Sin x 4 sin y 4 x 4 y 4

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• [PDF File]triple int16 8 - University of Notre Dame

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  (b) R is the region bounded by y = −x + 4, y = x + 1, and y = x/3 − 4/3 and the transformation is x = 1 2 (u+v), y = 1 2 (u− v) Soln: (a) Plug the transformation into the equation for the ellipse. (u 2)2 + (3v)2 36 = 1 u2 4 + 9v2 36 = 1 u 2+v = 4 After the transformation we had a disk of radius 2 in the uv-plane. (b) Plugging in the ...

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• [PDF File]SOLUTIONS - UCSD Mathematics | Home

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  subject to the constraint 2x2 +(y 1)2 18: Solution: We check for the critical points in the interior f x = 2x;f y = 2(y+1) =)(0; 1) is a critical point : The second derivative test f xx = 2;f yy = 2;f xy = 0 shows this a local minimum with

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• [PDF File]Partial Derivatives Examples And A Quick Review of ...

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  = 30y 2(x +y3)9 (Note: Chain rule again, and second term has no y) 3. If z = f(x,y) = xexy, then the partial derivatives are ∂z ∂x = exy +xyexy (Note: Product rule (and chain rule in the second term) ∂z ∂y = x2exy (Note: No product rule, but we did need the chain rule) 4. If w = f(x,y,z) = y x+y+z, then the partial derivatives are ∂w ...

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• [PDF File]Graphing sine and cosine worksheet key

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  y: sin 4x y: 4 cos x y: 3 sin —x cos 5x 2 sin x 4 cos 5x Give the amplitude and period of each function graphed below. Then write an equation of each graph. 12. —2 It 13. —2 It f'. ZIT Give the amplitude and period of each function. Then graph of the function over the interval —27t x 21t.

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• [PDF File]Partial Differentiation

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  4. Let f(x,y) = sin(x − y). Determine the equations and shapes of the cross-sections when x = 0, y = 0, x = y, and describe the level curves. Use a three-dimensional graphing tool to

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• [PDF File]R R D Solution

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  x2 +y2 = 1 and x2 +y2 = 4; Solution: Z Z R (x+y)dA = Z 3π/2 π/2 Z 2 1 (rcosθ +rsinθ)rdrdθ = Z 3π/2 π/2 (sinθ +cosθ)dθ Z 2 1 r2dr = − 14 3. (c) R R R cos(x 2 + y2)dA where R is the region that lies above the x-axis within the circle x2 +y2 = 9; Solution: Z Z R cos(x2 +y2)dA = Z π 0 Z 3 0 cos(r2)rdrdθ = (Z π 0 dθ)(Z 3 0 rcos(r2)dr ...

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• [PDF File]858 Chapter 15: Multiple Integrals

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  (sin x + cos y) dx dy L-1 L /2 0 y sin x dx dy FIGURE 15.6 The double integral gives the volume under this surface over the rectangular region R (Example 1). 4R ƒ(x, y) dA R 1 2 1 50 z x –1 z 5 100 2 6x2y y 100 FIGURE 15.7 The double integral gives the volume under this surface over the rectangular region R

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• [PDF File]HOMEWORK#10 SOLUTION

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  At t = 0 and y = 2 cm, E = 4 V/m: 4 = x x 10-2 — Hence, which gives and — COS 0.84 rad ... (7t x 107t—0.2zx) i— sin(z x —0.2zx) —ý 1207t 207t cos(z x 107t—0.2zx) (A/m), 62.83 (Q). and kc x3x108 Rx 1010 (rad/s). Pmblem 7.8 An RHC-polarized wave with a modulus of 2 (Vim) is traveling in free space in the negative *direction. Write ...

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• [PDF File]Review for Exam 2. Section 14 - Michigan State University

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  Review for Exam 2. I Sections 13.1, 13.3. 14.1-14.7. I 50 minutes. I 5 problems, similar to homework problems. I No calculators, no notes, no books, no phones. I No green book needed. Section 14.7 Example (a) Find all the critical points of f (x,y) = 12xy − 2x3 − 3y2. (b) For each critical point of f , determine whether f has a local

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• [PDF File]Student’s Solutions Manual

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  4 CHAPTER 1. WHAT IS A DIFFERENTIAL EQUATION? The quadratic term factors yield two more roots, m2 = −2, m3 = 1/2, and two more solutions y2 = e−2x and y 3 = e x/2. These three solutions can be combined, as in Exercise 5, to produce a

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• [PDF File]Trigonometric Identities - Miami

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  [sin(x+ y) + sin(x y)] Sum-to-Product Formulas sinx+ siny= 2sin x+y 2 cos x y 2 sinx siny= 2sin x y 2 cos x+y 2 cosx+ cosy= 2cos x+y 2 cos x y 2 cosx cosy= 2sin x+y 2 sin x y 2 The Law of Sines sinA a = sinB b = sinC c Suppose you are given two sides, a;band the angle Aopposite the side A. The

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• [PDF File]Math 241 Homework 12 Solutions

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  5. The base of a solid is the region between the curve y = 22 sin x and the interval 30, p 4 on the x-axis. The cross-sections perpen-dicular to the x-axis are a. equilateral triangles with bases running from the x-axis to the curve as shown in the accompanying figure. 0 p y ! 2" sin x x y b. squares with bases running from the x-axis to the ...

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• [PDF File]Integrals in cylindrical, spherical coordinates (Sect. 15 ...

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  x2 + y2. Solution: (x = ρ sin(φ)cos(θ), y = ρ sin(φ)sin(θ), z = ρ cos(φ).) 2 y 1/ 2 x x + y = 1/22 2 z z = 1- x - y2 2 z = x + y2 The top surface is the sphere ρ = 1. The bottom surface is the cone: ρ cos(φ) = q ρ2 sin2(φ) cos(φ) = sin(φ), so the cone is φ = π 4. Hence: R = n (ρ,φ,θ) : θ ∈ [0,2π], φ ∈ h 0, π 4 i, ρ ...

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• [PDF File]Assignment 4 (MATH 215, Q1) - University of Alberta

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  4. Evaluate the integral by making an appropriate change of variables. (a) ZZ E x2 dxdy,where E is the elliptical region {(x,y) : x2/4+ y2/9 ≤ 1}. Solution. Let T be the mapping given by x = 2u and y = 3v. The Jacobian of T is

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• [PDF File]Math 311 - Spring 2014 Solutions to Assignment # 4 ...

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  Math 311 - Spring 2014 Solutions to Assignment # 4 Completion Date: Friday May 16, 2014 Question 1. [p 77, #1 (a)] Apply the theorem in Sec. 22 to verify that the function

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• [PDF File]DIFERENSIAL - UNY

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  1. y = sin x maka dy/dx = cos x 2. y = cos x dy/dx = -sin x 3. y = lg x dy/dx = sec2x 4. y = cotg x dy/dx = -cosec2 x 5. y = sec x dy/dx = sec x tg x 6. y = cosec x dy/dx = -cosec x ctg x 7. y = sinh x dy/dx = cosh x ...

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• [PDF File]Restricted Sine Function.

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  Evaluating sin 1 x. Example Evaluate sin 1 p 2 using the graph above. I We see that the point p 1 2; ˇ 4 is on the graph of y = sin 1 x. I Therefore sin:1 p 1 2 = ˇ 4 Example Evaluate sin 1(p 3=2) and sin 1(p 3=2). I sin 1(p 3=2) = y is the same statement as: y is an angle between ˇ 2 and 2 with siny = p 3=2. I Consulting our unit circle, we ...

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• Problem Bank 7: Partial Differential Equation

  if g(0,y) = 0. One possible choice is therefore g(x,y) = x, which gives u(x,y) = x+ 1 4 (x2 +y2) 4. Initial–boundary value problem Supposea metal rod of length L has an initial temperatureof sin π Lx and the temperaturesat its left and right ends are both fixed at 0 degree Celsius. What would be the initial–boundary

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• [PDF File]TRIGONOMETRY LAWS AND IDENTITIES

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  y =sin(x) x y ⇡ 2 ⇡ 3⇡ 2 2⇡ 1 1 y = cos(x) x y ⇡ 2 ⇡ 3⇡ 2 2⇡ 1 1 y = tan(x) x y 0 30 60 90 120 150 180 210 240 270 300 330 360 135 45 225 315 ⇡ 6 ⇡ 4 ⇡ 3 ⇡ 2 2 3 3 5 ⇡ 7⇡ 6 5⇡ 4 4⇡ 3 3⇡ 2 5⇡ 3 7⇡ 4 11⇡ 6 2⇡ ⇣p 3 2, 1 ⌘ ⇣p 2 2, p 2 ⌘ ⇣ 1 2, p 3 2 ⌘ ⇣ p 3 1 ⌘ ⇣ p 2 p 2 ⌘ ⇣ 1, p 3 ...

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• [PDF File]Maximum and Minimum Values

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  • fy(x,y)=12x24y2 Set both of these partial derivatives to zero. • y =(1/4)x2 • x = 2y2 Next, we solve the system of equations. y =(1/4)(2y2)2 =) y = y4 If y =1,thenx = 2. If y =0,thenx =0. Thecriticalpoints (2,1) and (0,0). We now calculate the second derivatives to classify the critical point. • fxx(x,y)=6x • fyy(x,y)=48y • fxy(x ...

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