, −2 sin(t)⟩ √ ◮ r′ (t) = 29 ◮ r′′ (t) = ⟨−2 sin(t), 0, −2 cos(t)⟩ . . . . . .
- 98. Parametric Surfaces Learning objectives for Section 10.5 ◮ To understand and be able to apply the concept of a parametric surface. A parametric surface may be deﬁned by a vector equation, r(t) = x(u, v) i + y(u, v) j + z(u, v) k or by a set of parametric equations x = x (u , v ) y = y(u, v) z = z(u, v), where u and v are variables with a domain D contained in R2 . ◮ To understand and be able to represent surfaces of revolutions parametrically. . . . . . .
- 99. Surfaces with easy parametrizations ◮ graph ◮ plane ◮ sphere ◮ surface of revolution . . . . . .
- 100. Parametrizing a graph Example Parametrize the surface described by z = x 2 + y2 , −2 ≤ x ≤ 2 , −3 ≤ y ≤ 3 . . . . . .
- 101. Parametrizing a graph Example Parametrize the surface described by z = x 2 + y2 , −2 ≤ x ≤ 2 , −3 ≤ y ≤ 3 Solution Let x = u, y = v, z = u 2 + v2 on the domain −2 ≤ u ≤ 2, −3 ≤ v ≤ 3. . . . . . .
- 102. parametrizing a sphere Example Find a parametrization for the unit sphere. . . . . . .
- 103. parametrizing a sphere Example Find a parametrization for the unit sphere. Solution Use spherical coordinates. Let: x = cos θ sin φ, y = sin θ sin φ, z = cos φ The domain of parametrization is 0 ≤ θ ≤ 2π , 0 ≤ φ ≤ π . . . . . . .
- 104. Parametrizing a surface of revolution Example Find a parametrization of the surface described by x2 − y 2 + z 2 = 1 , −3 ≤ y ≤ 3 . . . . . .
- 105. Parametrizing a surface of revolution Example Find a parametrization of the surface described by x2 − y 2 + z 2 = 1 , −3 ≤ y ≤ 3 Solution The surface is the graph of z2 − y2 = 1 revolved around the y-axis. So let √ √ x= 1+y 2 cos θ, y = u, z = 1 + y2 sin θ The domain is −3 ≤ u ≤ 3, 0 ≤ θ ≤ 2π . . . . . . .
- 106. Outline Vectors and the Geometry of Derivatives and Integrals of Space Vector Functions Three-Dimensional Arc Length (not Curvature) Coordinate Systems Motion in Space: Velocity and Vectors Acceleration The Dot Product Parametric Surfaces The Cross Product Partial Derivatives Equations of Lines and Planes Functions of Several Variables Functions and surfaces Utility Functions and Cylindrical and Spherical indifference curves Coordinates Limits and Continuity Vector Functions Partial Derivatives Vector Functions and Space Tangent Planes and Linear Curves Approximations . . . . . .
- 107. Functions of Several Variables Learning objectives for Section 11.1 ◮ To understand functions of several variables and be able to represent these functions using level sets. . . . . . .
- 108. A contour plot is a topographic map of a graph √ Consider the graph z = x2 + y2 Intersect the cone with planes z = c and what do you get? Circles. A contour plot shows evenly spaced circles. 3 2 1 0 4 3 -1 2 2 1 0 -2 0 -2 0 -2 -3 -3 -2 -1 0 1 2 3 2 . . . . . .
- 109. The paraboloid Example Graph z = x2 + y2 . 3 2 1 0 15 10 -1 2 5 0 0 -2 -2 0 -2 -3 -3 -2 -1 0 1 2 3 2 . . . . . .
- 110. The hyperbolic paraboloid Example Graph z = x2 − y2 . 3 2 1 0 5 -1 0 2 -5 -2 0 -2 0 -2 -3 -3 -2 -1 0 1 2 3 2 . . . . . .
- 111. Plotting a Cobb-Douglas function Example Plot z = x1/2 y1/2 . 3 2.5 2 1.5 3 2 3 1 1 2 0 0 0.5 1 1 2 0 0 0.5 1 1.5 2 2.5 3 30 . . . . . .
- 112. Utility Functions and indifference curves ◮ If u is a utility function, a level curve of u is a curve along which all points have the same u value. ◮ We also know this as an indifference curve . . . . . .
- 113. Limits and Continuity Learning objectives for Section 11.2 ◮ To understand and be able to apply the concept of a limit of a function of several variables. ◮ To understand and be able to apply the deﬁnition of continuity for a function of several variables. . . . . . .
- 114. Deﬁnition We write lim f(x, y) = L (x,y)→(a,b) and we say that the limit of f(x, y) as (x, y) approaches (a, b) is L if we can make the values of f(x, y) as close to L as we like by taking the point (x, y) to be sufﬁciently close to (a, b). . . . . . .
- 115. easy limits ◮ lim x=a (x,y)→(a,b) ◮ lim y=b (x,y)→(a,b) ◮ lim c=c (x,y)→(a,b) . . . . . .
- 116. Like regular limits, limits of multivariable functions can be ◮ added ◮ subtracted ◮ multiplied ◮ composed ◮ divided, provided the limit of the denominator is not zero. . . . . . .
- 117. Limit of a Polynomial Example Find lim (x5 + 4x3 y − 5xy2 ) (x,y)→(5,−2) . . . . . .
- 118. Limit of a Polynomial Example Find lim (x5 + 4x3 y − 5xy2 ) (x,y)→(5,−2) Solution lim (x5 + 4x3 y − 5xy2 ) = (5)5 + 4(5)3 (−2) − 5(5)(−2)2 (x,y)→(5,−2) = 3125 + 4(125)(−2) − 5(5)(4) = 2025. . . . . . .
- 119. Limit of a Rational Expression Example Compute x2 lim . (x,y)→(1,2) x2 + y2 . . . . . .
- 120. Limit of a Rational Expression Example Compute x2 lim . (x,y)→(1,2) x2 + y2 Solution x2 (1)2 lim = (x,y)→(1,2) x2 + y2 (1)2 + (2)2 1 = 5 . . . . . .
- 121. What can go wrong? The only real problem is a limit where the denominator goes to zero. ◮ If the numerator goes to some number and the denominator goes to zero then the quotient cannot have a limit. . . . . . .
- 122. What can go wrong? The only real problem is a limit where the denominator goes to zero. ◮ If the numerator goes to some number and the denominator goes to zero then the quotient cannot have a limit. ◮ If on the other hand the numerator and denominator both go to zero we have no clue. . . . . . .
- 123. Showing a limit doesn’t exist Theorem Suppose lim f(x, y) = L. Then the limit of f as (x, y) → (a, b) is L (x,y)→(a,b) along all paths through (a, b). There are two contrapositives to this statement: ◮ If there is a path through (a, b) along which the limit does not exist, the two-dimensional limit does not exist ◮ If there are two paths through (a, b) along which the limits exist but disagree, the two-dimensional limit does not exist . . . . . .
- 124. Continuity Deﬁnition A function f of two variables is called continuous at (a, b) if lim f(x, y) = f(a, b). (x,y)→(a,b) We say f is continuous on D if f is continuous at every point (a, b) in D. . . . . . .
- 125. Partial Derivatives Learning objectives for Section 11.3 ◮ To understand and be able to apply the deﬁnition of a partial derivative. ◮ To be able to compute partial derivatives. ◮ To understand and be able to apply Clairaut’s Theorem. If f is deﬁned on a disk D that contains the point (a, b) and the functions fxy and fyx are continuous on D, then fxy (a, b) = fyx (a, b). ◮ To understand the idea of a partial differential equation, and to be able to verify solutions to partial differential equations. . . . . . .
- 126. Deﬁnition Let f(x, y) be a function of two variables. We deﬁne the partial ∂f ∂f derivatives and at a point (a, b) as ∂x ∂y ∂f f (a + h, b) − f (a, b) (a, b) = lim ∂x h→0 h ∂f f (a, b + h) − f (a, b) (a, b) = lim ∂y h→0 h In other words, we temporarily treat the other variable as constant and differentiate the resulting one-variable function as in Calculus I. . . . . . .
- 127. Example Let f(x, y) = x3 − 3xy2 . Find its partial derivatives. . . . . . .
- 128. Example Let f(x, y) = x3 − 3xy2 . Find its partial derivatives. Solution ∂f When ﬁnding , we hold y constant. So ∂x ∂f ∂ = 3x2 − (3y2 ) (x) = 3x2 − 3y2 ∂x ∂x Similarly, ∂f = 0 − 3x(2y) = −6xy ∂y . . . . . .
- 129. Second derivatives If f(x, y) is a function of two variables, each of its partial derivatives are function of two variables, and we can hope that they are differentiable, too. So we deﬁne the second partial derivatives. ( ) ∂2f ∂ ∂f = = fxx ∂ x2 ∂x ∂x ( ) ∂2f ∂ ∂f = = fxy ∂y ∂x ∂y ∂x ( ) ∂2f ∂ ∂f = = fyx ∂x ∂y ∂x ∂y ( ) ∂2f ∂ ∂f = = fyy ∂ y2 ∂y ∂y . . . . . .
- 130. Don’t worry about the mixed partials The “mixed partials” bookkeeping may seem scary. However, we are saved by: Theorem (Clairaut’s Theorem/Young’s Theorem) If f is deﬁned near (a, b) and fxy and fyx are continuous at (a, b), then fxy (a, b) = fyx (a, b). The upshot is that we needn’t worry about the ordering. . . . . . .
- 131. Example (Continued) Let f(x, y) = x3 − 3xy2 . Find the second derivatives of f. . . . . . .
- 132. Example (Continued) Let f(x, y) = x3 − 3xy2 . Find the second derivatives of f. Solution We have fxx = (3x2 − 3y2 )x = 6x fxy = (3x2 − 3y2 )y = −6y fyx = (−6xy)x = −6y fyy = (−6xy)y = −6x Notice that fyx = fxy , as predicted by Clairaut (everything is a polynomial here so there are no concerns about continuity). The fact that fxx = fyy is a coincidence. . . . . . .
- 133. Partial Differential Equations Deﬁnition A partial differential equation (PDE) is a differential equation for a function of more than one variable. So the derivatives involved are partial derivatives. . . . . . .
- 134. Partial Differential Equations Deﬁnition A partial differential equation (PDE) is a differential equation for a function of more than one variable. So the derivatives involved are partial derivatives. Example Let f(x, y) = x3 − 3xy2 . Show that f satisﬁes the Laplace equation ∂2f ∂2f + 2 =0 ∂ x2 ∂ y . . . . . .
- 135. Partial Differential Equations Deﬁnition A partial differential equation (PDE) is a differential equation for a function of more than one variable. So the derivatives involved are partial derivatives. Example Let f(x, y) = x3 − 3xy2 . Show that f satisﬁes the Laplace equation ∂2f ∂2f + 2 =0 ∂ x2 ∂ y Solution We already showed that fxx = 6x and fyy = −6x. So fxx + fyy = 0. . . . . . .
- 136. Tangent Planes and Linear Approximations I Learning objectives for Section 11.4 ◮ To understand the concept of a tangent plane to a surface z = f(x, y) and to be able to compute the equation of tangent planes ◮ To understand and be able to ﬁnd a linear approximation to a function z = f(x, y). . . . . . .
- 137. Tangent Planes and Linear Approximations II Learning objectives for Section 11.4 ◮ To understand the concept of a tangent plane to a parametrically deﬁned surface r(u, v) = x(u, v) i + y(u, v) j + z(u, v) k and to be able to compute the equation of tangent planes. . . . . . .
- 138. Tangent Planes and Linear Approximations III Learning objectives for Section 11.4 ◮ To understand and be able to ﬁnd the differential to a function z = f(x, y), ∂z ∂z dz = dx + dy ∂x ∂y To be able to use the differential to estimate maximum error. . . . . . .
- 139. Deﬁnition If f is a function and P0 = (x0 , y0 , z0 = f(x0 , y0 ) a point on its graph, then ◮ The linearization of f near (x0 , y0 ) is the function ∂f ∂f L(x, y) = f(x0 , y0 ) + (x0 , y0 )(x − x0 ) + (x0 , y0 )(y − y0 ) ∂x ∂y ◮ The tangent plane to the graph at P0 is the graph of L The linearization is the best possible linear approximation to a function at a point. . . . . . .
- 140. Example Find an equation of the tangent plane to the surface z = y cos(x − y) at (2, 2, 2). . . . . . .
- 141. Example Find an equation of the tangent plane to the surface z = y cos(x − y) at (2, 2, 2). Solution We have ∂z ∂z = −y sin(x − y), = cos(x − y) + y sin(x − y) ∂x ∂y So ∂z ∂z = 0, =1 ∂x (2,2) ∂y (2,2) Therefore the equation of the tangent plane is z = 2 + 0(x − 2) + 1(y − 2) = y . . . . . .
- 142. Example The number of units of output per day at a factory is [ ]−1/2 1 −2 9 P(x, y) = 150 x + y−2 , 10 10 where x denotes capital investment (in units of $1000), and y denotes the total number of hours (in units of 10) the work force is employed per day. Suppose that currently, capital investment is $50,000 and the total number of working hours per day is 500. Estimate the change in output if capital investment is increased by $5000 and the number of working hours is decreased by 10 per day. . . . . . .
- 143. Solution ( )[ ] ( ) ∂P 1 1 −2 9 −2 −3/2 −2 −3 (x, y) = 150 − x + y x ∂x 2 10 10 10 [ ] 1 −2 9 −2 −3/2 −3 = 15 x + y x 10 10 ∂P (50, 50) = 15 ∂x ( )[ ]− 3 / 2 ( ) ∂P 1 1 −2 9 9 (x, y) = 150 − x + y−2 (−2)y−3 ∂y 2 10 10 10 [ ] 1 −2 9 −2 −3/2 −3 = 15 x + y x 10 10 ∂P (50, 50) = 135 ∂y So the linear approximation is L = 7500 + 15(x − 50) + 135(y − 50) . . . . . .
- 144. Solution, continued So the linear approximation is L = 7500 + 15(x − 50) + 135(y − 50) If ∆x = 5 and ∆y = −1, then L = 7500 + 15 · 5 + 135 · (−1) = 7440 The actual value is P(55, 49) ≈ 7427 13 So we are off by ≈ 1.75% 7427 . . . . . .
- 145. Tangent planes to parametrized surfaces ◮ If r(u, v) parametrizes a surface S, and P0 = r(u0 , v0 ), then the vector r(u0 + h, v0 ) − r(u0 , v0 ) ru (u0 , v0 ) = lim h→0 h is tangent to S at P. ◮ If r(u, v) = x(u, v)i + y(u, v)j + z(u, v)k, then ∂x ∂y ∂z ru (u0 , v0 ) = (u0 , v0 )i + (u0 , v0 )j + (u0 , v0 )k ∂u ∂u ∂u ◮ We have another tangent vector ∂x ∂y ∂z rv (u0 , v0 ) = (u0 , v0 )i + (u0 , v0 )j + (u0 , v0 )k ∂v ∂v ∂v ◮ Their cross product is normal to the tangent plane. . . . . . .
- 146. Example Find an equation of the tangent plane to the parametric surface r(u, v) = u2 i + 2u sin v j + u cos v k at the point u = 1 and v = 0. . . . . . .
- 147. Example Find an equation of the tangent plane to the parametric surface r(u, v) = u2 i + 2u sin v j + u cos v k at the point u = 1 and v = 0. Solution The point we are ﬁnding the tangent plane at is r(1, 0) = ⟨1, 0, 1⟩. The tangent plane is spanned by the two vectors ru (1, 0) and rv (1, 0): ru (u, v) = ⟨2u, 2 sin v, cos v⟩ =⇒ ru (1, 0) = ⟨2, 0, 1⟩ rv (u, v) = ⟨0, 2u cos v, −u sin v⟩ =⇒ rv (1, 0) = ⟨0, 2, 0⟩ So a normal vector to the tangent plane is ru (1, 0) × rv (1, 0) = ⟨2, 0, 1⟩ × ⟨0, 2, 0⟩ = ⟨−2, 0, 4⟩ This means an equation for the tangent plane is −2(x − 1) + 4(z − 1) = 0. . . . . . .
- 148. Differentials This is really just another way to express linear approximation. Deﬁne new variables dx = ∆x, dy = ∆y, dz. Then the equation for the tangent plane through (x0 , y0 , z0 = f(x0 , y0 )) is ∂f ∂f dz = dx + dy ∂x ∂y And the concept that this is a good linear approximation is expressed as ∆z = f(x + dx, y + dy) − f(x, y) ≈ dx when dx and dy are “small enough” . . . . . .
- 149. Example If z = 5x2 + y2 and (x, y) changes from (1, 2) to (1.05, 2.1), compare the values of ∆z and dz. . . . . . .
- 150. Example If z = 5x2 + y2 and (x, y) changes from (1, 2) to (1.05, 2.1), compare the values of ∆z and dz. Solution We have dz = 10x dx + 2y dy When x = 1, y = 2, dx = 0.05, and dy = 0.1, we get dz = 0.9. On the other hand z(1.05, 2.1) = 9.9225 So ∆z = 0.9225. The difference is 0.0225, or 2.4%. . . . . . .

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