Solutions to Quiz 2

1. True or False? If the statement is true supply a \underline{short} explanation. If it is false provide a \underline{counterexample}. (2pts each)

a) If gradf(a, b)=gradg(a, b) for all points (a, b), then f = g.

False, if f(x, y) = g(x, y) + c, for any nonzero constant c, then gradf =gradg for all points (a, b), but the functions are different.

b) There is no function f with ∇f = ⟨x^{2} − y^{2}, −2xy⟩.

False, f(x, y) = x^{3}/3 − xy^{2} has the given gradient.

c) fu⃗(a, b) = f_{x}(a, b)u_{1} + f_{y}(a, b)u_{2} for all vectors u⃗ = ⟨u_{1}, u_{2}⟩.

False, in general this is only true for unit vectors. A counterexample is f(x, y) = 2x; then f_{3i⃗}(a, b) = f_{x}(a, b) = 2 and not 3f_{x}(a, b) = 6.

d) If f is differentiable at (a, b), f_{x}(a, b) = 0 and f_{y}(a, b) = 0, then all directional derivatives of f at (a, b) are zero.

This is true, since fu⃗(a, b) = u_{1}f_{x}(a, b) + u_{2}f_{y}(a, b) = 0 for all unit vectors u⃗ = ⟨u_{1}, u_{2}⟩.

e) gradf · u⃗ = 0 where u⃗ is tangent to the level curves of f at the point where gradf is computed.

This is true, since gradf is perpendicular to the level curve at (a, b) and u⃗ is tangent to the level curve at (a, b); thus gradf and u⃗ are perpendicular to each other and their dot product has to be 0.

2. Let V (x, y, z) represent the \underline{volume} of a rectangular box with length x, width y and height z.

a) Compute ∂V/∂x, ∂V/∂y and ∂V/∂z. (2pts)

V = xyz, so ∂V/∂x = yz, ∂V/∂y = xz and ∂V/∂z = xy

Now suppose that the dimensions of the box are x = 1cm, y = 3cm and z = 5cm at the instant t = 0. Also at the instant t = 0 both x and y are increasing at a rate of 2cm/sec while z decreases at 4cm/sec.

b) Determine the rate at which the volume is changing at t = 0. (3pts)

Since we look for dV/dt, we have to use the chain rule. The tree has V on top and then branches out to x, y and z and each one of those only depends on t:

\frac{dV}{dt} = \frac{∂V}{∂x} \frac{dx}{dt} + \frac{∂V}{∂y} \frac{dy}{dt} + \frac{∂V}{∂z} \frac{dz}{dt} = yz \frac{dx}{dt} + xz \frac{dy}{dt} + xy \frac{dx}{dt}

= (yz)(2) + (xz)(2) + (xy)(−4) = 15 · 2 + 5 · 2 + 3 · (−4) = 28.

3. Suppose that the quadratic approximation to the function f(x, y) near the origin is

of the form

f(x, y) ≈ A + Bx + Cy + Dx^{2} + Exy + F y^{2}.

If the level curves of the function are as below, what are the signs (positive, negative, or zero) of A, B, C, D, E and F? Explain. (5pts)

A = f(0, 0) = 0 (at the origin z = 0), B = f_{x}(0, 0) < 0 (f decreases in xdirection), C = f_{y}(0, 0) < 0 (f decreases in y-direction), D = .5f_{xx}(0, 0) > 0 (f is concave up in x-direction), D = .5f_{yy}(0, 0) > 0 (f is concave up in ydirection), F = f_{xy}(0, 0) > 0 (the slope of f in x-direction gets bigger (smaller negative number) as we move in positive y-direction