Calculus II, Spring 2016

Midterm 2

May 10, 2016 Show all details.

1. (10 pts)True or False?

If f(x, y) is differentiable at (0, 0), then f(x, y) is continuous at (0, 0). Answer. True. (2 pts) Since f is differentiable at (0, 0), both f_{x}(0, 0) and f_{y}(0, 0) exist (2 pts) and f(x, y)−f(0, 0) = f_{x}(0, 0)x+f_{y}(0, 0)y +ϵ_{1}x+ϵ_{2}y where ϵ_{1}, ϵ_{2} → 0 as x, y → 0 (4 pts) ⇒ |f(x, y)−f(0, 0)| = |f_{x}(0, 0)||x|+|f_{y}(0, 0)||y|+|ϵ_{1}||x|+|ϵ_{2}||y| → 0 as (x, y) → (0, 0). (2pts)

2. (10 pts)Evaluate

\frac{d}{dy} ∫_{1}^{2+y^{2}} \frac{cos(xy)}{x}dx.

Answer. Let G(y, z) = ∫_{1}^{z} \frac{cos(xy)}{x} dx. (2 pts) Then we need to compute

(G(y, 2+y^{2}))′ = G_{y}(y, 2+y^{2})+G_{z}(y, 2+y^{2})·2y = − ∫_{1}^{2+y^{2}} sin(xy) dx+\frac{cos((2 + y^{2})y)}{2 + y^{2}}·2y (4 pts)

= \frac{1}{y}[cos((2 + y^{2})y) − cos y] + \frac{cos((2 + y^{2})y)}{2 + y^{2}} · 2y. (4pts)

3. (10 pts)Find the equation of plane normal to the following curve at (1, −1, 1)

\{ \begin{matrix}x^{2} + 2y^{2} + 3z^{2} = 6 \\ x + y + z = 1 \end{matrix}

Answer. First compute the two gradients at (1, −1, 1)

(2x, 4y, 6z)_{(1,−1,1)} = (2, −4, 6) (2 pts)

and

(1, 1, 1)|_{(1,−1,1)} = (1, 1, 1). (2 pts)

The normal vector of the plane is parallel to the outer product of these two gradients:

n = | \begin{matrix}i & j & k \\ 1 & 1 & 1 \\ 1 & −2 & 3 \end{matrix} | = (5, −2, −3). (2 pts)

Therefore, the equation of the plane is 5(x − 1) − 2(y + 1) − 3(z − 1) = 0. (4 pts)

4. (10 pts)Show that, for all a ∈ ℝ, the point (x_{0}, y_{0}) = (0, 0) is a critical point of the

function

f_{a}(x, y) = (a − 1)(x + y)^{2} + (a + 1)(x − 2y)^{2}.

For what values of a is the point (0, 0) a local minimum, a local maximum and a saddle point, respectively? Answer.

Method.1 For a ≥ 1, f_{a}(x, y) = (a − 1)(x + y)^{2} + (a + 1)(x − 2y)^{2} is a sum of two squares. And (0, 0) admits the minimum 0. (2 pts) For −1 < a < 1, f_{a}(x, y) = (a + 1)(x − 2y)^{2} − (1 − a)(x + y)^{2} is a difference of two squares. And (0, 0) is a saddle point since f_{a}(1, −1) = 9(a + 1) > 0 and f_{a}(2, 1) = −9(1 − a) < 0. (6 pts) For a ≤ −1, f_{a}(x, y) = −[(1 − a)(x + y)^{2} + (−1 − a)(x − 2y)^{2}] is a negative sum of two squares. And (0, 0) admits the maximum 0. (2 pts)

Method.2 First compute the gradient at (0, 0)

∇f_{a}(0, 0) = (2(a−1)(x+y)+2(a+1)(x−2y), 2(a−1)(x+y)+2(a+1)(x−2y)(−2))|_{(0,0)} = (0, 0),

and thus (0, 0) is a critical point. (2 pts)

Now compute the second derivatives (2 pts)

(f_{a})_{xx}(0, 0) = 2(a − 1) + 2(a + 1) = 4a,
(f_{a})_{yy}(0, 0) = 2(a − 1) + 8(a + 1) = 10a + 6 = 2(5a + 3),

and

(f_{a})_{xy}(0, 0) = 2(a − 1) − 4(a + 1) = −2(a + 3).

A local minimum occurs as (f_{a})_{xx} > 0 and (f_{a})_{xx}(f_{a})_{yy} − (f_{a})_{xy}^{2} > 0, i.e.

a > 0, 8a(5a + 3) − 4(a + 3)^{2} = 36(a^{2} − 1) > 0
⇔ a > 1. (2 pts)

A local maximum occurs as (f_{a})_{xx} < 0 and (f_{a})_{xx}(f_{a})_{yy} − (f_{a})_{xy}^{2} > 0, i.e.

a > 0, 8a(5a + 3) − 4(a + 3)^{2} = 36(a^{2} − 1) > 0
⇔ a < −1. (2 pts)

A saddle point occurs as (f_{a})_{xx}(f_{a})_{yy} − (f_{a})_{xy}^{2} < 0, i.e.

8a(5a + 3) − 4(a + 3)^{2} = 36(a^{2} − 1) < 0
⇔ −1 < a < 1. (2 pts)

For a = 1, f_{a}(x, y) = 2(x − 2y)^{2}. And thus (0, 0) admits a local minimum. For a = −1, f_{a}(x, y) = −2(x + y)^{2}. And this (0, 0) admits a local maximum.

Method.1

Figure 1: Method 1 for problem 5, The gradient analysis.

Getting the two equalities a = ±1 correctly: (extra 2 pts).

5. (10 pts)Find absolute maxima and minima of f(x, y) = x^{2} + xy + y^{2} − 6x + 2 on the rectangular 0 ≤ x ≤ 5, −3 ≤ y ≤ 3. Answer.

First find the gradient:

∇f(x, y) = (2x + y − 6, x + 2y). (2 pts)

Therefore one can plot the gradients as in figure 1. From the plot it is easy to see that f(4, −2) = −10 is indeed local and absolute minima. Moreover, the local maximum consists of the four corners. Upon comparing the values of f on the four corners, it follows that the absolute maxima is f(5, 3) = 21.

Method.2 First find the interior critical points:

∇f(x, y) = (2x+y−6, x+2y) = (0, 0) ⇒ (x, y) = (4, −2), f(4, −2) = −10. (2 pts)

For x = 0, f(0, y) = y^{2} + 2, −3 ≤ y ≤ 3. Compute (f(0, y))′ = 2y = 0 ⇒ y = 0, and thus we compare these values

f(0, 0) = 2, f(0, −3) = f(0, 3) = 11. (1 pt)

For x = 5, f(5, y) = y^{2} + 5y − 3, −3 ≤ y ≤ 3. Compute (f(5, y))′ = 2y + 5 = 0 ⇒ y = −5/2, and thus we compare these values

f(5, −3) = −9, f(5, 3) = 21, f(5, −5/2) = −\frac{37}{4}. (1 pt)

For y = −3, f(x, −3) = x^{2} − 9x + 11, 0 ≤ x ≤ 5. Compute (f(x, −3))′ = 2x − 9 = 0 ⇒ x = 9/2, and thus we compare these values

f(0, −3) = 11, f(9/2, −3) = −\frac{37}{4}, f(5, −3) = −9. (1 pt)

For y = 3, f(x, 3) = x^{2} − 3x + 11, 0 ≤ x ≤ 5. Compute (f(x, 3))′ = 2x − 3 = 0 ⇒ x = 3/2, and thus we compare these values

f(0, 3) = 11, f(3/2, 3) = \frac{35}{4}, f(5, 3) = 21. (1 pt)

Therefore, the absolute maxima is f(5, 3) = 21 and the absolute minimum is f(4, −2) = −10. (4 pts)

6. (10 pts)Give an example of a constraint optimization problem that, upon applying the method of Lagrangian multipliers, results in a system of 5 equations with 5 unknowns (and write down the equations). Need not solve it. Answer. Find the points on the intersection curve of these two surfaces x^{2}+2y^{2}+3z^{2} = 1 and z = 0 which admit the shortest distance from the origin. Let f(x, y, z) = x^{2} +y^{2} +z^{2}, g_{1}(x, y, z) = x^{2} +2y^{2} +3z^{2} −1, and g_{2}(x, y, z) = z. (5 pts) Then applying the method of Lagrangian multipliers result in this system

\{ \begin{matrix}2x = 2λ_{1}x \\ 2y = 4λ_{1}y \\ 2z = 6λ_{1}z + λ_{2} \\ x^{2} + 2y^{2} + 3z^{2} − 1 = 0 \\ z = 0. (5 pts) \end{matrix}

7. (10 pts)Use Taylor’s formula to find the quadratic approximation of f(x, y, z) =

\frac{1}{1 − x − y + z} near the origin.

Answer. Find all of the first and second derivatives first.

f_{x}(0, 0, 0) = f_{y}(0, 0, 0) = −f_{z}(0, 0, 0) = \frac{1}{(1 − x − y + z)^{2}} |_{(0,0,0)} = 1, (2 pts)

and

f_{xx}(0, 0, 0) = f_{yy}(0, 0, 0) = f_{xy}(0, 0, 0) = −f_{zz}(0, 0, 0) = −f_{yz}(0, 0, 0)

= −f_{zx}(0, 0, 0) = \frac{2}{(1 − x − y + z)^{3}} = 2. (2 pts)

Therefore, the gradratic approximation is

Q(x, y, z) = f(0, 0, 0) + f_{x}(0, 0, 0)x + f_{y}(0, 0, 0)y + f_{z}(0, 0, 0)z

+\frac{1}{2}[f_{xx}(0, 0, 0)x^{2} + f_{yy}(0, 0, 0)y^{2} + f_{zz}(0, 0, 0)z^{2}

+2f_{xy}(0, 0, 0)xy + 2f_{yz}(0, 0, 0)yz + 2f_{zx}(0, 0, 0)zx] (4 pts)

= 1 + x + y − z + x^{2} + y^{2} − z^{2} + 2xy − 2yz − 2zx. (2 pts)

8. (10 pts)Evaluate (\frac{∂u}{∂x})_{y} at (x, y, z, w) = (1, 1, 1, 1) where u(x, y, z, w) = x^{2}+y^{2}+z^{2}+w^{2} with the constraint x + y + z + w = 4 and x − y + z − w = 0. Answer. First we compute z_{x} and w_{z} at (1, 1, 1, 1).

1 + z_{x} + w_{x} = 0 (2 pts)
1 + z_{x} − w_{x} = 0 (2 pts)
⇒ z_{x} = −1, w_{x} = 0. (2 pts)

Now we compute (\frac{∂u}{∂x})_{y} at (1, 1, 1, 1).

(\frac{∂u}{∂x})_{y} = 2x + 2zz_{x} + 2ww_{x}|_{(1,1,1,1)} (2 pts) = 2 − 2 = 0. (2 pts)

9. (10 pts)Evaluate

∫_{0}^{2} ∫_{y}^{2} x^{2} cos(xy) dxdy

Answer.

∫_{0}^{2} ∫_{y}^{2} x^{2} cos(xy) dxdy = ∫_{0}^{2} ∫_{0}^{x} x^{2} cos(xy) dydx (4 pts)
= ∫_{0}^{2} x sin(xy)|_{0}^{x} dx
= ∫_{0}^{2} x sin(x^{2}) dx (3 pts)
= −\frac{1}{2} cos(x^{2})|_{0}^{2} = −\frac{1}{2}(cos 4 − 1). (3 pts)

10. (10 pts)Change

∫_{√\overline{2}}^{2} ∫_{√\overline{4−y^{2}}}^{y} dxdy

into an equivalent polar integral and evaluate the polar integral. Answer.

∫_{√\overline{2}}^{2} ∫_{√\overline{4−y^{2}}}^{y} dxdy = ∫_{π/4}^{π/2} ∫_{2}^{2 csc θ} r drdθ (2 pts)

= ∫_{π/4}^{π/2} \frac{r^{2}}{2} |_{2}^{2 csc θ} dθ (2 pts)
= ∫_{π/4}^{π/2}(2 csc^{2} θ − 2) dθ (2 pts)

= (−2 cot θ − 2θ)|_{π/4}^{π/2} (2 pts) = 2 − \frac{π}{2} . (2 pts)

11. (10 pts)Is ∫_{1}^{∞} sin \frac{1}{x^{2}} convergent? Explain.

Answer.Yes. (2 pts)

∵ lim_{x→∞} \frac{sin \frac{1}{x^{2}}}{\frac{1}{x^{2}}} = 1 ∈ (0, ∞) (4 pts) and ∫_{1}^{∞} \frac{1}{x^{2}} dx converges. (4 pts)

∴ we can conclude from Limit Comparison Test that the improper integral converges.

12. (10 pts)Evaluate ∑_{n=0}^{∞} \frac{x^{n}}{n + 2} on |x| < 1 using computational rules of power series.

Answer. The value is 1/2 as x = 0. (2 pts) Consider x ≠ 0.

∑_{n=0}^{∞} \frac{x^{n}}{n + 2} = x^{−2} (∑_{n=0}^{∞} \frac{x^{n+2}}{n + 2})

= x^{−2} (∑_{n=0}^{∞} ∫_{0}^{x} t^{n+1} dt) (2 pts)

= x^{−2} (∫_{0}^{x} ∑_{n=0}^{∞} t^{n+1} dt)

= x^{−2} (∫_{0}^{x} \frac{t}{1 − t} dt) (2 pts)
= x^{−2} (∫_{0}^{x} −1 + \frac{1}{1 − t} dt)
= x^{−2} (−x − ln(1 − x)) (2 pts)
= −\frac{1}{x} − \frac{ln(1 − x)}{x^{2}}. (2 pts)