Exam. There are 6 problems. Your 5 best answers count. Please pay attention to the presentation of your work! Best 5

Similar documents
Later in this chapter, we are going to use vector functions to describe the motion of planets and other objects through space.

Vectors, dot product, and cross product

SOLUTIONS TO SECOND PRACTICE EXAM Math 21a, Spring 2003

MA CALCULUS II Friday, December 09, 2011 FINAL EXAM. Closed Book - No calculators! PART I Each question is worth 4 points.

Practice problems for Exam 1. a b = (2) 2 + (4) 2 + ( 3) 2 = 29

Vector Functions & Space Curves MATH 2110Q

MA 126 CALCULUS II Wednesday, December 10, 2014 FINAL EXAM. Closed book - Calculators and One Index Card are allowed! PART I

0, such that. all. for all. 0, there exists. Name: Continuity. Limits and. calculus. the definitionn. satisfying. limit. However, is the limit of its

Tangent and Normal Vector - (11.5)

Department of Mathematical and Statistical Sciences University of Alberta

Analytic Geometry MAT 1035

Analytic Geometry MAT 1035

(1) Recap of Differential Calculus and Integral Calculus (2) Preview of Calculus in three dimensional space (3) Tools for Calculus 3

Solution. The relationship between cartesian coordinates (x, y) and polar coordinates (r, θ) is given by. (x, y) = (r cos θ, r sin θ).

DEPARTMENT OF MATHEMATICS AND STATISTICS UNIVERSITY OF MASSACHUSETTS. MATH 233 SOME SOLUTIONS TO EXAM 1 Fall 2018

1.1 Single Variable Calculus versus Multivariable Calculus Rectangular Coordinate Systems... 4

Exercise. Exercise 1.1. MA112 Section : Prepared by Dr.Archara Pacheenburawana 1

3 = arccos. A a and b are parallel, B a and b are perpendicular, C a and b are normalized, or D this is always true.

MATH 12 CLASS 5 NOTES, SEP

Honors Calculus Homework 1, due 9/8/5

Solutions to old Exam 3 problems

Name: ID: Math 233 Exam 1. Page 1

8. Find r a! r b. a) r a = [3, 2, 7], r b = [ 1, 4, 5] b) r a = [ 5, 6, 7], r b = [2, 7, 4]

SCORE. Exam 3. MA 114 Exam 3 Fall 2016

Page Problem Score Max Score a 8 12b a b 10 14c 6 6

Multiple Choice. 1.(6 pts) Find symmetric equations of the line L passing through the point (2, 5, 1) and perpendicular to the plane x + 3y z = 9.

MATH Final Review

Exam 3 Solutions. Multiple Choice Questions

AP Calculus BC - Problem Solving Drill 19: Parametric Functions and Polar Functions

Topic 2-2: Derivatives of Vector Functions. Textbook: Section 13.2, 13.4

Math 114, Section 003 Fall 2011 Practice Exam 1 with Solutions

Math 165 Final Exam worksheet solutions

Tangent and Normal Vector - (11.5)

Mathematics 2203, Test 1 - Solutions

Practice problems. 1. Given a = 3i 2j and b = 2i + j. Write c = i + j in terms of a and b.

MTH Calculus with Analytic Geom I TEST 1

Lecture 4: Partial and Directional derivatives, Differentiability

Worksheet 1.7: Introduction to Vector Functions - Position

Calculus III (MAC )

SCORE. Exam 3. MA 114 Exam 3 Fall 2016

MA 351 Fall 2007 Exam #1 Review Solutions 1

AP Calculus (BC) Chapter 10 Test No Calculator Section. Name: Date: Period:

Worksheet 1.8: Geometry of Vector Derivatives

FINAL EXAM CALCULUS 2. Name PRACTICE EXAM

AP Calculus BC Summer Assignment. Please show all work either in the margins or on separate paper. No credit will be given without supporting work.

Without fully opening the exam, check that you have pages 1 through 11.

In mathematics (and physics...) such quantities are called vectors and are represented by a directed line segment.

MA 162 FINAL EXAM PRACTICE PROBLEMS Spring Find the angle between the vectors v = 2i + 2j + k and w = 2i + 2j k. C.

Study guide for Exam 1. by William H. Meeks III October 26, 2012

Math 210, Exam 1, Practice Fall 2009 Problem 1 Solution

Math 3c Solutions: Exam 1 Fall Graph by eliiminating the parameter; be sure to write the equation you get when you eliminate the parameter.

Curriculum Catalog

Math 116 Practice for Exam 2

Math 190 (Calculus II) Final Review

MATH 251 Examination II April 7, 2014 FORM A. Name: Student Number: Section:

Math 234. What you should know on day one. August 28, You should be able to use general principles like. x = cos t, y = sin t, 0 t π.

Exam 1 Review SOLUTIONS

Parametric Equations, Vectors, and Vector Valued Functions. Different parametric equations can yield the same curve:

D. Correct! This is the correct answer. It is found by dy/dx = (dy/dt)/(dx/dt).

Exam 1. (2x + 1) 2 9. lim. (rearranging) (x 1 implies x 1, thus x 1 0

MAT 272 Test 1 Review. 1. Let P = (1,1) and Q = (2,3). Find the unit vector u that has the same

APPM 2350, Summer 2018: Exam 1 June 15, 2018

Unit Speed Curves. Recall that a curve Α is said to be a unit speed curve if

Pre-Calculus EOC Review 2016

Final Exam Review Part I: Unit IV Material

3. Interpret the graph of x = 1 in the contexts of (a) a number line (b) 2-space (c) 3-space

Parametric Functions and Vector Functions (BC Only)

Fall Exam 4: 8&11-11/14/13 - Write all responses on separate paper. Show your work for credit.

Math 112 (Calculus I) Final Exam

MATH 32A: MIDTERM 2 REVIEW. sin 2 u du z(t) = sin 2 t + cos 2 2

Math 107. Rumbos Fall Solutions to Review Problems for Exam 1

UNIVERSITY OF REGINA Department of Mathematics and Statistics. Calculus I Mathematics 110. Final Exam, Winter 2013 (April 25 th )

MTHE 227 Problem Set 2 Solutions

Chapter 10 Conics, Parametric Equations, and Polar Coordinates Conics and Calculus

Math 113 Winter 2005 Departmental Final Exam

Math 114: Make-up Final Exam. Instructions:

Math 53: Worksheet 9 Solutions

TMTA Calculus and Advanced Topics Test 2010

Things you should have learned in Calculus II

MATH 2433 Homework 1

OHSx XM521 Multivariable Differential Calculus: Homework Solutions 13.1

HOMEWORK 2 SOLUTIONS

9/27/2017 FIRST HOURLY PRACTICE 5 Math 21a, Fall Name:

Page Problem Score Max Score a 8 12b a b 10 14c 6 6

AP Calculus AB. Free-Response Questions

Math 206 First Midterm February 1, 2012

MAC Module 5 Vectors in 2-Space and 3-Space II

Final exam (practice 1) UCLA: Math 32B, Spring 2018

MA 126 CALCULUS II Wednesday, December 14, 2016 FINAL EXAM. Closed book - Calculators and One Index Card are allowed! PART I

Math 116 Second Midterm November 14, 2012

Chapter 12 Review Vector. MATH 126 (Section 9.5) Vector and Scalar The University of Kansas 1 / 30

TARGET QUARTERLY MATHS MATERIAL

Introduction to Vector Functions

Detailed objectives are given in each of the sections listed below. 1. Cartesian Space Coordinates. 2. Displacements, Forces, Velocities and Vectors

MATH 1020 WORKSHEET 12.1 & 12.2 Vectors in the Plane


MATH H53 : Mid-Term-1

MTH 111, Math for Architects, Exam I, Summer 2013

AP Calculus AB 2015 Free-Response Questions

Transcription:

Department of Mathematical Sciences Instructor: Daiva Pucinskaite Calculus III June, 06 Name: Exam There are 6 problems. Your 5 best answers count. Please pay attention to the presentation of your work! 4 5 6 Best 5

. Consider the curve given by the parametric equations x = t, y = t t. a. Which of the following points is on this curve 0,0, 7,4, 7,? A point a,b is on the curve given by the parametric equations if there exists t such that a = t, b = t t. If 0,0 is on the curve, then 0 = t implies t = ±. Since 0 ± ±, the point 0,0 is not on the curve. If 7,4 is on the curve, then 7 = t implies t = ±. Since 4 = and 4, we have for t = The point 7,4 is on the curve. 7 = t, 4 = t t. If 7, is on the curve, then 7 = t implies t = ±. Since ± ±, the point 7, is not on the curve. b. Find an equation of the line tangent of the curve at the point determined in a. A line given by y = cx+m is the line tangent at the point 7,4, if c is the slope of this line at 7 }{{} x, 4 }{{} y, i.e. c = y x and the point 7,4 is on this line, i.e. 4 = c 7+m. Since x t = t and y t = t, we have that the slope of the tangent line at 7,4 is c = y x = = 6 6 = The equation 4 = 9 7+m implies m = 4 = 9, thus the line tangent at 7,4 is y = 9 x

. Consider the points P =,,, Q =,,5, R =,6,. State the equation of the plane P through the points P, Q, R. A normal vector of the plane containing the points P, Q, R is the cross product of the vectors PQ =,, 5 = 0,0, andpr =, 6, = 0, 4,0 : 0,0, 0, 4,0 = 0 4 0,0,0 0 0 0 0,,0 + 0 0 0 4 0,0, =,0,0 An equation of the plane passing through the point P =,, and orthogonal to the vector,0,0 is x +0y +0z 0 = 0 = x =. Determine the point which lies not on the plane P: P =,,, Q =,,, R =,0,0. A point a,b,c is on the plane x + 0y + 0z 0 = 0, if a + 0b +0c 0 = 0, i.e. a,b,c is on the plane if a =. P =,, is on the plane, because the first component of this point is. Q =,, is not on the plane, because the first component of this point is not. R =,0,0 is on the plane, because the first component of this point is. Find an equation of the plane T passing through the point determined in and parallel to the plane P. Since the plane given by x = is orthogonal to the vector,0,0, we have that each plane parallel to the plane x = is orthogonal to,0,0. An equation of the plane T passing through the point Q =,, and orthogonal to the vector,0,0 is x +0y +0z = 0 = x =. 4 Exercise for Extra Credit points Find the distance between the planes T and P. The equations x =, and x = represent the planes parallel to the xz-plane and resp. units from it. Thus the distance between the planes is.

. a. Which geometric object is defined as the set of all points x,y in a plane such that 4x 9y = 6? Specify the location of the foci, and sketch a graph of this object. Describe the relationship between the foci and a point x,y which satisfies this equation. 4x 9y = 6 = 4 6 x + 9 6 y = = x 9 + y 4 = = x + y = 4x 9y = 6 represents an ellipse x a + y b =, where a =, and b = with the foci at c,0 and c,0 where c = a b = 9 4 = 5 5,0 0, x,y 5,0 - -0-9 -8-7 -6-5 -4 - - - 0 4 5 6 7 8 9 0-0,- Let x,y be a point of the ellipse given by x + y =. If d is the distance from the focus at 5,0 to the point at x,y and d is the distance from the focus at 5,0 to x,y, then d +d = a = = 6. b. Which geometric object is defined as the set of all points x,y in a plane such that 4x +9y = 6? Specify the location of the foci, and sketch a graph of this object.

4x +9y = 6 = 4 6 x + 9 6 y = = x 9 y 4 = = x y = 4x 9y = 6 represents a hyperbola x a y b =, where a =, and b = with the foci at c,0 and c,0 where c = a +b = 9+4 =,0,0 - -0-9 -8-7 -6-5 -4 - - - 0 4 5 6 7 8 9 0 - c. Exercise for Extra Credit points Which geometric object is defined as the set of all points x,y in a plane with the following property: If l is the distance form x,y and 5,0, and l is the distance form x,y and 5,0, then l l = 8 or l l = 8 Write an equation, and sketch a graph of this object. Bythedefinitionahyperbolaisthesetofallpointsx,yinaplanesuchthatthedifference of the distances from two fixed points foci is constant. If 5,0 and 5,0 are the foci ±c,0 = ±5,0 l is the distance form x,y and 5,0, and l is the distance form x,y and 5,0, such that l l = a = 8 i.e. a = 4 then the equation of the hyperbola is x 4 + y b =, where c = a +b, i.e. 5 = 4 +b implies b = 5 6 = 9 =. Thus an equation of the hyperbola is x 4 + y =.

4. Consider the following vectors in R Find the cross product v w. v = 0,, and w =,0,. v w = 0,0,0 0 0,,0 + 0 0,0, =, 4, All vectors in R orthogonal to v and w are scalar multiple of the vector v w. Explain why this is true. If a,b,c 0 is orthogonal to v and w, then a,b,c is a normal vector of the plane P containing the vectors v and w. Since v w =, 4, is also a normal vector of the plane P the vectors, 4, and a,b,c are parallel. Thus a,b,c is a scalar multiple of, 4,. Find a vector u perpendicular to v and w, such that the volume of the box determined by v, w, and u is 58. Since u is perpendicular to v and w, the volume of the box determined by v, w, and u is the volume of the box determined by v, w, and u = the area of the parallelogram spanned by v, w, The area of the parallelogram spanned by v and w is v w =, 4, = + 4 + = 9. the length of u From we know that u is a scalar multiple of v w, thus there exist c such that u = c, 4, = c, 4c,c. Since the length of u is u = c + 4c +c = c 9 = c 9 we obtain that if c = 58 9 the volume of the box determined by v, w, and u =, so c = ±. For the vector u we have = 9 c 9 = c 9 = 58 u =, 4, = 6, 8,4, or u =, 4, = 6,8, 4.

5. An object moves on the helix cost, sint, t, for t 0. a Let f be a differentiable function with ft 0. Explain why the position function cosft, sinft, ft describes motion along the helix. The position function rt = cos ft, sin ft, ft describes motion along the helix, if each point of the curve of rt, i.e. the point of the form cosft, sinft, ft is on the helix given by cost, sint, t The curve of the position function cost, sint, t, for t 0 contains the points of the form cos t, sin t, t, where t is a non-negative real number. By the assumption the real number t = ft is non-negative for each t. The point cosft, sinft, ft = cost, sint, t is of the form described above for all t 0. Therefore each point of the curve of the position function cosft, sinft, ft is on the helix given by cost, sint, t. b Find a position function r that describes the motion along the helix if it occurs with the speed. Find the length of the curve for 0 t, and for t. A position function that describes the motion along the helix is of the form rt = cosft, sinft, ft Themagnitudeofthevectorr t = cosft, sinft, ft = sinft f t, cosft f t, f t gives the speed of the object at the time t. We have r t = sinft f t, cosft f t, f t = sinft f t +cosft f t +f t = f t sin ft+cos ft +f }{{} t = f t +f t = f t = f t = if f t =, thus f t =, and consequently ft = function rt = cos t, sin t, dt = t. The position t

describes the motion along the helix where it occurs with the speed. b b b Since the length of the curve for a t b is r t dt = dt = t = a b we a a a have the length the length of the curve = dt = t =, and of the curve = dt = t =. for 0 t 0 0 for t c Find a position function r that describes the motion along the helix if it occurs with the speed t. Find the length of the curve for 0 t, and for t. A position function that describes the motion along the helix is of the form rt = cosft, sinft, ft The magnitude of the vector r t = sinft f t, cosft f t, f t gives the speed of the object at the time t. We have r t = sinft f t +cosft f t +f t = f t sin ft+cos ft+f t = f t +f t = f t = f t = t if f t = t, thus f t = t, and consequently ft = t = t. The position function t t rt = cos, sin, t t dt = t dt = describes the motion along the helix where it occurs with the speed t. The length of the curve for a t b is have the length of the curve for 0 t = 0 b a =, and r t dt = b a the length of the curve for t t dt = t b a = d Exercise for Extra Credit points Interpret the results from b and c. The position functions = b a we = 7

cos t, sin t, t, and t t cos, sin, t describe the motion along the same helix with speeds, and t respectively. The particle moving along the helix given by cos t, sin t, t is at time 0,, at the positions,0,0 = cos cos, sin 6 cos 6, sin 0, sin,, 6 0, 0 Since the speed is constant, the distance along the helix between two points of the same time interval is the same. The time intervals from t = 0 to t =, and from t = to t = is the same, namely 0 = =, therefore the distance along the helix from,0,0 to cos, sin, as well as from cos, sin, to cos 6, sin 6, 6 is the same, namely.