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by: Ben Morrison

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Calc 2 Exam 2 Study Guide Math 1020

Marketplace > Rensselaer Polytechnic Institute > Math > Math 1020 > Calc 2 Exam 2 Study Guide
Ben Morrison
RPI

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This study guide goes over the basic information you should know from each section.
COURSE
Calculus II
PROF.
Gina I. Kucinski
TYPE
Study Guide
PAGES
10
WORDS
CONCEPTS
Math, Calc, Calc 2, Study Guide, exam
KARMA
50 ?

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This 10 page Study Guide was uploaded by Ben Morrison on Friday March 25, 2016. The Study Guide belongs to Math 1020 at Rensselaer Polytechnic Institute taught by Gina I. Kucinski in Spring 2016. Since its upload, it has received 30 views. For similar materials see Calculus II in Math at Rensselaer Polytechnic Institute.

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Date Created: 03/25/16
Calc II Exam 2: Study Guide Sections covered in this exam: 11.1 – Parametric Equations 11.2 – Arc Length and Speed 11.3 – Polar Coordinates 11.4 – Areas and Arc Length in Polar Coordinates 12.1 / 12.2 – Vectors in 2D and 3D 12.2 – Vectors in 3 Dimensions 12.3 – The Dot (Scalar) Product 11.1 Parametric Equations Let C be a curve in the plane. A particle’s motion along C can be described by giving its coordinates as functions of time t. x=f(t), y=g(t) The equations above are called parametric equations and t is a parameter for these equations. As t varies, (x, y) varies and traces the parametric curve C. Using substitution to express in form y=f(x): If x=t+3 and y=4t, solve for t so that t=x-3. x=t+3  t=x-3 Then, substitute t into y=4t. y=4t  y=4(x-3)  y=4x-12 This is called eliminating the parameter. Graphing the curve: Let’s use the same example to show how to graph a parametric curve. Plug in values of t that will give you enough information to sketch the curve. For this equation, I would start at t=-3 so that x starts at 0. When t=-3: x=t+3  x=-3+3=0 y=4(0)-12=-12 When t=-2: x=t+3  x=-2+3=1 y=4(1)-12=-8 When t=-1: x=t+3  x=-1+3=2 y=4(2)-12=-4 When t=0: x=t+3  x=0+3=3 y=4(3)-12=0 You can use these values to sketch the parametric curve. Parametric Equations and the Unit Circle: Sometimes you can use the unit circle equation to eliminate the parameter: x=cos(t), y=sin(t) Eliminate parameter: (cos(t))^2 + (sin(t))^2 = 1  x^2 + y^2 = 1 (unit circle) Other forms: If the equation is given in the form (x-h)^2 + (y-k)^2 = R^2: (x-h)^2 = (R^2)(cos(t))^2  x = h + Rcos(t) (y-k)^2 = (R^2)(sin(t))^2  y = k + Rsin(t) 0 ≤ t ≤ 2π If the equation is given in the form (a,b) through slope m: t = x-a  x = a+t y = b + mt ­∞ ≤ t ≤ ∞ If a segment is being joined from P(a,b) to Q(c,d): x = a + t(c – a) y = b + t(d – b) 0 ≤ t ≤ 1 Finding equation of tangent line to curve C: The slope of the tangent line to a parametric curve is represented by the equation dy/dx = (dy/dt) / (dx/dt) Example: Let’s find an equation of the tangent line to C if x=t^2, y=t^3 – 3t, and t = 2: dy/dx = (dy/dt) / (dx/dt) = (3t^2 – 3) / 2t = (3(2^2)-3) / 2(2) = 9/4 In this problem, the slope of the tangent line is 9/4. Now, plug t into the x and y equations to find a point on the line: When t=2, x = 2^2 = 4, y = (2^3) – 3(2) = 8-6 = 2 Slope = 9/4, point = (4,2) Equation: y-2 = (9/4)(x-4) 11.2 Arc Length and Speed Arc Length Equation b dx 2 dy 2 S = ∫ √(dt( ) dt dt a Speed Equation S’(t) = √ f't)+g't)2 Surface Area Equation b 2π∫ yt)√x't)+y (t) S = a dt 11.3 Polar Coordinates Plotting polar coordinates on a graph: θ + : measured counter-clockwise - θ : measured clockwise Example: (2, π/6) Equivalent polar coordinates: (-2, 7π/6) (2, -11π/6) (-2, -5π/6) Converting Cartesian Point to Polar Point Given the polar point (x,y), use the unit circle equation to find R: 2 2 2 x +y =R θ Then, use the x and y conversion equations to find: x=R cos(θ) y=Rsin(θ) Example: x +y =R2 √2 (-1,1)   1+1=R^2  R = x=R cos(θ)-1 = √2 cos θ  cos θ = -1/ √2 y=Rsin(θ) 1= √2 sin θ sin θ = 1/ √2 θ √2 θ √2 θ=3π/4 If cos = -1/ and sin = 1/ , θ¿ √2,3π/4 (R, = ( ) Converting Polar Point to Cartesian Point Example: Convert (2, π/3) to a cartesian point. R = 2 θ=π/3 x=R cos(θ)  x = 2cos(π/3)  x = 2(1/2)  x = 1 y=Rsin(θ)  y = 2sin(π/3)  y = 2( √3/2¿  y = √3 (x, y) = (1, √3¿ 11.4 Areas and Arc Length in Polar Coordinates To sketch a graph of a polar equation, first sketch as a cartesian curve, then create a table of values from that graph. Use those points to sketch the graph in unit circle form. To find the area, use the equation: β 1 2 A= ∫ 2(R )dθ α 12.1 / 12.2 Vectors in 2D and 3D: A vector has a magnitude and a direction. It’s represented as a directed line segment (such as PQ ) Two vectors are equivalent if they have the same direction and same length. V 1isatranslateof V 2if V1isequivalent¿V2butV 2hasadifferentintial pointthanV 1 Writing a directed line segment in component form: For PQwhereP= x1,y1,z1 ∧Q= (x2,y2,z2 : <x2-x1, y2-y1, z2-z1> Solving for length/magnitude of a vector: ⃗ For V=¿ v1,v2, v3>: ‖ V‖= √1 +v2 +v3 2 Vector addition and multiplication: ⃗ ⃗ For V=¿ v1,v2, v3>¿U=¿u1,u2,u3>: u+v⃗=¿u1+v1,u2+v2,u3+v3>¿ For a scalar number K: k v=¿kv1, kv2,kv3>¿ If k > 0, kv⃗ has the same direction as v kv⃗ ⃗ If k < 0, has the opposite direction as If k = 0, kv⃗ is 0 12.2 Vectors in 3 Dimensions A 2D plane has 4 quadrants: A 3D plane is divided into 8 octants. The first octant is where x, y, z > 0. Finding the distance between P1 and P2 in 3D: P1(x1, y1, z1) P2(x2, y2, z2) 2 2 2 |1P 2 | |2B + P| | | 2 2 2 ¿|P2B | |1A + | |AB | 2 2 2 ¿(z2−z1 ) +(x2−x1 )+ y2−y1 ) 2 2 2 P1P 2 |= √(x2−x1 )+ y2−y1 )+(z2−z1 ) Finding the center and radius of a sphere: Complete the square to get an equation of this form: 2 2 2 2 x−x0 )+ (y−y0 ) +(z−z0 )=R Center: (x0, y0, z0), Radius: R 12.3 The Dot (Scalar) Product Ia=¿a1,a2,a3>¿b=¿b1,b2,b3>,the productof ⃗a∧bis: a1b1 + a2b2 + a3b3 I⃗,b≠0∧θistheanglebetween⃗a∧bwhere0≤θ≤π,then: ⃗×b ‖ ‖‖ ‖ ¿ −1 θ=cos ¿ ⃗ ⃗ ⃗ ⃗ v1is∥¿v2if v1=kv2 v1is perpendicular ¿v 2if v1 x v2=0

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