Math 3C — Exam #1 Sample Laney College, Spring 2011 Fred Bourgoin 1. (14 points) Consider the function f (x, y) = x2 + 3y 2 + 1. (a) Sketch a contour diagram for f with at least 3 level curves. (b) Sketch the graph of f . (c) Find a function g(x, y, z) such that the graph of f is a level surface of g. (Don’t forget to indicate which level surface of g it is.) 2. (12 points) Consider the points P = (3, 2, −5) and Q = (−1, 3, −2) in R3 . −−→ (a) Write the vector P Q in terms of ~i, ~j, and ~k. −−→ (b) Find the magnitude of QP . (c) What is the distance between P and Q? 3. (10 points) Suppose P and Q are as in the preceding problem, and let R = (2, 6, 4). (a) What is the area of the triangle △P QR? (b) Find an equation for the plane through P , Q, and R. 4. (8 points) Find the angle between the planes z = 2x − y + 3 and x − 3y + 2z = 5. p 5. (8 points) Sketch the domain of f (x, y) = 1 + x − y 2 . 6. (10 points) Shortly after takeoff, a plane is climbing northwest through still air at an airspeed of 200 km/hr, and rising at a rate of 300 m/min. Resolve its velocity vector into components. The x-axis points east, the y-axis points north, and the z-axis points up. 7. (10 points) The following table gives values of a function h(x, y) at 20 points. 1 3 x 5 7 9 −3 y −2 −4 −2 −3 −5 −6 −7 −1 0 1 3 0 2 −3 −1 1 −5 −3 −1 −1 −4 −2 0 (a) Could h be a linear function? Why, or why not? (b) Find a possible expression for h. 1 8. (6 points) Describe the level surfaces of f (x, y, z) = p 1 x2 + 2y 2 + z 2 . 9. (16 points) Let ~u = 2~i + ~j − 3~k, ~v = ~i − ~j + 2~k, and w ~ = ~i − 5~j − ~k. (a) Compute 2~u − 3~v . (b) Are two of the vectors perpendicular? Which ones? Justify. ~ 2 such that w ~ =w ~1 + w ~ 2, w ~ 1 is parallel to ~v , and w ~2 (c) Find vectors w ~ 1 and w is perpendicular to ~v . 10. (6 points) Find the volume of the parallelepiped defined by the vectors ~u = 2~i + ~j − 3~k , ~v = ~i − ~j + 2~k , and w ~ = ~i − 5~j − ~k of the preceding problem. EC. True/False (No justification required) (a) If ~u · ~v < 0, then the angle between ~u and ~v is greater than π/2. (b) If the contours of g(x, y) are concentric circles, then the graph of g is a cone. (c) It is never true that ~v × w ~ =w ~ × ~v . (d) The sphere x2 + y 2 + z 2 = 10 intersects the plane x = 10. (e) For any vectors ~u and ~v , we have (~u + ~v ) · (~u − ~v ) = k~uk2 − k~v k2 . 2 Math 3C — Exam #1 Sample Solutions Laney College, Spring 2011 Fred Bourgoin 1. (14 points) Consider the function f (x, y) = x2 + 3y 2 + 1. (a) Sketch a contour diagram for f with at least 3 level curves. Solution. Setting f (x, y) = c yields x2 c−1 + y2 = , 3 3 so your contour diagram should consist of concentric ellipses, closer to one another as you move away from the origin. Shown here are the contours c = 2, 3, 4, 5, 6 (from the inside out). (b) Sketch the graph of f . Solution. The graph is a paraboloid with vertex (0, 0, 1). 3 (c) Find a function g(x, y, z) such that the graph of f is a level surface of g. (Don’t forget to indicate which level surface of g it is.) Solution. Let g(x, y, z) = z − (x2 + 3y 2 + 1). Then the graph of f (x, y) is also the graph of the level surface g(x, y, z) = c = 0. 2. (12 points) Consider the points P = (3, 2, −5) and Q = (−1, 3, −2) in R3 . −−→ (a) Write the vector P Q in terms of ~i, ~j, and ~k. −−→ Solution. P Q = (−1 − 3)~i + (3 − 2)~j + (−2 + 5)~k = −4~i + ~j + 3~k. −−→ (b) Find the magnitude of QP . p √ −−→ −−→ Solution. kQP k = kP Qk = (−4)2 + (1)2 + (3)2 = 26. (c) What is the distance between P and Q? √ −−→ Solution. The distance between P and Q is the magnitude of P Q, so 26. 3. (10 points) Suppose P and Q are as in the preceding problem, and let R = (2, 6, 4). (a) What is the area of the triangle △P QR? −→ Solution. Since P R = −~i + 4~j + 9~k, we have ~i ~j ~k −−→ −→ P Q × P R = −4 1 3 = −3~i + 33~j − 15~k . −1 4 9 −−→ The area of △P QR is half the area of the parallelogram with edges P Q and −→ P R, so √ √ 1323 21 3 1p = . (−3)2 + (33)2 + (−15)2 = 2 2 2 (b) Find an equation for the plane through P , Q, and R. −−→ −→ Solution. The vector P Q × P R is normal to the plane, so the equation is of the form −3x + 33y − 15z = c. To find c, plug in one of the points; you get −3x + 33y − 15z = 132 . 4. (8 points) Find the angle between the planes z = 2x − y + 3 and x − 3y + 2z = 5. Solution. The angle between the planes is the same as the angle between vectors that are normal to them. Rewriting the first equation as 2x − y − z = −3, we see that ~u = 2~i − ~j − ~k is a normal vector to the first plane. A vector that is normal to the second plane is ~v = ~i − 3~j + 2~k. The angle between ~u and ~v is ~u · ~v 3 3 −1 −1 −1 √ √ √ θ = cos = cos = cos ≈ 1.24 rad. k~uk k~v k 6 14 2 21 p 5. (8 points) Sketch the domain of f (x, y) = 1 + x − y 2 . Solution. We need to have 1 + x − y 2 ≥ 0; that is, x ≥ y 2 − 1. This is the region “inside” the sideways parabola x = y 2 − 1 (and the parabola itself). 4 6. (10 points) Shortly after takeoff, a plane is climbing northwest through still air at an airspeed of 200 km/hr, and rising at a rate of 300 m/min. Resolve its velocity vector into components. The x-axis points east, the y-axis points north, and the z-axis points up. Solution. 300 m/min is 18 km/hr. The plane’s ground speed is p p 2002 − 182 = 39,676 ≈ 199 km/hr. Since it is flying 45◦ north of west, its velocity is √ √ p 2~ p 2~ ~v = − 39,676 i + 39,676 j + 18 ~k 2 2 p p = − 19,838~i + 19,838 ~j + 18 ~k . 7. (10 points) The following table gives values of a function h(x, y) at 20 points. 1 3 x 5 7 9 −3 y −2 −4 −2 −3 −5 −6 −7 −1 0 1 3 0 2 −3 −1 1 −5 −3 −1 −1 −4 −2 0 (a) Could h be a linear function? Why, or why not? Solution. Yes, since both m = ∆z ∆x (b) Find a possible expression for h. = − 12 and n = ∆z ∆y = 2 are constant. Solution. If the function is linear, then it is of the form h(x, y) = − 12 x + 2y + c. To find c, notice that f (1, 0) = 3; so c = 72 . Hence, a possible expression for h is h(x, y) − 12 x + 2y + 72 . 8. (6 points) Describe the level surfaces of f (x, y, z) = p Solution. Setting f (x, y, z) = c yields x2 z2 1 + y2 + = 2. 2 2 2c 5 1 x2 + 2y 2 + z 2 . These are ellipsoids. 9. (16 points) Let ~u = 2~i + ~j − 3~k, ~v = ~i − ~j + 2~k, and w ~ = ~i − 5~j − ~k. (a) Compute 2~u − 3~v . Solution. 2~u − 3~v = 2(2~i + ~j − 3~k) − 3(~i − ~j + 2~k) = ~i + 5~j. (b) Are two of the vectors perpendicular? Which ones? Justify. Solution. Yes, ~u and w ~ are perpendicular since ~u · w ~ = 2 − 5 + 3 = 0. (c) Find vectors w ~ 1 and w ~ 2 such that w ~ =w ~1 + w ~ 2, w ~ 1 is parallel to ~v , and w ~2 is perpendicular to ~v . Solution. First replace ~v with a unit vector in the same direction: ~v ′ = ~v 1 = √ (~i − ~j + 2~k) . k~v k 6 Then 2 ~ ~ (i − j + 2~k) 3 1 and w ~2 = w ~ −w ~ 1 = (~i − 13~j − 7~k) . 3 ~ · ~v ′ ) ~v ′ = w ~ 1 = (w 10. (6 points) Find the volume of the parallelepiped defined by the vectors ~u = 2~i + ~j − 3~k , ~v = ~i − ~j + 2~k , and w ~ = ~i − 5~j − ~k of the preceding problem. Solution. The volume is |(~u × ~v ) · w|. ~ ~i ~j ~k ~u × ~v = 2 1 −3 = −~i − 7~j − 3~k 1 −1 2 so |(~u × ~v ) · w| ~ = |(−~i − 7~j − 3~k) · (~i − 5~j − ~k)| = 37 . EC. True/False (No justification required) (a) If ~u · ~v < 0, then the angle between ~u and ~v is greater than π/2. Solution. True. ~u · ~v = k~uk k~v k cos θ < 0 =⇒ cos θ < 0 =⇒ π < θ < π. 2 (b) If the contours of g(x, y) are concentric circles, then the graph of g is a cone. Solution. False. Let g(x, y) = x2 + y 2 . Its graph is not a cone even though its level curves are concentric circles. (c) It is never true that ~v × w ~ =w ~ × ~v . Solution. False. If ~v = w ~ = ~0, then ~v × w ~ =w ~ × ~v . 6 (d) The sphere x2 + y 2 + z 2 = 10 intersects the plane x = 10. Solution. False. To look for the intersection, plug 10 in the equation of the sphere for x: 102 + y 2 + z 2 = 10 =⇒ y 2 + z 2 = −90 , which is impossible; they don’t intersect. (e) For any vectors ~u and ~v , we have (~u + ~v ) · (~u − ~v ) = k~uk2 − k~v k2 . Solution. True. Using the properties of the dot product, (~u + ~v ) · (~u − ~v ) = ~u · (~u − ~v ) + ~v · (~u − ~v ) = ~u · ~u − ~u · ~v + ~v · ~u − ~v · ~v = k~uk2 − k~v k2 . 7