By applying the numbers to the above equation,

\(\nabla f(x,y) = (2x,-2y)\).

\(\nabla f (2,0) = (4,0), ∇f(0,2) = ⟨0,−4⟩ \)

$$∇f(2,2) = ⟨4,−4⟩, ∇f(2,1) = ⟨4,−2⟩$$

$$∇f (−3, 2) = ⟨−3, −4⟩, ∇f (−2, −4) = ⟨−4, 8⟩$$

$$∇f(0,0) = ⟨0,0⟩$$

According to the definition, $$\nabla f(x,y) = (fx(x,y),fy(x,y)$$. Therefore \(\nabla f(x,y) = (2x,-2y)\)

$$fx(x,y) = sin(x-y)$$

(a) \(fx(x,y) = cos(x-y)\). When \(fx(-4,-4) = cos(0) = 1\)

(b) We need to use chain rule for fy(x,y) \(fy(x,y) = cos(x-y)*(-1) = -cos(x-y)\). When \(fy(-4,-4) =-cos(0) = -1\)

For \(CT\), we think S and D as a constant.

So \(CT = 4.6-0.11T+0.00087T^2-0.01S+0.35\)

For \(CS\), we think T and D as a constant. Therefore, \(CS = 1.34-0.01T\)

For \(CD\), we treat T and S as a constant.

So \(CD=0.016\) since everything becomes zero except \(0.016D\)

The scalar equation of the plane is \(2(x-0)+-1(y-2)+1(z-4) =0\) with \(n=<2,-1,1>\) and \(p1 = (0,2,4)\).

To find a parametric equations for the line \(L\), we need to know what the point \(P = (x0,y0,z0)\) and the vector \(v = <a,b,c>\). \(P = OP1 = <1,2,-1>\) and \(v = P2P1 = <3,1,1>\). Thus the parametric equations of the line \(L\) are \(x(t) = x0+at = 1+3t, y(t) = y0+bt = 2+t, z(t) = z0+ct = -1 +t\)

There is also another way to solve the 3X3 matrix. It is called a Basketweave Method. It is more easier to memorize. You can easily remember as a fishing net.