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Exercise 8.5 · Q2

Q.Let z(x,y)=x2y+3xy4, x,y∈Rz(x,y)=x^2y+3xy^4,\ x,y\in\mathbb R. Find the linear approximation for zz at (2,−1)(2,-1).

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✓ Free question

Evaluate z,zx,zyz,z_x,z_y at (2,−1)(2,-1) and substitute into the linear-approximation formula.

Step 1. Evaluate zz at (2,−1)(2,-1). z(2,−1)=(2)2(−1)+3(2)(−1)4=4(−1)+3(2)(1)=−4+6=2z(2,-1)=(2)^2(-1)+3(2)(-1)^4=4(-1)+3(2)(1)=-4+6=2.

Step 2. Partial derivatives. zx=2xy+3y4z_x=2xy+3y^4, so zx(2,−1)=2(2)(−1)+3(1)=−4+3=−1z_x(2,-1)=2(2)(-1)+3(1)=-4+3=-1.

zy=x2+12xy3z_y=x^2+12xy^3, so zy(2,−1)=4+12(2)(−1)3=4+12(2)(−1)=4−24=−20z_y(2,-1)=4+12(2)(-1)^3=4+12(2)(-1)=4-24=-20.

Step 3. Substitute into L(x,y)=z(2,−1)+zx(2,−1)(x−2)+zy(2,−1)(y+1)L(x,y)=z(2,-1)+z_x(2,-1)(x-2)+z_y(2,-1)(y+1).

L(x,y)=2+(−1)(x−2)+(−20)(y+1)=2−x+2−20y−20=−x−20y−16.L(x,y) = 2+(-1)(x-2)+(-20)(y+1) = 2-x+2-20y-20 = -x-20y-16.

✓Final answer

The linear approximation is L(x,y)=−x−20y−16L(x,y)=\boxed{-x-20y-16}

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