Winter 2012 Math 255. Review Sheet for First Midterm Exam Solutions

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1 Winter 1 Math 55 Review Sheet for First Midterm Exam Solutions 1. Describe the motion of a particle with position x,y): a) x = sin π t, y = cosπt, 1 t, and b) x = cost, y = tant, t π/4. a) Using the general relation sin θ = 1 cosθ)/, we find that the particle moves along the parabola y = x + 1 starting at 1, 1) and ending at,1). b)usingthegeneralrelation1+tan θ = sec θ, wefindthattheparticle moves along y = 4/x ) 1 starting at,) and ending at,).. Match the parametric equations with Curve I left), Curve II center), and Curve III right). a) x = t+coscost), y = tancost), b) x = t+sin4t, y = tansint), c) x = t t, y = t. a) III, b) II, c) I. The parametric equations x = 4cosθ, y = sinθ, θ π give an ellipse. Find the Cartesian equation for this ellipse. x 4 + y = 1. 1 Math Math Math Math

2 4. Find an equation of the tangent to the curve, x = cscθ, y = cotθ at point, ) by two methods: a) without eliminating the parameter, and b) by first eliminating the parameter. a) The point, ) arises from θ = π/6. Note that dy dx = dy/dθ dx/dθ = 1/sin θ cosθ/sin θ = secθ. Therefore, the slope of the tangent when θ = π/6 is dy/dx = /. The equation of the tangent at, ) is y = 1 x 1). 1) b) Notethat y = x 1. Thepoint, ) isonthe curvey = x 1. By taking the derivative, we obtain dy dx = x x 1. Therefore, the slope of the tangent at, ) is dy/dx = /. The equation of the tangent at, ) is given by Eq. 1). 5. Find the points on the curve x = t +t 7t, y = t t +1 where the tangent is horizontal or vertical. Note that dx dt = t+7)t 1), dy dt = tt+4). The tangent is horizontal when dy/dt = dx/dt ). This happens when t = 4/,. The corresponding points are 84 7, 5 ) and,1). 7 The tangent is vertical when dx/dt = dy/dt ). This happens when t = 7/, 1. The corresponding points are 9 7, 76 ) and 4, ). 7 Math Math Math Math

3 6. Using the parametric equations of the ellipse, x /4 + y / = 1, find the area that it encloses. The parametric equations are given by x = 4cosθ, y = sinθ, θ π. The area A is calculated as A = 4 4 = 48 = 1π. ydx = 4 π/ π/ sinθ 4sinθ)dθ 1 cosθ dθ = 48 π/ 1 dθ 7. Find the area of the region enclosed by the deltoid x = cosθ+cosθ, y = sinθ sinθ, θ π. The area A is calculated as A = ydx / 1 / ydx ) = Note that dx/dθ = sinθ sinθ. where we used A = = = π, π π yθ) dx π/ dθ dθ sin θ +sinθsinθ ) dθ π 1+cosθ cosθ cosθ)dθ sin θ = 1 cosθ, sinθsinθ = The latter relation is obtained by π/ cosθ cosθ. ) yθ) dx dθ dθ cosa B) cosa+b) = cosacosb +sinasinb) cosacosb sinasinb) = sinasinb. Note) The parameter θ is not the angle Θ in polar coordinates although both move from to π. Indeed θ is different from Θ except for θ = nπ/ Math Math Math Math

4 n =,±1,±,...). Therefore, we cannot calculate the area as A = = π π 1 π 1 [ r dθ = cosθ +cosθ) +sinθ sinθ) ] dθ 1 5+4cosθ)dθ = 5π. 8. Find the length of the deltoid x = cosθ +cosθ, y = sinθ sinθ, θ π. The length L is calculated as L = = π π = = = 16. dx ) ) dy + dθ dθ dθ sinθ sinθ) +cosθ cosθ) dθ π π 1 cosθdθ sin θ ) dθ 9. Find the area of the region that lies both inside the circle x /) + y = 9/4 and inside the cardioid r = 1+cosθ. Use the fact that the circle is given by r = cosθ in polar coordinates.) The two curves intersect when cosθ = 1+cosθ, which gives θ = π/ and 5π/. The area A is obtained as [ A = 9 π/ ] 4 π 1 π/ 1 cosθ) dθ 1+cosθ) dθ. 4 Math Math Math Math

5 Note that π/ cosθ) dθ = 9 Thus, π/ 1+cosθ) dθ = = 9 π/ π π/ = π cosθ)dθ ) +, 4 1+cosθ + 1+cosθ [ A = 9 4 π π π 9 ] = π. ) dθ 1. Identify the curve by finding a Cartesian equation for the curve r = 4sinθ +cosθ. We first obtain x = rcosθ = cosθ +sinθ +, y = rsinθ = cosθ + sinθ +. From the above two equations, we obtain cosθ = x 4y + 7 ), sinθ = 4x+y 1). 5 5 Then, we have x = 9 4 cos θ +4sin θ +6cosθsinθ + 9 cosθ +6sinθ + 9 4, y = 4cos θ sin θ 6cosθsinθ 8cosθ +6sinθ +4. Finally, we obtain x +y = x+4y. This is a circle with center /,) and radius 5/: x ) 5 +y ) ) =. 5 Math Math Math Math

6 11. Find a polar equation for the curve represented by the Cartesian equation, x = 5. Since x = rcosθ, we obtain rcosθ = Sketch the curve x 4 +y 4 +xyxy 1) =. The equation is rewritten as Note that We obtain x +y ) = xy. x = rcosθ, y = rsinθ, x +y = r. r = ± sinθ. Thus, r is defined in θ [,π/] and θ [π,π/]. The curve is sketched as follows using r = sinθ θ π/, π θ π/). 6 Math Math Math Math

7 1. Find the points of intersection of two cardioids C 1 : r = 1+sinθ and C : r = 1 cosθ. Find the slopes of two tangents at r,θ) = 1+ 1, 4 ). π Two cardioids intersect at r,θ) = 1+ 1, 1 4π), 1, 7 4π) in addition to the origin. For C 1, we have x = rcosθ = cosθ + 1 At θ = 4π, the slope is obtained as For C, we have dy dx = dy/dθ dx/dθ 1 cosθ sinθ, y = rsinθ = sinθ +. = cosθ +sinθ sinθ +cosθ = 1+. x = rcosθ = cosθ 1+cosθ, y = rsinθ = sinθ 1 sinθ. At θ = 4π, the slope is obtained as dy dx = dy/dθ dx/dθ = cosθ cosθ sinθ +sinθ = Sketch the curve r = cos θ, and find the area it encloses. The curve is sketched as follows. 7 Math Math Math Math

8 The area A is calculated as follows. A = = 1 = 1 8 π π π π 1 1 r dθ = cos4 θdθ ) 1+cosθ dθ = 1 8 dθ = 8 π. π +cosθ + 1 ) cos4θ dθ 15. Find all the points of intersection of r = cosθ and r = sinθ. At the points of intersection, we have cosθ = sinθ. Therefore, θ = π/4+nπ, where n is an integer n Z). It is enough if we consider θ in the range [,π]. We obtain θ = π 1, 4 π, 17 1 π. For these values of θ, we have r = 1/4. Let us check the case of r = because two curves can take different values of θ if they intersect at the origin. The curve r = cosθ reaches the origin when θ = π 6, π, 5 6 π, 7 6 π, π, and 11 6 π. The curve r = sinθ reaches the origin when θ =, π, π, π, 4 π, and 5 π. Therefore, the two curves also intersect at the origin. In polar coordinates, the points of intersection are obtained as π ) ) r,θ) =,θ), 1 4,, 1 4, 1 4 π, 1 4, 17 1 π ). 16. Find the length of the polar curve r = θ for θ π. The length L is calculated as follows. L = π r + We introduce t as θ = sinht. We obtain L = sinh 1 π) ) dr π dθ = θ dθ +1dθ. cosh tdt = = sinh 1 π)+π 1+π. 8 sinh 1 π) 1+cosht dt = Math Math Math Math

9 Here, we used cosh sinh 1 π) ) = 1+sinh sinh 1 π) ) = 1+π. If we put sinh 1 π) = lnu, we obtain π = u 1 u) / and thus u = π + 1+π. So, the solution is also written as L = ln π + ) 1+π +π 1+π. 17. Find the surface area generated by rotating the curve r = cosθ about the line θ =. The curve is sketched as follows. We have r = cosθ π/4 θ π/4, π/4 θ 5π/4). We note that the line θ = is the x-axis, and x = rcosθ and y = rsinθ. The surface S 9 Math Math Math Math

10 is calculated as S = = = = 1π π/4 π/4 π/4 dx ) πy + dθ πx π/4 = 6 π. r + ) dy dθ dθ ) dr dθ dθ π cosθcosθ cos θdθ cosθ + sinθ cosθ ) dθ 18. Find an equation of a sphere if one of its diameters has endpoints 1,, ) and 1,1,). The diameter of the sphere, which is the distance between two endpoints, is 1+1) +1+) ++) = 7. The center of the sphere, which is the midpoint between two endpoints, is 1+1, +1, + ) =, 1 ),. Therefore, the equation of the sphere is given by x + y + ) 1 +z = Describe in words the region of R represented by the inequality, x + y +z x+4y +. The equation is rewritten as x 1) +y ) +z 5. Therefore, the region consists of the points whose distance from the point 1,,) is at most 5. 1 Math Math Math Math

11 . Find the volume of the solid that lies inside both of the spheres x + y +z x y +1 = and x +y +z +z =. Two equations are rewritten as x 1) +y 1) +z = 1, x +y +z +1) = 1. The radius of each sphere is 1. The distance between the centers of two spheres is =. 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 ) are circles X +Y = 1 and is then calculated as follows. V = = = π X + 1 πy dx + πy dx πy dx ) 1 X + dx ] 1 [ = π 1 X X X = π ) ) + 8 = π ). 8 ) +Y = 1. The volume V 1 4 )) 8 1. Find a, a+b, a b when a = i+j k and b =,4,. a = 14, a+b = 5,5,1, a b =, 1, Math Math Math Math

12 . A boy walks due west on the deck of a ship at 4mi/h. The ship is moving south at a speed of mi/h. Find the speed of the boy relative to the surface of the water. The speed vmi/h is calculated as v = 4i j = 14 = A -lb weight hangs from two strings. The strings, fastened at different heights, make angles of 6 and 45 with the horizontal. Find the tension in each string and the magnitude of each tension. Two tension vectors are written as T 1 = T 1 cos6 i+ T 1 sin6 j, T = T cos45 i+ T sin45 j. We have T 1 cos6 = T cos45, T 1 sin6 + T sin45 =. From the first equation, we obtain T = 1/ ) T 1. Then, we plug this relation into the second equation, and obtain T 1 = 6 1+, T = 1+. The tension vectors are obtained as T 1 = 1+ i+ ) j, T = 1+ i+j). 4. 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 = x 1,y 1 ), Q = x,y ), R = x,y ), and S = x 4,y 4 ). The vertices of the quadrilateral formed by joining the midpoints of the segments of the quadrilateral PQRS are then obtained as p = x 1 +x, y 1+y ), 1 Math Math Math Math

13 ). We con- ), u = q = x +x, y +y ), r = x +x 4, y +y 4 ), and s = x4 +x 1 sider vectors u 1 = pq = x x 1, y y 1 ), u = qr = x 4 x rs = x1 x ), and u4 = sp = x x 4, y 4+y 1, y 4 y, y 1 y, y y 4 ). We see that u1 and u are parallel, and u and u 4 are also parallel. Therefore, the quadrilateral pqrs is a parallelogram. 5. Which of the following expressions are meaningful? Which are meaningless? Explain. a) a b c, b) ab, c) ab c), d) a b, e) a b c, and f) a+b. a) meaningless, b) meaningless, c) meaningful, d) meaningful, e) meaningless, f) meaningful. 6. Find the angle between vectors a =,1, 1 and b =,,5. Find an exact expression.) Let θ be the angle. Therefore, θ = cos 1 1/ 51). cosθ = a b a b = Determine whether the given vectors are orthogonal, parallel, or neither. a) u = 1, 16,1, v = 4,6,1, b) u = 18i 1j + k, v = 1i+14j 5k, and c) u = 1, 16,1, v = 18i 1j+k. a) orthogonal, b) parallel, c) neither. 8. For vectors a =,1 and b =, 4, find orth a b The vector projection is calculated as proj a b = a b a a = 5,1. 1 Math Math Math Math

14 Then the orthogonal projection is calculated as 4 orth a b = b proj a b =, 4 5, = , Find the distance from the point 1,) to the line x y +7 =. Let us, for example, consider the y-intercept of the line. By putting x =, we see that the line passes,7/). Indeed, the equation for the line is rewritten as x y 7/) =. This is equivalently expressed as, x,y 7/ =. Therefore, the line is perpendicular to the vector a =,. We consider the displacement vector b from,7/) to the point 1,), and take the scalar projection onto a. comp a b = a b a =, 1, 1/ 9+4 = 1. The distance is comp a b = / 1.. Find a vector equation of a sphere with the radius and the center at 9,,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 PRQ makes the right angle. This relation is expressed as r a) r b) =. This is a vector equation of the sphere. Since the radius is and the center is at C9,,4), a and b are given as a = OC u, b = OC +u, where u is a vector with the magnitude. For example, if we take u =,,), we obtain r 7,,4 ) r 11,,4 ) =. 14 Math Math Math Math

15 1. By the protonation of ammonia NH, 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 Use cos 1 1/) 19.5.) In Cartesian coordinates, we place hydrogen atoms at 1,,),,1,),,,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/, 1/, 1/). The bond angle θ is the angle between displacement vectors u = 1,, 1/,1/,1/ = 1/, 1/, 1/ and v =,1, 1/,1/,1/ = 1/,1/, 1/. cosθ = u v u v = 1. Therefore, θ = cos 1 1/) For a = ti + 1 t)j + k and b = ti + j + t )k, find the cross product a b and verify that it is orthogonal to both a and b. a b = t,1 t,1 t,1,t = t +t )i+ t +4t)j+t t)k. Furthermore, we have a a b) = t,1 t,1 t +t, t +4t,t t =, b a b) = t,1,t t +t, t +4t,t t =.. 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 Math Math Math

16 4. Find the area of the parallelogram with vertices A, 1,1), B1,,), C5,6,), and D6,5,). We obtain AB = 1,1,, AC =,7,1, and AD = 4,6, 1. Note that AB + AD = AC. The area A is calculated as A = AB AD = 1,7, 1 = Find the volume of the parallelepiped determined by vectors a =,,5, b = 1,1,, and c = 4,6, 1. The volume V is calculated as V = a b c) =,,5 1,7, 1 = 55 = Verify that the vectors a = 5i + 5j k, b = 4i + 6j k, and c = 6i+4j 5k are coplannar. The scalar triple product is calculated as a b c) = 5,5, 6,14, =. Therefore, a, b, and c are coplannar. 7. Determine whether the points A5,6,), B, 1,1), C9,11, 4), and D1,,) lie in the same plane. For vectors a = AB =, 7, 1, b = AC = 4,5, 6, and c = AD = 4, 6,1, the scalar triple product is calculated as a b c) =, 7, 1 1,, 4 = 4. Therefore, A, B, C, and D don t lie in the same plane. 16 Math Math Math Math

17 8. Find the distance from the point P,1,4) to the plane through the points A1,1,), B,,), and C,,). The normal vector n of the plane spanned by AB and AC is n = AB AC = 1,1,, 1, = 5,7, 1. Then, the distance d is calculated as d = n AP n = 5)+7) 14) = Prove a b c) = a c)b a b)c. a b c) = a 1,a,a b c b c,b c 1 b 1 c,b 1 c b c 1 = a b 1 c b c 1 ) a b c 1 b 1 c ),a b c b c ) a 1 b 1 c b c 1 ), a 1 b c 1 b 1 c ) a b c b c ) = [a c +a c )b 1 a b +a b )c 1 ]i+[a 1 c 1 +a c )b a 1 b 1 +a b )c ]j +[a 1 c 1 +a c )b a 1 b 1 +a b )c ]k = [a 1 c 1 +a c +a c )b 1 a 1 b 1 +a b +a b )c 1 ]i +[a 1 c 1 +a c +a c )b a 1 b 1 +a b +a b )c ]j +[a 1 c 1 +a c +a c )b a 1 b 1 +a b +a b )c ]k = a c)b a b)c. 4. Suppose that a. a) If a b = a, does it follow that a = b? b) If a b = b a, does it follow that b =? a) No. In this case, proj a b = a. But, in general, b itself is different from a. b) Yes. 17 Math Math Math Math

18 41. Find parametric equations and symmetric equations for the line of intersection of the planes x+y +z = and x+y = 1. We eliminate x in the system { x+y +z =, x+y = 1, and obtain z = y 5)/ 4). Similarly by eliminating y, we obtain z = x+)/. Therefore, we can write x+)/ = y 5)/ 4) = z, or x+ = y 5 4 The parametric equations are obtained as = z. ) x = +t, y = 5 4t, z = t. Note) The line of intersection can also be obtained as follows. The normal vectors of the planes are n 1 = 1,1, and n =,1,. The line of intersection is parallel to n 1 n =,4, 1 and passes through,5,) in the xy-plane. Therefore, the symmetric equations are written as Eq. ). 4. Find an equation of the plane through the point 1, 8,) and perpendicular to the line x = 1+t, y = 5t, z = t. The normal vector is, 5,1. The equation of the plane is x 1) 5y +8)+z ) = or x 5y +z 44 =. 4. Find symmetric equations for the line of intersection of the planes x y + z = 8 and x + 5y + z =. Let θ denote the angle between the planes. Find cosθ. We eliminate x in the system { x y +z = 8, x+5y +z =, 18 Math Math Math Math

19 and obtain z = y + 1). Similarly by eliminating y, we obtain z = /8)x 7). Therefore, we have /8)x 7) = y+1) = z. Symmetric equations are obtained as The normal vectors of the planes are Hence, We obtain x 7 8 = y +1 = z. ) n 1 = 1, 1,, n = 1,5,1. cosθ = n 1 n n 1 n = 11) 15)+1) = 1. cosθ = 1. Note) The line of intersection can also be obtained as follows. The line is parallel to n 1 n = 16,,6 and passes through 7, 1,) in the xy-plane. Therefore, the symmetric equations are written as Eq. ). 44. Find the point at which lines,1, 1 + t 1, 1, and x 1 = y = z+ intersect. Then, find an equation for the plane that contains these lines. Parametric equations of two lines are obtained as x = +t, y = 1 t, z = 1+t, x = 1+s, y = s, z = +s Therefore, we have t = and s = 1 at the point of intersection. Two lines intersect at, 1, 4). The normal vector n is calculated as n = 1, 1, 1,1,1 = 4,,. The point, 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+y+1)+z+4) =, or x+y +z +5 =. 19 Math Math Math Math

20 45. Identify the surface x 4y 4z +8y 16z = 16. The equation is rewritten as y 1) +z +) x = 1. 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. 46. Find an equation for the surface obtained by rotating the line z = x, y = about the z-axis. The surface is a cone whose equation is z = x +y. 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 Px,y,z), the distance to the y-axis is x +z and the distance to the xz-plane is y. Therefore, the equation for P is x +z = y 4. Thus, the surface is a cone. The trace in the y = k plane is x + z = k/), which is a circle with radius k/. The trace in the x = k plane is y /k) z /k = 1, which is a hyperbola. The trace in the z = plane is y /k) x /k = 1, which is also a hyperbola. 48. Change, π/6, π/4) from spherical to cylindrical coordinates. In spherical coordinates, we have ρ,θ,φ) =, π/6,π/4). r = ρsinφ = sin π 4 ) =, z = ρcosφ = cos π 4 ) =. Math Math Math Math

21 49. Identify the surface y = y +z and write the equation both in cylindrical coordinates and in spherical coordinates. First of all, the equation is written as follows in Cartesian coordinates. y 1 ) +z = 1 4. Therefore, the surface is a circular cylinder whose axis is parallel to the x- axis and its traces are circles. In particular, the trace in the plane x = is a circle whose center is, 1/, ). This circular cylinder is given in cylindrical coordinates as rsinθ 1 ) +z = 1 4, and given in spherical coordinates as ρsinφsinθ 1 ) +ρ cos φ = A solid lies above the cone z = x +y and below the sphere x + y +z = z. Write a description of the solid in terms of inequalities involving spherical coordinates. In Cartesian coordinates, the solid is expressed as z > x +y, x +y + z 1 ) < ) 1. Using x = ρsinφcosθ, y = ρsinφcosθ, and z = ρcosφ ρ and φ π), we obtain < φ < π 4, < ρ < cosφ. 51. Find a vector perpendicular to the plane through the points A1, 4, ), B,, 1), and C,4,). Then, find the area of triangle ABC. Two vectors in the plane are AB =, 4, 1 and AC = 1,,. Their cross product is perpendicular to the plane., 4, 1 1,, = 1, 7,4. 1 Math Math Math Math

22 TheareaofthetriangleABC ishalfoftheareaoftheparallelogramspanned by the vectors AB and AC. The area is 1, 4, 1 1,, = 1 9 1) + 7) +4 =. Math Math Math Math