Math 21a Midterm I Review

Slide 1: Midterm I Review Math 21a March 5, 2008 . . Image: Flickr user Mr. Theklan . . . . . .

Slide 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 . . . . . .

Slide 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 . . . . . .

Slide 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 . . . . . . .

Slide 5: Axes in three-dimensional space z . . z y . . y . x . x . x . y . . y x . z . . z . . . . . .

Slide 6: The right-hand rule z . x . . . y . . . . . . .

Slide 7: Distance in space Example Find the distance between the points P1 (3, 2, 1) and P2 (4, 4, 4). . . . . . .

Slide 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 . . . . . .

Slide 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. . . . . . .

Slide 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 . . . . . .

Slide 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). . . . . . .

Slide 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. . . . . . .

Slide 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 . . . . . . .

Slide 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 . . . . . .

Slide 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 . . . . . .

Slide 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 . . . . . .

Slide 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 . . . . . .

Slide 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 . . . . . . .

Slide 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 ⟩ . . . . . .

Slide 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 ⟩ . . . . . .

Slide 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. . . . . . .

Slide 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 . . . . . .

Slide 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. . . . . . .

Slide 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. . . . . . .

Slide 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. . . . . . .

Slide 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. . . . . .

Slide 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 . . . . . .

Slide 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 . . . . . . .

Slide 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| . . . . . .

Slide 30: Other uses of the dot product ◮ similarity ◮ Work W=F·d . . . . . .

Slide 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. . . . . . .

Slide 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 ⟩ . . . . . .

Slide 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 . . . . . .

Slide 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 . . . . . .

Slide 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 . . . . . .

Slide 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)| . . . . . .

Slide 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⟩ . . . . . .

Slide 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. . . . . . .

Slide 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 . . . . . .

Slide 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. . . . . . .

Slide 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 . . . . . .

Slide 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). . . . . . .

Slide 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 . . . . . .

Slide 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 . . . . . .

Slide 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 . . . . . . .

Slide 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). . . . . . .

Slide 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. . . . . . .

Slide 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 . . . . . .

Slide 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. . . . . . .

Slide 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 . . . . . .

Slide 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. . . . . . .

Slide 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 } . . . . . .

Slide 53: Example Example √ Find the domain and range of f(x, y) = xy. . . . . . .

Slide 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). . . . . . .

Slide 55: Solution continued Here is a sketch of the domain: . . . . . . .

Slide 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 . . . . . .

Slide 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 . . . . . .

Slide 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 . . . . . .

Slide 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 . . . . . .

Slide 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 . . . . . .

Slide 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 . . . . . .

Slide 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 . . . . . .

Slide 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 . . . . . .

Slide 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 . . . . . . .

Slide 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 . . . . . . .

Slide 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 . . . . . . .

Slide 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 . . . . . .

Slide 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 . . . . . .

Slide 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 . . . . . .

Slide 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 . . . . . .

Slide 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 . . . . . .

Slide 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)⟩ . . . . . .

Slide 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. . . . . . .

Slide 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 . . . . . .

Slide 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. . . . . . .

Slide 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 . . . . . .

Slide 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 . . . . . .

Slide 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) . . . . . .

Slide 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 . . . . . .

Slide 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). . . . . . .

Slide 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 . . . . . .

Slide 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 . . . . . .

Slide 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 . . . . . .

Slide 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 . . . . . .

Slide 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 . . . . . .

Slide 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 . . . . . .

Slide 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 . . . . . .

Slide 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 . . . . . .

Slide 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 . . . . . .

Slide 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 . . . . . . .

Slide 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 . . . . . .

Slide 92: Example Example Find the length of the parabola y = x2 from x = 0 to x = 1. . . . . . .

Slide 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 . . . . . .

Slide 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. . . . . . .

Slide 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). . . . . . .

Slide 96: Example Find the velocity, acceleration, and speed of a particle with position function r(t) = ⟨2 sin t, 5t, 2 cos t⟩ . . . . . .

Slide 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)⟩ . . . . . .

Slide 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. . . . . . .

Slide 99: Surfaces with easy parametrizations ◮ graph ◮ plane ◮ sphere ◮ surface of revolution . . . . . .

Slide 100: Parametrizing a graph Example Parametrize the surface described by z = x 2 + y2 , −2 ≤ x ≤ 2 , −3 ≤ y ≤ 3 . . . . . .

Slide 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. . . . . . .

Slide 102: parametrizing a sphere Example Find a parametrization for the unit sphere. . . . . . .

Slide 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 . . . . . . .

Slide 104: Parametrizing a surface of revolution Example Find a parametrization of the surface described by x2 − y 2 + z 2 = 1 , −3 ≤ y ≤ 3 . . . . . .

Slide 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 . . . . . . .

Slide 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 Fu