Contents. Chapter 1: The Three-Dimensional Coordinate System Section 1.1: Rectangular Coordinates... 1

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1 Contents Chapter 1: The Three-Dimensional Coordinate System... 1 Section 1.1: Rectangular Coordinates... 1 Subsection 1.1.1: Distance and Spheres... 2 Chapter 2: Vectors... 5 Section 2.1: Vectors... 5 Section 2.2: The Dot Product... 8 Section 2.3: Angles Between Vectors Section 2.4: Projections Section 2.5: The Cross Product Chapter 3: Geometry in Three Dimensions Section 3.1: Planes Section 3.2: Lines Section 3.3: Cylinders Section 3.4: Quadric Surfaces Chapter 4: Vector Functions Section 4.1: Vector-valued Functions Section 4.2: Limits of Vector Functions Section 4.3: Continuity of Vector Functions Section 4.4: Derivatives of Vector Functions Section 4.5: Tangent Lines of Vector Curves Section 4.6: Integrals of Vector Functions Section 4.7: Arc Length i

2 ii Section 4.8: Curvature Section 4.9: Normal and Binormal Vectors Section 4.10: Velocity and Acceleration Vectors Chapter 5: Functions of Several Variables Section 5.1: Functions and Level Curves Section 5.2: Limits and Continuity of Functions of Several Variables Subsection 5.2.1: Limits Subsection 5.2.2: Continuity Section 5.3: Partial Derivatives Section 5.4: Local Extrema Section 5.5: Absolute Extrema Chapter 6: Derivatives Section 6.1: Tangent Planes, Linear Approximations, and Differentials Subsection 6.1.1: Tangent Planes Subsection 6.1.2: Linear Approximations Subsection 6.1.3: Differentials Section 6.2: The Derivative Section 6.3: The Chain Rule Section 6.4: Gradients and Directional Derivatives Subsection 6.4.1: The Gradient Subsection 6.4.2: Directional Derivatives Section 6.5: Tangent Planes to Level Surfaces Section 6.6: Lagrange Multipliers... 62

3 Chapter 7: Multiple Integrals Section 7.1: Double Integrals Section 7.2: Fubini s Theorem Section 7.3: Double Integrals over General Regions Section 7.4: Double Integrals with Polar Coordinates Section 7.5: Cylindrical Coordinates Section 7.6: Spherical Coordinates Section 7.7: Triple Integrals Section 7.8: Triple Integrals with Cylindrical Coordinates Section 7.9: Triple Integrals with Spherical Coordinates Chapter 8: Vector Fields Section 8.1: Vector Fields Section 8.2: Line Integrals Section 8.3: The Fundamental Theorem for Line Integrals Section 8.4: Green s Theorem Section 8.5: Curl and Divergence Section 8.6: Surface Area Section 8.7: Surface Integrals iii

4 Chapter 1: The Three-Dimensional Coordinate System 1 Section 1.1: Rectangular Coordinates Example 1: Plot the following points. a. (3,1,4) b. (3,1,-4) c. (-3,1,-4)

5 2 Subsection 1.1.1: Distance and Spheres Theorem 2 (Distance Formula): The distance between the points (x 1,y 1,z 1 ) and (x 2,y 2,z 2 ) is (x 1 x 2 ) 2 +(y 1 y 2 ) 2 +(z 1 z 2 ) 2. Example 3: Find the distance between the given pair of points. a. (1,-1,3) and (-1,6,2) Example 4: Let T be the triangle with vertices (1,-2,1),(-2,1,4), and (2,1,3). a. Find the perimeter of T. b. Is T a right triangle?

6 Theorem 5 (Equation of a Sphere): The equation of the sphere with center (h,k,l) and radius r is (x h) 2 +(y k) 2 +(z l) 2 = r 2. Example 6: For each of the following, find the equation of the sphere being described. a. Centered at (1,-2,4) with the point (2,1,-3) on its surface. 3 Example 7: For each of the following, find the center and radius of the sphere. a. x 2 +y 2 +z 2 4x+2y +6z = -11

7 4 Note 8: Discuss projections. Example 9: All three projections of (1,2,3). Example 10: Page 814: Parts of 10. Example 11: Page 814: even. Exercises Page 814: 1 43 odd.

8 5 Chapter 2: Vectors Section 2.1: Vectors 1,4 Notation 12: The set of all vectors in R n will be denoted V n. Definition 13: Let v = v 1,v 2,v 3 be a vector. The vector with initial point (0,0,0) and terminal point (v 1,v 2,v 3 ) is called the position vector representation of v. Example 14: Find the terminal point of the vector v = 2,0,-1 if the initial point is (1,-3,4). Definition 15: Let v = v 1,v 2,...,v n and w = w 1,w 2,...,w n be vectors. Then v +w = v 1 +w 1,v 2 +w 2,...,v n +w n. Definition 16: Let v = v 1,v 2,...,v n and w = w 1,w 2,...,w n be vectors. Then v w = v 1 w 1,v 2 w 2,...,v n w n. Example 17: Let v = -1,3 and w = 2,7. a. v +w = b. v w =

9 6 Note 18: The geometric interpretation of vector addition in R 2 is illustrated below. v 2 w 2 w v v v 1 w 1 w v +w Example 19: Let v = 2,3 and w = 3,-1. Draw vectors v, w, v +w, and v w. Definition 20: Let v = v 1,v 2,...,v n be a vector and let a R. Then av = av 1,av 2,...,av n. Example 21: Let v = -1,5 and calculate 4v. Example 22: Let v = 4,8,-3 and w = -1,3,2. 2v w =

10 7 Notation 23: The vector (-1)v is usually written -v. Theorem 24: Let u,v,w V n and a,b R. Then (i) v +w = w+v for all v,w V n ; (ii) (u+v)+w = u+(v +w) for all u,v,w R n ; (iii) there is a unique vector 0 R n such that v +0 = 0+v = v; (iv) for each v R n, 1v = v; (v) v +(-v) = 0; (vi) a(v +w) = av +aw; (vii) (a+b)v = av +bv; (viii) (ab)v = a(bv). Definition 25: Two vectors v and w are parallel if there exists a real number α such that αv = w. Example 26: Are the following parallel? a. v = 1,0,4 and w = 3,0,8? Standard Basis Vectors V 2 V 3 i = 1,0 j = 0,1 i = 1,0,0 j = 0,1,0 Exercises k = 0,0,1 Page 822: 1 17 odd.

11 8 Section 2.2: The Dot Product Definition 27: Let v = v 1,v 2,...,v n and w = w 1,w 2,...,w n. Then v w = v 1 w 1 +v 2 w 2 + +v n w n is called the dot product or inner product of v and w. Example 28: Let v = 2,0,-1 and w = 1,-1,4. Then v w = Theorem 29: If v and w are vectors in R n and a is a scalar, then a(v w) = (av) w = v (aw). Proof: Let v = v 1,v 2,...,v n and w = w 1,w 2,...,w n be vectors in R n and let a R. Consider a(v w) = a(v 1 w 1 +v 2 w 2 +v n w n ) a(v w) = a(v 1 w 1 +v 2 w 2 +v n w n ) = av 1 w 1 +av 2 w 2 +av n w n = (av 1 )w 1 +(av 2 )w 2 +(av n )w n = (av) w. = v 1 aw 1 +v 2 aw 2 +v n aw n = v 1 (aw 1 )+v 2 (aw 2 ) +v n (aw n ) = v (aw). Also, Theorem 30: Let u,v,w R n and a R, then (i) v w = w v; (ii) u (v +w) = u v +u w; (iii) (av) w = a(v w) = v (aw); (iv) 0 w = w 0 = 0. Definition 31: The norm or length or magnitude of a vector v is v v and is denoted v or v. Example 32: Let v = 2,0,-1 and w = 1,-1,4. a. v b. w

12 9 Definition 33: A vector with norm 1 is called a unit vector. Proposition 34: If v is a nonzero vector, then 1 v is a unit vector in the same direction v as v. Example 35: For each of the following, find a unit vector in the same direction as the given vector. a. Let v = 6,8. b. v = -1,4,6 Note 36: See example 7 on page 821. Example 37: Page 823: 36. Page 822: odd. Page 830: 1 7 odd. Exercises

13 10 Section 2.3: Angles Between Vectors Theorem 38: If u and v are unit vectors, then u v = cosθ where θ is the angle between u and v. Proof: By the law of cosines, we have v w 2 = v 2 + w 2 2 v w cosθ (v w) (v w) = 2 2cosθ w v w v v v w w v+w w = 2 2cosθ θ v v 2 2v w + w 2 = 2 2cosθ 2 2v w = 2 2cosθ v w = cosθ. Corollary 39: If u and v are unit vectors, then u v 1. Corollary 40 (Cosine Formula): If v and w are nonzero vectors, then v w v w = cosθ where θ is the angle between v and w. Proof: By the previous theorem, v w v w = v v w w = cosθ. Corollary 41: Two vectors v and w are orthogonal if and only if v w = 0. Example 42: Find the angle between the vectors 0, 1 2,0 and 4 2 2,0, Example 43: Find a nonzero vector orthogonal to 3,2,4.

14 Corollary 44 (Cauchy-Buniakowsky-Schwarz Inequality): If v and w are any two vectors, then v w v w. Proof: By the Cosine Formula, v w v w = cosθ v w = cosθ v w v w v w = cosθ. So we have v w = cosθ v w v w 1 v w v w v w. 11 Corollary 45 (The Triangle Inequality): v +w v + w. If v and w are any two vectors, then Proof: First consider v +w 2 = (v +w) (v +w) = (v +w) v +(v +w) w = v v +w v +v w +w w = v 2 +2v w + w 2 v 2 +2 v w + w 2 v 2 +2 v w + w 2 = ( v + w ) 2. = v v +2v w +w w Since v +w 2 ( v + w ) 2, v +w ( v + w ).

15 12 Definition 46: The direction angles of a nonzero vector are the angles (between 0 and π) that a vector makes with i, j, and k. These angles are usually denoted α, β, and γ, respectively. The cosines of these angles are called the direction cosines. Note 47: Let v = v 1,v 2,v 3. Then cosα = v i v i = v i v = v 1 v, cosβ = v j v j = v j v = v 2 v, and cosγ = v k v k = v k v = v 3 v. Therefore, cos 2 α+cos 2 β +cos 2 γ = = ( ) v i 2 ( ) v j 2 ( ) v k v i v j v k ( v1 v ) 2 + ( v2 v ) 2 + ( v3 v = 1 v 2 [(v 1) 2 +(v 2 ) 2 +(v 3 ) 2 ] ) 2 = 1 v 2(v v) = v 2 v 2 = 1. Also, since v 1 = v cosα, v 2 = v cosβ, v 3 = v cosγ, v = v cosα, v cosβ, v cosγ = v cosα,cosβ,cosγ and so v = cosα,cosβ,cosγ. v Exercises Page 830: 9 37 odd, 38.

16 13 Section 2.4: Projections Definition 48: Suppose that v and w are two vectors in V n. (i)thescalar projection ofwonto v orthecomponent ofw along v iscomp v w = v w v. (ii) The vector projection of w onto v is proj v w = Note 49: ( ) ( ) v w v v w v v = v 2 v. (i) (ii) w w θ proj v w v proj v w α θ v proj v w w = cosθ proj v w w = cosα proj v w w = v w v w proj v w = v w v proj v w = comp v w proj v w w = (-v) w v w proj v w = -(v w) v proj v w = -comp v w (iii) If 0 < θ < π 2, then proj v w has the same direction as v and comp v w = proj v w. (iv) If π 2 < θ < π, then proj v w has the opposite direction as v and comp v w = - proj v w. (v) If θ = 0 or θ = π, then proj v w = w. (vi) If θ = π 2, then proj v w = 0. Example 50: Let v = 1,-3,5 and w = 2,-1,4. a. comp v w b. proj v w c. Find the angle between v and w.

17 14 Theorem 51: Let v be a nonzero vector and let w be any vector. Then there exist unique vectors w 1 and w 2 such that w 1 is parallel to v, w 2 is orthogonal to v, and w = w 1 +w 2. Furthermore, w 1 = proj v w and w 2 = w proj v w. Proof: First, note that proj v w +(w proj v w) = w. Since proj v w = ( ) v w v 2 v, proj v w is parallel to v by Definition 25. To see that w proj v w is orthogonal to v, note that (w proj v w) v = [ ( ) ] v w w v 2 v v = w v [( ) ] v w v 2 v v = w v = w v = w v w v = 0. ( ) v w v 2 (v v) ( ) v w v 2 v 2 It remains to show that proj v w and w proj v w are unique. So suppose that x and y are vectors such that such that x + y = w, x is parallel to v, and y is orthogonal to v. By Theorem 25, there is α R such that x = αv. Since y is orthogonal to v, 0 = y v 0 = (w x) v 0 = w v x v x v = w v (αv) v = w v α(v v) = w v α v 2 = w v α = w v v 2. So x = αv = ( w v v 2 ) v = projv w and y = w x = w proj v w. Example 52: A wagon is pulled 100 ft by exerting a force of 10 lb on the handle at an angle of π with the horizontal. How much work is done? 3 Page 830: odd (omit 45). Exercises

18 Section 2.5: The Cross Product 15 Definition 53: Let v = v 1,v 2,v 3,,w = w 1,w 2,w 3, V 3. The cross product of v and w is the vector v w = v 2 w 3 v 3 w 2,v 3 w 1 v 1 w 3,v 1 w 2 v 2 w 1. Note 54: One can use determinants to easily remember the formula. i j k v 1 v 2 v 3 w 1 w 2 w 3 v w = v 2 w 3 w 2 v 3, (v 1 w 3 w 1 v 3 ),v 1 w 2 w 1 v 2 Example 55: Let v = 1,-2,1 and w = -3,0,2. a. Calculate v w. b. Calculate w v. Example 56: If possible, find vectors u, v, and w such that (u v) w u (v w). If this is not possible, prove that it is not.

19 16 Theorem 57: Suppose that u,v,w V 3 and α R. Then (i) u v = -(v u); (ii) u 0 = 0; (iii) u u = 0; (iv) α(u v) = (αu) v = u (αv); (v) u (v +w) = (u v)+(u w); (vi) (v +w) u = (v u)+(w u); (vii) u (v w) = (u w)v (u v)w; (viii) (u v) w = (u w)v (v w)u; (ix) u (v w) = (u v) w; (x) u v 2 = ( u v ) 2 (u v) 2. Proof: Theorem 58: If v and w are nonzero vectors, then between v and w. Proof: By the previous theorem, v w 2 = ( v w ) 2 (v w) 2 v w 2 = ( v w ) 2 ( v w cosθ) 2 v w 2 = ( v w ) 2 (1 cosθ) 2 v w 2 = ( v w ) 2 sin 2 θ v w 2 ( v w ) 2 = sin2 θ v w v w = sinθ. v w v w = sinθ where θ is the angle Corollary 59: If v,w R 3, then v and w are parallel if and only if v w = 0. Proof: First, suppose that v and w are parallel. Then there exists α R such that w = αv. Then v w = v (αv) = α(v v) = 0. Now suppose that v w = 0. Then 0 = v w = v w sinθ. So either v = 0, w = 0, or sinθ = 0 which means that either v = 0, w = 0, or θ = 0, or θ = π. In any case, v and w are parallel. Note 60: If u,v,w R 3, then u (v w) =. u 1 u 2 u 3 v 1 v 2 v 3 w 1 w 2 w 3

20 Theorem 61: The volume of the parallelpiped determined by the vectors u, v, and w is 17 u (v w) =. u 1 u 2 u 3 v 1 v 2 v 3 w 1 w 2 w 3 Example 62: Findthevolumeoftheparallelpipeddetermined bythevectorsu = 1,2,-4, v = 0,1,0, and w = 1,8,6. Theorem 63: For any two vectors v and w, v w is orthogonal to both v and w. Proof: Consider v (v w) = v 1 v 2 v 3 v 1 v 2 v 3 w 1 w 2 w 3 = v 1 (v 2 w 3 w 2 v 3 ) v 2 (v 1 w 3 w 1 v 3 )+v 3 (v 1 w 2 w 1 v 2 ) = v 1 v 2 w 3 v 1 v 3 w 2 v 1 v 2 w 3 +v 2 v 3 w 1 +v 1 v 3 w 2 v 2 v 3 w 1 = 0. Likewise, w (v w) = 0. Note 64: The torque vector is the cross product of the position and force vectors. τ = r F τ = r F τ = r F sinθ Example 65: Page 839: 40. Page 838: 1 43 odd. Exercises

21 18 Chapter 3: Geometry in Three Dimensions Section 3.1: Planes Definition 66: Let P be a plane and n a vector. If n is orthogonal to every line in the plane, then n is called a normal vector to P. Theorem 67: Theequationoftheplanecontainingthepoint(x 0,y 0,z 0 )withnormalvector n = a,b,c is a(x x 0 )+b(y y 0 )+c(z z 0 ) = 0. Example 68: For each of the following, give the equation of the plane being described. a. Contains the point (1,-1,0) and has normal vector 1,-2,-1. b. Contains the points (1,1,0), (2,-1,1), and (2,-1,3). Definition 69: Suppose that two planes P 1 and P 2 have normal vectors n 1 and n 2, respectively. (i) If n 1 and n 2 are parallel, then P 1 and P 2 are parallel. (ii) If n 1 and n 2 are orthogonal, then P 1 and P 2 are orthogonal.

22 Theorem 70: Suppose that two planes P 1 and P 2 have normal vectors n 1 and n 2, respectively. Then the following are equivalent: (i) P 1 and P 2 are parallel; (ii) n 1 = αn 2 for some α R; (iii) n 1 n 2 = 0. Theorem 71: Suppose that two planes P 1 and P 2 have normal vectors n 1 and n 2, respectively. Then P 1 and P 2 are orthogonal if and only if n 1 n 2 = 0. Theorem 72: The shortest distance between the plane with equation ax+by+cz = d and a point (x 0,y 0,z 0,) that does not lie on the plane is ax 0 +by 0 +cz 0 d a2 +b 2 +c Page 848: odd, 51, 53, 55, 61, 63, 67. Exercises Section 3.2: Lines Definition 73: Let l be the line containing the point (x 0,y 0,z 0 ) in the direction of the vector v = a,b,c. (i) The parametric equations for l are x = x 0 +ta, y = y 0 +tb, z = z 0 +tc. (ii) The symmetric equations for l are x x 0 a = y y 0 b = z z 0. c (iii) The vector equation for l is r = r 0 +tv where r 0 = x 0,y 0,z 0. Example 74: For each of the following, give an equation of the line with the given properties. a. Contains the points (1,-3,4) and (2,0,1).

23 20 Example 75: Does the line with parametric equations x = 1+2t, y = -2+t, and z = 5 4t intersect the plane x+2y +z = 4? If so, where? Example 76: Find an equation of the line where the following planes intersect. 2x y +3z = 1-3x+y +5z = 7 Exercises Page 848: 1 21 odd, 35, 37, 45, 47, 49, 57, 59, 69, 71, 73.

24 Section 3.3: Cylinders 21 Definition 77: Let C be a curve contained in a plane P and l be a line not contained in P. Also, let L be the set of all lines that are parallel to l and intersect C. (i) The cylinder generated by C and l is the set of all points that lie on some line contained in L. (ii) The curve C is called the directrix of the cylinder. (iii) The line l is called the generating line or the generator of the cylinder. (iv) Any line in L is called a ruling of the cylinder. Note 78: The directrix is not necessarily a circle although it can be. Example 79: x 2 +y 2 = 9 Example 80: 9x 2 +4y 2 = 36 Definition 81: Quadric Surfaces (i) ellipsoid: x 2 a 2 + y2 b 2 + z2 c 2 = 1; (Page 854) Section 3.4: Quadric Surfaces (ii) cone: x 2 a 2 + y2 b 2 = z2 c 2 ; (Page 854) (iii) elliptic paraboloid: x2 + y2 = z ; (Page 854) a 2 b 2 c (iv) hyperboloid of one sheet: x 2 a 2 + y2 b 2 z2 c 2 = 1; (Page 854) (v) hyperboloid of two sheets: - x2 a 2 y2 b 2 + z2 c 2 = 1; (Page 854) (vi) hyperbolic paraboloid: x2 y2 = z ; (Page 854) a 2 b 2 c Example 82: For each of the following, find the surface in the planes x = k, y = k, and z = k. a. x 2 +y 2 +3z 2 = 1 b. 2x 2 y 2 +2z 2 = 1

25 22 Example 83: Sketch each of the following. a. y 2 = 3x 2 +2z 2 b. y 2 = 2x 2 3z 2 c. 10x 2 +3y 2 +z 2 = 1 Exercises Page 856: 1 19 odd, 21 28, odd, odd.

26 Chapter 4: Vector Functions 23 Section 4.1: Vector-valued Functions Example 84: For each of the following, sketch the curve of the given vector equation. Draw arrows along the curve in the direction of increasing t. a. f(t) = t 4,t 2 b. f(t) = t 2,t 3 c. f(t) = t 2 2,t d. f(t) = t 3,t

27 24 e. f(t) = cost,sint,t Exercises Page 869: 7 19 odd, 21 26, odd, 47. Section 4.2: Limits of Vector Functions Definition 85: Suppose f : R R 3 and c R. Then lim t c f(t) = L if for every ε > 0 there is a δ > 0 such that f(t) L < ε whenever 0 < t c < δ. Theorem 86: Suppose f : R R 3 is defined by f(t) = x(t),y(t),z(t) and c R. Then lim f(t) = lim x(t),limy(t),limz(t) provided each limit exists. x c x c x c x c

28 25 Example 87: Calculate the following limits. a. lim t 5π 6 sint,cost,t 2 b. lim t 0 sint,t 2,tlnt c. lim sint,t 2, sint t 0 t Exercises Page 869: 1 6. Section 4.3: Continuity of Vector Functions Definition 88: Suppose f : R R 3 and c R. Then f is continuous at c if for every ε > 0 there is a δ > 0 such that f(t) f(c) < ε whenever 0 < t c < δ. Theorem 89: Suppose f : R R 3 is defined by f(t) = x(t),y(t),z(t) and c R. If x, y, and z are all continuous at c, then f is continuous at c. Theorem 90: A vector-valued function f is continuous at a number c if and only if the following three conditions are satisfied: (i) f(c) exists; (ii) lim x c f(x) exists; (iii) lim x c f(x) = f(c).

29 26 Section 4.4: Derivatives of Vector Functions Definition 91: Supposef : R V 3. Thenthederivativeoff isf f(t+h) f(t) (t)= lim h 0 h if the limit exists. Theorem 92: Suppose f : R R 3 is defined by f(t) = x(t),y(t),z(t). If x, y, and z are all differentiable functions, then f is a differentiable function and f (t) = x (t),y (t),z (t). Example 93: For each of the following, find the derivative of the given vector function. a. r(t) = sint,cost,t 2 Theorem 94: Suppose that f : R R 3, g : R R 3, and s : R R are all differentiable. Then (i) (f +g) (t) = f (t)+g (t); (ii) (f g) (t) = f (t) g (t); (iii) (sf) (t) = s (t)f(t)+s(t)f (t); (iv) (f g) (t) = f (t) g(t)+f(t) g (t); (v) (f g) (t) = f (t) g(t)+f(t) g (t); (vi) (f s) (t) = f [s(t)] s (t). Example 95: Differentiate. a. r(t) = sint,cost,t 2 Exercises Page 876: 9 15 odd, 47, 48.

30 Section 4.5: Tangent Lines of Vector Curves 27 Example 96: For each of the following, find parametric equations for the line that is tangent to the given curve at the specified point. a. r(t) = t 4,t 2 +1,4 ; t = 2 b. r(t) = t,cosπt,e t ; t = 1 Note 97: The unit tangent vector is T(t) = r (t) r (t). Example 98: Let r(t) = 2sint, 5t,2cost. Find the unit tangent vector T(t). Page 876: 1 8, odd. Exercises

31 28 Section 4.6: Integrals of Vector Functions Definition 99: Suppose f : R R 3 is defined by f(t) = x(t),y(t),z(t). If x, y, and z b b b b are all integrable on [a,b], then f(t)dt = x(t) dt, y(t) dt, z(t) dt. Example 100: Integrate. a a a a a. 2 0 ( 3i 5t 2 j ) dt b. [(2t)i+(sint)j +(e t sint)k ] dt Exercises Page 876:

32 29 Note 101: Recall that arc length is s(t) = Then t a s (t) = r (t). [x (q)] 2 +[y (q)] 2 +[z (q)] 2 dq = Section 4.7: Arc Length t a r (q) dq. Example 102: Let C be the curve defined by the vector function f(t) = cost,sint, 3t. Parameterize the curve with respect to arc length. Page 884: 1 6, Exercises

33 30 Section 4.8: Curvature Definition 103: Suppose that a curve is parameterized by arc length s and T is the unit tangent vector. Then the curvature is κ = dt ds. Theorem 104: Suppose that a curve is defined by the vector function r(t) and T is the unit tangent vector. Then the curvature of the curve at the point r(t) is κ(t) = T (t) r (t). Proof: First, note that by the chain rule, T (t) = dt ds = T (t) ds dt. So κ = dt ds = T (t) ds = T (t) ds dt dt = T (t) r (t). Example 105: Find the curvature of each of the following. a. r(t) = cost,sint,0 b. r(t) = 2cost,2sint,0 Theorem 106: Suppose that a curve is defined by the vector function r(t). Then the curvature of the curve at the point r(t) is κ(t) = r (t) r (t) r (t) 3. Page 884: odd, 42, 43, 45. Exercises

34 Section 4.9: Normal and Binormal Vectors 31 Note 107: (i) N(t) = T (t) T (t) ; (ii) B(t) = T(t) N(t). Example 108: Let C be the curve defined by r(t) = 2sint, 5t,2cost. a. Find the unit tangent vector T(t). b. Find the unit normal vector N(t). c. Find the binormal vector B(t). d. Find the curvature κ(t).

35 32 Example 109: Let r(t) = 3sint,4,3cost. a. Find the unit tangent vector T(t). b. Find the curvature κ(t). c. Find the unit normal vector N(t). d. Find the binormal vector B(t). Page 884: 17 20, 47, 48, 49, 55. Exercises

36 Section 4.10: Velocity and Acceleration Vectors 33 Definition 110: Suppose that a point moves through space in such a way that its position at time t is (x(t),y(t),z(t)) and define s : R V 3 by s(t) = x(t),y(t),z(t). (i) The velocity of the point at time t is v(t) = s (t) provided s (t) exists. (ii) The speed of the point at time t is s(t) = v(t) = s (t) provided s (t) exists. (iii) The acceleration of the point at time t is a(t) = v (t) = s (t) provided s (t) exists. Example 111: The motion of an object is described by s(t) = e t,t 2,sint. a. Find the velocity of the object at time t. b. Find the speed of the object at time t. c. Find the acceleration of the object at time t.

37 34 Example 112: A projectile is launched with an initial speed of 100 feet per second at an angle of π to the horizontal. Assume that the only force acting on the object is gravity. 4 a. Find the initial velocity vector. b. Find the vector functions that describe velocity and motion. c. Find the maximum height. d. Find the horizontal range. e. Find the speed of impact.

38 Theorem 113: Suppose that a point moves through space in such a way that its position at time t is (x(t),y(t),z(t)) and define s : R V 3 by s(t) = x(t),y(t),z(t). Then the acceleration at time t is a(t) = v (t)t(t)+v(t) T (t) N(t). Proof: 35 Definition 114: Suppose that a point moves through space in such a way that its position at time t is (x(t),y(t),z(t)) and define s : R V 3 by s(t) = x(t),y(t),z(t). (i) The tangential component of acceleration is v (t) and is denoted a T (t). (ii) The normal component of acceleration is v (t) T (t) and is denoted a N (t). Exercises Page 894: 3 31 odd (omit 17), odd.

39 36 Chapter 5: Functions of Several Variables Section 5.1: Functions and Level Curves Example 115: Find the domain and range of each of the following functions. a. f(x,y) = x 2 y 2 b. f(x,y) = x+y c. g(x,y,z) = x 2 +y 2 +z 2 d. h(x,y,z) = x 2 +y 2 +z Example 116: For each of the following, sketch the level curve for the given value of k. a. f(x,y) = x 2 +y 2 ; k = 1 b. f(x,y) = x 2 +y 2 ; k = 4 c. f(x,y) = x 2 +y 2 ; k = 0

40 d. f(x,y) = x 2 y 2 ; k = 1 e. f(x,y) = x 2 y 2 ; k = -1 f. f(x,y) = x 2 y 2 ; k = 0 37 Example 117: For each of the following, describe the surface curve for the given value of k. a. f(x,y,z) = x 2 +y 2 +z 2 ; k = 1 b. f(x,y,z) = x 2 +y 2 +z 2 ; k = 9 c. g(x,y,z) = x+y z; k = 1 d. g(x,y,z) = x+y z; k = 4 Example 118: Page 912: 24, 26, 28, 32. Page 912: 1 31 odd, Exercises

41 38 Section 5.2: Limits and Continuity of Functions of Several Variables Subsection 5.2.1: Limits Definition 119: The Euclidean norm on R n is the real-valued function on R n defined by x = x 2 1 +x2 2 x2 n for all x = (x 1,x 2,...,x n) R n. Definition 120: The Euclidean metric on R n is the real-valued function on R n R n defined by d(x,y) = (x 1 y 1 ) 2 +(x 2 y 2 ) 2 +(x n y n ) 2 for all x = (x 1,x 2,...,x n ),y = (y 1,y 2,...,y n ), R n. Definition 121: Let x R n and ε > 0. Then the ball about x of radius ε is {y R n : d(x,y) < ε}. Notation 122: Let x R n and ε > 0. Then the ball about x of radius ε is sometimes written B(x, ε). Definition 123: Let f : R n R, c R n, and L R. Then the limit as x approaches c of f(x) is L if for every ε > 0 there exists δ > 0 such that f(x) B(L,ε) whenever x B(c,δ)\{c}. Theorem 124: Let f : R n R, c R n, and L R. Then the limit as x approaches c of f(x) is L if and only if for every ε > 0 there exists δ > 0 such that d(f(x),l) < ε whenever 0 < d(x,c) < δ. Theorem 125: If f : R n R is a polynomial, then lim x c f(x) = f(c). Corollary 126: If f : R n R is a rational function, then lim x c f(x) = f(c) if f(c) exists. Example 127: For each of the following, find the limit or prove that it does not exist. ( a. lim 2x 3 y +xy 2 1 ) xy b. lim (x,y) (-1,2) (x,y) (0,0) x 2 +y 2

42 x c. lim (x,y) (0,0) y 39 x 2 y 4 d. lim (x,y) (0,0) x 2 +y 4

43 40 Subsection 5.2.2: Continuity Definition 128: Let f : R n R and c dom(f). Then f is continuous at c if for every ε > 0 there exists δ > 0 such that f(x) B(f(c),ε) whenever x B(c,δ). Theorem 129: Let f : R n R and c dom(f). Then f is continuous at c if and only if for every ε > 0 there exists δ > 0 such that d(f(x),f(c)) < ε whenever d(x,c) < δ. Theorem 130: Let f : R n R and c R n. Then f is continuous at c if and only if the following three conditions are satisfied: (i) f(c) exists; (ii) lim x c f(x) exists; (iii) lim x c f(x) = f(c). sinθ Example 131: Use the definition of limit and the fact that lim θ 0 θ sin(x 2 +y 2 ) lim = 1. (x,y) (0,0) x 2 +y 2 = 1 to prove that Page 923: 1, 5 21 odd, 25, odd. Exercises

44 Section 5.3: Partial Derivatives 41 Definition 132: Let z = f(x,y) be a function of two variables. Then the partial derivative of f with respect to x is f x (x,y) = z x = lim h 0 f(x+h,y) f(x,y), if the limit exists. h Likewise, the partial derivative of f with respect to y is f y (x,y) = z y = lim h 0 f(x,y +h) f(x,y), if the limit exists. h Example 133: f(x,y) = 4x 3 y 2 12x 2 y 4 +x+2y a. f x (x,y) = b. f y (x,y) = Example 134: f(x,y) = x2 y 2 +8x 2 +4y 2 x 2 +y +1 a. f x (x,y) = b. f y (x,y) =

45 42 Definition 135: Let z = f(x 1,x 2,...,x n ) be a function of n variables. Then the partial derivative of f with respect to x i is f xi (x 1,x 2,...,x n ) = z f(x = lim 1,...,x,x +h,x,...,x i 1 i i+1 n) f(x 1,...,x i,...,x n ), x i h 0 h if the limit exists. Example 136: f(x,y,z) = xe yz +sinxy a. f x (x,y,z) = b. f y (x,y,z) = c. f z (x,y,z) = Definition 137: Let z = f(x,y) be a function of two variables such that both partial derivatives exist. Then the second order partial derivatives are: ( f xx (x,y) = (f x ) x (x,y) = z ) x x = 2 z f = lim x (x+h,y) f x (x,y), if the limit exists; x 2 h 0 h ( f xy (x,y) = (f x ) y (x,y) = z ) y x = 2 z = lim y x h 0 ( f yx (x,y) = (f y ) x (x,y) = z ) x y = 2 z = lim x y h 0 f x (x,y +h) f x (x,y), if the limit exists; h f y (x+h,y) f y (x,y), if the limit exists; h ( f yy (x,y) = (f y ) y (x,y) = z ) y y = 2 z f = lim y (x,y +h) f y (x,y), if the limit exists. y 2 h 0 h

46 43 Example 138: f(x,y) = x 3 y 2 +xy 3 +xy +x Example 139: f(x,y) = xe y +xy Theorem 140 (Clairaut s Theorem): Let z = f(x,y) be a function of two variables. If f, f x, f y, f xy, and f yx are continuous on an open set U R R, then f xy (u,v) = f yx (u,v) for all (u,v) U. Page 935: odd, 99. Exercises

47 44 Section 5.4: Local Extrema Definition 141: Let f be a function of two variables. Then f has a local maximum at (a,b) if there exists r > 0 such that f(x,y) f(a,b) for all (x,y) in the interior of the circle (x a) 2 +(x b) 2 = r 2. Definition 142: Let f be a function of two variables. Then f has a local minimum at (a,b) if there exists r > 0 such that f(a,b) f(x,y) for all (x,y) in the interior of the circle (x a) 2 +(x b) 2 = r 2. Definition 143: Let z = f(x,y) be a function of two variables. Then a point (a,b) dom(f) is a critical point of f if one of the following is true: (i) f x (a,b) = 0 and f y (a,b) = 0. (ii) f x (a,b) or f y (a,b) does not exist. Theorem 144: Let f be a function of two variables. If f has a local extrema at (a,b) then (a,b) is a critical point of f. Definition 145: If (a,b) is a critical point for a function f and f does not have a local extrema at (a,b), then (a,b) is called a saddle point of f. Theorem 146 (The Second Partials Test): Let f be a function of two variables that hascontinuoussecondorderpartialderivativesandlet(a,b) dom(f)suchthatf x (a,b) = 0 and f y (a,b) = 0. Also, let D(a,b) = f xx (a,b)f yy (a,b) [f xy (a,b)] 2. If D(a,b) > 0 and f xx (a,b) < 0, then f has a local maximum at (a,b). If D(a,b) > 0 and f xx (a,b) > 0, then f has a local minimum at (a,b). If D(a,b) < 0, then f has a saddle point at (a,b). Example 147: For each of the following, find all local extrema and saddle points. a. g(x,y) = x 3 y +3x 2 +y

48 45 b. f(x,y) = 1 2 x2 y 2 +x y +1 Exercises Page 977: 5 19 odd. Section 5.5: Absolute Extrema Theorem 148 (Extreme Value Theorem): Let f be continuous on a closed bounded subset of R 2. Then f attains both an absolute maximum and an absolute minimum. Fact 149: Let f be continuous on a closed bounded subset E of R 2. Then the absolute extrema (guaranteed by the Extreme Value Theorem) occur at either the critical points of f or the boundary points of E.

49 46 Example 150: Let g(x,y) = x 3 y +3x 2 +y. Find the absolute extrema of g on the region bounded by the triangle with vertices (0,0), (1,0), and (1,1).

50 Example 151: Let f(x,y) = x 2 y 2 xy. Find maximum and minimum of f on the region defined by (i) x 2 +y 2 1; and (ii) y 0. 47

51 48 Exercises Page 977: odd, odd.

52 Chapter 6: Derivatives 49 Section 6.1: Tangent Planes, Linear Approximations, and Differentials Subsection 6.1.1: Tangent Planes Theorem 152: Suppose f : R 2 R has continuous partial derivatives. An equation of the plane that is tangent to the surface z = f(x,y) at the point (a,b,f(a,b)) is z f(a,b) = f x (a,b)(x a)+f y (a,b)(y b). Example 153: For each of the following, give the equation of the plane that is tangent to the given surface at the specified point. a. z = 2x 2 y +y 3 ; (1,2,12) b. z = xe xy ; (1,0,1) Subsection 6.1.2: Linear Approximations Definition 154: Let f : R n R and suppose c R n such that the first order partial derivatives f 1,f 2,...,f n all exist at c. The linear approximation or linearization of f at c is the function L(x) = f(c)+f 1 (c)(x 1 c 1 )+f 2 (c)(x 2 c 2 ) +f n (c)(x n c n ).

53 50 Subsection 6.1.3: Differentials Definition 155: Let z = f (x 1,...,x n ) where f : R n R. If each x i is incremented by x i, then the corresponding increment of the dependent variable z is z = f (x 1 + x 1,...x n + x n ) f (x 1,...,x n ). Definition 156: Let z = f (x 1,...,x n ) where f : R n R. If all first order partial derivatives exist, then define the differentials dx 1, dx 2,..., dx n, dz as follows: (i) dx i = x i for each i, 1 i n; (ii) dz = f 1 (x 1,x 2,...,x n )dx 1 +f 2 (x 1,x 2,...,x n )dx 2 +f n (x 1,x 2,...,x n )dx n. Example 157: Differentials of previous examples. Page 946: 1 5 odd, odd. Exercises

54 Section 6.2: The Derivative 51 Definition 158: Let f : R n R and c R n. Also, suppose that the first order partial derivatives f x1, f x2,..., f xn all exist on an open set containing c. Then f is differentiable at c if ) f (c 1 +h 1,...,c n +h n ) f (c 1,...,c n ) (f x1 (c) h 1 + +f xn (c) h n lim = 0. (h 1,...,h n ) (0,...,0) (h 1,...,h n ) Definition 159: Let f : R n R and c R n. Also, suppose that the first order partial derivatives f 1, f 2,..., f n all exist on an open set containing c. If f is differentiable at c, then the derivative of f at c is the function f c : Rn R defined by f c (x) = f 1(c) x 1 +f 2 (c) x 2 + +f n (c) x n. Example 160: Letf(x,y) = x 2 +2y. Showthatf isdifferentiableat(a,b)forall(a,b) R 2 and find f (a,b) (x,y).

55 52 Example 161: Prove that f(x,y) = 2x+y 2 +1 is differentiable for all (a,b) R 2 and find f (a,b) (x,y).

56 Example 162: Let g(x,y,z) = x 2 +y 2 +z 3. Show that g is differentiable at (a,b,c) for all (a,b,c) R 3 and find g (a,b,c) (x,y,z). 53

57 54 Theorem 163: Let f : R n R and c R n. If f is differentiable at c, then f is continuous at c. Theorem 164: Let f : R n R and c R n. If the first order partial derivatives of f are continuous on an open set containing c, then f is differentiable at c. Page 946: 1 6, 11 16, 43, 44. Exercises For the list of functions above, find the derivative.

58 Section 6.3: The Chain Rule 55 Theorem 165 (The Chain Rule): Suppose that x 1 : R R, x 2 : R R,..., x n : R R, and f : R n R are all differentiable. Define g : R R by g(t) = f (x 1 (t),x 2 (t),...,x n (t)). Then g is differentiable and g (t) = f 1 (x 1 (t),x 2 (t),...,x n (t)) x 1 (t)+ +f n(x 1 (t),x 2 (t),...,x n (t)) x n (t) Example 166: Let z = e x y 2 +y where x = t 2 +1 and y = sint. Find dz dt. Example 167: Let z = sinx+cosy where x = e t and y = t a. Use the chain rule to find dz dt. b. Express z as function of t and then find dz dt.

59 56 Question 168: What if x and y are multivariable functions? Example 169: Let z = e x y 2 +y where x = st+2 and y = s 2 +t. Find z s. Example 170: Let w = x 2 y xyz +zsinx where x = rs 2 t, y = st, and z = re s. w x = 2xy yz +zcosx w = y x2 xz w = -xy +sinx z a. w r b. w s c. w t Page 954: 1 47 odd. Exercises

60 Section 6.4: Gradients and Directional Derivatives 57 Subsection 6.4.1: The Gradient Definition 171: Let u 1 = 1,0,0,...,0, u 2 = 0,1,0,...,0,..., and u n = 0,0,0,...,1 be the standard unit vectors of V n. Also, suppose that c R n and f : R n R such that all first order partial derivatives exist at c. Then the gradient vector of f at c is the vector f 1 (c),f 2 (c),...,f n (c) = f 1 (c)u 1 +f 2 (c)u 2 + +f n (c)u n. Notation 172: The gradient of f at c is denoted f(c). Example 173: For each of the following, find the gradient of the given function at the indicated point. f(x,y,z) = x 2 y +xz +y 3 ; (1,-2,0) Definition 174: Let u 1 = 1,0,0,...,0, u 2 = 0,1,0,...,0,..., and u n = 0,0,0,...,1 be the standard unit vectors of V n. Also, suppose that f : R n R such that all first order partial derivatives exist at c. Then the gradient of f is the vector-valued function f : V n R n defined by f(x) = f 1 (x),f 2 (x),...,f n (x) = f 1 (x)u 1 +f 2 (x)u 2 + +f n (x)u n.

61 58 Subsection 6.4.2: Directional Derivatives Definition 175: Let f : R n R, c R n, and v V n. The directed derivative of f at c directed by v is f (c lim 1 +hv 1,c 2 +hv 2,...,c 2 +hv 2 ) f(c) h 0 h provided the limit exists. In the case that v is a unit vector, the above limit is called directional derivative of f at c in the direction of v. Notation 176: The directed derivative of f at c directed by v is denoted D v f(c). Theorem 177: If f : R n R is differentiable at c, then the directed derivative of f at c directed by v exists for all v V n. Furthermore, D v f(c) = f 1 (c)v 1 +f 2 (c)v 2 +f n (c)v n = f(c) v. Example 178: For each of the following, find the directional derivative of the function at the indicated point in the direction of the given vector. a. g(x,y) = x 2 y +xy +y +6, (-1,2), u = 1, b. f(x,y,z) = xy 3 +2yz e x, (0,1,2), u = 1,0, c. f(x,y,z) = 2x 3 z +2z xz, (1,-6,4), k

62 Theorem 179: Let f : R n R, c R n, and u V n be a unit vector. Then D u f(c) is the rate of change of f in the direction of u at c. Theorem 180: Let f : R n R be differentiable at c R n. Then the maximum rate of change of f at c is f(c) and occurs in the direction of f(c). Corollary 181: Let f : R n R be differentiable at c R n. Then the minimum rate of change of f at c is - f(c) and occurs in the direction of - f(c). Example 182: Let f(x,y) = x 3 y 2 +x a. Find the rate of change of f at (1,1) in the direction of 4,5. 59 b. In what direction does the maximum rate of change of f occur at (1,1)? c. What is the maximum rate of change of f at (1,1)? Page 967: 5 35 odd. Exercises

63 60 Section 6.5: Tangent Planes to Level Surfaces Definition 183: Let f(x,y,z) be a differentiable function of three variables such that x, y, and z are differentiable functions of t. Then the tangent plane to the level surface f(x,y,z) = k at the point (x 0,y 0,z 0 ) is the plane that passes through the point (x 0,y 0,z 0 ) with normal vector f(x 0,y 0,z 0 ). The equation of this plane is f x (x 0,y 0,z 0 )(x x 0 )+f y (x 0,y 0,z 0 )(y y 0 )+f z (x 0,y 0,z 0 )(z z 0 ) = 0. The normal line to the surface at (x 0,y 0,z 0 ) is the line passing through (x 0,y 0,z 0 ) and perpendicular to the tangent plane. The normal line is given by the parametric equations x = x 0 +f x (x 0,y 0,z 0 )t y = y 0 +f y (x 0,y 0,z 0 )t z = z 0 +f z (x 0,y 0,z 0 )t. Example 184: Let f(x,y,z) = x 3 z+x 2 y+yz where x, y, and z are differentiable functions of t. a. Find an equation for the tangent plane to the level surface f(x,y,z) = 11 at the point (1,2,3). b. Find an equation for the normal line to the level surface f(x,y,z) = 11 at the point (1,2,3).

64 61 Example 185: Let f(x,y) = xe y +xy where x = t 2 and y = sinπt. a. Find an equation for the tangent plane to the surface z = f(x,y) where t = 2. b. Find an equation for the normal line to the surface z = f(x,y) where t = 2. Page 967: odd, odd. Exercises

65 62 Section 6.6: Lagrange Multipliers Theorem 186 (Lagrange s Theorem): Let f : R n R and g : R n R such that (i) f and g have continuous first order partial derivatives; (ii) f has a local extremum at c R n subject to the constraint g(x) = 0; (iii) g(c) 0. Then there exists λ R such that f(c) = λ g(c). Definition 187: In the above theorem, λ is called the Lagrange multiplier. Example 188: Maximize x 2 +4y +6z subject to x 2 +y 2 +z 2 = 1.

66 Example 189: Maximize andminimize f(x,y,z) = 4x+12y+20z subject to theconstraint x 2 +y 2 +z 2 =

67 64 Example 190: Maximize f(x,y,z) = xyz subject to 9x 2 +36y 2 +4z 2 = 36.

68 Theorem 191 (Lagrange s Theorem): Let f : R n R, g : R n R, and h : R n R such that (i) f, g, and h have continuous first order partial derivatives; (ii) f has a local extremum at c subject to the constraints g(x) = 0 and h(x) = 0; (iii) g(c) 0 and h(c) 0; (iv) g(c) and h(c) are not parallel. Then there exist λ,µ R such that f(c) = λ g(c)+µ h(c). Example 192: Maximize and minimize f(x,y,z) = 2x+y 2 z subject to x 2 +y 2 = 1 and x+z = Page 987: 3 43 odd. Exercises

69 66 Chapter 7: Multiple Integrals Section 7.1: Double Integrals Definition 193: Suppose that R = [a,b] [c,d] and f : R R. Then f is integrable on R if there is a number L such that for all ε > 0 there is a δ > 0 such that whenever a = x 0 < x 1 < x 2 < < x m 1 < x m = b, c = y 0 < y 1 < y 2 < < y n 1 < y n = d, x i x i 1 < δ for all i, 1 i m, y j y j 1 < δ for all j, 1 j n, and ( ) [ ] [ ] r ij,s ij xi 1,x i yj 1,y j for n m all i,j, 1 i m, 1 j n, then f ( ) ( ) ( ) r ij,s xi ij x yj i 1 y j 1 L < ε. j=1 i=1 The number L is called the integral of f over R. Notation 194: f(x,y)da Page 1005: 1 4 R Exercises Section 7.2: Fubini s Theorem Theorem 195 (Fubini s Theorem): Let R = [a,b] [c,d] and suppose f : R R is d b b d integrable on R. Then f(x,y)da = f(x,y)dxdy = f(x,y)dydx. R ( Example 196: Let R = [0,2] [0,4] and calculate 3x 2 y +xy ) da. c a R a c

70 Example 197: Let R = [2,3] [1,2] and calculate R ( x 2 y +2x y ) da. 67 Theorem 198: Suppose that R is a closed bounded region in R 2 and f : R R is continuous and f(x,y) 0 for all (x,y) R. Then the volume of the solid that lies above ( ) R and below the graph of f is f(x,y) da. R Example 199: Findthevolumeofthesolidboundedbytheellipticparaboloid3x 2 +y 2 +z = 25, the planes x = 2 and y = 3, and the coordinate planes. Page 1011: 1 31 odd. Exercises

71 68 Section 7.3: Double Integrals over General Regions Definition 200: Let R R 2. Then R is said to be of type I if there exists a closed interval [a,b] and two continuous functions g 1 and g 2 such that R = {(x,y) R 2 : a x b,g 1 (x) y g 2 (x)}. Definition 201: Let R R 2. Then R is said to be of type II if there exists a closed interval [c,d] and two continuous functions h 1 and h 2 such that R = {(x,y) R 2 : c y d,h 1 (y) x h 2 (y)}. Theorem 202: Let R R 2 be a type I region and f : R R be continuous. Then b g2 (x) f(x,y)da = f(x,y)dydxwhere g 1 and g 2 are continuous real-valued functions R a g 1 (x) such that R = {(x,y) R 2 : a x b,g 1 (x) y g 2 (x)}. Theorem 203: Let R R 2 be a type II region and f : R R be continuous. Then d h2 (y) f(x,y)da = f(x,y)dxdy whereh 1 andh 2 arecontinuousreal-valuedfunctions R c h 1 (y) such that R = {(x,y) R 2 : c y d,g 1 (y) x g 2 (y)}. Example 204: Let R be the region bounded by x = 0, x = 4, and x 2 = y and calculate ( x 2 +xy ) da. R

72 Example 205: Let R be the region bounded by the graphs of y = x 2 2x 5 and y = x+5 and calculate xda. R 69 Example 206: Let R = {0 y 3,y x 3} and calculate e x2 da. R Exercises Page 1019: 1 16, odd, 35, 37, odd.

73 70 Section 7.4: Double Integrals with Polar Coordinates ( Example 207: Let R be the unit circle and evaluate x 2 +y 2 +1 ) da. R Theorem 208: Let f : R 2 R be continuous on the polar region R = {(r,θ) : 0 a r b,α θ β,α β α+2π}. Then R f(x,y)da = β b α a f(rcosθ,rsinθ)rdrdθ. ( Example 209: Let R be the unit circle and evaluate x 2 +y 2 +1 ) da. R

74 71 Example 210: Let R = {(x,y) : x 2 +y 2 4}. ( x 2 +y 2 +x ) da R Theorem 211: Let f : R 2 R be continuous on the polar region R = {(r,θ) : α θ β,h 1 (θ) r h 2 (θ)}. Then R f(x,y)da = β h2 (θ) α h 1 (θ) f(rcosθ,rsinθ)rdrdθ. Example 212: Find the area enclosed by one loop of the eight-leaved rose r = sin4θ.

75 72 Example 213: Find the volume of the region inside the sphere x 2 + y 2 + z 2 = 81 and outside of the cylinder x 2 +y 2 = 9. Note 214: It is easier to find the volume of the solid inside of both regions first. Page 1026: 1 6, 7 35 odd. Exercises

76 Section 7.5: Cylindrical Coordinates 73 Definition 215: The cylindrical coordinate representation of the point with rectangular coordinate representation (x, y, z) is (r, θ, z) where (r, θ) is the rectangular coordinate representation of (x, y) in the xy-plane. Note 216: Recall: (i) x = rcosθ; (iii) y x = tanθ; (iv) x2 +y 2 = r 2. (ii) y = rsinθ; Example 217: Convert each of the following points from rectangular to cylindrical coordinates. a. ( 2, 2,3 ) b. (0,1,-5) Example 218: Convert each of the following points from cylindrical to rectangular coordinates. a. ( 3, π 3,1) b. ( -2, π 4,6)

77 74 Example 219: Convert each of the following rectangular equations to cylindrical equations. a. y = x b. x 2 +y 2 = 9 c. x 2 +y 2 = z Example 220: Convert each of the following cylindrical equations to rectangular equations. a. θ = π 6 b. r = 5 c. z = r 2 Page 1055: Exercises

78 75 Section 7.6: Spherical Coordinates Definition 221: The spherical coordinate representation of the point with rectangular coordinate representation (x, y, z) and cylindrical representation (r, θ, z) is (ρ, θ, φ) where ρ is the distance the point is from (0,0,0) and φ is the angle between the positive z-axis and the the segment joining (0,0,0) to the point. Proposition 222: (i) x = rcosθ = ρsinφcosθ; (ii) y = rsinθ = ρsinφsinθ; (iii) z = ρcosφ; (iv) ρ 2 = r 2 +z 2 = x 2 +y 2 +z 2. Example 223: For each of the following, give the coordinates of the point in the the other two coordinate systems. a. Rectangular: ( 2,2 3,4 ) b. Cylindrical: ( 3, π 4,4) c. Spherical: ( 2, π 4, π 6 ) Page 1061: Exercises

79 76 Section 7.7: Triple Integrals Definition 224: Suppose that R = [a,b] [c,d] [u,v] and f : R R. Then f is integrable on R if there is a number L such that for all ε > 0 there is a δ > 0 such that whenever a = x 0 < x 1 < x 2 < < x l 1 < x l = b, c = y 0 < y 1 < y 2 < < y m 1 < y m = d, u = z 0 < z 1 < z 2 < < z n 1 < z n = v, x i x i 1 < δ for all i, 1 i l, y j y j 1 < δ for all j, 1 j m, z k z k 1 < δ for all k, 1 k n, and ( ) [ r ijk,s ijk,t ijk,x ] [ xi 1 i,y ] [ yj 1 j,z ] zk 1 k for all i,j,k, 1 i l, 1 j m, 1 k n, then n m l f ( ) ( ) ( ) ( ) r ijk,s ijk,t xi ijk x yj i 1 y zk j 1 z k 1 L < ε. The number k=1 j=1 i=1 L is called the integral of f over R. Notation 225: R f(x,y,z)dv Theorem 226 (Fubini s Theorem): LetR = [a,b] [c,d] [u,v]andsupposef : R 3 R is integrable on R. Then R f(x,y,z)dv = = = v d b u c a v b d u a c d v b c u a f(x,y)dxdydz f(x,y)dydxdz f(x,y)dxdzdy = = = d b v c a u b v d a u c b d v a c u f(x,y)dzdxdy f(x,y)dydzdx f(x,y)dzdydx. Example 227: Let R = [0,1] [1,3] [-1,0] and calculate ( x 2 y +x 2 z +yz ) dv. R

80 Definition 228: Let R R 3 and let P be the projection of R onto the xy-plane. Then R is said to be of type I if there exists two continuous functions g 1 : P R and g 2 : P R such that R = {(x,y,z) R 3 : (x,y) P,g 1 (x,y) z g 2 (x,y)}. Definition 229: Let R R 3 and let P be the projection of R onto the yz-plane. Then R is said to be of type II if there exists two continuous functions h 1 : P R and h 2 : P R such that R = {(x,y,z) R 3 : (y,z) P,h 1 (y,z) x h 2 (y,z)}. Definition 230: Let R R 3 and let P be the projection of R onto the xz-plane. Then R is said to be of type III if there exists two continuous functions k 1 : P R and k 2 : P R such that R = {(x,y,z) R 3 : (x,z) P,k 1 (x,z) y k 2 (x,z)}. Theorem 231: Let R = {(x,y,z) R 3 : (x,y) P,g 1 (x,y) z g 2 (x,y)} be a type I region where P is the projection of R onto the xy-plane. If f : R 3 R is integrable over [ ] g2 (x,y) R, then f(x,y,z)dv = f(x,y,z)dz da. R P g 1 (x,y) Theorem 232: Let R = {(x,y,z) R 3 : (y,z) P,h 1 (y,z) x h 2 (y,z)} be a type II region where P is the projection of R onto the yz-plane. If f : R 3 R is integrable over [ ] h2 (y,z) R, then f(x,y,z)dv = f(x,y,z)dx da. R P h 1 (y,z) Theorem 233: Let R = {(x,y,z) R 3 : (x,z) P,k 1 (x,z) y k 2 (x,z)} be a type III region where P is the projection of R onto the xz-plane. If f : R 3 R is integrable over [ ] k2 (x,z) R, then f(x,y,z)dv = f(x,y,z)dy da. R P k 1 (x,z) Example 234: Use a triple integral to find the volume of the solid tetrahedron bounded by x = 0, y = 0, z = 0, and x+y +z = 2. 77

81 78 Example 235: Let R be the solid tetrahedron with vertices (0,0,0), (3,0,0), (0,4,0), and (0,0,5). ( ) 108x+216 dv R Page 1049: 1 35 odd. Exercises

82 79 Section 7.8: Triple Integrals with Cylindrical Coordinates Note 236: For a type I region f(x,y,z)dv R = P [ ] g2 (x,y) f(x,y,z)dz da g 1 (x,y) = = β h2 (θ) g2 (rcosθ,rsinθ) α h 1 (θ) g 1 (rcosθ,rsinθ) β h2 (θ) g2 (rcosθ,rsinθ) α h 1 (θ) g 1 (rcosθ,rsinθ) (f(rcosθ,rsinθ,z)dz)rdrdθ rf(rcosθ,rsinθ,z)dzdrdθ Example 237: Let R = {(x,y,z) R 3 : x 2 + y 2 4,0 z 5} and calculate 2e x2 +y 2 dv. R

83 80 Example 238: 3 9 x x 2 y 2 z 2 dzdydx Exercises Page 1055: odd, 29.

84 Section 7.9: Triple Integrals with Spherical Coordinates 81 Theorem 239: Let f be continuous on the spherical wedge R = {(ρ,θ,φ) : 0 a ρ b,α θ β,β α+2π,γ φ δ,δ γ +π}. Then f(x,y,z)dv = R δ β b γ α a f(ρsinφcosθ,ρsinφsinθ,ρcosφ)ρ 2 sinφdρdθdφ. Example 240: Let R be the unit sphere and calculate x2 +y 2 +z 2 dv. R Example 241: Use a triple integral to find the volume of the region inside the sphere x 2 +y 2 +z 2 = 4 and outside the sphere x 2 +y 2 +z 2 = 1.

85 82 Example 242: Page 1046: 22, 24. Exercises Page 1061: odd, 39, 41.

86 Chapter 8: Vector Fields 83 Section 8.1: Vector Fields Definition 243: Suppose that A R n and F : A V n. Then F is called a vector field in R n. (i) If n = 2, F is called a vector field in the plane. (ii) If n = 3, F is called a vector field in space. Example 244: Define F : R 2 R 2 by F(a,b) = - a,b. Sketch the graph. Example 245: Define F : R 2 R 2 by F(a,b) = 1,1. Sketch the graph. Definition 246: Suppose that f : R n R. Then f : R n V n is a vector field called the gradient vector field. Definition 247: If F : R n R n is a vector field and f : R n R such that F = f, then F is called a conservative vector field and f is called a potential function for F. Example 248: A particle moves in velocity field V (x,y,z) = x+z,y 2,2z. At time t = 2 the particle is located at (1,4,-3). Estimate its position at time t = Example 249: Page 1085: Exercises Page 1085: 1 9 odd, 15 18, 21, 23, 25,

87 84 Section 8.2: Line Integrals Theorem 250: Suppose that x : [a,b] R, y : [a,b] R, and f : R 2 are continuous. b Then f(x,y)ds = f(x(t),y(t)) [x (t)] 2 +[y (t)] 2 dt. C a Theorem 251: Suppose that x : [a,b] R, y : [a,b] R, z : [a,b] R, and f : R 3 are b continuous. Then f(x,y,z)ds = f(x(t),y(t),z(t)) [x (t)] 2 +[y (t)] 2 +[z (t)] 2 dt. C a Example 252: Let C be the part of the circle centered at (0,0) of radius 4 that lies in the first quadrant and calculate x 3 yds. C Note 253: The line integrals along C with respect to x and y are calculated as follows: (i) f(x,y)dx = b C a (ii) f(x,y)dy = b C a f(x(t),y(t))x (t)dt; f(x(t),y(t))y (t)dt. Example 254: Consider C : x = t 2, y = t, z = t, 1 t 4 xdx+y 3 dy +zdz C

88 Definition 255: Suppose that F : C V n is a continuous vector field and C is a smooth curve parameterized by r : [a,b] R n. The the line integral of F along C is F dr = b a F(r(t)) r (t)dt. Example 256: Let F(x,y,z) = 2xy,x 2 +z,y +z and C be the curve defined by r(t) = t 3,t 2,t for 0 t 1. Compute the following integral. C F dr C 85 Page 1096: 1 15 odd, 19, 21. Exercises

89 86 Section 8.3: The Fundamental Theorem for Line Integrals Theorem 257 (The Fundamental Theorem for Line Integrals): Suppose that C is a smooth curve defined by the vector function r : [a,b] R n (n = 2,3). Also, suppose that f : C R is differentiable and f is continuous on C. Then F dr = f(r(b)) f(r(a)). Example 258: Let F(x,y,z) = 2xy,x 2 +z,y +z and C be the curve defined by r(t) = t 3,t 2,t for 0 t 1. Compute the following integral. C F dr C Definition 259: Suppose that F : D V n is a continuous vector field. Then the line integral F dr is independent of path if F dr = F dr whenever C 1 and C 2 are C C 1 C 2 two paths in D with the same terminal points. Definition 260: A curve in R n is closed if its initial point and terminal point are the same. Theorem 261: Suppose that F : D V n is a continuous vector field. Then independent of path if and only if F dr = 0 for every closed path C in D. C C F dr is Definition 262: A set A R n is (path) connected if whenever a,b A, there is a path from a to b. Theorem 263: Suppose that F : D V n is a continuous vector field and D is open and connected. If F dr is independent of path in D, then F is a conservative vector field. C Theorem 264: Suppose that F(x,y) : D R 2 where F(x,y) = p(x,y),q(x,y is a conservative vector field and p and q have continuous first-order partial derivatives on D. Then p = q on D. y x