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Exercise 8.5 · Q1

Q.If w(x,y)=x3−3xy+2y2, x,y∈Rw(x,y)=x^3-3xy+2y^2,\ x,y\in\mathbb R, find the linear approximation for ww at (1,−1)(1,-1).

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✓ Free question

Evaluate w,wx,wyw,w_x,w_y at (1,−1)(1,-1) and substitute into L(x,y)=w(x0,y0)+wx(x0,y0)(x−x0)+wy(x0,y0)(y−y0)L(x,y)=w(x_0,y_0)+w_x(x_0,y_0)(x-x_0)+w_y(x_0,y_0)(y-y_0).

Step 1. Evaluate ww at (1,−1)(1,-1). w(1,−1)=13−3(1)(−1)+2(−1)2=1+3+2=6w(1,-1)=1^3-3(1)(-1)+2(-1)^2=1+3+2=6.

Step 2. Partial derivatives. wx=3x2−3yw_x=3x^2-3y, so wx(1,−1)=3(1)−3(−1)=3+3=6w_x(1,-1)=3(1)-3(-1)=3+3=6.  wy=−3x+4y\ w_y=-3x+4y, so wy(1,−1)=−3(1)+4(−1)=−3−4=−7w_y(1,-1)=-3(1)+4(-1)=-3-4=-7.

Step 3. Substitute into the linear-approximation formula (12).

L(x,y)=6+6(x−1)+(−7)(y−(−1))=6+6x−6−7y−7=6x−7y−7.L(x,y) = 6 + 6(x-1) + (-7)(y-(-1)) = 6+6x-6-7y-7 = 6x-7y-7.

✓Final answer

The linear approximation is L(x,y)=6x−7y−7L(x,y)=\boxed{6x-7y-7}

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