This is a Calculus 3 problem. Please explain c) clearly, no cursive writing. The function \( f(x, y) \) is defined as:\[f(x, y) = \frac{-x^2 y^2}{x^2 + y^2}\]This expression represents a two-variable function. The numerator \(-x^2 y^2\) indicates that the function value depends on the square of both \(x\) and \(y\), multiplied together and made negative. The denominator \(x^2 + y^2\) represents a sum of squares, which is often associated with radial symmetry from the origin in polar coordinates. The function might exhibit interesting behaviors, such as rotational symmetry, and is undefined when both \(x\) and \(y\) are zero due to the division by zero. c) Find the second degree polynomial approximation to \( f \) at \( (a, b) = (1, -2) \)\[ T_2(x, y) = \] [blank box for student input]

Name: This is a Calculus 3 problem. Please explain c) clearly, no cursive wr
Uploaded: 2024-03-05T11:52:21.000Z
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Description: This is a Calculus 3 problem. Please explain c) clearly, no cursive writing. The function \( f(x, y) \) is defined as:\[f(x, y) = \frac{-x^2 y^2}{x^2 + y^2}\]This expression represents a two-variable function. The numerator \(-x^2 y^2\) indicates tha

Given :- The second degree polynomial approximation to f at a,b = 1,-2 f x,y = -x2y2x2 +y2 To determine :- f x,y = -x2y2x2 +y2 f 1,-2 = -12×2212 +22 =-45 Now, differentiating with respect to x and y, fx x,y = -y2x2 +y2.2x-x2.2xx2 +y22 fx x,y = -y22x3 +2xy2-2x3x2 +y22 fxx,y = -2xy4x2 +y22fx1,-2 = -2×1×-2412+-222 =-3225 fy x,y = -x2x2 +y2.2y-y2.2yx2 +y22 fy x,y = -x22y3 +2x2y-2y3x2 +y22 fyx,y = -2x4yx2 +y22fy1,-2 = -2×-2×1412+-222 =425fxx x,y = -2y4x2 +y22.1-x.2x2 +y2.2xx2 +y24 fxx x,y = -2y4x2 +y2x2 +y2-4x2x2 +y24 fxxx,y = -2y4y2 -3x2x2 +y23fxx1,-2 = -2×-24×-22 -3×1212+-224 =-32125 fyy x,y = -2x4x2 +y22.1-y.2x2 +y2.2yx2 +y24 fyy x,y = -2x4x2 +y2y2 +x2-4y2x2 +y24 fyyx,y = -2x4x2 -3y2x2 +y23fyy1,-2 = -2×14×12 -3×-2212+-223 =22125 fxy x,y = -2xx2 +y22.4y3-y4.2x2 +y2.2yx2 +y24 fxy x,y = -2xx2 +y2y2 +x2.4y3-4y5x2 +y24 fxyx,y = -8x3 y3x2 +y23fxy1,-2 = -8×13×-2312+-223 =64125 By Taylor's Theorem, fx,y =fa,b +x-a fxa,b + y-b fy a,b +12! x-a2 fxxa,b +2x-a y-b fxy a,b +y-b2 fyya,b fx,y =f1,-2 +x-1 fx1,-2 + y+2 fy 1,-2 +12! x-12 fxx1,-2 +2x-1 y+2 fxy 1,-2 +y+22 fyy1,-2 fx,y =-45+x-1 ×-3225+ y+2 ×425 +12! x-12 ×-32125 +2x-1 y+2 × 64125+y+22 ×22125fx,y =-45-3225 x-1 +425 y+2 +12! x-12 ×-32125 +2x-1 y+2 × 64125+y+22 ×22125