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 55- -- Math 55- -- Math 55- -- Math
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 55- -- Math 55- -- Math 55- -- Math
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 55- -- Math 55- -- Math 55- -- Math
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 55- -- Math 55- -- Math 55- -- Math
Note that π/ cosθ) dθ = 9 Thus, π/ 1+cosθ) dθ = = 9 π/ π π/ = π + 9 8. 1+cosθ)dθ ) +, 4 1+cosθ + 1+cosθ [ A = 9 4 π π + 9 8 π 9 ] = 5 8 4 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 + 9 4 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 55- -- Math 55- -- Math 55- -- Math
11. Find a polar equation for the curve represented by the Cartesian equation, x = 5. Since x = rcosθ, we obtain rcosθ = 5. 1. 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 55- -- Math 55- -- Math 55- -- Math
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 = 1 1+. 14. Sketch the curve r = cos θ, and find the area it encloses. The curve is sketched as follows. 7 Math 55- -- Math 55- -- Math 55- -- Math
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 55- -- Math 55- -- Math 55- -- Math
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 55- -- Math 55- -- Math 55- -- Math
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 = 49. 19. 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 55- -- Math 55- -- Math 55- -- Math
. 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 1 +1 +1 =. 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 1+ 1 1 X + 1 πy dx + πy dx πy dx ) 1 X + dx ] 1 [ = π 1 X X + 1 4 X = π 1 1 + 5 ) + 8 7 ) + 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, 1. 11 Math 55- -- Math 55- -- Math 55- -- Math
. 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 = 4 65.. 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 55- -- Math 55- -- Math 55- -- Math
). 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 = 1 51. 7. 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 55- -- Math 55- -- Math 55- -- Math
Then the orthogonal projection is calculated as 4 orth a b = b proj a b =, 4 5, = 5 11 5,. 5 9. 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 55- -- Math 55- -- Math 55- -- Math
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 19.5. 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/) 1.916 19.5.. 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 55- -- Math 55- -- Math 55- -- Math
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 = 18. 5. 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 = 55. 6. 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 55- -- Math 55- -- Math 55- -- Math
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) = 6 5+49+1 5. 9. 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 55- -- Math 55- -- Math 55- -- Math
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 55- -- Math 55- -- Math 55- -- Math
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+1+9 1+5+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 55- -- Math 55- -- Math 55- -- Math
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 55- -- Math 55- -- Math 55- -- Math
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 = 1 4. 5. 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 55- -- Math 55- -- Math 55- -- Math
TheareaofthetriangleABC ishalfoftheareaoftheparallelogramspanned by the vectors AB and AC. The area is 1, 4, 1 1,, = 1 9 1) + 7) +4 =. Math 55- -- Math 55- -- Math 55- -- Math