Math 214 Advanced Calculus II Midterm Exam I

March 17, 2001 Ali Sinan Sertöz

1) Give an example of a function f : ℝ^{2} → ℝ such that f_{x} and f_{y} exist everywhere on ℝ^{2} but the function f itself is not differentiable at (0, 0). Prove however that if you further assume that f_{x} and f_{y} are continuous at the origin, then f has to be differentiable there.

Solution: Such an example is given in Example 2 on page 291. The proof that continuity of the partials implying differentiability is the content of Theorem 6.9 on page 292.

2) \begin{matrix}Let f : ℝ^{3} → ℝ be a C^{2} function. Let a, b and c be points in ℝ^{3} and \\ define H(x) = (\frac{∂^{2}f}{∂x_{i}∂x_{j}} (x))_{1≤i,j≤3} for x ∈ ℝ^{3}. We have the following data: \\ ∇f(a) = 0, ∇f(b) = 0, ∇f(c) = 0 and \\ H(a) = ( \begin{matrix}−1 & 0 & 0 \\ 0 & −13 & 5 \\ 0 & 5 & −2 \end{matrix} ) , & H(b) = ( \begin{matrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & −2 \end{matrix} ) , H(c) = ( \begin{matrix}3 & 2 & 1 \\ 2 & 3 & 0 \\ 1 & 0 & 1 \end{matrix} ) . \\ With this much information can you decide which of these points a, b \\ and c are min/max or saddle points for f? If so, classify these points \\ accordingly. \end{matrix}

Solution: The criteria to make such decisions is given in Theorem 6.22 on page 321. For the matrix H = (h_{ij})_{1≤i,j≤3} define H_{1} = h_{11}, H_{2} = h_{11}h_{22} − h_{12}h_{21} and H_{3} = det(H). For the first matrix we have H_{1} = −1, H_{2} = 13, H_{3} = −1. Then this is a negative definite matrix and satisfies part (ii) of Theorem 6.22. Thus the point a gives local maximum for f. The second matrix clearly satisfies the condition (iii) of this theorem so the point b gives a saddle point. For the third matrix H_{1} = 3, H_{2} = 5, H_{3} = 2. This is then a positive definite matrix and satisfies condition (i) of the theorem and thus c gives a local minimum point for f.

3) Consider the functions f : ℝ^{3} → ℝ^{2} and g : ℝ^{2} → ℝ^{3} given by f(u, v, w) = (u^{3} + v^{2} + 2u + 2v + w, u^{7} + v^{6} + w^{8} + u + 2v − w), g(x, y) = (x^{5} + xy + y^{3} + x + 2y, x^{5}y^{7} + y^{7} + 3x + 4y, x^{21} + y^{30} + 3x + 2y). Calculate D(f ∘ g)(0, 0).

Solution: Note that g(0, 0) = (0, 0, 0). Let f = (f_{1}, f_{2}) and g = (g_{1}, g_{2}, g_{3}).

Then

Df(0, 0, 0) = ( \begin{matrix}\frac{∂f_{1}}{∂u} (0, 0, 0) & \frac{∂f_{1}}{∂v} (0, 0, 0) & \frac{∂f_{1}}{∂w} (0, 0, 0) \\ \frac{∂f_{2}}{∂u} (0, 0, 0) & \frac{∂f_{2}}{∂v} (0, 0, 0) & \frac{∂f_{2}}{∂w} (0, 0, 0) \end{matrix} )

= ( \begin{matrix}2 & 2 & 1 \\ 1 & 2 & −1 \end{matrix} ) = A

and

Dg(0, 0) = ( \begin{matrix}\frac{∂g_{1}}{∂x} (0, 0) & \frac{∂g_{1}}{∂y} (0, 0) \\ \frac{∂g_{2}}{∂x} (0, 0) & \frac{∂g_{2}}{∂y} (0, 0) \\ \frac{∂g_{3}}{∂x} (0, 0) & \frac{∂g_{3}}{∂y} (0, 0) \end{matrix} ) = ( \begin{matrix}1 & 2 \\ 3 & 4 \\ 3 & 2 \end{matrix} ) = B.

Now D(f ∘ g)(0, 0) = AB = ( \begin{matrix}11 & 14 \\ 4 & 8 \end{matrix} ) .

The theory behind this is in Theorem 6.13 on page 298.

4) Let F (y) = ∫_{0}^{∞} e^{−x^{2}y^{2}}dx, y ∈ [1, 100]. Calculate F′(1) and justify your answer. (You may want to be reminded that ∫_{0}^{∞} e^{−t^{2}}dt = \sqrt{π}/2)

Solution: First assume that F′(y) converges uniformly on [1, 100]. Then we can apply Theorem 6.6 on page 284 and differentiate under the integral sign:

F′(1) = ∫_{0}^{∞}(−2yx^{2}e^{−x^{2}y^{2}})|_{y=1}dx
= −2 ∫_{0}^{∞} x^{2}e^{−x^{2}}dx
= xe^{−x^{2}}|_{x=0}^{x=∞} − ∫_{0}^{∞} e^{−x^{2}}dx
= −\frac{\sqrt{π}}{2},

where we used integration by parts with u = x and dv = xe^{−x^{2}}dx for integration and used the hint in the last line. Now for the justification observe that when y ∈ [1, 100], | − 2yx^{2}e^{−x^{2}y^{2}}| = 2|yx^{2}| |e^{−x^{2}y^{2}}| ≤ 2|100x^{2}| |e^{−x^{2}}| = 200x^{2}e^{−x^{2}} And this function is improperly integrable on [0, ∞) as we showed above. Thus by Weierstrass M-Test, on page 283, the integral F′(y) converges uniformly. This satisfies the main requirement of Theorem 6.6 on page 284 and differentiation under the integral sign is justified.

5) \begin{matrix}Prove that there exist functions u(x, y), v(x, y) and w(x, y) and r > 0 \\ such that u, v, w are continuously differentiable and satisfy the equations \\ u^{5} + xv^{2} − y + w & = & 0 \\ v^{5} + yu^{2} − x + w & = & 0 \\ w^{4} + y^{5} − x^{4} & = & 1 \\ on B_{r}(1, 1), and u(1, 1) = 1, v(1, 1) = 1, w(1, 1) = −1. \end{matrix} Solution: This is Exercise 5 on page 318, and I solved it in detail in class. Here we basically use the Implicit Function Theorem on page 315.