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MATH 4389 Final: Calculus_3

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MATH 4389

Calculus 3 Notes Cartesian Space Coordinates Ex: Give the equation of a plane that is parallel to the xz-plane that passes through the point (-1, 3, -2). Distance Formula: dP (,P) ( =x )x (− y)+y( 2− z) z+ − 1 2 2 1 2 1 2 1 xx+y+yz+ z Midpoint Formula: ⎜ ⎟1 2 1 2, , ⎝ ⎠ 2 2 2 2 2 2 Equation of a Sphere: (xa− ) (+b −zc ) (r − ) = 2 Example: Give the equation of the sphere that has A and B as the endpoints of a diameter. A (2, 0, -1) B (2, 1, 3) Vectors A vector is an ordered triple (in space) where addition and multiplication by scalars holds. Vectors have a direction and a length (magnitude or norm). Properties of vectors: Commutative: a + b = b + a Associative: (a + b) + c = a + (b + c) The zero vector 0 = (0,0,0) (note: a⋅00 = ) Vectors can be multiplied by a scalar: if a = aa,,a , then 2a = 2a ,2a ,2a ( 1 2 3 ) ( 1 2 3) 2 2 2 The norm of a vector a = (a1 2 3 ) is a =+ +aa1 2 3 Two vectors are parallel if a = α b for some real number α . If α >0, then a and b have the same direction. If α <0, then a and b have opposite directions. Unit Vectors are vectors of norm 1. a ua= u aas direction a a Find the unit vector for a = (3,4,-2) There are 3 special unit vectors: i = (1,0,0) j = (0,1,0) k = (0,0,1) All vectors can be represented by a linear combination of these: = a i + a j + a k ( a1 2 3 ) 1 2 3 Dot Product Given a = (a , a , a ) and b = (b , b , b ), a b = (a )(b ) + (a )(b ) + (a )(b ) 1 2 3 1 2 3 1 1 2 2 3 3 Note that this gives an answer that is NOT a vector. The dot product gives an answer that is a scalar. Example: a = 6i + 5j + 4k, b = i + 3j a b = The angle between two vectors is found with this formula: cosθ = aib a b π Find all the numbers x for which the angle between c = xi + j + k and d = i + xj + k is 3 Projection of a on b proj a = (aiu )u where aiu = comp a b b b b b Given a = 4i + 3j, b = i - 3j + 2k Find the component of a in the b direction. Find the projection of a in the b direction. The angles αβ ,,γ that a vector makes with unit vectors i, j, and k are called direction angles of a. A unit vector with these direction angles is: cosαi+cosβj+cosγ k Find the direction angles of a = i - 3 k Cross Product If vectors a and b are NOT parallel, they form two adjacent sides of a parallelogram: b a a b is the vector perpendicular to this plane. So, if a || b then a b = 0 The area of this parallelogram is A= || a b || × The area of the triangle with sides a and b is A= || a b || /2 × for a = (a1 2 3 ) and b = (b12 3 ), a b = Example F ind the area of triangle PQR. P(2, 0, -3), Q(3, -1, 0), R(2, 3, -2) Volume of a Parallelepiped: V= |(a × b) i c | How can we tell if 3 vectors are coplanar? Lines Vector form: r(t)= (x i+ y j+ 0 k)+t(0 i+d j0d k) 1 2 3 x(t)= x +t0 1 Scalar form: y(t)= y +td 0 2 z(t)= z +td 0 3 x − x y − y z − z Symmetric form: 0 = 0 = 0 d 1 d 2 d 3 Examples: 1) Find a vector parameterization for the line that passes through P(3, 2, 3) and is parallel to the line r(t) = (i + j - k) + t(2i + 3j - 3k). 2) Give the symmetric form for the line that passes through P(2, 2, 1) and Q(2, -2, -3). So, if we have two lines, we can determine if they are parallel, coincident, skew, or intersecting. Parallel – Coincident – Skew – Intersect – Determine whether the lines l and l are parallel, coincident, skew, or intersecting. If they intersect, 1 2 find the point of intersection: To find the angle between two lines: cosθ = u •u d D Distance from a point to a line: P P ×d d(P,l) = 0 1 d Find the distance from P(2, 0, 2) to the line through P (3, -1, 1) parallel to 0 - 2j - 2k. Planes Let N = Ai + Bj + Ck be the nonzero vector perpendicular to a plane: The equation of the plane containing P(x ,0y ,0z )0 with normal N = Ai + Bj + Ck is A(x− x )+ 0(y− y )+C(z− z0)= 0 0 Angle between two planes: cosθ = u •u N2 N2 | Ax + By +Cz + D| Distance from a point to a plane d(P,℘1= 1 1 1 A + B +C 2 2 Examples: 1) Find an equation for the plane which passes through the point P(2, 4, 0) and is perpendicular to i - 3j + 2k. 2) Find an equation for the plane which passes through the point P(3, -2, 3) and is parallel to the plane: 3) Find an equation for the plane which passes through the point P(1, 3, 4) and contains the line: 4) Find an equation in x, y, z for the plane that passes through the given points. P(1, -1, 2), Q(3, -1, 3), R(2, 0, 2) Two planes will be parallel if their norms are scalar multiples of each other. If two planes are not parallel, then they will intersect in a line. The line of intersection will have a direction vector equal to the cross product of their norms. 5) Find a set of scalar parametric equations for the line formed by the two intersecting planes. and Vector Functions Vector functions in R 3are in the form: fkj)fttf+()+ () () where f , 1 ,2and f ar3 the 1 2 3 component functions. The scalar functions would take the form: f1 102+03t 0 t y = (t+ft z d t=+ () We can think of the vector f1 2 if3tk+()ft () or (f1 2 3t ,t () ) as the position of a moving particle at time t. In order to sketch the graph of a vector function, it is easier to look at it in terms of the scalar vectors: x(t) =f =t1 =2yt) 3 ( t) z( t) f ( t) The domain of a vector function is the set of all t’s for which all the component functions are defined. It is the largest possible interval for which all three components are defined. L t Lft Lft f f lif mfL (t)i==== lim ( 1 1 2 2 ,3lim3 ( ) , lim ( ) tt >0 0 0 0 tt −> tt −> tt −> So, if we can find the limit, we can find the derivative: A vector function f is differentiable at t if f)(f+)( t− lim exists. h−>0 h f)(f+)( t− And if the limit exists then f '(t) = h−>0 h ′ ′ ′ Or rather, given fj)fttf+(1 2 3 () () , ′ (t) = f 1t)i+ f (t)2+ f (t)k 3 Note, if any component of f(t) is not differentiable, then f '(t)DNE. Integration: b b b b fjt)td = f+tdt+) f tt () f tt () ∫∫ ∫ ⎜⎜⎟⎟⎟∫ 1 2 3 a a ⎝⎝⎠⎠⎠ a Properties of the derivatives: Let r and u be differentiable vector-valued functions of t, and let f be a differentiable real-valued function of t, and let c be a scalar. d 1. [cr(t) ] cr (t′ dt d 2. [r(t)±u(t) = r](t)±′ (t) ′ dt d 3. dt [ f(t)r(t) ] f(t)r (t)′ f (t)r′t) 4. d r(t)⋅u(t) = r(t)⋅u (t)+r (t)⋅u(t) dt [ ] ′ ′ 5. d [r(t)×u(t) = r]t)×u (t)+r (′)×u(t) ′ dt d 6. [r(f(t)) =]r (f(′))f (t) ′ dt 7.If r(t)⋅r(t)= c,then r(t)⋅r (t)= 0′ Curves ! The!parameterization!of!the!curve!represented!by!the!vector6valued!function!! r(t)! ▯x(t)!i#▯▯y(t)!j#▯▯z(t)!k###! is!smooth!on!an!open!interval!I!if! f ′ ′ , and h ′ !are!continuous!on!I!and! r ′t)≠ 0 !for!any!value!of! t!on!the!interval!I.! ! Let!C!be!a!smooth!curve!represented!by!r!on!an!open!interval!I.!!The!unit#tangent#vector#T(t)!at! r'(t) t!is!defined!to!be! T(t) = ,r'(t) ≠ 0 r'(t) ! A!tangent!line!to!!a!curve!at!a!point!is!the!line!passing!through!the!given!point!that!is!parallel!to! the!tangent!vector.! Now if T'0 (t) = then the unit tangent vector does not change direction. If T'0(t) ≠ , then we can find the principal normal vector: T'(t) N(t) = T'(t) The plane determined by the unit tangent vector and the principal normal vector is called the osculating plane. Intersecting Curves: r1'(t)ir '2u) cosθ = r1'(t) r '2u) Example: Find the point at which the curves intersect and find the angle of intersection. 3t 2 r1ite=+ j4sin+( πt / 2) ( −1 ) 2 r2i) k=+ +4u ( 2− ) Arc Length Recall from calc II: (9.8) Arc length and speed: b 2 2 L(c) = ∫ [x' t ] [+ y' t) ] dt a b L(c) = ∫ 1+ f ' x( )]2dx a β L(c) = [ρ θ ) ] + ρ' θ( )]2dθ α Speed along a curve: t 2 2 Path (x t , y t takes from 0 to t: s = ∫ [x' u) ] + y' u( )] du 0 So, v t = s' t = [x' t ] + y' t) ]2 Now apply this to a path C traced out by xt= () , yt= () and ztz= () for tb∈ [ ] and we have: L(C) =+ + x [ t′ ′ ] [ t () z] [d t ] ∫a Or b L(C) = ( ∫a) t t Speed: Let r(t) = x(t) i + y(t) j + z(t) k, t∈ [a, b] be a continuously differentiable curve. If s is the length of the curve from the tip of r(a) to the tip of r(t), then the speed is: 2 2 2 ds = ++⎡⎤ ⎡ ⎤ ⎡⎤ dy dz dt ⎣⎦ ⎣ ⎦ ⎣⎦ dt dt Note: a curve is said to be a unit speed curve if r ( ) is a unit vector for all t Curvilinear Motion; Curvature We can describe the position of a moving object at time t by a radius vector r(t). As t ranges over a time interval I , the object traces out some path C : r(t) = x(t) i + y(t) j + z(t) k, t ∈ I If r is twice differentiable, then we can find rv( ) ( ) - velocity and r v( )t== ′( ) ( ) - acceleration Also, v ( )t = ′( ) is the speed at time t . The magnitude of the velocity vector is thus the rate of change of arc distance with respect to time. This is why we call it the speed of the object. The Curvature of a Plane Curve In this figure we have a curve through point P with tangent line l that intersects the x-axis at angle φ dφ Then κ = (the magnitude of the rate of change ds of the angle per unit of arc length) is called the curvature. If we have the tangent vector T, then κ = dT ds y′′ If a curve is given by yx= () , then κ = 2 3/2 ⎡1+ y( )′ ⎣ ⎦ Now, if we have a curve given parametrically, r( )txy + ( ) ( ) and Note: • Along a straight line, the curvature is constantly zero. • Along a circle of radius r the curvature is constantly l/r. 1 Also, the reciprocal of the curvature is called the radius of curvature: ρ = κ And the point at the distance ρ from the curve in the direction of the principal normal is called the center of curvature. The Curvature of a Space Curve A space curve bends in two ways. It bends in the osculating plane (the plane of the unit tangent T and the principal normal N) and it bends away from that plane. The first form of bending is measured by the rate at which the unit tangent T changes direction. The second form of bending is measured by the rate at which the vector T × N changes direction. We will concentrate here on the first form of bending, the bending in the osculating plane. The measure of this is called curvature. If the space curve is given in terms of a parameter t, C : r(t) = x(t) i + y(t) j + z(t) k, t∈ ,[ ] Then the curvature is: κ = = tTd / ds/ dt 3 2 The vector function r(t) = (cos t + t sin t)i + (sin t − t cos t)j + 2 t k determines a curve C in space. (a) Find the unit tangent vector T(t) and the principal normal vector N. (b) Find the curvature, κ of C. Partial Derivatives The partial derivative of f with respect to x is the function fx obtained by differentiating f with respect to x, treating y as a constant. The partial derivative of f with respect to y is the function fy obtained by differentiating f with respect to y, treating x as a constant. Examples: Find the partial derivatives: 1. fx(y, ) 3x= 2 y yx + xy 2 ⎛ ⎞ x 2. fx(y, ) e =+ n l ⎜ xy+ ⎝ ⎠ Higher order derivatives: In the case of a function of three variables you can look for three first partials fx y z and there are NINE second partials: ff f,,f,f,f,f,f,f, xy yx xz zx yz zy xx yy zz Example: DIFFERENTIABILITY AND GRADIENT We say that f is differentiable at x if there exists a vector y such that f (x + h) − f (x) = y i h + o(h). g(h) We will say that g(h) is o(h) if h→0 =0 h Let f be differentiable at x. The gradient of f at x is the unique vector ∇ f (x) such that f (x + h) − f (x) = ∇ f (x) i h + o(h). NOTE: The gradient is a VECTOR!!!! Ex: Find the gradient of fx(, ) 2 e=is+x )y 2 Gradients and Directional Derivatives Properties of gradients: Directional Derivatives: f 'u gives the directional derivative of f in the direction u. In other words, fu′ gives the rate of change of f in the direction of u. f ' = ∇f (x)iu u Example: 2 2 ⎛ 1 ⎞ ⎛ 3 ⎞ Find the directional derivative for fx(, ) x =+ y3 at Q ⎜1, ⎟ towards R ⎜, ⎟ . ⎝ 2 ⎠ ⎝ 2 ⎠ Note that the directional derivative in a direction u is the component of the gradient vector in that direction. Important: Example: Find a unit vector in the direction in which f increases most rapidly at P and give the rate of change of f in that direction; find a unit vector in the direction in which f decreases most rapidly at P and give the rate of change of f in that direction. fx(, ) y =e1 ,ta22 x Example: Find the rate of change of f with respect to t along the given curve. f (x, y) = x y, r(t) = e i + e j t −t Other chain rules: If Then And if Then The Gradient as a Normal; Tangent Lines and Tangent Planes Suppose that f (x, y) is a non-constant function that is continuously differentiable. That means f is differentiable and its gradient ∇ f is continuous. We saw last week that at each point in the domain ∇ f (if ≠ 0) points in the direction of the most rapid increase of f. Also, a
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