Practice Problems For Final Math 5B

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Find a vector that is perpendicular to the vector (1,2,3) with the same length. Also ..... Practice Problem: Find the tangent plane to the surface x z. + z y. = 2 at (2,1,1) . .... they are all local minimums by the second derivative test. f(x, y) evaluated at ...
Practice Problems For Final

Math 5B

Chapter 1: Vectors, Matrices, and Applications Practice Problem: Write an equation for the plane that contains the points (2, 0, −3), (−4, −5, 2), and (0, 3, −4) in the form ax + by + cz = d. Solution: Let ~v = (−4, −5, 2) − (2, 0, −3) = (−6, −5, 5) and w ~ = (0, 3, −4) − (2, 0, −3) = (−2, 3, −1). Then ~ ~ ~ ~v × w ~ = i(5 − 15) − j(6 + 10) + k(−18 − 10) = (−10, −16, −28). We can choose ~n to be any vector in the same direction as ~v × w ~ so let ~n = (5, 8, 14). Then the plane has the form 5x + 8y + 14z = d. Substituting the point (2, 0, −3) for (x, y, z) and solving for d gives d = 10 + 0 + (−42) = −32. So the plane has the equation 5x + 8y + 14z = −32 . Practice Problem: Find a parametric form for the line passing through the point (1, 2) in the direction (3, 4), which we will call c1 (t). Set c1 (t) equal to (x, y) and eliminate t to get the line into y = mx + b form. Now find a different parametrization c2 (t) of the same line such that c2 (0) = (−2, −2) and c2 (2) = (−5, −6). Solution: c1 (t) = (1, 2) + t(3, 4) = (1 + 3t, 2 + 4t). Setting (x, y) = (1 + 3t, 2 + 4t) yields x = 1 + 3t and y = 2 + 4t. Solving the former equation for t yields t = (x − 1)/3. Substituting this into the second equation then gives us y = 43 x + 23 . Let c2 (t) = p + t~v . c2 will then be a parametrization of the same line given by c1 if p is a point on the same line and ~v is in the same direction as (3, 4) (i.e. some scalar multiple of (3, 4)). Since c2 (0) = (−2, −2) we will choose p = (−2, −2) (you can check that this point indeed lies on the line parametrized by c1 ). Then c2 (2) = (−2, −2) + 2~v = (−5, −6), so we get that ~v = (−3/2, −2), which , −2) is a different parametrization of is indeed a scalar multiple of (3, 4). So c2 (t) = (−2, −2) + t( −3 2 the line parametrized by c1 . Practice Problem: Suppose an object is traveling due east at a speed of 5 m/s. A wind is blowing in a northwestern direction with a constant force of 2N. How much work is done by the wind over a span of 30 s. Solution: Let the positive x and y directions correspond to due east and due north respectively. Since the object is traveling due east at 5 m/s for 30 s, then the object’s displacement vector has direction (1, 0) and magnitude 5 m/s × 30 s = 150 m. So d~ = (150, 0). Meanwhile the wind has direction (cos(3π/4), sin(π/4)) = (− √12 , √12 ) and magnitude 2N, so F = (− √22 , √22 ). Thus the work done by the √ √ = −150 2 joules. wind is F · d = − 300 2

Practice Problem: Find a vector that is perpendicular to the vector (1, 2, 3) with the same length. Also, find a plane perpendicular to (1, 2, 3) that passes through the point (3, 2, 1). Solution: A vector ~v is perpendicular to (1, 2, 3) if ~v · (1, 2, 3) = 0. There are infinite number of possibilities to choose from, but one possible choice for ~v is (2, −1, 0). However, we want this vector to have length √ √ √ 14 k(1, 2, 3)k = 14. Since k(2, −1, 0)k = 5, we need to rescale our vector to be √ (2, −1, 0) . 5 If (1, 2, 3) is a normal vector to a plane then the plane will have the form x + 2y + 3z = d. Since the plane passes through the point (3, 2, 1), we substitute these values for x, y, and z to get 3 + 4 + 3 = 10 = d so our plane equation is x + 2y + 3x = 10 . Practice Problem: Write an equation for the plane that contains the point (1, 0, 3) and the line (−3, −2, −2) + t(1, 2, −1) in the form ax + by + cz = d. Solution: Since the plane contains the line (−3, −2, −2) + t(1, 2, −1) we know that one tangent vector to the plane is ~v = (1, 2, −1). We can get a second tangent vector by finding the vector between (−3, −2, −2) and (1, 0, 3). So let w ~ = (4, 2, 5). Then ~v × w ~ = ~i(10 + 2) − ~j(5 + 4) + ~k(2 − 8) = (12, −9, −6). So we can choose ~n = (4, −3, −2) and our plane has the form 4x − 3y − 2z = d. Plugging in (1, 0, 3) for (x, y, z) and solving for d yields 4x − 3y − 2z = −2 . Practice Problem: Find the minimum distance between the point (3, −3, −3) and the plane 2x + y − z = 3. Solution: The point in the plane closest to (3, −3, −3) lies on a line that is perpendicular to the plane and passes through (3, −3, −3). Since (2, 1, −1) is a normal vector to the plane, we will use it as the direction of this line. Thus a parametrized form of the line is c(t) = (3, −3, −3) + t(2, 1, −1) = (3 + 2t, −3 + t, −3 − t). We substitute this into the plane equation to find its intersection with the plane and get: 2(3 + 2t) + 1(−3 + t) − (−3 − t) = 6 + 6t = 3 ⇒ t = − 12 . So the point in the plane closest to (3, −3, −3) is c(− 12 ) = (2, − 27 , − 25 ). The distance between the point and the plane is thus q q 1 2 1 2 3 2 1 + ( 2 ) + (− 2 ) = . 2

Practice Problem: Find the minimum distance between the point (4, 2, −3) and the line (1, 0, 2) + t(−1, −1, 2). Solution: Let ~v (t) represent the vector from the point (4, 2, −3) and the line (1, 0, 2) + t(−1, −1, 2) = (1 − t, −t, 2 + 2t) at any t ∈ R. So ~v (t) = (4, 2, −3) − (1 − t, −t, 2 + 2t) = (3 + t, 2 + t, −5 − 2t). We want to find the t such that ~v (t) is perpendicular to the line, which is when ~v (t) · (−1, −1, 2) = 0. (3 + t, 2 + t, −5 − 2t) · (−1, −1, 2) = −15 − 6t = 0 ⇒ t = − 52 . So the length of ~v (− 25 ) should represent the minimum distance from (4, 2, −3) and the line. 1 k~v (−5/2)|| = k(1/2, −1/2, 0)k = √ . 2 Practice Problem: Find the vector projection of (3, 2) onto (−1, −1). Then find the area of the triangle with one side the vector (3, 2) and another side the result of this projection. Solution: −5 (−1, −1) = ( 52 , 52 ) . proj(−1,−1) (3, 2) = 2 Then the area of the triangle with sides (3, 2) and ( 52 , 25 ) is one half the area of the parallelogram with sides (3, 2) and ( 52 , 52 ). So the area of the triangle is 21 k(3, 2, 0) × ( 25 , 52 , 0)k = 12 k(0, 0, 25 )k = Note that the two vectors were converted to lie in R3 so that we could take the cross product.

5 4

.

Chapter 2: Calculus of Functions of Several Variables Practice Problem: Find the domain and range of the following functions: 1. f (x, y) =

2y 2 x2 + y 2

y x p 3. f (x, y) = xy 2 2. f (x, y) =

4. f (x, y) = e−x

2

5. F~ (x, y) = (ln(x), ln(y)) 6. f (x, y) = arctan(x/y) Solution: 2y 2 x2 + y 2 Domain = R2 − (0, 0)

1. f (x, y) =

Range = [0, 2] since if y = 0 then f (x, 0) = 0 and if y 6= 0 then f (x, y) is any value in (0, 2]. y x Domain = R2 except for the y-axis

2. f (x, y) =

Range = R since if x = 1 then f (1, y) = y which has range R. p 3. f (x, y) = xy 2 Domain = (x, y) such that x ≥ 0 Range = all non-negative real numbers since f (1, y) = |y| and negative values are impossible. 2

4. f (x, y) = e−x Domain = R2 Range = (0, 1] as f (0) = 1 and as x increases or decreases from 0, f (x, y) approaches 0. 5. F~ (x, y) = (ln(x), ln(y)) Domain = (x, y) such that x ≥ 0 and y ≥ 0 Range = R2 since the range of ln(x) is all of R. 6. f (x, y) = arctan(x/y) Domain = R2 except for the x-axis Range = [−π/2, π/2] as f (x, 1) = arctan(x) which has range [−π/2, π/2].

Practice Problem: 3x − 6y Find lim . (x,y)→(0,0) sin(x − 2y) Solution: 3u , a single-variable limit. As both the numerator Let u = x − 2y. Then the limit becomes lim u→0 sin u 3u and denominator are equal to 0 when u is 0, we can apply L’Hopital’s rule to get lim = u→0 sin(u) 3 lim = 3. u→0 cos(u) Practice Problem: x Find lim . (x,y)→(0,0) x + y 2 Solution: 0 If we approach (0,0) along x = 0, then the limit becomes lim 2 = 0. If we approach (0,0) along y→0 y x y = 0, the limit becomes lim = 1. Since the limit is different depending on the direction we x→0 x approach (0,0), the limit does not exist . Practice Problem: √ xy Find lim . 2 (x,y)→(0,0) x + y 0 = 0 and similarly the limit is y→0 y zero when we approach along y = 0. However, when we approach along x = y, we get that the limit x 1 is lim 2 = lim = 1. Thus the limit does not exist . x→0 x + x x→0 2x + 1 Solution: If we approach (0,0) along x = 0 the limit becomes lim

Practice Problem: Find lim (x + y)e−1/(x+y) . (x,y)→(0,0)

Solution: If we approach (0,0) along y = 0 we get lim xe−1/x . This single-variable limit depends x→0

on the direction x approaches 0. If x approaches 0 from a positive direction then lim xe−1/x = + x →0

−1/x

( lim x)( lim e + + x →0

x →0

) = (0)(0) = 0.

If x approaches 0 from a negative direction, lim e−1/x = ∞ so we cannot calculate the limit in − x →0

the same manner. This time let x = −1/n. So x approaching 0 from a negative direction is equivalent −1 n −en −1/x to n approaching infinity. We then get lim xe = lim e = lim by L’Hopital’s rule, n→∞ n n→∞ 1 x− →0 which has limit −∞. Thus since from one direction the limit is 0 and the other it is −∞, the limit does not exist .

Practice Problem: x−y Find lim (x,y)→(1,1) x2 − y 1−y = 1. If we approach (1,1) along y→1 1 − y

Solution: If we approach (1,1) along x = 1 we get lim

x−1 1 1 = lim = . Since the limit is different when approaching (1,1) from 2 x→1 x − 1 x→1 x + 1 2 two different directions, the limit does not exist .

y = 1 we get lim

Practice Problem: If the radius of a cylinder is decreasing at a rate of 2 cm/s and the height is increasing at a rate of 5 cm/s, what is the rate that the volume of the cylinder is changing when the radius is 4 cm and its height is 3 cm? ∂h ∂V ∂r Solution: The volume of a cylinder is given by V (r, h) = πr2 h. Then = π2rh + πr2 . ∂t ∂t ∂t ∂h ∂V ∂r 2 , and we get = π2(4)(3)(−2) + π(4) (5) = 32π So entering the given information for r, h, ∂t ∂t ∂t 3 cm /s. Practice Problem: Find the tangent plane to the surface f (x, y) = 3x2 − 2y 2 + 5 at (1, 1). Solution: Using the linear approximation method, we get z = L(1,1) (x, y) = f (1, 1)+(6x)|(1,1) (x− 1) + (−4y)|(1,1) (y − 1) = 6 + 6(x − 1) − 4(y − 1) = 6x − 4y + 4. So the plane is 6x − 4y − z = −4 . Using the gradient method, we set F (x, y, z) = 3x2 − 2y 2 − z, so our surface is a level curve of F at -5. Also, when (x, y) = (1, 1) we get that z = f (1, 1) = 6. Then ∇F (1, 1, 6) = (6x, −4y, −1)|(1,1,6) = (6, −4, −1). So our plane has the form 6x − 4y − z = d. Substituting (1,1,6) for (x, y, z) and solving for d results in the plane being 6x − 4y − z = −4 . Practice Problem: x z Find the tangent plane to the surface + = 2 at (2, 1, 1). z y Solution: In this case we cannot solve nicely for z and put the surface in a form where we can x z take a linear approximation. So instead we will use the gradient mehtod. Set F (x, y, z) = + so z y 1 z x 1  the surface is the level curve of F at 2. Then ∇F (2, 1, 1) = ,− ,− + = (1, −1, −1). z y 2 z 2 y (2,1,1) So our plane has the form x − y − z = d. Substituting in (2, 1, 1) and solving for d gives us the plane x−y−z =0.

Practice Problem: Find the directional derivative of f (x, y) = 3x2 − 2y 2 at the point (2, 4) in the direction of (−3, 4). What is the direction in which f is increasing most rapidly at (2, 4)? What is this maximum rate of increase?  3 4 (−3, 4) = − , is the unit vector in the direction of (−3, 4). Then Solution: ~u = k(−3, 4)k 5 5  3 4 36 64 D~u f (2, 4) = ∇f (2, 4)·~u = (6x, −4y)|(2,4) ·~u = (12, −16)· − , =− − = −20 . The direction 5 5 5 5 that f is increasing the most rapidly at (2, 4) is the direction of ∇f (2, 4) = (12, −16) A unit vector in the direction of (12, −16) is (3/5, −4/5) so the maximum rate of increase is (12, −16) · (3/5, −4/5) = 20 .

Chapter 3: Vector-Valued Funcitons of One Variable Practice Problem: Find a parametric equation for the curve that is the intersection of the cylinder x2 + y 2 = 4 and the surface xz = y. Solution: The easiest way to parametrize x2 + y 2 = 4 is to set x = 2 cos(t) and y = 2 sin(t). Then substituting into xz = y and solving for z yields z = tan(t). Therefore, a possible parametrization of this curve of intersection is ~c(t) = (2 cos(t), 2 sin(t), tan(t)) . Note that there are many other ways to parametrize this curve. Practice Problem: Find a parametric equation ~c(t) for the parabola 2x + y 2 = 3 such that ~c(0) = (1, 1). Then SET UP the integral that would give the length of the parabola between (1, 1) and (−3, 3). Solution: We again have a lot of ways to parametrize this curve. Since it is easier to solve for 3 − t2 x in the parabola equation, I’ll let y = t and then solving for x gives us x = . So one way to 2  3 − t2  parametrize this curve is , t . However, this does not give us that ~c(0) = (1, 1). So instead 2   3 − (t + 1)2 , t + 1 . So now ~c(0) = (1, 1). We also find that replace the t with t + 1 to get ~c(t) = 2 ~c(2) = (−3, 3). So the arc length between (1, 1) and (−3, 3) involves finding the arc length of this parametrization forZ t ∈ [0, 2] and we get the following integral: Z 2 2 R2√ kc~0 (t)kdt = k(−t − 1, 1)kdt = 0 t2 + 2t + 2dt 0

0

Practice Problem: Find a parametric equation ~c(t) for the ellipse 4x2 + y 2 = 9 such that ~c(0) = (0, 3) and ~c( π2 ) = (− 23 , 0). Solution: If we let y = 3 cos(t) or y = 3 sin(t) then solving for x is nice. Since ~c(0) = (0, 3) we decide to let y = 3 cos(t). Then substituting y = 3 cos(t) and solving for x, we find that x = 3 3 sin(t) or x = − sin(t). Since ~c( π2 ) = (− 32 , 0), we choose the latter. Thus the parametrization is 4 4  3  ~c(t) = − sin(t), 3 cos(t) . 4 Practice Problem: Find a parametrization of the curve x3 = 8y 2 and use it to find the length of this curve for x ∈ [0, 2]. Solution: One possible parametrization ofZ this curve is ~c(t) = (2t2 , t3 ). ThenZ x = 0 corresponds Z 1 1 1√ to t = 0 and x = 2 corresponds to t = 1. So kc~0 (t)kdt = k(4t, 3t2 )kdt = 16t2 + 9t4 dt = 0 0 0 Z 1 √ 1 1 1 61 t 16 + 9t2 dt = . (16 + 9t2 )3/2 = (125 − 64) = 27 27 27 0 0

Practice Problem: √ Let ~c(t) = (et , e−t , 2t). Find the arc length for t ∈ [0, 1]. Z 1 Z 1 Z 1√ Z 1p √ t −t Solution: kc~0 (t)kdt = k(e , −e , 2)kdt = e2t + e−2t + 2dt = (et + e−t )2 dt = 0 0 0 0 Z 1 et + e−t dt = (et − e−t |10 = e − e−1 0

Practice Problem: Let ~c(t) = (cos(t), sin(t), t). Reparametrize by arc length. Solution: Z t function is Z t√ Z t The arc length √ 0 kc~ (t)kdt = k(− sin(t), cos(t), 1)kdt = s(t) = 2dt = 2t. So t = √s2 . 0 0  s  s   0 s  √ Therefore, the new parametrization is ~c(s) = cos , sin √ , √ . 2 2 2

Chapter 4: Scalar and Vector Fields Practice Problem: 1 and the origin. Find the shortest distance between the surface z = xy Solution: 1 Let f (x, y) represent the squared distance between a point on z = (i.e. a point of the form xy     1 1 2 1 x, y, ) and the origin. So f (x, y) = x2 + y 2 + = x2 + y 2 + 2 2 . The reason why we xy xy xy are using the squared distance is that the function will have the same critical ponts as the standard distance function (since distance is non-negative) but we don’t have to deal with any square roots.  1 2  2 Then ∇f (x, y) = 2x − 3 2 , 2y − 2 3 . Setting the first coordinate to zero, we get that x4 = 2 xy xy y 1 and similarly setting the second coordinate to zero we get y 4 = 2 . Solving for y 2 in the first equation x 1 1 1 2 gives us y = 4 . Substituting this into the second equation yields 8 = 2 ⇒ x6 = 1 ⇒ x = ±1. If x x x x = 1 or x = −1 we get that y is equal to 1 or −1 as well. So we end up with four critical points: (1, 1), (1, −1), (−1, 1), and (−1, −1).  6   4 2 12 12 20 6  D(x, y) = 2 + 4 2 2 + 2 4 − 3 3 = 4 + 4 2 + 2 4 + 6 6 . xy xy xy xy xy xy Since all of the powers are even, the discriminant of all the critical points will be the same: D(1, 1) = 4 + 12 + 12 + 20 = 48 > 0. Furthermore, fxx = 8 > 0 for all four of the critical points so they are all local minimums by the second derivative test. f (x, y) evaluated at each of the critical points is 3. However, since f (x, y) represents the squared distance, the minimum distance from the √ surface to the origin must then be 3 . Practice Problem: Find and classify all of the critical points of f (x, y) = x2 y + xy 2 − 4x. Solution: We again look for critical points: ∇f (x, y) = (2xy + y 2 − 4, x2 + 2xy) = (0, 0). Setting the second coordinate to zero gives us x2 + 2xy = 0 ⇒ x(x + 2y) = 0. So x = 0 or x = −2y. Substituting x = 0 into the first coordinate gives us y 2 − 4 = 0, so y = 2 or −2. If we substitute x = −2y into the first coordinate we get 2(−2y)y + y 2 − 4 = −3y 2 − 4 = 0 which has no real solutions. So our only critical points are (0, 2) and (0, −2) . D(x, y) = (2y)(2x) − (2x + 2y)2 = 4xy − 4x2 − 8xy − 4y 2 = −4x2 − 4xy − 4y 2 . So D(0, 2) and D(0, −2) are both negative and are therefore both saddle points . Practice Problem: Find the absolute extreme values of the function f (x, y) = x − y on the disk D = {(x, y)|x2 + y 2 ≤ 1}. Solution: ∇f (x, y) = (1, −1). So there are no critical points. A parametrization of the boundary is ~c(t) = (cos(t), sin(t)). So (f ◦ ~c)(t)) = cos(t) − sin(t) and so  3π  √ 3π 7π =− 2 (f ◦ ~c)0 (t) = − sin(t) − cos(t). Then − sin(t) − cos(t) = 0 when t = and . (f ◦ ~c) 4 4 4  7π  √ and (f ◦ ~c) = 2. There are no end points of the boundary to check. 4 √ √ Thus the global maximum over D is 2 and the global minimum is − 2 .

Practice Problem: Let f (x, y) = xy. Find global maximum and minimum over the region bounded by x2 + y 2 = 4. Solution: ∇f (x, y) = (y, x). So (0, 0) is the only critical point on the interior and f (0, 0) = 0. A parametrization of the boundary is ~c(t) = (2 cos(t), 2 sin(t)). (f ◦ ~c)(t) = 4 cos(t) sin(t), so (f ◦~c)0 (t) = −4 sin2 (t) + 4 cos2 (t) = 4(cos(t) − sin(t))(cos(t) + sin(t)). So the critical points of (f ◦~c)(t) π 3π 5π 7π π 5π 3π 7π are , , , and . (f ◦ ~c)( ) = (f ◦ ~c)( ) = 2 and (f ◦ ~c)( ) = (f ◦ ~c)( ) = −2. 4 4 4 4 4 4 4 4 So the global maximum and global minimum over x2 + y 2 = 4 are 2 and −2 respectively. Practice Problem: √ Let f (x, y) = ln(x − y + 1). Find global maximum and minimum over the region bounded by y = x, y = 0, and x = 1.  −1  1 , . So there are no critical points on the interior. Solution: ∇f (x, y) = x−y+1 x−y+1 There are three boundary components that we need to parametrize. Let c~1 (t) = (t2 , t) for t ∈ [0, 1] √ 2t − 1 be the parametrization of y = x. Then (f ◦ c~1 )(t) = ln(t2 − t + 1) and (f ◦ c~1 )0 (t) = 2 . So t −t+1 the critical point along this boundary component is t = 12 . (f ◦ c~1 )( 21 ) = ln( 34 ). Let c~2 (t) = (t, 0) for t ∈ [0, 1] represent the boundary of y = 0. Then (f ◦ c~2 )(t) = ln(t + 1) and 1 (f ◦ c~2 )0 (t) = which has no critical points. Similarly, when we let c~3 (t) = (1, t) for t ∈ [0, 1] we t+1 −1 which also has no critical points. get (f ◦ c~3 )(t) = ln(−t + 2) and (f ◦ c~3 )0 (t) = −t + 2 Last of all, we need to check the endpoints of the boundary. f (0, 0) = ln(1) = 0, f (1, 0) = ln(2), and f (1, 1) = ln(1) = 0. So the global maximum is ln(2) and the global minimum is ln( 34 ) (since ln( 34 ) < 0). Practice Problem: Find the absolute minimum and absolute maximum of the function f (x, y) = sin(x + y) over the π π rectangle D = {(x, y)|0 ≤ x ≤ , ≤ y ≤ π}. 2 2 Solution: This is a special case since f (x, y) is a bounded function - it clearly has a maximum of 1 and of −1,and the only question is whether f (x, y) attains this min and maxover D.  aπminimum  π π π 0, is in D and f 0, = sin = 1, so the global maximum is 1 . Furthermore, , π is 2 2  2 2 π  3π also in D and f , π = sin = −1. Therefore the global minimum is −1 and we have avoided 2 2 the need to take a gradient or parametrize the boundary. Practice Problem: Calculate the divergence and curl of F~ (x, y, z) = (x2 , yz, exz ). Solution: div F~ = 2x + z + xexz . curlF~ = ~i(0 − y) − ~j(zexz − 0) + ~k(0 − 0) = (−y, −zexz , 0) .

Chapter 5: Integration Along Paths Practice Problem: Z Compute the path integral of zexy ds where ~c is the line segment from (0, 0, 0) to (3, 2, 1). ~c

Solution: xy 6t2 Let ~c(t) = (3t, 2t, t) √ for t ∈ [0, 1]. Then if we let f (x, y, z) = ze then (f ◦ ~c)(t) = te and kc~0 (t)k = k(3, 2, 1)k = 14. √ Z Z 1 √ √  1 6t2 1 14 6 6t2 f ds = te 14dt = 14 e = (e − 1) . 12 12 0 ~c 0 Practice Problem: Z √ Compute the path integral of f ds where f (x, y) = x + y and ~c(t) is the straight line from (0, 0) ~c

to (1, 1). Solution: √ √ Let ~c(t) = (t, t) for t ∈ [0, 1]. Then (f ◦ ~c)(t) = 2t and kc~0 (t)k = k(1, 1)k = 2. Z Z 1√ √ 1 2 4 3/2 So f ds = 2t 2dt = 2 t = . 3 3 0 ~c 0 Practice Problem: Z Compute the path integral of (x2 + y 2 + xy)ds where ~c is the semicircle of radius 1 centered at the ~c

origin above the y-axis.

Solution: Let ~c(t) = (cos(t), sin(t)) for t ∈ [0, π] and let f (x, y) = x2 + y 2 + xy. Then (f ◦ ~c)(t) = cos2 (t) + sin2 (t) Z+ cos(t) sin(t) kc~0 (t)k = k(− sin(t), cos(t))k = 1. Z π = 1 + cos(t) sin(t) and  1 2 π So f ds = 1 + cos(t) sin(t)dt = t + sin (t) = π . 2 0 ~c 0 Practice Problem: Z Compute the path integral of F~ ·ds where F~ (x, y, z) = (y−2z, x+z, y−2x) and ~c(t) = (t, t2 −1, 6−t), for t ∈ [0, 2].

~c

Solution: Is F~ a gradient vector field? We could take the curl to be sure, and indeed curlF~ = ~i(1 − 1) − ~j(−2 + 2) + ~k(1 − 1) = (0, 0, 0). So we want to find an f such that ∇f = F~ and we get that Z Z f (x, y, z) = xy − 2xz + yz. So by the Fundamental Theorem of Calculus, F~ · ds = ∇f · ds = ~c

~c

f (c(2)) − f (c(0)) = f (2, 3, 4) − f (0, −1, 6) = 2 − (−6) = 8 . 2 If we didn’t see it was a gradient vector field, Z we couldZ instead find that (f ◦ ~c)(t) = (t + 2t − 2

13, 6, t2 −2t−1) and c~0 (t) = (1, 2t, −1) to get that Z 2 (16t − 12)dt = (8t2 − 12t|20 = 8 . 0

F~ ·ds =

~c

(t2 +2t−13, 6, t2 −2t−1)·(1, 2t, −1)dt =

0

Practice Problem: Z Compute the path integral of F~ · ds where F~ (x, y) = (xy, 2x − y) and ~c(t) is the line from (0, 0) to ~c

(1, 2). Solution: It turns out this F~ is not a gradient vector field and we can verify this since the scalar curl of F~ is 2 − x 6= 0. Z 2 0 ~ Let ~c(t) = (t, 2t) for t ∈ [0, 1]. Then (f ◦ ~c)(t) = (2t , 0) and c (t) = (1, 2). So F~ · ds = ~ c Z 1 Z 1  2 1 2 2 (2t , 0) · (1, 2)dt = 2t2 dt = t3 = . 3 0 3 0 0 Practice Problem: Suppose that ∇f (x, y, z) = (y 2 z cos(x), 2yz sin(x), y 2 sin(x)). If f (0, 1, 2) = 5, find f ( π2 , 1, 2). Solution: There is a couple of ways to go about this problem. One way is to try to figure out what f (x, y, z) is based on its gradient. Working backwards you find that f (x, y, z) = y 2 z sin(x) + C, where C is some constant. Usually we don’t need to worry about the constant because it gets cancelled out when we are integrating gradient vector fields, but here it is important. Since f (0, 1, 2) = 5, we get that C = 5. Therefore, f ( π2 , 1, 2) = 7. R Another method is to let F~ = ∇f (x, y, z) and then find ~c F~ · ds along some path ~c(t) between (0, 1, 2) and (π/2, 1, 2), since by the Fundamental Theorem of Calculus, the result of this integral should be f (π/2, 1, 2) − f (0, 1, 2). So let ~c(t) = ( π2 t, 1, 2) for t ∈ [0, 1]. Then (f ◦ ~c)(t) = (2 cos( π2 t), 4 sin( π2 t), sin( π2 t)) and c~0 (t) = ( π2 , 0, 0). 1  R R1 R1 So F~ · ds = (2 cos( π t), 4 sin( π t), sin( π t)) · ( π , 0, 0)dt = π cos( π t)dt = 2 sin( π t) = 2. So ~c

0

2

2

2

2

0

f (π/2, 1, 2) − f (0, 1, 2) = 2 ⇒ f (π/2, 1, 2) = 2 + f (0, 1, 2) = 2 + 5 = 7 .

2

2

0

Chapter 6: Double and Triple Integrals Practice Problem: RR pπ pπ pπ 2 ), and ( , 2 ). Evaluate cos(y )dA where D is the triangle with verticles (0, 0), (0, 2 2 D Solution: This triangle is a type 3. However, this integral is much easier if we integrate with respect p to x first. So we will consider this triangle a type 2. As a type 2, we have that 0 ≤ x ≤ y and 0 ≤ y ≤ π2 . So the double integral becomes: √π R √ π2 R √ π2 R √ π2 R y 2 1 2 2 y 2 2 cos(y )dxdy = (x cos(y )| dy = y cos(y )dy = ( = 12 . sin(y )| 0 0 0 0 0 2 0 Practice Problem: Find the volume of the three-dimensional solid bounded by the plane 3x + 2y + z = 6 in the first octant. Solution: Let z = f (x, y) = 6 − 3x − 2y. Then we will find the correct volume if we integrate f (x, y) over the region in the xy-plane bounded by x = 0, y = 0, and where this plane intersects the xy-plane. We can find the latter by setting z = 0 to get y = 3 − 32 x. So the region we are integrating over is the triangle with vertices (0, 0), (0, 3), and (2, 0). It does not really matter in which order we integrate, so lets treat this triangle as a type 1. So 0 ≤ y ≤ 3 − 32 x and 0 ≤ x ≤ 2. So our integral is: 2 3− 3 x   R 2 R 3− 3 x R2 R2 2 9 2 3 3 9 2 2 2 x dx = 9x− x + x (6−3x−2y)dydx = dx = 6y−3xy−y 9−9x+ = 4 2 4 0 0 0 0 0

0

6. Practice Problem: √ Find the area of the region bounded by y = x2 and y = x. Solution: √ Let’s integrate this region as a type 1, so 0 ≤ x ≤ 1 and x2 ≤ y ≤ x.  2 3 1 1 2 1 R 1 √x R 1 R √x R1 √ 1 2 1dydx = 0 (y|x2 dx = 0 ( x − x )dx = x 2 − x3 = − = . 0 x2 3 3 0 3 3 3 Practice Problem: RR Evaluate xydA where D is the region bounded by 1 ≤ x2 + y 2 ≤ 4 in the first quadrant. D Solution: Given D it would appear this integral would be easier using polar coordinates, in which case our bounds are 1 ≤ r ≤ 2 and 0 ≤ θ ≤ π2 . xy = r2 cos(θ) sin(θ). So our double integral becomes: 2 π  15 R π R2 3 Rπ 1 4 R π 15 2 2 2 2 2 r cos(θ) sin(θ)drdθ = r cos(θ) sin(θ) dθ = cos(θ) sin(θ)dθ = sin (θ) = 0 1 0 0 4 4 8 1 0 15 . 8

Practice Problem: RRR Evaluate xdV where W is the region bounded by the plane 3x + 2y + z = 6 in the first octant. W Solution: Lets integrate with respect to z first. Since z is bounded by the plane z = 0 and z = 6 − 3x − 2y, we have that 0 ≤ z ≤ 6 − 3x − 2y. Then we have reduced this problem to a double integral, where we want to integrate x and y over the xy-cross-sectional area of this solid. As we saw in an earlier practice problem, this area is a triangle and is bounded by the functions x = 0, y = 0, and the line y = 3 − 23 x (which we get by setting z = 0). So if we again treat this triangle as a type 2, we get the following triple integral: R 2 R 3− 3 x R 2 R 3− 3 x R 6−3x−2y R 2 R 3− 3 x 6−3x−2y 2 2 2 xdzdydx = (xz| dydx = (6x − 3x2 − 2xy)dydx 0 0 0  0 0 0 0 0 2  3 R2 R2 3− x 9 4 = 0 (6xy − 3x2 y − xy 2 |0 2 dx = 0 9x − 9x2 + 94 x3 dx = 92 x2 − 3x3 + 16 x = 3 0

Practice Problem: RRR Evaluate zdV where W is the cone bounded by z 2 = x2 + y 2 and z = 4. W Solution: One way to calculate this integral is with cylindrical coordinates. Clearly 0 ≤ θ ≤ 2π, since this cone can be thought of as a triangle rotated around the z-axis. To find this triangle, we can think about the cross-section of this cone with the xz-plane. This triangle has vertices (0, 0), (−4, 4), and (4, 4). But the triangle that is being rotated around the z-axis is really half of this triangle, the one with vertices (0, 0), (0, 4), and (4, 4). From this picture we can determine that 0 ≤ z ≤ 4 and 0 ≤ r ≤ z (note we could have alternatively chosen 0 ≤ r ≤ 4 and r ≤ z ≤ 4). So our triple integral becomes: R 2π R 2π  1 4 4 R 2π R 4 R z R 2π R 4 1 3 (z )dzdθ = z dθ = 0 32dθ = 64π . (zr)drdzdθ = 8 0 0 0 0 0 0 2 0

Practice Problem: RRR Evaluate zdV where W is the three-dimensional solid that is bounded by the sphere x2 + y 2 + W z 2 = 4 and above the xy-plane. Solution: In this case spherical coordinates seems to be the best choice. The bounds would then be 0 ≤ ρ ≤ 2, 0 ≤ φ ≤ π2 , and 0 ≤ θ ≤ 2π. z = ρ cos(φ). So our triple integral becomes: π R 2π R π R 2 3 R 2π R π R 2π R 2π 2 2 2 2 ρ cos(φ) sin(φ)dρdφdθ = 4 cos(φ) sin(φ)dφdθ = (2 sin (φ)| dθ = 2dθ = 4π . 0 0 0 0 0 0 0 0 Practice Problem: What is the volume of the three-dimensional solid bounded by the sphere x2 + y 2 + z 2 = 9 and the planes z = 1 and z = 2? Solution: Although there is a sphere involved, the resulting solid looks closer to a cylinder, so we will use cylindrical coordinates. Again θ is going to vary from 0 to 2π and so it’s really down to figuring out the relationship between r and z. If we say 1 ≤ z ≤ 2 then we need to figure out how r depends on z. Substituting r2 = x2 + y 2 into the sphere √ equation above and solving for r yields √ 2 r = 9 − z , giving us the upper bound on r. So 0 ≤ r ≤ 9 − z 2 . This gives the following triple integral: 2 R 2π R 2 R √9−z2 R 2π R 2 1 R 2π 1  R 2π 20π 1 3 2 rdrdzdθ = (9 − z )dzdθ = 9z − z dθ = 0 10 dθ = . 3 3 0 1 0 0 1 2 0 2 3 1

Chapter 7: Integration Over Surfaces, Properties, Applications Practice Problem: RR Compute f dS where f (x, y, z) = (x2 + y 2 )z and S is the hemisphere x2 + y 2 + z 2 = 4 above the S xy-plane. Solution: Spherical or cylindrical-type coordinates make the most sense for parametrizing a hemisphere and given the function f I’m going to go with √ a cylindrical aproach. So let ~r(u, v) = (u cos(v), u sin(v), 4 − u2 ) where 0 ≤ v ≤ 2π and 0 ≤ u ≤ 2. I got this parametrization by using cylindrical coordinates with u playing the role of r, v the role of θ, and then by substituting r2 for x2 + y 2 in the sphere equation you can solve for z. ∂~r  −u  ∂~r and Then = cos(v), sin(v), √ = (−u sin(v), u cos(v), 0). ∂u ∂v 4 − u2  u2 cos(v) u2 sin(v)  ~ So N (u, v) = √ ,√ ,u . 4r − u2 4 − u2 r r 2 4 2 4 4 u cos (v) (v) u 4u2 u sin 2 = 2 = ~ (u, v)k = Thus kN + + u + u . 2 4 − u2 4 − u2 4 − u2 √4 − u √ ~ (u, v)k = u2 4u2 = 2u3 . fR~(~r(u, v)) = u2 4 −R u2 . So f~(~r(u, v))kN R 2π 2 2π (2u3 )dudv = 0 8dv = 16π . 0 0 Practice Problem: Compute the surface area of the cone z 2 = x2 + y 2 that is bounded above by z = h. Solution: Again cylindrical coordinates gets the nod for the parametrization and we get that ~r(u, v) = ~ (u cos(v), u sin(v), √ u) where 0 ≤ v ≤ 2π and 0 ≤ u ≤ h. Then N (u, v) = (−u cos(v), −u sin(v), u) and ~ (u, v)k = 2u. kN √ R 2π R h √ R 2π h2 √ ( dv = 2u)dudv = 2πh2 . 0 0 0 2 Practice Problem: RR Compute f dS where f (x, y, z) = x2 + y 2 + z 2 and S is the hemisphere x2 + y 2 + z 2 = 1 and x ≤ 0. S Solution: In this case, spherical coordinates are the best way to parameterize this surface since f can be written nicely in terms of spherical coordinates. So let ~r(u, v) = (sin(u) cos(v), sin(u) sin(v), cos(u)) where 0 ≤ u ≤ π and − π2 ≤ θ ≤ π2 . Then ~ (u, v) = (sin2 (u) cos(v), sin2 (u) sin(v), cos(u) sin(u)) and kN ~ (u, v)k = sin(u). N f (~r(u, v)) = sin2 (u) cos2 (v) + sin2 (u) sin2 (v) + cos2 (u) = 1. R π2 R π R π2 sin(u)dudv = π − − π 2dv = 2π . 0 2

2

Practice Problem: RR Compute F~ · dS where F~ (x, y, z) = (x, y, z) and S is the cylinder x = y 2 + z 2 bounded by x = 0 S and x = 2 oriented with normal vector pointing outwards. Solution: Since this is a cylinder whose “height” is in the x-direction, we will parameterize the surface using cylindrical coordinates but swapping the roles of x and z. So ~r(u, v) = (v, cos(u), sin(u)) where 0 ≤ v ≤ 2 and 0 ≤ u ≤ 2π (so v is the length in the x-direction and u is the angle of rotation ~ (u, v) = (0, cos(u), sin(u)). To check if this normal vector is pointing around the x-axis). Then N ~ (0, 0) = (0, 1, 0). So the normal outwards, let’s test the point (u, v) = (0, 0). ~r(0, 0) = (0, 1, 0) and N vector at (0,1,0) is (0,1,0) and thus the normal vector is pointing outwards (since the vector is pointing ~ away R 2 the x-axis). F (~r(u, v)) = (v, cos(u), sin(u)).R 2π R 2 R 2πfrom (v, cos(u), sin(u)) · (0, cos(u), sin(u))dvdu = 0 0 1dvdu = 4π . 0 0 Practice Problem: Compute the downwards flux of F~ (x, y, z) = (x, y, z) across the plane x + y + z = 3 in the first octant. Solution: The easiest way to parameterize this plane is to set u = x and v = y and then solving for z. So let ~r(u, v) = (u, v, 3 − u − v). Then since we are in the first octant, we have that 0 ≤ u and 0 ≤ v. To find the upper bounds, we see that 3 − u − v hits the uv-plane along the line v = 3 − u. So we get the bounds 0 ≤ u ≤ 3 and 0 ≤ v ≤ 3 − u. ~ (u, v) = (1, 1, 1) which is pointing upwards. So to find the downwards flux we will need to Then N multiply our answer by −1. ~ v)) = (u, v, 3 − u − v), so RF 3(~rR(u, R 3 R 3−u R3 3−u . (u, v, 3 − u − v) · (1, 1, 1)dvdu = 3dvdu = (9 − 3u)du = 27 2 0 0 0 0 0 . So our answer is − 27 2 Practice Problem: Compute the inwards flux of F~ (x, y, z) = (x, y, z) across the sphere x2 + y 2 + z 2 = 1. Solution: Here we will use a parameterization based on spherical coordinates with ρ = 1. So let ~r(u, v) = (sin(u) cos(v), sin(u) sin(v), cos(u)) where 0 ≤ u ≤ π and 0 ≤ v ≤ 2π. Then ~ (u, v) = (sin2 (u) cos(v), sin2 (u) sin(v), cos(u) sin(u)). N ~ points inwards or outwards, so lets check (u, v) = ( π , 0) (we We now need to check whether N 2 would check (0,0), but our parameterization is actually not smooth at that point, see example 7.7 on ~ ( π , 0) = (1, 0, 0). So the normal vector is page 441 for a similar example). ~r( π2 , 0) = (1, 0, 0) and N 2 pointing outwards (away from the origin). ~ (~r(u, v)) = (sin(u) cos(v), sin(u) sin(v), cos(u)) RF 2π Rπ (sin(u) cos(v), sin(u) sin(v), cos(u)) · (sin2 (u) cos(v), sin2 (u) sin(v), cos(u) sin(u))dudv 0R 0R R 2π R π R 2π 2π π = 0 0 sin3 (u) cos2 (v)+sin3 (u) sin2 (v)+cos2 (u) sin(u)dudv = 0 0 sin(u)dudv = 0 2dv = 4π. So our answer is −4π . Note that it makes sense that 4π is the result of the above integral since the vector field F~ (x, y, z) = (x, y, z) emanates out from the origin and so the net volume passing through the sphere per unit time should be equal to the surface area of the sphere, which is 4π.

Practice Problem: RR Compute F~ · dS where S is the paraboloid y = 4 − x2 − z 2 with normal vector pointing in the S positive y-direction and F~ (x, y, z) = (0, 3, 0). Solution: There is actually a quick way to find the solution to this problem without having take any integrals. If we imagine F~ to be describing the velocity of a liquid that fills all of R3 , then remember that this integral is telling us the net volume of liquid that passes through S in a unit of time. F~ (x, y, z) = (0, 3, 0) is telling us that this liquid is moving in the positive y-direction at a rate of 3 units per unit time. Furthermore, if you picture yourself at the origin and looking in the positive y-direction, S appears to be just a circle of radius 2. So this cross-sectional area of S is really the entire surface area that the liquid is passing through. So the volume of water passing through S per unit time is just the area of this circle (4π) times the velocity of water (3), giving us 12π . Of course, we can also back this answer up with the integral. Let ~r(u, v) = (u cos(v), 4 − 2 u , u sin(v)) (basically cylindrical coordinates but swapping the roles of y and z) with 0 ≤ v ≤ 2π ∂~r ∂~r ~ (u, v) = = (cos(v), −2u, sin(v)) and = (−u sin(v), 0, u cos(v)) so N and 0 ≤ u ≤ 2. Then ∂u ∂v (−2u2 cos(v), −u, −2u2 sin(v)). Note that due to the −u in the second coordinate, this normal vector is pointing in the negative y-direction on S, so we will need to multiply the result of this integral by ~ (u, v) = −3u. −1.R F~ (~rR(u, v)) = (0, 3,R0) so F~ (~r(u, v)) · N 2π 2 2π −3ududv = 0 −6dv = −12π. 0 0 So we take the negative of this result and get 12π .