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Are P and P 0 subspaces of R3?Solve each differential equation 2)show that 5xy^2 sin (y)= sin (x^2 1) is an implicite solution to the differential equation dy/dx=2xcos (x^21)5y^2/10xycos (y) 3) find value for k for which y= e^kx is a solution of the differential equation y"11y'28y=0 4)A tank contains 480 gallons of water in which 60First, there is a solution since the feasible set is compact Lagrange multipliers gives yz 2λx = 0, xz 4λy = 0, xy 6λz = 0 Multiply these equations by x,y,z Method to find the extremal values of xyz subject to x^22y^23z^2=a Method to find the extremal values of

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(x^2 y^2-1)^3=x^2y^3 meaning-1 This gives points on the circle with vertical tangents at (3,2), and (1,2) 12The relation x2 2xy y2 6x 10y 29 = 0 de nes a parabola (a)Find the points where the curve has a horizontal tangent line (b)Find the points where the curve has a vertical tangent line (a)Use implicit di erentiation, then set dy dx = 0Choose one of the equations and solve it for x by isolating x on the left hand side of the equal sign 2x=2y2 Add 2y to both sides of the equation x=\frac {1} {2}\left (2y2\right) Divide both sides by 2 x=y1 Multiply \frac {1} {2} times 22y y13y=3 Substitute y1 for x

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2xy9x^2 (2yx^21) (dy)/ (dx)=0, y (0)=3 \square!More Examples •Example 3 Find the extreme values of f(x,y)=x 22y2 on the disk x2y ≤1 •Solution Compare the values of f at the critical points with values at the 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

0 y 2 Solution We look for the critical points in the interiorIf x^2 y^2 = 25, what is the value of d^2y/dx^2 at point (4,3)?1 = {(x,y)x2 y2 2} is closed because it includes its boundary while D 2 = {(x,y)x2 y2 < 2} is not closed because it does not D 1 D 2 To find the absolute maximum and absolute minimum, follow these steps 1 Find the the critical points of f on D 2 Find the extreme values of f

Answer For any real number r, the plane x 2y z = r is parallel to P, since all such planes have a common normal vector i2jk = 1 2 1Sin (x)cos (y)=05 2x−3y=1 cos (x^2)=y (x−3) (x3)=y^2 y=x^2 If you don't include an equals sign, it will assume you mean " =0 " It has not been well tested, so have fun with it, but don't trust it If it gives you problems, let me know Note it may take a few seconds to finish, because it has to do lots of calculationsAnswer (1 of 3) Solution The given equation is 3x2y4z and the values are x=1, y=2, z=5 Let we substitute the values in the given equation as, 3(1)2(2)4(5)→3–4=19 The final result is 19 Solution The given equation is 3x2y4z and the values are x=1, y=2, z=5 Let we substitute the values in the given equation as, 3(1)2(2)4(5)→3–4=19

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2x3y=2 Geometric figure Straight Line Slope = 1333/00 = 0667 xintercept = 2/2 = 1 yintercept = 2/3 = Rearrange Rearrange the equation by subtracting what is to the How do you graph \displaystyle{2}{x}{2}{y}={2} by plotting points?Solve your math problems using our free math solver with stepbystep solutions Our math solver supports basic math, prealgebra, algebra, trigonometry, calculus and moreTheorem 36 Let F be any partition of the set S Define a relation on S by x R y iff there is a set in F which contains both x and y Then R is an equivalence relation and the equivalence classes of R are the sets of F Pf Since F is a partition, for each x in S there is one (and only one) set of F

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Example Let F R2!R3 be the function F(x;y) = (x 2y;sin(x);ey) = (F 1(x;y);F 2(x;y);F 3(x;y)) Its derivative is a linear transformation DF(x;y) R2!R3 The matrix of the linear transformation DF(x;y) is DF(x;y) = 2 6 4 @F 1 @x @F 1 @y @F 2 @x @F 2 @y @F 3 @x @F 3 @y 3 7 5= 2 4 1 2 cos(x) 0 0 ey 3 5 Notice that (for example) DF(1;1) is aGraph y=2(x1)^23 Find the properties of the given parabola Tap for more steps Use the vertex form, , to determine the values of , , and Since the value of is negative, the parabola opens down Opens Down Find the vertex Find , the distance from the vertex to the focusF(x,y) = (2x)(2y) = 4xy, 0 < x < 1,0 < y < 1 (b) Draw the line x y = 1 2, indicating the origin and coordinates x and y Then indicate the area of xy < 1 2 in the support of f(x,y) (Draw by yourself) (c) Find Pr{X Y < 1 2} Pr{X Y < 1 2} = Z 1 2 0 Z 1 2 −y 0 4xydxdy = Z 1 2 0 2yx2 1 2 −y 0 dy = Z 1 2 0 2y(1 2 −y)2dy = Z 1 2 0

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Convert to Radical Form x^ (3/2) x3 2 x 3 2 If n n is a positive integer that is greater than x x and a a is a real number or a factor, then ax n = n√ax a x n = a x n ax n = n√ax a x n = a x n Use the rule to convert x3 2 x 3 2 to a radical, where a = a =, x = x =, and n = n = √x3 x 3It is best to apply implicit differentiation Differentiating y with respect to x yieldsProblem 2 Determine the global max and min of the function f(x;y) = x2 2x2y2 2y2xy over the compact region 1 x 1;

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Directrix x = 17 4 x = 17 4 Select a few x x values, and plug them into the equation to find the corresponding y y values The x x values should be selected around the vertex Tap for more steps Substitute the x x value 2 2 into f ( x) = 1 √ 4 − x f ( x) = 1 4 x In this case, the point is ( 2, ) ( 2, )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 x2Answer (1 of 4) First we separate equation \qquad \dfrac{dy}{dx} 2y = 3 \qquad \dfrac{dy}{dx} = 2y 3 \qquad \dfrac{dy}{2y3} = dx Next we integrate both sides

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For example , if I separate it so that its partial x (2x3) = 2 partial y (2y2) = 2 2=2, so its exact BUT why can't I go partial x (2y2) = 0 partial y (2x3) = 0 0=0 so thats also exactThe solution set is obviously symmetric with respect to the y axis Therefore we may assume x ≥ 0 In the domain { ( x, y) ∈ R 2 x ≥ 0 } the equation is equivalent with x 2 y 2 − 1 = x 2 / 3 y , which can easily be solved for y y = 1 2 ( x 2 / 3 ± x 4 / 3 4 ( 1 − x 2))2 How do I know which variable gets differentiated?

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Y0= xy 2y x 2 xy 3y x 3;Finding the derivative when you can't solve for y You may like to read Introduction to Derivatives and Derivative Rules first Implicit vs Explicit A function can be explicit or implicit Explicit "y = some function of x"When we know x we can calculate y directly The inner stationary points has zero partial derivatives $$\begin{cases} f'_\lambda = x y^2 z^3 1 = 0\\ f'_x = z^2 2xy \lambda = 0\\ f'_y = x^2 2yz 2\lambda y = 0\\ f'_z = y^2 2zx 3\lambda z^2 = 0 \end{cases}$$ After the excluding of parameter $\lambda$ get the system $$\begin{cases} x y^2 z^3 1 = 0\\ x^2 2yz = 2y(z^2

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First we need to calculate fx (x,y) and fy (x,y) For fx (x,y) we differentiate with respect to x ==> fx (x,y) = 3x^2 (2y^3)x Now substitute with (2,1) ==> fx (2,1)= 3 (4) 2 (1)2 = 124= 16 Use the method of separation of variables if x ≠ 0 and y ≠ 0 (note that y = 0 is a stationary solution) then x = − ( 1 y 2) y 3 ⋅ y ′ = ( − 1 y 3 − 1 y) ⋅ y ′ which implies that x 2 2 = ∫ x d x = ∫ ( − 1 y 3 − 1 y) d y = 1 2 y 2 − ln ⁡ y C Therefore a solution y ( x) satisfies the equation x 2 = 1 y ( x) 2 − lnThe two parts of this product have useful meaning \((ba)(dc)\) is of course the area of the rectangle, and the double sum adds up \(mn\) terms of the form \(f(x_j,y_i)\Delta x\Delta y\text{,}\) which is the height of the surface at a point multiplied by the area of one of the small rectangles into which we have divided the large rectangle

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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 and y = u1 On the RHS(a) Suppose that X and Y are independent random variables with Var(X) = 1, Var(Y) = 2 Find Var(1−2X 3Y) 8 pts Solution This is an exercise in using the properties of a variance Var(1−2X 3Y) = 0(−2)2 Var(X)32 Var(Y) = 4·19·2 = 22 22x2y=3 Geometric figure Straight Line Slope = 1 xintercept = 3/2 = yintercept = 3/2 = Rearrange Rearrange the equation by subtracting what is to the right of the

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It is usually best to see how we use these two facts to find a potential function in an example or two Example 2 Determine if the following vector fields are conservative and find a potential function for the vector field if it is conservative →F = (2x3y4 x)→i (2x4y3 y)→j F → = ( 2 x 3 y 4 x) i → ( 2 x 4 y 3 y) j →Simplify ( (3x^ (3/2)y^3)/ (x^2y^ (1/2)))^2 ( 3x3 2 y3 x2y−1 2)−2 ( 3 x 3 2 y 3 x 2 y 1 2) 2 Move x3 2 x 3 2 to the denominator using the negative exponent rule bn = 1 b−n b n = 1 b n ( 3y3 x2y−1 2x−3 2)−2 ( 3 y 3 x 2 y 1 2 x 3 2) 2 Multiply x2 x 2 by x−3 2 xThis problem has been solved!

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I wondered the same thing The answer is, not really It is still always going to be implicit Given x = r*cos(θ) and y = r*sin(θ), the equation posted reduces to (r 2 1) 3 = r 5 cos 2 (θ) sin 3 (θ) Thats about as simple as I could get itCheck to see if f y is equal to N f y = −x 2g0(y) = 6y −x2 3 so that g0(y) = 6y2 3 That gives g(y) = 2y3 3y Put this back in to get the full solution, f(x,y) = c x3 −x2y 2x2y3 3y = C 3 Problem 4 (2xy2 2y)(2x2y 2x)dy dx = 0 Check for "exactness" $\begingroup$ Unsure if this approach will solve the problem The approach might failThe premise is of form $$\frac{E^2}{F^2} \frac{G^2}{H^2} = 1,$$ which resembles an equation of the form $\sin^2(\theta) \cos^2(\theta) = 1$

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 1 How do I separate the terms?Get stepbystep solutions from expert tutors as fast as 1530 minutes Your first 5 questions are on us!This is the meaning of x1 and x2 in the formula We need to verify that L satisfies the two properties in the definition of linear function For any x,y∈ R2, we have L(xy) = L x1 y1 x2 y2 = (x1 y1) 4(x2 y2) 3(x1 y1) −(x2 y2) (x2 y2) (In the formula, x1 y1 plays the role of x1 and x2 y2 plays the role of x2) = (x1 4x2) (y1

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Algebra Calculator is a calculator that gives stepbystep help on algebra problems See More Examples » x3=5 1/3 1/4 y=x^21 Disclaimer This calculator is not perfect Please use at your own risk, and please alert us if something isn't working Thank youRewrite the equation as − 3 2 x 2 = 0 3 2 x 2 = 0 − 3 2 x 2 = 0 3 2 x 2 = 0 Add 3 2 3 2 to both sides of the equation x 2 = 3 2 x 2 = 3 2 Since the expression on each side of the equation has the same denominator, the numerators must be equal x = 3 x = 3 Multiply both sides of the equation by 2 21 Problem 216 Let P be the plane in 3space with equation x 2y z = 6 What is the equation of the plane P 0 through the origin parallel to P?

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X2 2x 3 = 0 (x 3)(x 1) = 0 x= 3;Solve the given inequalities x – 2y ≤ 2, x y ≥ 3, –2x y ≤ 4, x ≥ 0, y ≥ 0 graphically in two – dimensional plane asked Jul 22 in Linear Equations by KumarArun ( 146k points) pair of linear equations in two variablesEXAMPLE root(3,x^7)=root(3,x^6*x)=root(3,x^6)root(3,x)=x^2root(3,x) The cases when there are fractions in the radicand and radicals in the denominator of a fraction will be discussed later EXAMPLE Put root(2^3x^5) in standard form

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SOLUTION Using independence and the facts that EX= 1 2, E(X2) = 1 3, EY = 1 and E(Y2) = 2 gives the answers for (a)(c) as E(XY) = 3 2, E(XY) = 1 2, and E(X Y)2 = E(X2) 2EXEY E(Y2) = 4 3 For part (d) we must also compute E(e2Y) = Z 1 0 e2ye ydy= Z 1 0 eydy= 1 Therefore, E(X2e2Y) = 1 Problem 7 (p 367 #6) Let Xand Y be independent standard normal variables Find 3386 Chapter 15 Multiple Integration c y 1 y 2 y 3 d a x 1 x 2 x 3 x 4 x 5 b ∆x ∆y Figure 1511 A rectangular subdivision of a,b×c,d Using sigma notation, we can rewrite the approximation 1 mnAnswer (1 of 8) Let k = 2^x = 3^y = 6^{z} x = \frac{\ln k}{\ln 2} y = \frac{\ln k}{\ln 3} z = \frac{\ln k}{\ln 6} \begin{align*} \frac{1}{x} \frac{1}{y} \frac

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Example Factorize 9x 2 12xy 4y 2 Solution Step 1 Identify which identity can be applied in the expression We can apply (x y) 2 = x 2 2xy y 2 Step 2 Rearrange the expression so that it can appear in the form of the above identity 9x 2 12xy 4y 2 = (3x) 2 2 × 3x × 2y (2y) 2 Step 3 Once the expression is arranged in the form of the identity, write its factors31 Expectation The mean, expected value, or expectation of a random variable X is written as E(X) or µ 1),(x 2,y 2),,(x N,y N) If X and Y are dependent, the value x i might affect the value y i, and vice versa, so we have to keep the observations together in their pairings As the number of pairs N tends to infinity, the average

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