X 3 xy 2 y 3
[PDF File]Quiz 6 Problem 1. Solution.
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We also have D(x,y) = f xxf yy − f2 xy = 36xy − 81. Next, D(0,0) = −81 < 0, therefore (0,0) is a saddle point. f xx(3,3) = 18 > 0 and D(3,3) = 243 > 0, therefore (3,3) is a point of relative minimum. Problem 2. Find the absolute maximum and the absolute minimum of f(x,y) = x2 +2y2 −x on the closed disk x2 +y2 ≤ 4. Solution. We start ...
[PDF File]Examples: Joint Densities and Joint Mass Functions
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f(x,y)dydx = Z 1 0 Z 2 0 (cx2 + xy 3)dydx = 2c 3 + 1 3, so c = 1. (b). Draw a picture of the support set (a 1-by-2 rectangle), and intersect it with the set {(x,y) : x + y ≥ 1}, which is the region above the line y = 1 − x. See figure above, right. To compute the probability, we double integrate the joint density over this subset of the ...
[PDF File]Unit #5 - Implicit Di erentiation, Related Rates Implicit ...
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x2 xy+ y2 = 3 x2 x( x) + ( x)2 = 3 x2 + x2 + x2 = 3 x2 = 1x = 1 Since x= 1 is the point we started at, x= +1 must be the other intersection. At that point, y= x, so the point is (x;y) = (1; 1). Below is a graph of the scenario. The curve with equation 2y3 + y2 y5 = x4 2x3 +x2 has been likened to a bouncing wagon (graph it to see why). Find the ...
[PDF File]微積分学 II 演習問題
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第1回の演習問題の解答 1. (1) f: (xy) 2 R 2 xy̸= 1 2}! (1 ; 1 2) [(2;1, g: 1 ; 1 2) [(2;1R をf(xy) = xy, g(t) = cos(ˇt)1+2t で定めれば, f, gはともに連続関数だから, lim (xy)!(21) cos(ˇxy)1+2xy = lim (xy)!(21) g(f(xy)) = g lim (y )!(21) f(x y) g(f(2 1)) = g(2) = 1 5. (2) f: R2!R, g: R ! R をf(xy) = xy, g(t) = 8
[PDF File]ASSIGNMENT 5 SOLUTION
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We can nd these points by solving x2+xy+y2 = 0. Completing the square for x gives (x + 1 2 y) 2 + 3 4 y 2 = 0, and this happens only when (x;y) = (0;0). So f is indeed continuous away from the origin, i.e., the only possibile discontinuity is at the origin. [Note:
[PDF File]Implicit equations: an example
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The equation x2 xy + y2 = 3 describes an ellipse. Let’s nd out what the derivative dy dx is by implicit di erentiation: 2x (y + xy0) + 2yy0= 0 ( x+ 2y)y0= y 2x y0= y 2x 2y x = 2x y x 2y We may as well nd the second derivative d2y dx2. Here is one way: y00= 2x y x 2y 0 = (2 0y )(x 2y) (2x y)(1 2y0) (x 22y) = 3 xy0 y (x 22y) = 3 x 2x y x 2y y
[PDF File]AP CALCULUS AB 2015 SCORING GUIDELINES
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y xy: 3: −= 2. It can be shown that : 2. 3: dy y dx y x = ... y x; 2; −= In part (c) the student correctly differentiates , dy dx so the first 2 points were earned . Title: ap15_calculus_ab_q6 Author: ETS Subject: calculus_ab_q6 Created Date: 8/17/2015 12:16:34 PM ...
[PDF File]Section 7.4: Lagrange Multipliers and Constrained Optimization
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Find the extrema of F(x,y) = x2y − ln(x) subject to 0 = g(x,y) := 8x +3y. 3 Solution We solve y = −8 3 x. Set f(x) = F(x, −8 3 x) = −8 3 x 3 − ln(x). Differentiating we have f0(x) = −8x2 − 1 x. Setting f 0(x) = 0, we must solve x3 = −1 8, or x = −1 2. Differentiating again, f00(x) = −16x + 1 x2 so that f 00(−1 2) = 12 ...
[PDF File]Solution to Math 2433 Calculus III Term Exam. #3
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2 Z x+6 x2 xdydx= Z 3 2 x x+ 6 x2 dx = Z 3 22 x2 + 6x x3 dx= 1 3 x3 + 3x2 1 4 x4 = 9 + 27 81 4 + 8 3 12 + 4 = 125 12 2. Let T be the solid bounded by the paraboloid z= 4 x2 y2 and below by the xy-plane. Find the volume of T. (Hint, use polar coordinates). Answer The intersection of z= 4 2x 22y and xyplane is 0 = 4 x2 y;i.e. x2 +y = 4: In polar ...
[PDF File]Chapter 10 Differential Equations
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so particular solution in this case is y= 3 2 x2 +C= (i) y= 3 2 x2 − 1 2 (ii) y= 3 2 x2 + 1 2 (iii) y= 3 2 x2 + 3 2 (f) Graphs of dy dx = 3x. There are/is (i) one (ii) many curves/graphs associated with differential equation dy dx = 3x, as shown in the picture below. y x y = (3/2)x - 1/2 2 y = (3/2)x + 3/2 2 y = (3/2)x + 1/2 2 (1,2) (1,3 ...
[PDF File]Joint and Marginal Distributions
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Exercise 2. Find 1. P{X = Y} 2. P{X +Y ≤ 3}. 3. P{XY = 0}. 4. P{X = 3}. As before, the mass function has two basic properties. • f X,Y (x,y) ≥ 0 for all x and y. 1 ...
[PDF File]Maximum and Minimum Values
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D is the enclosed triangular region with vertices (0,0),(0,2), and (4,0). f(x,y)=x+y xy Let’s first draw a picture of D to help us visualize everything. 141 of 155. Multivariate Calculus; Fall 2013 S. Jamshidi x =0 y =0 y = x 2 +2 First, we find the critical points on D. We begin by finding the partials and setting them equal
[PDF File]Lecture 16: Harmonic Functions - Furman
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v(x,y) = 3x2y −y3 +ϕ(x) for some function ϕ of x. It now follows from the second equation that −6xy = −v x(x,y) = −(6xy +ϕ0(x)), and so ϕ0(x) = 0. Hence for any real number c, the function v(x,y) = 3x2y −y3 +c is a harmonic conjugate of u. 3
[PDF File]Homework 5: Solutions
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3.1.32: f(x;y)=œ xy(x2−y2) x2+y2 (x;y)≠(0;0) 0 (x;y)=(0;0) Note: fis continuous, (by computing lim(x;y)→(0;0) of the formula above, e.g. using polar coorinates). (a) Find f x and f y when (x;y)≠(0;0): Away from (0;0);fcan be di erentiated using the formula de ning it, as: @f @x (x;y)= (x2 +y2)[y(x2 −y2)+2x2y]−2x2y(x2 −y2) (x 2+y ...
[PDF File]Partial Derivatives Examples And A Quick Review of ...
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1. Given x4 +y4 = 3, find dy dx. ANSWER: Differentiating with respect to x (and treating y as a function of x) gives 4x3 +4y3 dy dx = 0 (Note the chain rule in the derivative of y4) Now we solve for dy dx, which gives dy dx = −x3 y3. Note that we get both x’s and y’s in the answer, but at least we get some answer. 2. Given y3 −x2y − ...
[PDF File]triple int16 8 - University of Notre Dame
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(a) R is the ellipse x2 + y2 36 = 1 and the transformation is x = u 2, y = 3v. (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
[PDF File]Chain Rule and Total Differentials
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Chain Rule and Total Differentials 1. Find the total differential of w = x. 3. yz + xy + z + 3 at (1, 2, 3). Answer: The total differential at the point (x
[PDF File]極値問題 解答 - 熊本大学
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極値問題 解答 1 (1) f(x,y) = x2 +xy +y2 −3x+2y +2 fx = 2x+y −3 = 0 fy = x+2y +2 = 0 を解いて，(x,y) = 8 3,− 7 3) を得る。 fxx = 2, fxy = 1, fyy = 2 より ∆ = fxxfyy −(fxy)2 = 2×2−12 = 4−1 = 3 > 0 となるので，上記の点でf は極値を取る。fxx = 2 > 0だからその極値は極小値である。 f (8 3,− 7 3) = − 13 3 以上により，f(x,y)は(x ...
[PDF File]Solution to Exercise 5 - CUHK Mathematics
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3. Consider the function f(x;y) = xy(x2 y2) x2 + y2; (x;y) 6= (0 ;0) ; and f(0;0) = 0. Show that f xy and f yxexist but are not equal at (0;0). Solution. For (x;y) 6= (0 ;0), f x= x4y+ 4x2y3 y5 (x2 + y2)2; f y= x5 4x3y2 xy4 (x2 + y2)2: Summer 2017 MATH2010 2 When (x;y) = (0;0), f x(0;0) = 0 and f y(0;0) = 0. We have f xy(0;0) = lim y!0 f
[PDF File]Lecture 9: Partial derivatives
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f(x,y) = x3 − 3xy2 is an example satisfying the Laplace equation. 7 The advection equation f t = f x is used to model transport in a wire. The function f(t,x) = e−(x+t)2 satisfy the advection equation. 8 The eiconal equation f2 x +f2 y = 1 is used to see the evolution of wave fronts in optics. The function f(x,y) = cos(x) +sin(y) satisfies ...
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