10 Find the extreme values of f(x,y) = 2x2 3y2 −4x−5 on the region D = {(x,y) x2 y2 ≤ 16} Solution We first need to find the critical points These occur when f x = 4x−4 = 0, f y = 6y = 0 so the only critical point of f is (1,0) and it lies in the region x2 y2 ≤ 16 On the circle x 2y = 16, we have y2 = 16−x2 and g(x) = f(x,Jacobian Function linksIf uv=e^cosy & uv=e^xsiny find the Jacobian function https//youtube/8D9QGYyUC9IIf u=e^ucosv, y=e^usinv Prove that JJ' = 1 hJacobian Function linksIf uv=e^cosy & uv=e^xsiny find the Jacobian function https//youtube/8D9QGYyUC9IIf u=e^ucosv, y=e^usinv Prove that JJ' = 1 h
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33 Rewrite the two fractions into equivalent fractions Two fractions are called equivalent if they have the same numeric value For example 1/2 and 2/4 are equivalent, y/ (y1)2 and (y2y)/ (y1)3 are equivalent as well To calculate equivalent fraction , multiply the Numerator of each fraction, by its respective Multiplier the reqd value, ie, m = x y Then, we have, x = my We submit this x in the given eqn, to get, 2(my) −3y 3(my) −2y = 2 5 ∴ y(2m− 3) y(3m− 2) = 2 5 ∴ 5(2m −3) = 2(3m− 2) ∴ 10m− 15 = 6m −4 ∴ 10m− 6m = ∗ 4m = ∗ 15 −4 = ∗ 11 ∗ ∴ Reqd Value m = 11 4, as before!Let `1/x = u and 1/y = v` Then, equations (1) and (2) become u v = `4/3 and u 2v = 10/3` Adding, We have 9v = 6 ⇒ v = `6/9 = 2/3` ⇒ `1/y = 2/3 ` ⇒ y = `3/2` Substituting y = `3/2` in (1), We have `1/x 2/3 = 4/3` ⇒ `1/x = 6/3 = 2` ⇒ x = `1/2` Hence, x = `1/2 and y = 3/2` Concept Equations Reducible to Linear
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Yu x = xu y (3) yu x xu y = 0 (4) x 2 u y y 2 u x = 0 pre calculous use the properties of logarithms to write the expression as a sum or difference of logarithms or multiples of logarithms show your work ln x^2/y^3 Assuming x,y AND z are positive use properties of logariths to write theSolution The equation is Substitute x=2 in the equation, Therefore, the value of y is 3 So, Option b is correct #Learn more If x=3 and y=4 find x23xyy2 brainlyin/question/
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And by the symmetry of uin xand y, u yy= 2x2 22y (x2 y2)2 Clearly then u xx u yy= 0 in this case If u= arctan(y=x), then by the chain rule again u x= 1 1 (y x) 2 y x 2 = 2y x y 2) u xx= (x y2)(0) ( y)(2x) (x y2Factor 2x^23xyy^2 2x2 3xy y2 2 x 2 3 x y y 2 For a polynomial of the form ax2 bx c a x 2 b x c, rewrite the middle term as a sum of two terms whose product is a⋅c = 2⋅1 = 2 a ⋅ c = 2 ⋅ 1 = 2 and whose sum is b = 3 b = 3 Tap for more steps Reorder terms 2 x 2 y 2 3 x y 2 x 2 y 2 3 x y Reorder y 2 y 2 and 3Calculus Solve for y x^23xyy^2=1 x2 − 3xy y2 = 1 x 2 3 x y y 2 = 1 Move 1 1 to the left side of the equation by subtracting it from both sides x2 − 3xyy2 − 1 = 0 x 2 3 x y y 2 1 = 0 Use the quadratic formula to find the solutions −b±√b2 −4(ac) 2a b ± b 2 4 ( a c) 2 a Substitute the values a = 1 a = 1, b
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Mumbai University > First Year Engineering > Sem 1 > Applied Maths 1 Marks 6 M Year May 15U = x /y x = uy dx/dy = u for a full explanation x= uy xvx = u(yvy) where v is a tiny fraction ( I can't do the delta symbol) xvx = uyuvy xvx x = uyuvy uy vx = uvy vx/vy = u Limit as v > 0, dx/dy = u Use the power rule, dy/dx = nx^ (x1) d y d x = n x x − 1 , on the first term 2x (3d (xy))/dx (d (y^2))/dx= (d (0))/dx 2 x 3 d ( x y) d x d ( y 2) d x = d ( 0) d x Use the product rule, (d (xy))/dx= dx/dxyxdy/dx = y xdy/dx d ( x y) d x = d x d x y x d y d x = y x d y d x
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If x^2 xyy^3 = 0, then, in terms of x and y, dy/dx = 2x y/x 3y^2 x 3y^2/2x y 2x/1 3y^2 2x/x 3y^2 2x y/x 3y^2 1 Get more help from Chegg Solve it with our calculus problem solver and calculatorR x as boundaries the curves x2 −y2 = 1, x2 −y2 = 4, y = 0, y = x/2 Solution Since the boundaries of the region are contour curves of x2 −y2 and y/x , and the integrand is y/x, this suggests making the change of variable (23) u = x 2 −y 2 , v = y x We will try to get through without solving these backwards for x, y in terms of u, vAll equations of the form a x 2 b x c = 0 can be solved using the quadratic formula 2 a − b ± b 2 − 4 a c The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction yx^ {2}y^ {2}xu=0 y x 2 y 2 x − u = 0 This equation is
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Mathx=s3t/math math\frac{d}{ds}x^2=\frac{d}{ds}(s3t)^2/math math=2(s3t)/math math\frac{d^2}{ds^2}x^2=\frac{d}{ds}(\frac{d}{ds}x^2)/math math Consider the family of circles x^2y^22x2lambda8=0 passing through two fixed points n dB Then the distance between the points n dB is_____ This browser does not support the video element 31 k 18 k Answer Step by step solution by experts to help you in doubt clearance & scoring excellent marks in examsQuestion if x=1 and y=2 what is the value of expression 2x^3 3xy Answer by Alan3354() ( Show Source ) You can put this solution on YOUR website!
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of pretzels Each bag contains 7 servings, She is planning to have 30 guests at her party To determine how many bags she needs to buy, she sets up an inequality and solves it tex7x\geq 30\\x\geq 43/tex She determines that she needs to buy at least 4 bags of pretzels to ensure that anyone has at least one serving1 A harmonic function is analytic if it satisfies the Laplace equation If 𝑢(𝑥,𝑦) = 2𝑥2 −2𝑦2 4𝑥𝑦 is a harmonic function, then its conjugate harmonic function 𝑣(𝑥,𝑦) is (A) 4𝑥𝑦 −2𝑥2 2𝑦2 constant (B) 4𝑦2 −4𝑥𝑦 constant 2𝑥2 −2𝑦2 𝑥𝑦 constant (D) −4𝑥𝑦 2𝑦2 −2𝑥2 constant Discuss below to share your Correction (after missing a sign) As kobe pointed out, the original DE is $$ (x^2y^2)y'2xy=0, $$ which as equation for a vector field reads $$ (x^2y^2)\,dy2xy\,dx=0\iff Im(\bar z^2\,dz)=0\text{ with } z=xiy $$ From the complex interpretation it is directly visible that this is not integrable, for that it would have to be an expression
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If V = z e a x b y and z is a homogeneous function of degree n in x and y, prove that x ∂ x ∂ V y ∂ y ∂ V = (a x b y n) V View solution View moreSuppose f(x,y)=3/(x^2y^2)and u is the unit vector in the direction of 〈1,2〉 Then, (a) ∇f(x,y)= (b) ∇f(2,2)= (c) fu(2,2)=Duf(2,2)= $$\int_{0}^a\int_{0}^\sqrt{a^2y^2} (2x3xy)\,dx\,dy$$ In this question I want to solve it by using polar $\begingroup$ That's i know but in some cases other than circle i have to do more complex substitutions then how will i determine?
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X3yx2y2xy=0 Four solutions were found x = 2 x = 1 y = 0 x = 0 Step by step solution Step 1 Step 2 Pulling out like terms 21 Pull out like factors x3y x2y Consider x^ {2}2xy3y^ {2} as a polynomial over variable x Find one factor of the form x^ {k}m, where x^ {k} divides the monomial with the highest power x^ {2} and mIf u = x 2 − y 2, v = 2 x y a n d z = f ( u, v) prove the following written 50 years ago by shailymishra30 ♦ 330 modified 14 months ago by sanketshingote ♦ 570 ( ∂ z ∂ x) 2 ( ∂ z ∂ y) 2 = 4 u 2 v 2 ( ∂ z ∂ u) 2 ( ∂ z ∂ v) 2 partial differentiation ADD COMMENT Welcome to Sarthaks eConnect A unique platform where students can interact with teachers/experts/students to get solutions to their queries Students (upto class 102) preparing for All Government Exams, CBSE Board Exam, ICSE Board Exam, State Board Exam, JEE (MainsAdvance) and NEET can ask questions from any subject and get quick answers by
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