Review Sheet for First Midterm Exam

Winter 2012 Review Sheet for First Midterm Exam Solutions 1. Describe the motion of a particle with position (x, y): (a) x = sin π2 t, y = cos πt, 1 ≤ t ≤ 2, and (b) x = 2 cos t, y = 3 tan t, 0 ≤ t ≤ π/4. .......................................................................... (a) Using the general relation sin2 θ = (1 − cos 2θ)/2, we find that the particle moves along the parabola y = −2x2 + 1 starting at (1, −1) and ending at (0, 1). (b) Using the general relation 1+tan2 θ = sec2 θ, we find that the p √ particle 2 moves along y = 3 (4/x ) − 1 starting at (2, 0) and ending at ( 2, 3). 2. Match the parametric equations with Curve I (left), Curve II (center), and Curve III (right). (a) x = t+cos(cos t), y = tan(cos t), (b) x = t+sin 4t, y = tan(sin t), (c) x = t3 − t, y = 3 − t2 . .......................................................................... (a) III, (b) II, (c) I 3. The parametric equations x = 4 cos θ, y = 3 sin θ, 0 ≤ θ ≤ 2π give an ellipse. Find the Cartesian equation for this ellipse. .......................................................................... x2 y 2 + 2 = 1. 42 3 1 Math 255-003 -- Math 255-003 -- Math 255-003 -- Math 2 4. Find√an equation of the tangent to the curve, x = csc θ, y = cot θ at point (2, 3) by two methods: (a) without eliminating the parameter, and (b) by first eliminating the parameter. . . . . . . . . . . . . . . . . . . . . .√ ..................................................... (a) The point (2, 3) arises from θ = π/6. Note that dy/dθ −1/ sin2 θ dy = = = sec θ. dx dx/dθ − cos θ/ sin2 θ √ Therefore, the slope of the tangent when θ = π/6 is dy/dx = 2/ 3. The √ equation of the tangent at (2, 3) is 1 (1) y = √ (2x − 1). 3 √ √ (b) Note that y 2 = x2 −1. The point (2, 3) is on the curve y = x2 − 1. By taking the derivative, we obtain x dy =√ . dx x2 − 1 √ √ Therefore, the slope of the tangent at (2, 3) is dy/dx = 2/ 3. The equa√ tion of the tangent at (2, 3) is given by Eq. (1). 5. Find the points on the curve x = t3 + 2t2 − 7t, y = −t3 − 2t2 + 1 where the tangent is horizontal or vertical. .......................................................................... Note that dx = (3t + 7)(t − 1), dt dy = −t(3t + 4). dt The tangent is horizontal when dy/dt = 0 (dx/dt 6= 0). This happens when t = −4/3, 0. The corresponding points are   284 5 ,− and (0, 1). 27 27 The tangent is vertical when dx/dt = 0 (dy/dt 6= 0). This happens when t = −7/3, 1. The corresponding points are   392 76 , and (−4, −2). 27 27 2 Math 255-003 -- Math 255-003 -- Math 255-003 -- Math 2 6. Using the parametric equations of the ellipse, x2 /42 + y 2 /32 = 1, find the area that it encloses. .......................................................................... The parametric equations are given by x = 4 cos θ, y = 3 sin θ, 0 ≤ θ ≤ 2π. The area A is calculated as Z 0 Z 4 3 sin θ(−4 sin θ)dθ ydx = 4 A = 4 π/2 0 = −48 = 12π. Z 0 π/2 1 − cos 2θ dθ = 48 2 Z π/2 0 1 dθ 2 7. Find the area of the region enclosed by the deltoid x = 2 cos θ + cos 2θ, y = 2 sin θ − sin 2θ, 0 ≤ θ ≤ 2π. .......................................................................... The area A is calculated as ! ! Z π Z −1 Z 0 Z 3 dx dx y(θ) dθ ydx = 2 A=2 y(θ) dθ − ydx − dθ dθ 2π/3 −3/2 2π/3 −3/2 Note that dx/dθ = −2 sin θ − 2 sin 2θ. Z π
2 sin2 θ + 2 sin θ sin 2θ dθ A = 2 Z0 π (1 + cos θ − cos 2θ − cos 3θ) dθ = 2 0 = 2π, where we used sin2 θ = 1 − cos 2θ , 2 sin θ sin 2θ = cos θ − cos 3θ . 2 The latter relation is obtained by cos(A − B) − cos(A + B) = (cos A cos B + sin A sin B) − (cos A cos B − sin A sin B) = 2 sin A sin B. (Note) The parameter θ is not the angle Θ in polar coordinates although both move from 0 to 2π. Indeed θ is different from Θ except for θ = nπ/3 3 Math 255-003 -- Math 255-003 -- Math 255-003 -- Math 2 (n = 0, ±1, ±2, . . . ). Therefore, we cannot calculate the area as A = = Z Z 2π 0 2π 0 Z 2π  1 1 2 (2 cos θ + cos 2θ)2 + (2 sin θ − sin 2θ)2 dθ r dθ = 2 2 0 1 (5 + 4 cos 3θ)dθ = 5π. 2 8. Find the length of the deltoid x = 2 cos θ + cos 2θ, y = 2 sin θ − sin 2θ, 0 ≤ θ ≤ 2π. .......................................................................... The length L is calculated as s Z 2π  2  2 dx dy L = + dθ dθ dθ 0 Z 2π p = (−2 sin θ − 2 sin 2θ)2 + (2 cos θ − 2 cos 2θ)2 dθ 0 √ Z 2π √ 1 − cos 3θdθ = 2 2 0 s   √ Z 2π 2 3 = 2 2 2 sin θ dθ 2 0 = 16. 9. Find the area of the region that lies both inside the circle (x − 3/2)2 + y 2 = 9/4 and inside the cardioid r = 1 + cos θ. (Use the fact that the circle is given by r = 3 cos θ in polar coordinates.) .......................................................................... The two curves intersect when 3 cos θ = 1 + cos θ, which gives θ = π/3 and 5π/3. The area A is obtained as "Z # Z π/3 π/3 9 1 1 A= π−2 (3 cos θ)2 dθ − (1 + cos θ)2 dθ . 4 2 2 0 0 4 Math 255-003 -- Math 255-003 -- Math 255-003 -- Math 2 Note that Z π/3 2 9 2 Z π/3 (1 + cos 2θ)dθ √ ! 9 π 3 = , + 2 3 4  Z π/3  Z π/3 1 + cos 2θ 2 1 + 2 cos θ + dθ (1 + cos θ) dθ = 2 0 0 √ π 9 3 = + . 2 8 (3 cos θ) dθ = 0 Thus, 0 " √ √ # 9 5 9 3 π 9 3 3 A= π− = π. π+ − − 4 2 8 2 8 4 10. Identify the curve by finding a Cartesian equation for the curve r = 4 sin θ + 3 cos θ. .......................................................................... We first obtain 3 3 x = r cos θ = cos 2θ + 2 sin 2θ + , 2 2 3 y = r sin θ = −2 cos 2θ + sin 2θ + 2. 2 From the above two equations, we obtain   7 2 2 (4x + 3y − 12) . 3x − 4y + , sin 2θ = cos 2θ = 25 2 25 Then, we have 9 9 9 cos2 2θ + 4 sin2 2θ + 6 cos 2θ sin 2θ + cos 2θ + 6 sin 2θ + , 4 2 4 9 2 2 2 y = 4 cos 2θ + sin 2θ − 6 cos 2θ sin 2θ − 8 cos 2θ + 6 sin 2θ + 4. 4 Finally, we obtain x2 + y 2 = 3x + 4y. x2 = This is a circle with center (3/2, 2) and radius 5/2:    2 3 2 5 2 x− + (y − 2) = . 2 2 5 Math 255-003 -- Math 255-003 -- Math 255-003 -- Math 2 11. Find a polar equation for the curve represented by the Cartesian equation, x = 5. .......................................................................... Since x = r cos θ, we obtain r cos θ = 5. 12. Sketch the curve x4 + y 4 + 2xy(xy − 1) = 0. .......................................................................... The equation is rewritten as (x2 + y 2 )2 = 2xy. Note that x = r cos θ, We obtain y = r sin θ, x2 + y 2 = r 2 . √ r = ± sin 2θ. Thus, r is defined √ in θ ∈ [0, π/2] and θ ∈ [π, 3π/2]. The curve is sketched as follows using r = sin 2θ (0 ≤ θ ≤ π/2, π ≤ θ ≤ 3π/2). 6 Math 255-003 -- Math 255-003 -- Math 255-003 -- Math 2 13. Find the points of intersection of two cardioids C1 : r =1 + sin θ and  C2 : r = 1 − cos θ. Find the slopes of two tangents at (r, θ) = 1 + √12 , 34 π . .......................................................................... Two cardioids intersect at (r, θ) = (1 + √12 , 43 π), (1 − √12 , 47 π) in addition to the origin. For C1 , we have x = r cos θ = cos θ + 1 sin 2θ, 2 y = r sin θ = sin θ + 1 − cos 2θ . 2 At θ = 34 π, the slope is obtained as √ dy dy/dθ cos θ + sin 2θ = = = 1 + 2. dx dx/dθ − sin θ + cos 2θ For C2 , we have x = r cos θ = cos θ − 1 + cos 2θ , 2 y = r sin θ = sin θ − 1 sin 2θ. 2 At θ = 43 π, the slope is obtained as dy dy/dθ cos θ − cos 2θ 1 √ . = = = dx dx/dθ − sin θ + sin 2θ 1+ 2 14. Sketch the curve r = cos2 θ, and find the area it encloses. .......................................................................... The curve is sketched as follows. 7 Math 255-003 -- Math 255-003 -- Math 255-003 -- Math 2 The area A is calculated as follows. Z 2π Z 2π 1 1 2 r dθ = cos4 θdθ A = 2 2 0 0     Z Z 1 2π 3 1 1 2π 1 + cos 2θ 2 dθ = + 2 cos 2θ + cos 4θ dθ = 2 0 2 8 0 2 2 Z 2π 3 3 1 dθ = π. = 8 0 2 8 √ 15. Find all the points of intersection of r = cos 3θ and r2 = sin 3θ. .......................................................................... At the points of intersection, we have 0 ≤ cos 3θ = sin 3θ. Therefore, 3θ = π/4 + 2nπ, where n is an integer (n ∈ Z). It is enough if we consider θ in the range [0, 2π]. We obtain θ= π , 12 3 π, 4 17 π. 12 For these values of θ, we have r = 2−1/4 . Let us check the case of r = 0 because two curves can√take different values of θ if they intersect at the origin. The curve r = cos 3θ reaches the origin when θ = π6 , π2 , 65 π, 76 π, 3 11 π 2 2 2 π, and 6 π. The curve r = sin 3θ reaches the origin when θ = 0, 3 , 3 π, π, 4 5 3 π, and 3 π. Therefore, the two curves also intersect at the origin. In polar coordinates, the points of intersection are obtained as      1 π −4 − 41 3 − 41 17 (r, θ) = (0, θ), 2 , , 2 , π , 2 , π . 12 4 12 16. Find the length of the polar curve r = 2θ for 0 ≤ θ ≤ π. .......................................................................... The length L is calculated as follows. s  2 Z πp Z π dr 2 r + θ2 + 1dθ. dθ = 2 L= dθ 0 0 We introduce t as θ = sinh t. We obtain Z sinh−1 (π) Z sinh−1 (π) 1 + cosh 2t dt = cosh2 tdt = 2 L = 2 2 0 0 p = sinh−1 (π) + π 1 + π 2 . 8 Math 255-003 -- Math 255-003 -- Math 255-003 -- Math 2 Here, we used cosh sinh −1 q
p (π) = 1 + sinh2 sinh−1 (π) = 1 + π 2 .

If we put sinh−1 (π) = ln u, we obtain π = u − u1 /2 and thus u = π + √ 1 + π 2 . So, the solution is also written as   p p L = ln π + 1 + π 2 + π 1 + π 2 . 17. Find the surface area generated by rotating the curve r2 = 3 cos 2θ about the line θ = 0. .......................................................................... The curve is sketched as follows. √ We have r = 3 cos 2θ (−π/4 ≤ θ ≤ π/4, 3π/4 ≤ θ ≤ 5π/4). We note that the line θ = 0 is the x-axis, and x = r cos θ and y = r sin θ. The surface S 9 Math 255-003 -- Math 255-003 -- Math 255-003 -- Math 2 is calculated as S = 2 = 2 = 2 Z Z Z π/4 2πy 0 s π/4 2πx 0 s dx dθ r2 + 2   dy dθ 2 dθ + dr dθ √ π/4 2π 3 cos 2θ cos θ 0 Z = 12π √ 0 = 6 2π. 2 dθ s 3 cos 2θ + 3 π/4  − sin 2θ √ cos 2θ 2 dθ cos θdθ 18. Find an equation of a sphere if one of its diameters has endpoints (−1, −2, −3) and (1, 1, 3). .......................................................................... The diameter of the sphere, which is the distance between two endpoints, is p (1 + 1)2 + (1 + 2)2 + (3 + 3)2 = 7. The center of the sphere, which is the midpoint between two endpoints, is     −1 + 1 −2 + 1 −3 + 3 1 , , = 0, − , 0 . 2 2 2 2 Therefore, the equation of the sphere is given by   1 2 2 + z 2 = 49. x + y+ 2 19. Describe in words the region of R3 represented by the inequality, x2 + y 2 + z 2 ≤ 2x + 4y + 20. .......................................................................... The equation is rewritten as (x − 1)2 + (y − 2)2 + z 2 ≤ 52 . Therefore, the region consists of the points whose distance from the point (1, 2, 0) is at most 5. 10 Math 255-003 -- Math 255-003 -- Math 255-003 -- Math 2 20. Find the volume of the solid that lies inside both of the spheres x2 + y 2 + z 2 − 2x − 2y + 1 = 0 and x2 + y 2 + z 2 + 2z = 0. .......................................................................... Two equations are rewritten as (x − 1)2 + (y − 1)2 + z 2 = 1, x2 + y 2 + (z + 1)2 = 1. The radius√of each sphere is √ 1. The distance between the centers of two spheres is 12 + 12 + 12 = 3. Let us define the X-axis so that it passes the centers of two spheres, and take the Y -axis so that it is perpendicular to the X-axis and it passes the midpoint between two centers. The Z-axis is drawn by the right-hand rule. In the XY -plane, traces of these two spheres   √ 2 √ 2 are circles X − 23 + Y 2 = 1 and X + 23 + Y 2 = 1. The volume V is then calculated as follows. V = Z 0 √ −1+ 23 √ = 2 Z = 2π 1− 1− 0 " 1− √ 3 2 πY 2 dX 0 3 2 0 Z πY 2 dX + Z πY 2 dX √  3 2 1 − √  √ !2 3  X+ dX 2 #1− √ 3 2 3 2 1 1 X + X = 2π − X 3 − 3 2 4 0 √ ! √ ! 13 5 3 3 7 3 = 2π − + + + − 12 8 2 8 √ ! 2 3 3 − . = 2π 3 8 √ !! 3 1 − 4 8 21. Find |a|, a + b, 2a − 3b when a = 3i + j − 2k and b = h2, 4, 3i. .......................................................................... |a| = √ 14, a + b = h5, 5, 1i, 2a − 3b = h0, −10, −13i. 11 Math 255-003 -- Math 255-003 -- Math 255-003 -- Math 2 22. A boy walks due west on the deck of a ship at 4 mi/h. The ship is moving south at a speed of 32 mi/h. Find the speed of the boy relative to the surface of the water. .......................................................................... The speed v mi/h is calculated as √ √ v = | − 4i − 32j| = 1040 = 4 65. 23. A 3-lb weight hangs from two strings. The strings, fastened at different heights, make angles of 60◦ and 45◦ with the horizontal. Find the tension in each string and the magnitude of each tension. .......................................................................... Two tension vectors are written as T1 = −|T1 | cos 60◦ i + |T1 | sin 60◦ j, T2 = |T2 | cos 45◦ i + |T2 | sin 45◦ j. We have |T1 | cos 60◦ = |T2 | cos 45◦ , |T1 | sin 60◦ + |T2 | sin 45◦ = 3. √ From the first equation, we obtain |T2 | = (1/ 2)|T1 |. Then, we plug this relation into the second equation, and obtain √ 6 3 2 √ , |T2 | = √ . |T1 | = 1+ 3 1+ 3 The tension vectors are obtained as  √  3 √ −i + 3j , T1 = 1+ 3 T2 = 3 √ (i + j) . 1+ 3 24. Use vectors to prove that the line segments joining the midpoints of the sides of any quadrilateral form a parallelogram. .......................................................................... Let us take the vertices of the quadrilateral in counterclockwise order as P = (x1 , y1 ), Q = (x2 , y2 ), R = (x3 , y3 ), and S = (x4 , y4 ). The vertices of the quadrilateral formed by joining the midpoints of the seg
2 y1 +y2 ments of the quadrilateral P QRS are then obtained as p = x1 +x , , 2 2 12 Math 255-003 -- Math 255-003 -- Math 255-003 -- Math 2


x3 +x4 y3 +y4 x4 +x1 y4 +y1 3 y2 +y3 , , r = , , and s = , . We conq = x2 +x 2 2 2 2 2 2

− → → x4 −x2 y4 −y2 1 y3 −y1 , , u = qr = , , u3 = sider vectors u1 = − pq = x3 −x 2 2 2 2 2

− → → − x2 −x4 y2 −y4 x1 −x3 y1 −y3 , and u4 = sp = . We see that u1 and u3 rs = 2 , 2 2 , 2 are parallel, and u2 and u4 are also parallel. Therefore, the quadrilateral pqrs is a parallelogram. 25. Which of the following expressions are meaningful? Which are meaningless? Explain. (a) a · b · c, (b) ab, (c) a(b · c), (d) |a|b, (e) |a · b| · c, and (f) |a + b|. .......................................................................... (a) meaningless, (b) meaningless, (c) meaningful, (d) meaningful, (e) meaningless, (f) meaningful. 26. Find the angle between vectors a = h2, 1, −1i and b = h0, 3, 5i. (Find an exact expression.) .......................................................................... Let θ be the angle. cos θ = √ Therefore, θ = cos−1 (−1/ 51). a·b 1 = −√ . |a||b| 51 27. Determine whether the given vectors are orthogonal, parallel, or neither. (a) u = h21, −16, 12i, v = h4, 6, 1i, (b) u = 18i − 12j + 30k, v = −21i + 14j − 35k, and (c) u = h21, −16, 12i, v = 18i − 12j + 30k. .......................................................................... (a) orthogonal, (b) parallel, (c) neither. 28. For vectors a = h2, 1i and b = h3, −4i, find ortha b .......................................................................... The vector projection is calculated as proja b = a·b 2 a = h2, 1i. 2 |a| 5 13 Math 255-003 -- Math 255-003 -- Math 255-003 -- Math 2 Then the orthogonal projection is calculated as     11 22 4 2 , = ,− . ortha b = b − proja b = h3, −4i − 5 5 5 5 29. Find the distance from the point (−1, 3) to the line 3x − 2y + 7 = 0. .......................................................................... Let us, for example, consider the y-intercept of the line. By putting x = 0, we see that the line passes (0, 7/2). Indeed, the equation for the line is rewritten as 3x − 2(y − 7/2) = 0. This is equivalently expressed as h3, −2i · hx, y − 7/2i = 0. Therefore, the line is perpendicular to the vector a = h3, −2i. We consider the displacement vector b from (0, 7/2) to the point (−1, 3), and take the scalar projection onto a. −2 a·b h3, −2i · h−1, −1/2i √ =√ . = |a| 9+4 13 √ The distance is | compa b| = 2/ 13. compa b = 30. Find a vector equation of a sphere with the radius 2 and the center at (9, 2, 4). .......................................................................... Let P and Q be endpoints of a diameter. Let R be any point on the −−→ −−→ −−→ sphere. We define position vectors a = OP , b = OQ, and r = OR. The angle ∠P RQ makes the right angle. This relation is expressed as (r − a) · (r − b) = 0. This is a vector equation of the sphere. Since the radius is 2 and the center is at C(9, 2, 4), a and b are given as −−→ a = OC − u, −−→ b = OC + u, where u is a vector with the magnitude 2. For example, if we take u = (2, 0, 0), we obtain (r − h7, 2, 4i) · (r − h11, 2, 4i) = 0. 14 Math 255-003 -- Math 255-003 -- Math 255-003 -- Math 2 31. By the protonation of ammonia NH3 , the ammonium cation NH+ 4 is obtained, in which four hydrogen atoms are at the vertices of a regular tetrahedron and a nitrogen atom is at the centroid. The bond angle is the angle formed by the H-N-H combination. Show that the bond angle is about 109.5◦ . (Use cos−1 (−1/3) ≃ 109.5◦ .) .......................................................................... In Cartesian coordinates, we place hydrogen atoms at (1, 0, 0), (0, 1, 0), (0, 0, 1), and (1, 1, 1) (here the unit of length is arbitrary). Then, these four points form a regular tetrahedron. Let us place the nitrogen atom at the centroid (1/2, 1/2, 1/2). The bond angle θ is the angle between displacement vectors u = h1, 0, 0i − h1/2, 1/2, 1/2i = h1/2, −1/2, −1/2i and v = h0, 1, 0i − h1/2, 1/2, 1/2i = h−1/2, 1/2, −1/2i. cos θ = u·v 1 =− . |u||v| 3 Therefore, θ = cos−1 (−1/3) ≃ 1.9106 ≃ 109.5◦ . 32. For a = ti + (1 − t)j + k and b = 2ti + j + (t − 2)k, find the cross product a × b and verify that it is orthogonal to both a and b. .......................................................................... a × b = ht, 1 − t, 1i × h2t, 1, t − 2i = (−t2 + 3t − 3)i + (−t2 + 4t)j + (2t2 − t)k. Furthermore, we have a · (a × b) = ht, 1 − t, 1i · h−t2 + 3t − 3, −t2 + 4t, 2t2 − ti = 0, b · (a × b) = h2t, 1, t − 2i · h−t2 + 3t − 3, −t2 + 4t, 2t2 − ti = 0. 33. State whether each expression is meaningful. If not, explain why. If so, state whether it is a vector or a scalar. (a) (a×b)×(c×d), (b) (a×b)·(c×d), and (c) (a × b) × (c · d). .......................................................................... (a) vector, (b) scalar, (c) meaningless (c · d is a scalar.) 15 Math 255-003 -- Math 255-003 -- Math 255-003 -- Math 2 34. Find the area of the parallelogram with vertices A(2, −1, 1), B(1, 0, 3), C(5, 6, 2), and D(6, 5, 0). .......................................................................... −−→ −→ −−→ We obtain AB = h−1, 1, 2i, AC = h3, 7, 1i, and AD = h4, 6, −1i. Note −−→ −−→ −→ that AB + AD = AC. The area A is calculated as √ −−→ −−→ A = |AB × AD| = |h−13, 7, −10i| = 318. 35. Find the volume of the parallelepiped determined by vectors a = h2, 3, 5i, b = h−1, 1, 2i, and c = h4, 6, −1i. .......................................................................... The volume V is calculated as V = |a · (b × c)| = |h2, 3, 5i · h−13, 7, −10i| = | − 55| = 55. 36. Verify that the vectors a = 5i + 5j − 3k, b = 4i + 6j − k, and c = 6i + 4j − 5k are coplannar. .......................................................................... The scalar triple product is calculated as a · (b × c) = h5, 5, −3i · h−26, 14, −20i = 0. Therefore, a, b, and c are coplannar. 37. Determine whether the points A(5, 6, 2), B(2, −1, 1), C(9, 11, −4), and D(1, 0, 3) lie in the same plane. .......................................................................... −−→ −→ For vectors a = AB = h−3, −7, −1i, b = AC = h4, 5, −6i, and c = −−→ AD = h−4, −6, 1i, the scalar triple product is calculated as a · (b × c) = h−3, −7, −1i · h−31, 20, −4i = −43 6= 0. Therefore, A, B, C, and D don’t lie in the same plane. 16 Math 255-003 -- Math 255-003 -- Math 255-003 -- Math 2 38. Find the distance from the point P (3, 1, 4) to the plane through the points A(1, 1, 0), B(0, 2, 2), and C(3, 0, 3). .......................................................................... −−→ −→ The normal vector n of the plane spanned by AB and AC is −−→ −→ n = AB × AC = h−1, 1, 2i × h2, −1, 3i = h5, 7, −1i. Then, the distance d is calculated as d= n −→ 5(2) + 7(0) − 1(4) 6 √ · AP = = √ . |n| 25 + 49 + 1 5 3 39. Prove a × (b × c) = (a · c)b − (a · b)c. .......................................................................... a × (b × c) = ha1 , a2 , a3 i × hb2 c3 − b3 c2 , b3 c1 − b1 c3 , b1 c2 − b2 c1 i = ha2 (b1 c2 − b2 c1 ) − a3 (b3 c1 − b1 c3 ), a3 (b2 c3 − b3 c2 ) − a1 (b1 c2 − b2 c1 ), a1 (b3 c1 − b1 c3 ) − a2 (b2 c3 − b3 c2 )i = [(a2 c2 + a3 c3 )b1 − (a2 b2 + a3 b3 )c1 ]i + [(a1 c1 + a3 c3 )b2 − (a1 b1 + a3 b3 )c2 ]j +[(a1 c1 + a2 c2 )b3 − (a1 b1 + a2 b2 )c3 ]k = [(a1 c1 + a2 c2 + a3 c3 )b1 − (a1 b1 + a2 b2 + a3 b3 )c1 ]i +[(a1 c1 + a2 c2 + a3 c3 )b2 − (a1 b1 + a2 b2 + a3 b3 )c2 ]j +[(a1 c1 + a2 c2 + a3 c3 )b3 − (a1 b1 + a2 b2 + a3 b3 )c3 ]k = (a · c)b − (a · b)c. 40. Suppose that a 6= 0. (a) If a · b = |a|2 , does it follow that a = b? (b) If a × b = b × a, does it follow that b = 0? .......................................................................... (a) No. In this case, proja b = a. But, in general, b itself is different from a. (b) Yes. 17 Math 255-003 -- Math 255-003 -- Math 255-003 -- Math 2 41. Find parametric equations and symmetric equations for the line of intersection of the planes x + y + 2z = 3 and 2x + y = 1. .......................................................................... We eliminate x in the system ( x + y + 2z = 3, 2x + y = 1, and obtain z = (y − 5)/(−4). Similarly by eliminating y, we obtain z = (x + 2)/2. Therefore, we can write (x + 2)/2 = (y − 5)/(−4) = z, or y−5 x+2 = = z. 2 −4 (2) The parametric equations are obtained as x = −2 + 2t, y = 5 − 4t, z = t. (Note) The line of intersection can also be obtained as follows. The normal vectors of the planes are n1 = h1, 1, 2i and n2 = h2, 1, 0i. The line of intersection is parallel to n1 × n2 = h−2, 4, −1i and passes through (−2, 5, 0) in the xy-plane. Therefore, the symmetric equations are written as Eq. (2). 42. Find an equation of the plane through the point (1, −8, 2) and perpendicular to the line x = 1 + 2t, y = 3 − 5t, z = t. .......................................................................... The normal vector is h2, −5, 1i. The equation of the plane is 2(x − 1) − 5(y + 8) + (z − 2) = 0 or 2x − 5y + z − 44 = 0. 43. Find symmetric equations for the line of intersection of the planes x − y + 3z = 8 and x + 5y + z = 2. Let θ denote the angle between the planes. Find | cos θ|. .......................................................................... We eliminate x in the system ( x − y + 3z = 8, x + 5y + z = 2, 18 Math 255-003 -- Math 255-003 -- Math 255-003 -- Math 2 and obtain z = 3(y + 1). Similarly by eliminating y, we obtain z = (−3/8)(x − 7). Therefore, we have (−3/8)(x − 7) = 3(y + 1) = z. Symmetric equations are obtained as z x−7 =y+1= . −8 3 (3) The normal vectors of the planes are n1 = h1, −1, 3i, n2 = h1, 5, 1i. Hence, cos θ = n1 · n 2 1(1) − 1(5) + 3(1) 1 √ =√ =− √ . |n1 ||n2 | 1 + 1 + 9 1 + 25 + 1 3 33 We obtain 1 | cos θ| = √ . 3 33 (Note) The line of intersection can also be obtained as follows. The line is parallel to n1 × n2 = h−16, 2, 6i and passes through (7, −1, 0) in the xy-plane. Therefore, the symmetric equations are written as Eq. (3). 44. Find the point at which lines h−2, 1, −10i + th1, −1, 3i and x − 1 = y = z + 3 intersect. Then, find an equation for the plane that contains these lines. .......................................................................... Parametric equations of two lines are obtained as x = −2 + t, x = 1 + s, y = 1 − t, y = s, z = −10 + 3t, z = −3 + s Therefore, we have t = 2 and s = −1 at the point of intersection. Two lines intersect at (0, −1, −4). The normal vector n is calculated as n = h1, −1, 3i × h1, 1, 1i = h−4, 2, 2i. The point (0, −1, −4) is a point on the plane (we can take any point on two lines). The equation of the plane is obtained as −4x+2(y +1)+2(z +4) = 0, or −2x + y + z + 5 = 0. 19 Math 255-003 -- Math 255-003 -- Math 255-003 -- Math 2 45. Identify the surface x2 − 4y 2 − 4z 2 + 8y − 16z = 16. .......................................................................... The equation is rewritten as x2 = 1. 22 Thus, the surface is a hyperboloid of one sheet with axis of symmetry the x-axis. Traces are circles in planes perpendicular to the x-axis and traces are hyperbolas in planes parallel to the x-axis. (y − 1)2 + (z + 2)2 − 46. Find an equation for the surface obtained by rotating the line z = 2x, y = 0 about the z-axis. .......................................................................... The surface is a cone whose equation is z2 = x2 + y 2 . 22 47. Find an equation for the surface consisting of all points P for which the distance from P to the y-axis is half the distance from P to the xz-plane. Identify the surface. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .√ ............................ For P (x, y, z), the distance to the y-axis is x2 + z 2 and the distance to the xz-plane is |y|. Therefore, the equation for P is y2 . 4 Thus, the surface is a cone. The trace in the y = k plane is x2 + z 2 = (k/2)2 , which is a circle with radius k/2. The trace in the x = k plane is y 2 /(2k)2 − z 2 /k 2 = 1, which is a hyperbola. The trace in the z = plane is y 2 /(2k)2 − x2 /k 2 = 1, which is also a hyperbola. x2 + z 2 = 48. Change (2, −π/6, 3π/4) from spherical to cylindrical coordinates. .......................................................................... In spherical coordinates, we have (ρ, θ, φ) = (2, −π/6, 3π/4).     √ √ 3π 3π r = ρ sin φ = 2 sin = 2, = − 2. z = ρ cos φ = 2 cos 4 4 20 Math 255-003 -- Math 255-003 -- Math 255-003 -- Math 2 49. Identify the surface y = y 2 + z 2 and write the equation both in cylindrical coordinates and in spherical coordinates. .......................................................................... First of all, the equation is written as follows in Cartesian coordinates.   1 1 2 + z2 = . y− 2 4 Therefore, the surface is a circular cylinder whose axis is parallel to the xaxis and its traces are circles. In particular, the trace in the plane x = 0 is a circle whose center is (0, 1/2, 0). This circular cylinder is given in cylindrical coordinates as   1 2 1 r sin θ − + z2 = , 2 4 and given in spherical coordinates as   1 1 2 + ρ2 cos2 φ = . ρ sin φ sin θ − 2 4 p 50. A solid lies above the cone z = x2 + y 2 and below the sphere x2 + y 2 +z 2 = z. Write a description of the solid in terms of inequalities involving spherical coordinates. .......................................................................... In Cartesian coordinates, the solid is expressed as  2  2 p 1 1 2 2 z > x2 + y 2 , x +y + z− < . 2 2 Using x = ρ sin φ cos θ, y = ρ sin φ cos θ, and z = ρ cos φ (ρ ≥ 0 and 0 ≤ φ ≤ π), we obtain π 0 < ρ < cos φ. 0<φ< , 4 51. Find a vector perpendicular to the plane through the points A(1, 4, 0), B(3, 0, −1), and C(2, 4, 3). Then, find the area of triangle ABC. .......................................................................... −−→ −→ Two vectors in the plane are AB = h2, −4, −1i and AC = h1, 0, 3i. Their cross product is perpendicular to the plane. h2, −4, −1i × h1, 0, 3i = h−12, −7, 4i. 21 Math 255-003 -- Math 255-003 -- Math 255-003 -- Math 2 The area of the triangle ABC is half of the area of the parallelogram spanned −−→ −→ by the vectors AB and AC. The area is √ 209 1p 1 2 2 2 (−12) + (−7) + 4 = |h2, −4, −1i × h1, 0, 3i| = . 2 2 2 22 Math 255-003 -- Math 255-003 -- Math 255-003 -- Math 2