SOLUTIONS TO PRACTICE EXAM 1

1. The following table of values was generated by a function f(x, y).

x\y  -1  0  1^{-1  -0.5  0.1  0.7}^{0  3.6  4.2  4.8} 1  7.7  8.3  8.9

a) Can f be a linear function? If so find its equation, if not explain why. It is a linear function, since all differences in x-direction are equal (4.1) and

all differences in y-direction are equal (0.6). So we also get that δz/δx = 4.1 and δz/δy = 0.6. Therefore f(x, y) = 4.1x + 0.6y + c and using f(0, 0) = 4.2 we get c = 4.2, so that f(x, y) = 4.1x + 0.6y + 4.2.

b) How do your answers to part a) change if 5 is added to every entry in the table?

The differences don’t change, so it is still a plane; the only thing that will change is c = 9.2, since the plane is lifted by 5: f(x, y) = 4.1x + 0.6y + 9.2.

2. For each of the following functions, decide whether it could be a bowl, a plate or neither. Consider a plate to be any fairly flat surface and a bowl to be anything that could hold water, assuming the positive z-axis is up.

a) z = x^{2} + y^{2}. Bowl.

b) z = 1 − x^{2} − y^{2}. Neither (Bowl opening downwards).

c) z = 3. Plate.

3. For which vectors u⃗, v⃗ and w⃗ are the following identities true? You don’t have to give any algebraic proofs, but provide brief explanations why you think your answer is true.

a) u⃗ · (v⃗ × w⃗) = (u⃗ · v⃗) × w⃗

For no choice of u⃗, v⃗, and w⃗, since u⃗ · v⃗ is a scalar and you can’t cross a scalar with a vector.

b) u⃗ · (v⃗ + w⃗) = (u⃗ · v⃗) + (u⃗ · w⃗)

For all u⃗, v⃗, and w⃗, since the dotproduct is distributive.

c) u⃗ × v⃗ = v⃗ × u⃗ u⃗ × v⃗ = −v⃗ × u⃗, so this is only true if v⃗ × u⃗ = 0. That means we need u⃗ = 0,

v⃗ = 0 or u⃗ and v⃗ are parallel. d) (u⃗ × u⃗) × v⃗ = u⃗ × (u⃗ × v⃗) u⃗ × u⃗ = 0, so the left side is 0. If u⃗ × v⃗ = 0, then the right side is also 0.

Otherwise u⃗ × v⃗ is a vector perpendicular to u⃗ and the right side will not be 0. So the identity is true if u⃗ = 0, v⃗ = 0 or u⃗ and v⃗ are parallel, like in c).

4. Same as Quiz 1#2.

5. Let f(x, y) = \frac{1}{3}x^{3} − x^{2}y + xy^{2} + 2y.

a) Find the plane tangent to f at the point (0,1,2).

f_{x}(x, y) = x^{2} − 2xy + y^{2} and f_{y}(x, y) = −x^{2} + 2xy + 2, so that f(0, 1) = 2, f_{x}(0, 1) = 1 and f_{y}(0, 1) = 2. Using the equation for the tangent plane we get: z = 1(x − 0) + 2(y − 1) + 2 or z = x + 2y.

b) Find a vector perpendicular to the graph of f at (0,1,2).

A vector perpendicular to the graph is the same as a vector perpendicular to the tangent plane. So the normal vector < 1, 2, −1 > of the plane is a valid answer to the question.