Mathematics 241, section 1 – Multivariable Calculus Solutions for Midterm Exam 2 November 5, 2010

I. All parts of this problem refer to the vector field

F(x, y) = (x^{2} − 2x, y^{2} + 2y).

A. (5) Find all critical points of F(x, y).

Solution: The critical points are the solutions of x^{2} − 2x = 0 and y^{2} + 2y = 0, so x = 0, 2 and y = 0, −2. This gives four critical points all together: (0, 0), (2, 0), (0, −2) and (2, −2).

B. (10) There are two vector fields plotted on the back of this sheet. Say which one shows F(x, y) and use that plot to classify each of the critical points as a source, sink, saddle, or center.

Solution: The correct plot is Vector Field Two, since Vector Field One apparently has only two critical points. Consulting the plot, we see that (0, 0) and (2, −2) are saddle points, (2, 0) is a source, and (0, −2) is a sink.

C. (10) Show that α(t) = (\frac{2}{1−2e^{2t}} , \frac{−2}{1−3e^{−2t}} ) is a flow line of F.

Solution: We must show that α′(t) = F(α(t)). The left side is

α′(t) = (\frac{8e^{2t}}{(1 − 2e^{2t})^{2}} , \frac{12e^{−2t}}{(1 − 3e^{−2t})^{2}} ) .

The right side is

F(α(t)) = ((\frac{2}{1 − 2e^{2t}} )^{2} − 2 (\frac{2}{1 − 2e^{2t}} ) , (\frac{−2}{1 − 3e^{−2t}} )^{2} + 2 (\frac{−2}{1 − 3e^{−2t}} ))

= (\frac{4 − 4(1 − 2e^{2t})}{(1 − 2e^{2t})^{2}}, \frac{4 − 4(1 − 3e^{−2t})}{(1 − 3e^{−2t})^{2}})
= (\frac{8e^{2t}}{(1 − 2e^{2t})^{2}} , \frac{12e^{−2t}}{(1 − 3e^{−2t})^{2}} ) .

This shows what we wanted.

D. (10) Show that any f(x, y) = \frac{x^{3}}{3} − x^{2} + \frac{y^{3}}{3} + y^{2} + c satisfies ∇f = F. What are the critical points of such an f? Classify each of them as a local maximum, local minimum, or saddle point. (Don’t “start over” here; use previous parts of this question as appropriate.)

Solutions: For any such f, we have \frac{∂f}{∂x} = x^{2} − 2x and \frac{∂f}{∂y} = y^{2} + 2y. So ∇f(x, y) = (x^{2} − 2x, y^{2} + 2y) = F(x, y). By properties of gradient vector fields, we know that f(x, y) has a local minimum at (2, 0), a local maximum at (0, −2), and saddle points at (0, 0) and (2, −2).

II. All parts of this problem refer to the function

f(x, y) = \frac{x^{3} − 3xy^{2}}{x^{2} + y^{2}} if (x, y) ≠ (0, 0) and f(0, 0) = 0.

A. (10) Compute \frac{∂f}{∂x}(x, y) and \frac{∂f}{∂y} (x, y) for (x, y) ≠ (0, 0).

Solution: By the quotient rule

\frac{∂f}{∂x} = \frac{(x^{2} + y^{2})(3x^{2} − 3y^{2}) − (x^{3} − 3xy^{2})(2x)}{(x^{2} + y^{2})^{2}}
= \frac{x^{4} + 6x^{2}y^{2} − 3y^{4}}{(x^{2} + y^{2})^{2}} .

and

\frac{∂f}{∂y} = \frac{(x^{2} + y^{2})(−6xy) − (x^{3} − 3xy^{2})(2y)}{(x^{2} + y^{2})^{2}}
= \frac{−8x^{3}y}{(x^{2} + y^{2})^{2}} .

B. (10) Do \frac{∂f}{∂x}(0, 0) and \frac{∂f}{∂y} (0, 0) exist? If so, find them; if not say why not.

Solution: The formulas from part A are not defined at (0, 0). However, this just says we need to use the limit definition to see whether the partial derivatives exist. We find

\frac{∂f}{∂x}(0, 0) = lim_{h→0} \frac{f(h, 0) − f(0, 0)}{h}
= lim_{h→0} \frac{h^{3}/h^{2} − 0}{h}
= 1

and

\frac{∂f}{∂y} (0, 0) = lim_{h→0} \frac{f(0, h) − f(0, 0)}{h}
= lim_{h→0} \frac{0/h^{2} − 0}{h}
= 0.

C. (10) Let m be arbitrary and compute lim_{t→0} f(t, mt) (the limit of the value of f along the line through the origin in the direction of the vector (1, m)). Does this prove anything about the existence of the limit lim_{(x,y)→(0,0)} f(x, y)? Explain your answer using the definition of the statement lim_{(x,y)→(0,0)} f(x, y) = L.

Solution: We have

f(t, mt) = \frac{(1 − 3m^{2})t^{3}}{(1 + m^{2})t^{2}} = (\frac{1 − 3m^{2}}{1 + m^{2}} ) t.

So as t → 0, the limit is zero. This does not show that lim_{(x,y)→(0,0)} f(x, y) = 0, though. For that, we would have to show that for all ε > 0, there exists r > 0 such that |f(x, y)| < ε for all (x, y) in the open ball B_{r}(0, 0), except possibly (0, 0).

D. (10 Extra Credit) Is f(x, y) differentiable at (0, 0)? Explain your answer.

Solution: The answer is no. To show differentiability, we would need to show that

lim_{(x,y)→(0,0)} \frac{f(x, y) − ℓ(x, y)}{\sqrt{x^{2} + y^{2}}} = 0,

where ℓ(x, y) is the linear approximation at (0, 0). Here from part B we know ℓ(x, y) = x, so

\frac{f(x, y) − ℓ(x, y)}{\sqrt{x^{2} + y^{2}}} = \frac{−4xy^{2}}{(x^{2} + y^{2})^{3/2}} .

If you let (x, y) → (0, 0) along the x or y axes, this is going to 0. However, along the line y = x this goes to the value −4/2^{3/2} ≠ 0. Hence the required limit does not exist. Since it is not equal to zero, f(x, y) is not differentiable.

III. All parts of this problem refer to f(x, y) = 4x^{2} − y^{2}.

A. (15) Sketch the contours of f(x, y) for c = −1, 0, 1 on the same set of axes.

Solution:

(These are hyperbolas; the c = −1 contour opens up and down; the c = 0 contour is the union of the two lines y = ±2x; the c = 1 contour opens left and right.)

B. (10) Consider

S = {(x, y) ∈ ℝ^{2} | −1 < f(x, y) < 0}.

Is S open, closed, or neither? Why? (An intuitive explanation referring to the definition of open or closed is OK here.)

Solution: S is open since none of the boundary points of S (the points on the two contours for c = −1 and c = 0) are included in S. This means that for any (x_{0}, y_{0}) ∈ S, there exists a ball B_{r}(x_{0}, y_{0}) ⊂ S for some r > 0.

C. (10) For which unit vectors u is the directional derivative D_{u}f(1, 2) = 0?

Solution: The best way is to recognize that f(1, 2) = 4 − 4 = 0. So (1, 2) is on the contour consisting of the two lines y = ±2x. We want u in the direction of the line with the plus sign to get the directional derivative equal to zero, so

u = ± (\frac{1}{\sqrt{5}}, \frac{2}{\sqrt{5}}) .