• 1. The (unique) critical point of the function is (b) (2,-1).

  Reason: Set and , that is

  

  Solving this linear system, we get a unique solution x=2, y=-1, which gives us the critical point (2,-1).

• 2. The distance of the point (3,0,0) to the paraboloid is (c) .

  Reason: Let (a,b, c) be the point on the paraboloid closest to (3,0,0). The square of the distance from a point (x,y,z) on the paraboloid to the point (3,0,0) is given by . So (a,b) is a point at which the function

  

  is minimized. We compute , and set them equal to zero to obtain

  

  From the second equation we have . But is always positive. So y=0. Substituting y=0 in the first equation, we get . The only real root of this equation is x=1. Substituting x=1 and y=0 into , we have z=1. Thus (1,0,1) is the point on the paraboloid closest to (3,0,0). The distance between these points is .

• 3. The partial derivatives of w=f(x,y) are given as

  

  Let be the polar coordinates. Then is (e) 0, and is (a) 1.

  Reason: Since and , we have , , and . In terms of polar coordinates, we have and . So, by the chain rule, we have

  

• 4. The gradient vector of the function at P=(1, 1, 2) is (d) (6,-2,-8). The directional derivative at P with is (e) (<-- Replace the 4 by 6 here).

  Reason: , which is (6, -2, -8) at (1,1,2). Also, a unit vector in the direction of is given by u where

  

  So, ( <-- Actually, the answer is 6 over the square root of 14).

• 5. Let be the gradient of a function f at a point P and let S be the level surface of f through P. Then

  (c) is a normal vector to S at P.

  A comment on (e): we may correct this statement by saying that (not ) is in the direction of the steepest descent.

• 6. The plane tangent to the level surface f=-3 of the scalar field at P=(1,1,2) is (b) 4(x-1)-(y-1)-4(z-2)=0.

  Reason: The normal direction of the plane is given by the vector at P, that is (4,-1, -4). Thus the equation of the tangent plane is

  

• 7. (Bonus) Consider two scalar fields f(x,y,z) and g(x, y,z). Suppose that f and g are functionally related in the sense that there is a function F(u,v) of two variables such that at each (u,v) and F(f(x,y,z),g(x,y,z))=0 for all (x,y,z). Prove that and are linearly dependent (that is, one is a scalar multiple of the other) at each (x,y,z).

  Proof. Taking partial derivatives on both sides of F(f(x,y,z), g(x, y,z))=0 and applying the chain rule, we have

  

  We can rewrite the above three identities in the vector form as . This shows that and are linearly dependent, since tells us that at least one of , is nonzero at each point.

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