7 Geometry and Elliptic Curves - UCI Mathematics
2 The homogenization of the hyperbola x2 y2 = 3 is the projective curve X2 Y2 = 3Z2 3 The homogenization of x2 + xy2 +2x +1 = 0 is X2Z + XY2 +2XZ2 + Z3 = 0 Definition 7 5 The real projective plane RP2 is defined via an equivalence relation on the set of non-zero points in R3: RP2 = R 3 ˘ where (X,Y, Z) ˘(A, B,C) X Y Z k A B C
180B Lecture Notes, W2011
2 Covariance and Correlation Suppose that (;P) is a probability space We say that X: R is in-tegrable if EjXj
C2 Trigonometr y: Trigonometric Equations PhysicsAndMathsTutor
(b) Solve, for 0 ≤x < 360°, 5sin 2x = 2cos 2x, giving your answers to 1 decimal place (5) (Total 6 marks) 2 (a) Show that the equation 5 sin x = 1 + 2 cos2 x can be written in the form 2 sin2 x + 5 sin x – 3 = 0 (2) (b) Solve, for 0 ≤x < 360°, 2 sin2 x + 5 sin x – 3 = 0 (4) (Total 6 marks) 3 (i) Solve, for –180° ≤ θ < 180°,
6 Gaussian Integers and other Rings
2 S = fmd: m 2Z[i]g, where S is defined as in Definition 6 4 3 Every common divisor of a and b divides d , whence d has the largest norm among all common divisors We leave the proof to the homework: as a hint, part 1 requires the use of the Division Algorithm
GCOC9: Lines and Angles 3 - JMAP
2) 70° 3) 110° 4) 180° 11 Angle A and angle B are complementary angles If m∠A =x, which expression represents the number of degrees in angle B? 1) x −180 2)180−x 3) x −90 4)90−x 12 If the measure of an angle is represented by 2 x, which expression represents the measure of its complement? 1) 180−2x 2) 90 −2x 3) 90 +2x 4) 88x
UNIVERSITY OF CAMBRIDGE INTERNATIONAL EXAMINATIONS
The diagram shows accurate graphs of y = sinx and y = cosx for 0° Y x Y 180° Use the graph to solve the equations (a) sinx – cosx = 0, Answer(a) x = [1] (b) sinx – cosx = 0 5 Answer(b) x = [2] 9 A fence is made from 32 identical pieces of wood, each of length 2 metres correct to the nearest centimetre
Math 180,Exam 2, Spring 2011 Problem 1Solution
Math 180,Exam 2, Spring 2011 Problem 2Solution 2 Sketch the graph of the function f(x) = x−2 −x2 following the steps below (a) Determine the domain of f and find all asymptotes (b) Find the intervals where the graph of f is increasing, decreasing, concave up and concave down
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A teacher asks one of her students to solve the equation 2 cos 2x + 3 = 0 for 0 ⩽ x ⩽ 180° The attempt is shown below 2cos2 = 3x 3 cos2 = 2 x 2 =cos1 3 2 x 2 =150x o x=75o w x o o oor =360 75 =295 so reject as out of range (a) Identify the mistake made by the student (1) (b) Write down the correct solutions to the equation (1)
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Math 180, Exam 2, Spring 2008 Problem 2 Solution 2 Let f(x) = x3 +3x2 − 9x− 11 (a) Find the critical points of f (b) Find the intervals on which f is increasing and the intervals on which f is decreasing
CBSE NCERT Solutions for Class 9 Mathematics Chapter 6
Let the common ratio between a and b be x, ∴a = 2x and b = 3x XY is a straight line, OM and OP rays stands on it Therefore, XOM + MOP + POY = 180 o ⇒b + a + POY + 180o ⇒3x + 2x + 90o= 180o ⇒5x = 90o ⇒x = 18o a = 2x ⇒= 2 × 18 ⇒= 36o b = 3x ⇒= 3 × 18 ⇒= 54o Now, MN is a straight line OX ray stands on it ∠b + ∠c = 180o
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Math 180, Exam 2, Spring 2008
Problem 1 Solution
1. Find the derivatives of the following functions: (do not simplify)
(a)f(x) =x2cos(x+ 1) (b)g(x) = sin(xlnx) (c)h(x) = tan1(2x2+ 1) (d)k(x) = (x+x3)1776Solution: (a) Use the Product Rule and the Chain Rule. f 0 (x) = [x2cos(x+ 1)]0 =x2[cos(x+ 1)]0+ (x2)0cos(x+ 1) =x2[sin(x+ 1)(x+ 1)0] + 2xcos(x+ 1) =x2sin(x+ 1) + 2xcos(x+ 1)(b) Use the Chain Rule and the Product Rule. g0(x) = [sin(xlnx)]
0 = cos(xlnx)(xlnx)0 = cos(xlnx)[x(lnx)0+ (x)0lnx] = cos(xlnx) x1x+ lnx = (1 + lnx)cos(xlnx) (c) Use the Chain Rule. tan1(2x2+ 1)0=11 + (2x2+ 1)2(2x2+ 1)0
11 + (2x2+ 1)2(4x)
(d) Use the Chain Rule. (x+x3)17760= 1776(x+x3)1775(x+x3)0 = 1776(x+x3)1775(1 + 3x2)1Math 180, Exam 2, Spring 2008
Problem 2 Solution
2. Letf(x) =x3+ 3x29x11.
(a) Find the critical points off. (b) Find the intervals on whichfis increasing and the intervals on whichfis decreasing. (c) Find the local minima and maxima off. Computexandf(x) for each local extremum. (d) Determine the intervals on whichfis concave up and the intervals on whichfis concave down. (e) Find the points of in ection off. (f) Sketch the graph off.Solution:
(a) The critical points off(x) are the values ofxfor which eitherf0(x) does not exist or f0(x) = 0. Sincef(x) is a polynomial,f0(x) exists for allx2Rso the only critical
points are solutions tof0(x) = 0. f0(x) = 0x3+ 3x29x110= 0
3x2+ 6x9 = 0
3(x2+ 2x3) = 0
3(x+ 3)(x1) = 0
x=3; x= 1Thus,x=3
andx= 1are the critical points off. (b) The domain offis (1;1). We now split the domain into the three intervals (1;3), (3;1), and (1;1). We then evaluatef0(x) at a test point in each in- terval to determine the intervals of monotonicity.IntervalTest Point,cf0(c)Sign off0(c)
(1;3)4f0(4) = 15+ (3;1)0f0(0) =9 (1;1)2f0(2) = 15+ 1 Using the table, we conclude thatfis increasing on (1;3)[(1;1) becausef0(x)>0 for allx2(1;3)[(1;1) andfis decreasing on (3;1) becausef0(x)<0 for
allx2(3;1). (c) Sincef0changes sign from + toatx=3 the First Derivative Test implies that f(3) = 16 is a local maximum and sincef0changes sign fromto + atx= 1 theFirst Derivative Test implies thatf(1) =16
is a local minimum. (d) To determine the intervals of concavity we start by finding solutions to the equation f00(x) = 0 and wheref00(x) does not exist. However, sincef(x) is a polynomial we
know thatf00(x) will exist for allx2R. The solutions tof00(x) = 0 are: f00(x) = 03x2+ 6x90= 0
6x+ 6 = 0
x=1 We now split the domain into the two intervals (1;1) and (1;1). We then evaluatef00(x) at a test point in each interval to determine the intervals of concavity.IntervalTest Point,cf0(c)Sign off0(c)
(1;1)2f00(2) =6 (1;1)0f00(0) = 6+ Using the table, we conclude thatfis concave down on (1;1) becausef00(x)<0 for allx2(1;1) andfis concave up on (1;1) becausef00(x)>0 for all x2(1;1). (e) The in ection points off(x) are the points wheref00(x) changes sign. We can see in the above table thatf00(x) changes sign atx=1. Therefore,x=1 is an in ection point. (f) -4 -3 -2 -1 1 2x -20 -16-12-8-448121620y
2Math 180, Exam 2, Spring 2008
Problem 3 Solution
3. Find an equation for the tangent line to the curve defined bythe equationx2y2+xy+y= 1
at the point (2;1). Solution: To findy0, the slope of the tangent line, we use implicit dierentiation. x2y2+xy+y= 1
(x2y2)0+ (xy)0+ (y)0= (1)0 (x2)(y2)0+ (y2)(x2)0+ [(x)(y)0+ (y)(x)0] +y0= 0(x2)(2yy0) + (y2)(2x)+ [(x)(y0) + (y)(1)] +y0= 02x2yy0+ 2xy2+xy0+y+y0= 0
2x2yy0+xy0+y0=2xy2y
y0(2x2y+x+ 1) =2xy2y
y0=2xy2y
2x2y+x+ 1
At the point (2;1), the value ofy0is:
y0(2;1) =2(2)(1)2(1)
2(2)2(1) + 2 + 1=35
An equation for the tangent line is then:
y+ 1 =3 5(x2) 1Math 180, Exam 2, Spring 2008
Problem 4 Solution
4. Find the minimum and the maximum values on the interval [0;5] for the function:
f(x) =x39x2+ 24x+ 1 Solution: The minimum and maximum values off(x) will occur at a critical point in the interval [0;5] or at one of the endpoints of the interval. The critical points are the values of xfor which eitherf0(x) = 0 orf0(x) does not exist. Sincef(x) is a polynomial,f0(x) exists for allx2R. Therefore, the only critical points are solutions tof0(x) = 0. f0(x) = 0x39x2+ 24x+ 10= 0
3x218x+ 24 = 0
3(x26x+ 8) = 0
3(x2)(x4) = 0
x= 2; x= 4 The critical pointsx= 2 andx= 4 both lie in [0;5]. Therefore, we check the value off(x) atx= 0;2;4, and 5. f(0) = 039(0)2+ 24(0) + 1 = 1 f(2) = 239(2)2+ 24(2) + 1 = 21 f(4) = 439(4)2+ 24(4) + 1 = 17 f(5) = 539(5)2+ 24(5) + 1 = 21The minimum value off(x) on [0;5] is 1
because it is the smallest of the above values of f. The maximum is 21 because it is the largest. 1Math 180, Exam 2, Spring 2008
Problem 5 Solution
5. You are designing a closed cardboard box (with a top, a bottom, and four sides). The
length of the box must be twice its width and the volume of the box must be 72 cubic inches. (a) Express the surface area of the box as a function of its width. (b) Determine the dimensions of the box that will use the least amount of cardboard.Solution:
(a) We begin by lettingxbe the width of the box,ybe the height, andzbe the length.The surface area then has the equation:
Surface Area = 2xy+ 2yz+ 2xz(1)
We are asked to write the surface area as a function of width,x. Therefore, we must find equations that relatextoyandzso that we can eliminate the latter two variables from the surface area equation. One of the constraints in the problem is that the length must be twice its width. This gives us the equation z= 2x(2) The other constraint in the problem is that the volume must be72 cubic inches. This gives us the equationVolume =xyz= 72 (3)
Plugging equation (2) into equation (3) and solving forywe get xyz= 72 xy(2x) = 72 y=36 x2(4) Finally, we write the surface area as a function ofxby plugging equations (2) and (4) into equation (1).Surface Area = 2xy+ 2yz+ 2xz
Surface Area = 2x36
x2 + 236x2 (2x) + 2x(2x)Surface Area =
216x+ 4x2(5) f(x) =216 x+ 4x2(6) 1 (b) The surface area represents the amount of cardboard necessary to construct the box. Therefore, we seek the value ofxthat minimizesf(x). The interval in the problem is (0;∞) becausexmust be nonnegative (it is a length quantity) and nonzero (the volume must be 72). The absolute minimum off(x) will occur either at a critical point off(x) in (0;∞) or it will not exist because the interval is open. The critical points off(x) are solutions tof?(x) = 0. f ?(x) = 0216 x+ 4x2 = 0 216
x2+ 8x= 0
8x3-216 = 0
x 3=216 8 x= 3Plugging this intof(x) we get:
f(3) =2163+ 4(3)2= 108
Taking the limits off(x) asxapproaches the endpoints we get: lim x→?+f(x) = lim x→?+ 216x+ 4x2 =∞+ 0 =∞ lim x→∞f(x) = limx→∞ 216
x+ 4x2 = 0 +∞=∞ both of which are larger than 108. We conclude that the cost isan absolute minimum atx= 3 and that the resulting surface area is 108. The last step isto find the corresponding values foryandzby pluggingx= 3 into equations (4) and (2). z= 2x= 2(3) = 6 y=36x2=3632= 4 2