Study Guide/Practice Exam 3 This study guide/practice exam covers only the material since exam 2. The final exam, however, is cumulative so you should be sure to thoroughly study earlier material. The distribution of content on this practice exam is not necessarily representative of the distribution of content on the actual exam. Some of the problems are significantly harder than what could appear on the final exam. (1) Let f (x, y) = x2 − sin(x) + cos(y). Find the first and second MacLaurin polynomials for f . (That is, the linear and quadratic approximations to f . Solution: The first Taylor polynomial is P1 (x, y) = 1 − x. The second is P2 (x, y) = 1 − x + x2 − (1/2)y 2 . (2) Let f (x, y) = x3 + x2 y − y 2 . Find and classify all critical points of f. Solution: We begin by calculating: ∇f (x, y) = (3x2 + 2xy, x2 − 2y) and   6x + 2y 2x Hf (x, y) = 2x −2 We have ∇f (x, y) = (0, 0) if (x, y) = (0, 0) or if (x, y) = (3, 9/2). These are our critical points. Plugging them into Hf (x, y) and using the second derivative test we discover that (0, 0) is a degenerate critical point and that (3, 9/2) is a saddle point. (** A degenerate critical point means that the 2nd derivative test is inconclusive **) (3) Let f (x, y) = ex of f .

2 +y 2

−e−(x

2 +y 2 )

. Find and classify all critical points

Solution: 2

2

2

2

fx (x, y) = 2x(ex +y + e−(x +y ) 2 2 2 2 fy (x, y) = 2y(ex +y + e−(x +y ) The only critical point of f is, therefore, (0, 0). Then 2

2

2

2

fxx (x, y) = 2(ex +y + e−(x +y ) ) + 4x2 ( something ) 2 2 2 2 fyy (x, y) = 2(ex +y + e−(x +y ) + 4y 2 ( something ) fyx (x, y) = 2x( something ) 1

2

  4 0 Consequently, Hf (0, 0) = . By the second derivative test, 0 4 (0, 0) is a minimum of f . (4) Find the point on the plane x − 2y + z = 1 that is closest to the point (1, 1, 0) and prove that you’ve found the point giving minimum distance. (Hint: minimize the square of the distance.) Solution: Let s(x, y, z) = (x − 1)2 + (y − 1)2 + z 2 . This is the square of the distance from a point (x, y, z) to the point (1, 1, 0). By setting z = 1 − x + 2y, we restrict to points on the plane. s(x, y, z) = (x − 1)2 + (y − 1)2 + (1 − x + 2y)2 . The partials are: sx = 2(x − 1) + 2(1 − x + 2y)(−1) = 4x + 4y − 4 sy = 2(y − 1) + 2(1 − x + 2y)(2) = −4x + 10y + 2 Setting equal to zero and solving, we find y = 7/3 and x = 19/3. The hessian is:   4 −4 Hs(x, y) = . −4 10 The determinant is 34. Since 4 > 0, the point (19/3, 7/3) minimizes s and hence also minimizes distance. 1 (5) Let g(x, y) = xy . Find the points on the graph of g which are closest to the origin in R3 . (Hint: Let s(x, y, z) = x2 +y 2 +z 2 be the square of the distance from (x, y, z) to the origin and set z = g(x, y).)

Solution: Let s(x, y) = x2 + y 2 + ∇s(x, y) = (2x −

2 y 2 x3

1 . x2 y 2

We have

, 2y −

2 y 3 x2

).

Setting ∇s(x, y) = (0, 0) and solving we get four critical points (1, 1), (−1, 1), (1, −1), and (−1, −1). Call these A, B, C, and D respectively.   2 + 6y −2 x−4 4y −3 x−3 Then Hs(x, y) = . Consequently: 4y −3 x−3 2 + 6y −4 x−2     8 4 8 −4 Hs(A) = Hs(B) = 4 8  −4 8 8 −4 8 4 Hs(C) = Hs(D) = −4 8 4 8 By the second derivative test, these are all minima.

3

(6) A company operates two plants which manufacture the same item and whose total cost functions are C1 = 8.5 + 0.03q12 C2 = 5.2 + 0.04q22 where q1 and q2 are the quantities produced by each plant. If the item costs p dollars then p = 60 − .04(q1 + q2 ). The goal is to find values for q1 and q2 which will maximize the company’s profit. Carefully set up and describe how you would solve this problem using multi-variable derivatives. You need not actually perform the calculations. Solution: Let P (q1 , q2 ) = (q1 + q2 )p − (C1 + C2 ). This is a function of q1 and q2 which represents the profit of the company. I would plug in the equations for p, C1 , and C2 and would find the critical points and then use the second derivative test to determine which critical points were maxima of P . (7) Let f (x, y) = x + y and let R be the rectangle in R2 with corners (1, 1), (1, 3), (2, 1) and (2, 3). (a) Subdivide R into four subrectangles RR of equal area and write down a sum which approximates f dA. R

Solution: Since the area of R is 2, the area of each subrectangle is ∆A = 2/4 = 1/2. Choose lower left corners for sample points (for example). Then we have sample points (1, 1), (3/2, 1), (2, 1), and (3/2, 2). The following sum approximates the double integral: ∆A(f (1, 1) + f (3/2, 1) + f (2, 1) + f (3/2, 2)) = (1/2)(1 + 1 + 3/2 + 1 + 2 + 1 + 3/2 + 2)

(b) Write

RR R

f dA as an iterated integral and solve.

4

Solution: RR

R3 R2

f dA =

(x + y) dx dy 2 (x2 /2 + yx) dy

1 1 R3

R

=

1 R3

=

1

(3/2) + y dy

1

= 7. (8) For the following functions fRRand regions R set up (but do not solve) an iterated integral equal to f dA. Your answer should be someR

thing that can be plugged into Mathematica to find the answer. (a) f (x, y) = x3 y and R is a disc of radius 1 centered at the point (1, −1). Solution: ZZ Z f dA =

2

Z √1−(x−1)2 −1 −

0

R

√

x3 y dy dx

1−(x−1)2 −1

(b) f (x, y) = sin(xy) and R is the triangular region with corners (0, 0), (2, 0), and (1, 5). Solution: ZZ

Z

5

Z

−y/5+2

sin(xy) dx dy

f dA = 0

R

y/5

(c) f (x, y) = x2 − y 2 and R is the region bounded by the graphs of y = x5 and y = x3 . ZZ R

Solution: Z 0Z f dA = −1

x5 2

2

Z

1

Z

x3

x − y dy dx +

x3

0

x2 − y 2 dy dx

x5

(9) Give an example of the following regions in R2 : (a) A Type I but not Type II region. (b) A Type II but not Type I region. (c) A Type III region

5

(d) A region that is not of Type I, II, or III. (10) Explain why Fubini’s theorem is true. Solution: Suppose that we want to find the volume between the graph of f (x, y) and a rectangle R = [a, b] × [c, d]. The area of an x-slice is Z d f (x, y) dx. A(x) = c

Subdivide the interval [a, b] into n subintervals each of length ∆x and pickPa point xi in each subinterval. The total volume isPapproximately A(xi )∆x. The total volume is equal to limn→∞ A(xi )∆x = Rb RR A(x) dx. Since the total volume is also R f dA we also have a ZZ Z b Z bZ d f dA = A(x) dx = f (x, y) dydx. R

a

a

c

(11) Describe two ways to define the double integral of a function f (x, y) over a non-rectangular region R in R2 . Solution: The first method is to subdivide the region R into n subrectangles, each of area ∆A. Pick a point xi in each subrectangle RR that is completely contained inside the region. Define f dA = R P limn→∞ f (xi )∆A. The second method is to choose a rectangle Q enclosing R and define n f (x) x ∈ R fb(x) = 0 x 6∈ R. Define

RR R

f dA =

RR Q

fbdA.

(12) Suppose that R is a rectangle and that for each n, R has been subdivided into n subrectangles each of area ∆n = Area(R)/n. Let xi be a point in the ith subrectangle. Suppose that f is a continuous function on R. Prove that the limit of the average values of f (xi ) is RR 1 equal to Area(R) f dA. R

Solution: See the solutions to HW 11. At most one problem similar to one of the remaining problems will be on the exam. (13) Carefully state the Change of Variables Theorem for double integrals.

6

Solution: Suppose that f : R2 → R is continuous and that x = x(u, v) and y = y(u, v) are a differentiable changeRRof coordinates. Let R ⊂ R2 be a region in the xy plane for which R f dA exists. b be the corresponding region in the u, v plane. Let fˆ be the Let R   xu xv composition of f with the coordinate change. Let J = . yu yv Then ZZ ZZ b fˆ | det J| dA.

f dA = R

b R

The theorem gives a way of converting an integral of f over a region in the xy plane to an integral over a region in the uv plane. The determinant of the Jacobian measures the infinitessimal change in area due to the coordinate change. It is necessary to include this so that the Riemann sum representing the integral on the left is equal to the Riemann sum representing the integral on the right. p (14) Let f (x, y) = y 4 − (x2 + y 2 ). Let R be the half disc defined by x2 + y 2 ≤ 1 and y ≥ 0.RRSet up an iterated integral in polar coordinates which is equal to f dA. R

Solution: In polar coordinates, we have y = r sin θ and r2 = x2 + y 2 . Thus, √ fˆ(r, θ) = r sin θ 4 − r2 . The half disc can be parameterized as 0 ≤ r ≤ 1 and 0 ≤ θ ≤ π. Thus ZZ Z πZ 1 √ f dA = r2 sin θ 4 − r2 dr dθ. R

0

0

(15) The following problems each give a function f , a region R, and a change of coordinates. For each, write down RR an iterated integral in the new coordinate system which equals f dA. R

(a) f (x, y) = x − y. R is the triangle with corners (0, 0), (3, 1) and (0, 2). The coordinate change is given by x = s−t, y = −s−t. Solution: We have fˆ(s, t) = 2s and det J = −2. Since the coordinate change is a linear coordinate change lines are taken to lines. Thus, we need only compute the image of the three corners of R in the st plane. To do this we solve for s and t in

7

terms of x and y, obtaining: s = x/2 − y/2 t = −x/2 − y/2 b is a triangular region with vertices (0, 0), (1, −2), and Thus R (−1, −1). Thus, we have RR RR b f dA = 4s dA R RRb1 R s R 1 R −2s = −1 −(s+1)/2−1 4s dt ds + 0 −(s+1)/2−1 4s dt ds √ (b) f (x, y) = 4x + y. R is the region bounded by x2 − y 2 = 1, x2 − y 2 = 4, y = 0, and y = x/2. The region appears in Figure 3. The coordinate change is given by x = r cosh θ, y = r sinh θ. It may help to remember the following facts: d/dθ cosh θ d/dθ sinh θ 2 cosh θ − sinh2 θ cosh θ sinh θ

= = = = =

sinh θ cosh θ 1 (eθ + e−θ )/2 (eθ − e−θ )/2

F IGURE 1 Solution: The new function is fˆ(r, θ) = (r cosh θ + r sinh θ)1/4 = r1/4 eθ/4 The new region is defined bounded by r2 r2 r sinh θ 2r sinh θ

= = = =

1 4 0 r cosh θ

The hyperbolas x2 − y 2 = 1 and x2 − y 2 = 4 each have two branches, if f > 0 we are on the right branches which is what

8

b we want. Thus, two of the curves bounding our new region R are r = 1 r = 2 b Consequently, our other Thus, r > 0 for all the points in R. b lines bounding R must be: sinh θ = 0 2(e − e−θ ) = eθ + e−θ θ

The first of these happens only if θ = 0 and the second only if b is the rectangle bounded by θ = ln(3)/2. Thus our region R r r θ θ

= = = =

1 2 0 (ln 3)/2

The Jacobian of the coordinate change is:   cosh θ r sinh θ J= sinh θ r cosh θ It has determinant: det J = r cosh2 θ − r sinh2 θ = r Hence, RR

f dA =

R

b fˆr dA R 2 R (ln3)/2

RR b R

=

1

0

r5/4 eθ/4 dθ dr

(c) Let f (x, y) = 1. Let R be the elliptical region Ax2 + Bxy + Cy 2 ≤ 1, where A, B, and C are positive constants such that C > B 2 /(4A2 ). Use the coordinate change
√ B s = x q+ 2A y A 2 t = y C−B . 4A RR Computing f dA will give the area of R. R

9

Solution: For convenience we let k = coordinate change can be rewritten as: x = √sA − y = t/k

p C − B 2 /(4A). The

Bt 2Ak

Thus, the determinant of the Jacobian is det J = region is determined as follows:

1 √ . k A

Our

Ax2 + Bxy + Cy 2 ≤ 1 2 Bt 2 Bt ) + B( √sA − 2Ak )( kt ) + Ct A( √sA − 2aK 2 ≤ 1 k  2  Bst Ct2 B 2 t2 B 2 t2 √ A sA − √Bts + + − 2 k2 2 + k2 ≤ 1 4A 2Ak AAk Ak s2 −

Bts √ Ak

+

B 2 t2 4Ak2 2 2

2 2

2

B t Ct + √Bst − 2Ak 2 + k2 ≤ 1 Ak 2 2 k s + (B −B + C)t2 ≤ k 2 4A 2A2 k 2 s2 + (− B + C)t2 ≤ k 2 4A k 2 s2 + k 2 t2 ≤ k 2 s2 + t2 ≤ 1

b is the disc of unit radius in the st plane. Thus the new region R Consequently, RR R

b f ( k√1 A ) dA b R RR b = k√1 A 1 dA

f dA =

RR

= =

1 √

k √

b R

π A

2π 4AC−B 2