X2 Y2 2y 1

Math 2263 Quiz 10 26 April, 12 Name 1 Evaluate RR S zdS, where S is the part of the plane 2x 2y z = 4 that lies in the rst octant Answer The x, y.

X2 y2 2y 1.  the answer is 6 2 (y1) for y=2 just means to replace the y with two So, it turns into 2 (21) 2 plus one is 3, so just 2 times 3 yw 👍. 1 day ago find stationary points of $ f(x,y) = y^4 (4x^2)(y^2) 2y^2 2x^2 1$ Ask Question Asked today Active today Viewed 31 times 1 = 0 \\ 4y^3(8x^24)y = 0 \end{cases}\implies \begin{cases} x = 0 \\ 4y^34y=0\end{cases}\implies (x,y) = (0,0), (0,1), (0,1)$ Thus those are the stationary points I hope it helps !. Y0= xy 2y x 2 xy 3y x 3;.

Answer (1 of 2) This almost looks like an EulerCauchy equation \qquad\qquad x^2 y’’ 2xy’ 2y = 0 So if we can find a solution to the equation above that is also a solution to \qquad\qquad y’’ = 0 then this solution will also be a solution to \qquad\qquad y’’ (x^2 y’’ 2xy’ 2y) =.  Explanation We can try isolate x from the first equation x = 1 2y substitute into the second (for x ) (1 2y)2 y2 = 29 1 4y 4y2 y2 = 29 5y2 4y −28 = 0 solve for y (using the Quadratic Formula) gives you y1,2 = −4 ± √16 560 10 = −4 ± 24 10. Reduce the equation to one of the standard forms, classify the surface, and sketch it x^2 y^2 z^2 4x 2y 2z 4 = 0 1) Reorganize the equation.

And uppersurface z = 2 q 1− x2 − y2 9 The projected region in the x−y is the the inside of the ellipse x2 y2 9 = 1 in the first quadrant, which may be described as a ysimple region in the 2D x − y plane n (x,y) 0 ≤ y ≤ 3 √ 1− x2,0 ≤ x ≤ 1 o So, the integral above is the same as Z 1 0 Z 3 √ 1−x2 0 Z 2 q 1−x2. Start with the inverse equation in explicit form Example y = sin −1 (x) Rewrite it in noninverse mode Example x = sin(y) Differentiate this function with respect to x on both sides Solve for dy/dx;. 1 y =2x 2 x =2y 3 x2 y2 =8 Notice that if one variable is zero, then the other is as well This violates equation (3), so we don’t need to consider it Let’s substitute (1) into (2) x =42x =) = ± 1 2 Plugging this value into equations (1) and (2) give us the following equation y = ±x We can then plug this into equation (3) Then 2 x2.

28 y dx/dyx=2y^2, y(1)=5 Ecuaciones lineales Alexander Estrada. 2 1 = λ2x λ2y = x y so 2y = x Sub into (15) to find 4y2y2 = 5 ⇒ y = ±1 Combining with 2y = x, we get the solutions (x,y) = (2,1) and (−2,−1) These are the same points we found in (c), and knowing their z values, we know that f(2,1) is a maximum while f(−2,−1) is a minimum on the constraint 29.  Show that 2 2 1 { ( )} x x y is an integrating factor of the differential equation ( ) 2 0 2 2 x y dx xydy asked in Mathematics by Så Y Äñ ( points) 0 votes.

2 Montrer que lescourbes intégrales onttoutesune asymptote quandxtendvers∞ 3. Nowwecanconsidertheothercase 1 2(x 2y) = 0,whichmeansthatx y2 = 1 Whatwillthis imply?. 0 y 2 Solution We look for the critical points in the interior.

SOLUTION 13 Begin with x 2 xy y 2 = 1 Differentiate both sides of the equation, getting D ( x 2 xy y 2) = D ( 1 ) , 2x ( xy' (1)y) 2 y y' = 0 , so that (Now solve for y' ) xy' 2 y y' = 2x y, (Factor out y' ) y' x 2y = 2 x y, and the first derivative as a. Y = ± p c(4x2)−2 (19) 2y(x1)dy = xdx, 2ydy= x x1 dx, Z 2ydy= Z x x1 dx, Resolvemos la integral del lado derecho Z x x1 dx = Z x1−1 x1 dx = Z µ.  Explanation Sort the x terms and y terms x2 −x y2 2y = −1 Now, complete the square for each variable x2 −x 1 4 y2 2y 1 = −1 1 4 1 Don't forget to balance both sides of the equation (If you add something to one side, add it to the other side as well) (x − 1 2)2 (y 1)2 = 1 4 This is in the standard form of a circle.

As a final step we can try to simplify more by substituting the original equation An. Extended Keyboard Examples Upload Random Compute answers using Wolfram's breakthrough technology & knowledgebase, relied on by millions of students & professionals For math, science, nutrition, history, geography, engineering, mathematics, linguistics, sports, finance, music.  phân tích đa thức thành nhân tử 9z 2 x 2 2xyy 2 1 Trả lời Cho tam giác ABC, trung tuyến AMGọi I là trung điểm của Am và D là giao điểm của BI và AC Chứng minh AC=3AD Bài 1Cho tam giác ABC, trung tuyến AMGọi I là trung điểm của Am và D là giao điểm của BI và AC a)Cm.

Vx = 2y ˚0(x) = uy = 2y 2 ) ˚(x) = 2xc ) f = x2 y2 2y i(2xy 2xc) where c is a real constant Let g = uiv We have v = 2xy y ) vy = 2x1 = ux) u = x2 x˚(y) ) uy = ˚0(y) = vx = 2y ) ˚(y) = y2 C ) g = x2 y2 xci(2xy y) where as before c is a real constant 3 Determine the domains where the following functions are holomorphic (i. 1 Every number is equal to itself x= xfor all x∈ R 2 Equalities can be “reversed” If x,y∈ Rand x= y, then y= x 3 You can “chain” equalities together If x,y,z∈ Rand x= yand y= z, then x= z These three properties are captured in the axioms for an equivalence relation Definition. Suppose we approach the origin along x = y2 Then f(x,y) = y2y2 y4 y4 = y4 2y4 = 1 2, so the limit is 1/2 Looking at figure 1421, it is apparent that there is a ridge above x = y2 Approaching the origin along a straight line, we go over the ridge and then drop.

ODEs Find the first four terms of the power series solution to the IVP y"2y'y=x, y(0)=0, y'(0)=1 To check our answer, we find the solution using th. The intersection of the two graphs is ( 2;. X2 x2 2y2 ≤ 1, we have the inequalities 0 ≤ x 2sin y x 22y ≤ sin2 y The limits of the outer two functions as (x,y) → (0,0) are both 0, and so the Squeeze Theorem tells us that lim (x,y)→(0,0) x2 sin2 y x2 2y2 = 0 The notion of the limit of a function of two variables readily extends to functions of three or more variables.

Area of the triangle determined by the line xy=3 and the bisector of angle between the lines x^2y^22y=1. Y(4) = 2 1 Rewriting the LHS in di erential form and factoring the RHS we get dy dx = (x 2)(y 1) (x 3)(y 1) 2 Separating the variables leads to y 1 y 1 dy = x 2 x 3 dx 3 To evaluate the integrals Z y 1 y 1 dy = Z x 2 x 3 dx we need usubstitution on both sides On the LHS, let u = y 1 and then du = dy.  x 2 – y 2 2y c – 1 = 0 Comparing this equation with Ax 2 2Hxy By 2 2Gx 2Fy C = 0, we get, A = 1, H = 0, B = 1, G = 0, F = 1, C = c – 1 Since the given equation represents a pair of lines, ← Prev Question.

If so, compute it If not, prove it lim (x,y)!(1,1) e xy cos(xy) Notice that the point (1, 1) is in the. 1) So the solution to the system of simultaneous equations is x = 2 and y = 1 We can also check the solution using algebraic methods Substitute equation ( 1) into ( 2) x = 2 y ∴ y = 2 ( 2 y) − 3 Then solve for y y − 4 y = − 3 − 3 y = − 3 ∴ y = 1. Cylinder z = (4−y2)1/2 below by the xy plane and the projection D of the solid onto the xyplane is the triangle with edges x = 2y, x = 0 and the intersection of the cylinder with the plane z = 0 which gives y 2 = 4 or y = 2 (first octant).

Substituting the value of x in equation (1), We have 5 (2) 2y = 2 ⇒ 10 2y = 2 ⇒ 2y = 10 2 ⇒ 2y = 8 ⇒ y = `8/2` ⇒ y = 4 Thus the values of x and y are x = 2 and y = 4 Concept Solving Exponential Equations. Di erentiating both sides with respect to x, y2 x 2y dy dx = 0 so dy dx = y2 2xy At the point (1 4;2), dy dx = 4, so we can use the point/slope formula to obtain the tangent line y= 4(x 1=4) 2 2Consider the circle de ned by x 2 y = 25 (a)Find the equations of the tangent lines to the circle where x= 4 (b)Find the equations of the normal. Again,wecanconsiderthesecondequation,pluginx2 y2 = 1,andseewhathappens (2y)ey2 x2 (x2 y2)ey2 x2(2y) = 0 turnsinto (2y)ey2 x2 ey2 x2(2y) = 0 ) 4yey2 x2 = 0 Once again, since 4ey2 x2 6= 0 , we see that y= 0 So what this tells us is that if x2 y2 = 1.

X 22y = (x2 2y2)(x2 2y2) x 2y2 = x2 22y Now we return to Step I and observe that (0,0) is in the domain of x2 2y2 Therefore, we can plug in Thus, lim (x,y)!(0,0) x4 4y2 x 22y =(0) 22(0) =0 Example 5212 Does the limit exist?. Now, y = x± q x2 −4(x2 −7) 2 = x± √ 28−3x2 2 Take the positive root since y(1) = 3 The restriction on x would be that 28−3x2 ≥ 0 Therefore, − s 28 3 < x < s 28 3 5 Problem 15 (xy2 bx2y)dx(xy)x2 dy = 0 First, for this to be exact. Answer (1 of 3) We have \dfrac{dy}{dx}=\dfrac{y}{x}\dfrac{x^2y^2–1}{x^2y^21} Put x=r\cos{t}, y=r\sin{t}, \text{ where } r \text{ is a function of } t \text.

Find the Symmetry x^2y^2xy=1 There are three types of symmetry 1 XAxis Symmetry 2 YAxis Symmetry 3 Origin Symmetry If exists on the graph, then the graph is symmetric about the 1 XAxis if exists on the graph 2 YAxis if exists on the graph 3 Origin if exists on the graph Check if the graph is symmetric about the xaxis by. Problem 2 Determine the global max and min of the function f(x;y) = x2 2x2y2 2y2xy over the compact region 1 x 1;. Divide 4 4 by − 4 4 Multiply − 1 1 by − 1 1 Add 0 0 and 1 1 Substitute the values of a a, d d, and e e into the vertex form a ( x d) 2 e a ( x d) 2 e Set x x equal to the new right side Use the vertex form, x = a ( y − k) 2 h x = a ( y k) 2 h, to determine the values of a a, h h, and k k.

Differential Equations Problems with Solutions By Prof Hernando Guzman Jaimes (University of Zulia Maracaibo, Venezuela). We can try to factor x 2 −2xy−y 2 but we must do some rearranging first Change signs y 2 2xy−x 2 = − 1 k 2 Replace − 1 k 2 by c y 2 2xy−x 2 = c. Z=\pm 5 is not a curve Of course, 2x^22y^2z^2=25 and x^2y^2=z^2 imply z=\pm 5, but don't forgot about x and y I'll show case z=5, z=5 is similar.

 You found $1z^2=xyyz−zxz^2=(yz)(xz)$ Similarly for $1x^2$ you get $(xz)(xy)$ And for $1y^2$ you get $(yz)(xy)$ Therefore, you can replace \begin{align} P= \frac{2x^2}{1x^2} \frac{2y^2}{1y^2}\frac{3z^2}{1z^2}\end{align}.  Section 43 Double Integrals over General Regions In the previous section we looked at double integrals over rectangular regions The problem with this is that most of the regions are not rectangular so we need to now look at the following double integral, ∬ D f (x,y) dA ∬ D f ( x, y) d A where D D is any region. Apply a linear substitution v' = t sin (2v t) 1/2, v (0) = pi/2 Solve a firstorder homogeneous equation through a substitution solve x y' = y* (log (x) log (y)) Make general substitutions solve 2 t^3 y' (t) = 1 sqrt (1 4 t^2 y (t)) y' (x) = (1x cos (y (x))) cot (y (x)) More examples.

F(x,y)dx = ˆR 1 0 4xydx = 2y if 0 ≤ y ≤ 1 0 otherwise (d) YES, X and Y are independent, since fX(x)fY (y) = ˆ 2x·2y = 4xy if 0 ≤ x ≤ 1 and 0 ≤ y ≤ 1 0otherwise is exactly the same as f(x,y), the joint density, for all x and y Example 4 X and Y are independent continuous random variables, each with pdf g(w) = ˆ 2w if 0 ≤ w. And now to solve for dy dx, we just have to divide both sides by 2y minus 2x minus 1 And we are left with we deserve a little bit of a drum roll at this point As you can see, the hardest part was really the algebra to solve for dy dx We get the derivative of y with respect to x is equal to 2y minus 2x plus 1 over 2y minus 2x minus 1. 1x 2 y′(x−1) y=x3−x2x1 On note yK(x) l’unique solution de cette équation telle que yK(0) = K On appelle courbe intégrale le graphe de yK(x)pourK∈R 1 Que direde latangente en x=1àune courbe intégrale?.

2Use cylindrical coordinates to evaluate the integral RRR E p x2 y2 dV, where E is the region E= f(x;y;z) 2R3 jx2 y2 1;y 0;4x z 6g The constraints x2 y2 1 and y 0 are equivalent to 0 r 1 and 0 ˇ In cylindrical coordinates, the constraint 4x z 6 becomes. Z= p 1 2x2 4y2 Then, the vector equation is obtained as r(x;y) = xi yj p 1 2x2 4y2k 176 Find a parametric representation for the surface which is the part of the elliptic paraboloid x y2 2z2 = 4 that lies in front of the plane x= 0 If you regard yand zas parameters, then the parametric equations are x= 4 y2 2z2;.