OHSx XM521 Multivariable Differential Calculus: Homework Solutions 13.1

OHSx XM521 Multivariable Differential Calculus: Homework Solutions 13.1 (37) If a bug walks on the sphere x 2 + y 2 + z 2 + 2x 2y 4z 3 = 0 how close and how far can it get from the origin? Solution: Complete the square of the sphere equation and group constants on one side, to get: (x + 1) 2 + (y 1) 2 + (z 2) 2 = 3 + 1 + 1 + 4 = 3 2. This is the equation of a sphere of radius 3 centered at ( 1, 1, 2). The distance, d, from the origin to a point P (x, y, z) is given by: d OP = x 2 + y 2 + z 2. It is possible to try to minimize and maximize this equation subject to the constraint that (x, y, z) must lie on the above sphere. But the techniques to do min-max problems with multiple variables have not been developed at this point in the course. Geometrically, the distance from the origin to the center of the sphere, C, is: d OC = ( 1) 2 + (1) 2 + (2) 2 = 6. Consider the triangle formed by the origin, the center of the sphere, and a point P (x, y, z) on the sphere. By the Law of Cosines: d 2 OP = d 2 OC + d 2 CP 2d OC d CP cos θ, where d CP = radius of sphere = 3 and θ = the angle formed by OCP. To maximize d OP, θ = π so that: d OP max = d 2 OC + d2 CP + 2d OCd CP = (d OC + d CP ) 2 = d OC + d CP = 6 + 3. This corresponds to moving from O to the center of sphere C and on the extension of line OC to P. The minimum corresponds to θ = 0, so that: d OP min = d 2 OC + d2 CP 2d OCd CP = (d OC d CP ) 2 = d OC d CP = 6 3. (41) As shown in the accompanying figure, a bowling ball of radius R is placed inside a box just large enough to hold it, and it is secured for shipping by a Styrofoam sphere into each corner of the box. Find the radius of the largest Styrofoam sphere that can be used. [HINT: Take the origin of a Cartesian coordinate system at a corner of the box with the coordinate axes along the edges.] Solution: A box just large enough to hold a sphere of radius R is a cube with edges of length 2R where the center of each face of the cube is tangent to the sphere. Place this 1

cube so that one corner of it is at the origin of a Cartesian coordinate system. The distance from the origin to the center of the sphere, C R = (R, R, R), is: d OCR = x 2 + y 2 + z 2 = R 2 + R 2 + R 2 = 3R. Let the radius of the Styrofoam sphere be r. Since this sphere will need to be placed in each corner of the cube, consider just the sphere in the corner of the origin. This little sphere must be tangent to the big sphere to hold it in place. It must also be tangent to the three faces of the cube in the origin corner. By the above analysis, the distance from the origin to the center of the Styrofoam sphere, C r = (r, r, r), is: d OCr = x 2 + y 2 + z 2 = r 2 + r 2 + r 2 = 3r. Let a be the distance from the origin to the little sphere along the line OC R so that from geometry: Also: d OCR = R + 2r + a = 3R. d OCr = r + a = 3r, so that a = ( 3 1)r. Substituting into the equation for d OCR : d OCR = R + 2r + ( 3 1)r = 3R, so that r( 3 + 1) = ( 3 1)R. Solving for r and rationalizing: r = ( 3 1) 2 R ( 3 + 1)( 3 1) = (2 3)R. (43) Show that for all values of θ and φ, the point lies on the sphere x 2 + y 2 + z 2 = a 2. (a sin φ cos θ, a sin φ sin θ, a cos φ) Solution: Let θ and φ be arbitrary. If x = a sin φ cos θ, y = a sin φ sin θ, and z = a cos φ, then using the trig identity sin 2 + cos 2 = 1, we have x 2 + y 2 + z 2 = (a sin φ cos θ) 2 + (a sin φ sin θ) 2 + (a cos φ) 2 = (a sin φ) 2 (sin 2 θ + cos 2 θ) + (a cos φ) 2 = (a sin φ) 2 + (a cos φ) 2 = a 2. Therefore, for all values of θ and φ, the point (x, y, z) of the given form satisfies x 2 +y 2 +z 2 = a 2. So all such points lie on this sphere. 2

OHSx XM521 Multivariable Differential Calculus: Homework Solutions 13.2 (38) Let r = x, y and r 0 = x 0, y 0. In each part, describe the set of all points (x, y) in 2-space that satisfy the stated condition (a) r r 0 = 1. Solution: In terms of components, r r 0 = x x 0, y y 0, and r r 0 = (x x 0 ) 2 + (y y 0 ) 2. So r r 0 = 1 if and only if (x x 0 ) 2 + (y y 0 ) 2 = 1. A point (x, y) satisfies this equation if and only if its distance to (x 0, y 0 ) is 1. Therefore, the set of points (x, y) such that r = x, y satisfies r r 0 = 1 is a circle of radius 1 centered at (x 0, y 0 ). (b) r r 0 1. Solution: By the same reasoning as in part (a), r r 0 1 if and only if (x x 0 ) 2 + (y y 0 ) 2 1. A point (x, y) satisfies this equation if and only its distance to (x 0, y 0 ) is less than or equal to 1. The set of such points is the disk (the circle and its interior) of radius 1 centered at (x 0, y 0 ). (c) r r 0 > 1. Solution: By the same reasoning as in part (a) and (b), r r 0 > 1 if and only if (x x 0 ) 2 + (y y 0 ) 2 > 1. A point (x, y) satisfies this equation if and only its distance to (x 0, y 0 ) is greater than 1. The set of such points consistns of those points lying strictly outside of the disk (the circle and its interior) of radius 1 centered at (x 0, y 0 ). (49) The accompanying figure shows a 250-lb traffic light supported by two flexible cables. The magnitude of the forces that the cables apply to the eye ring are called the cable tensions. Find the tensions in the cables if the traffic light is in static equillibrium (defined above exercise 47). Solution: Static equillibrium for a given point implies i F i = 0 where each F i is a force applied to the point and the sum is carried over all such forces F i. Use this equation to relate the forces on the eyering: F 1 + F 2 + F 250 = 0 where F 1 is the tension in the line making a 30 o angle with the eyering, F 2 is the tension in the line making a 45 o angle, and F 250 is the downward pointing force of the 250 lb traffic light. Locating the origin of an xy coordinate system at the eyering, look at the x, y component form of this vector equation: F 1 cos 150 o, sin 150 o + F 2 cos 45 o, sin 45 o + F 250 cos( 90 o ), sin( 90 o ) = 0, 0, where the identity v = v cos φ, sin φ has been used to find the various x, y components of the vectors. Simplifying the above expression gives: F 1 cos 150 o + F 2 cos 45 o + F 250 cos( 90 o ), F 1 sin 150 o + F 2 sin 45 o + F 250 sin( 90 o ) = 0, 0. Equating components gives the following two equations: F 1 cos 150 o + F 2 cos 45 o + F 250 cos( 90 o ) = 0; F 1 sin 150 o + F 2 sin 45 o + F 250 sin( 90 o ) = 0. 1

Substituting for various trignometric values, using F 250 = 250 lbs, and solving for F 1, F 2, gives: 3 F 1 + 2 F 2 = 0; F 1 + 2 F 2 = 2(250). Solving gives F 1 = 250( 3 1) 183 lbs, and F 2 225 lbs. (52) A vector w is said to be a linear combination of the vectors v 1, v 2, and v 3 if w can be expressed as w = c 1 v 1 + c 2 v 2 + c 3 v 3, where c 1, c 2, and c 3 are scalars. (a) Find scalars c 1, c 2, and c 3 to express 1, 1, 5 as a linear combination of the vectors v 1 = 1, 0, 1, v 2 = 3, 2, 0, and v 3 = 0, 1, 1. Solution: Substitute the various vectors into the vector equation w = c 1 v 1 + c 2 v 2 + c 3 v 3 to get: 1, 1, 5 = c 1 1, 0, 1 + c 2 3, 2, 0 + c 3 0, 1, 1. Multiply constants into vectors and add components on the right side to get: 1, 1, 5 = c 1 + 3c 2, 2c 2 + c 3, c 1 + c 3. Equate components on both sides to get following 3 equations in 3 unknowns: 1 = c 1 + 3c 2 ; 1 = 2c 2 + c 3 ; 5 = c 1 + c 3. Subtract the third equation from first to get: 6 = 3c 2 c 3. Add this equation to the second equation above to get: 5 = 5c 2. So c 2 = 1, implying 6 = 3 c 3, or c 3 = 3. Finally 5 = c 1 + 3, so c 1 = 2. (b) Show that the vector 2i + j k cannot be expressed as a linear combination of the vectors v 1 = i j, v 2 = 3i + k, and v 3 = 4i j + k. Solution: Suppose there are scalars c 1, c 2, c 3 such that 2, 1, 1 = c 1 1, 1, 0 + c 2 3, 0, 1 + c 3 4, 1, 1. Scalar multiply componentwise and add components on the right side to get: 2, 1, 1 = c 1 + 3c 2 + 4c 3, c 1 c 3, c 2 + c 3. Equate components on both sides to obtain the following 3 equations in 3 unknowns: 2 = c 1 + 3c 2 + 4c 3 ; 1 = c 1 c 3 ; 1 = c 2 + c 3. 2

Add the first and second equations together to get: 3 = 3c 2 + 3c 3. So 1 = c 2 + c 3. But from the third equation above 1 = c 2 + c 3. These two equations are inconsistent and will not have any solutions for c 2 and c 3. (The graphs of these two equations in the c 2 c 3 -plane are two parallel lines with no intersection point.) Hence 2i+j k is not a linear combination of v 1,v 2,v 3. (53) Use a theorem from plane geometry to show that if u and v are vectors in 2-space or 3-space, then u + v u + v, which is called the triangle inequality for vectors. Give some examples to illustrate this inequality. Solution: Consider the triangle formed by the vectors u, v, and u+v, where u and u+v have a common initial point, and the initial point of v is the terminal point of u. Let θ be the angle between the sides defined by u and v. (Note: As vectors, the angle φ between u and v is the angle formed between u and v when drawn with common initial point. So the angle θ defined above is the supplement of φ, although this won t matter.) Apply the Law of Cosines to this triangle to get: u + v 2 = u 2 + v 2 2 u v cos θ. (1) We now obtain a more useful bound on the right side of the above equation. Since cos θ 1, u 2 + v 2 2 u v cos θ u 2 + v 2 +2 u v cos θ u 2 + v 2 +2 u v. So, combining with (1) yields u + v 2 u 2 + v 2 + 2 u v = ( u + v ) 2. Since the terms being squared on both sides are nonnegative, we can take the square root of each side to obtain the desired inequality: u + v u + v. For an example of this relation, consider the vectors u = 1, 0, v = 0, 1, u + v = 1, 1, so that u = 1, v = 1, and u+v = 2. The inequality holds since 2 (1+1) = 2. (57) Use vectors to prove that the line segment joining the midpoint of two sides of a triangle is parallel to the third side and half as long. Solution: Let u, v, w be three vectors that overlay the sides u, v, and w of an arbitrary triangle, and assume that the terminal point of u is the initial point of v, the terminal point of v is the initial point of w, and the terminal point of w is the initial point of u. The midpoints of sides u and v can be represented as: m u = 0.5u; m v = u + 0.5v. The directed line segment from the midpoint of u to the midpoint of v is m v m u = 0.5(u + v). 3

Now u + v + w = 0 since tracing the three sides of a triangle makes a loop around a closed path resulting in 0 net displacement. Hence w = (u + v) and so m v m u = 0.5w. Since m v m u is a scalar multiple of w and m v m u = 0.5 w, the line segment joining the midpoint of two sides of a triangle is parallel to the third side and half as long. (58) Use vectors to prove that the midpoints of the sides of a quadrilateral are the vertices of a parallelogram. Solution: Let u, v, w, x be four vectors that overlay the sides u, v, w, and x of an arbitrary quadrilateral, and assume that each of the vectors u, v, w, x has terminal point equal to the initial point of the following vector. The midpoints of the various sides can be represented as: m u = 0.5u; m v = u + 0.5v; m w = u + v + 0.5w; m x = u + v + w + 0.5x. The line segments connecting the midpoints of sides u and v as well as w and x can be represented as the vector difference of the respective midpoint vectors: m v m u = 0.5(u + v); m x m w = 0.5(w + x). Now u + v + w + x = 0 since tracing the four sides of the quadrilateral makes a loop around a closed path resulting in 0 net displacement. Using this to simplify the above two equations results in m v m u = (m x m w ). Therefore these two line segments formed by joining the midpoints of the quadrilateral are parallel as well as equal in length. A similar argument applies to the line segments formed by joining the midpoints of u and x as well as v and w respectively, to show that these segments are also parallel and equal in length. Since the opposite sides are parallel and equal in length, the figure is a parallelogram. 4

OHSx XM521 Multivariable Differential Calculus: Homework Solutions 13.3 (14) True or False? If a b = a c and if a 0, then b = c. Justify your conclusion. Solution: False. Let a = 1, 1, 1, b = 1, 0, 0, and c = 0, 1, 0. a b = a c = 1 and a 0, but b c. (20) Show that two nonzero vectors v 1 and v 2 are orthogonal if and only if their direction cosines satisfy: cos α 1 cos α 2 + cos β 1 cos β 2 + cos γ 1 cos γ 2 = 0. Solution: Consider v 1 v 2 with v 1 and v 2 expressed in terms of their direction cosines: v 1 v 2 = ( v 1 (cos α 1 i + cos β 1 j + cos γ 1 k)) ( v 2 (cos α 2 i + cos β 2 j + cos γ 2 k)) = v 1 v 2 (cos α 1 cos α 2 + cos β 1 cos β 2 + cos γ 1 cos γ 2 ). If v 1 and v 2 are orthogonal and nonzero, then v 1 v 2 = 0, v 1 0, and v 2 0; so cos α 1 cos α 2 + cos β 1 cos β 2 + cos γ 1 cos γ 2 = 0 from the above equation for v 1 v 2. On the other hand, if cos α 1 cos α 2 + cos β 1 cos β 2 + cos γ 1 cos γ 2 = 0, and v 1 and v 2 are nonzero, then by the above equation v 1 v 2 = 0, implying that v 1 and v 2 must be orthogonal. (37) Let u and v be adjacent sides of a parallelogram. Use vectors to prove that the diagonals of the parallelogram are perpendicular if the sides are equal in length. Solution: Given u and v as the adjacent sides of a parallelogram (drawn with terminal point of u equal to the initial point of v), the diagonals of the parallelogram can be represented as u + v and u v. Since a b = b a and a a = a 2 for all vectors a and b, the dot product of the diagonals can be written as (u + v) (u v) = u u u v + v u v v = u u v v = u 2 v 2. So, if u = v, then (u+v) (u v) = 0, implying that these diagonals are perpendicular. (40) Prove u v = 1 4 u + v 2 1 4 u v 2. Solution: since a 2 = a a and b c = c b for all a, b, and c, u + v 2 = (u + v) (u + v) = u u + u v + v u + v v = u 2 + v 2 + 2u v. Similarly, u v 2 = u 2 + v 2 2u v. So as desired. 1 4 u + v 2 1 4 u v 2 = u 2 + v 2 + 2u v ( u 2 + v 2 2u v) 4 = 4u v 4 = u v, 1

(41) Show that if v 1, v 2, and v 3 are mutually perpendicular nonzero vectors in 3-space, and if a vector v in 3-space is expressed as: v = c 1 v 1 + c 2 v 2 + c 3 v 3, then the scalars c 1, c 2, and c 3 are given by the formulas: c i = (v v i )/ v i 2, i = 1, 2, 3. Solution: For v 1, consider: v v 1 = (c 1 v 1 + c 2 v 2 + c 3 v 3 ) v 1 = c 1 v 1 v 1 + c 2 v 2 v 1 + c 3 v 3 v 1. We know that v 1 v 1 = v 1 2, and v 1 v 2 = v 1 v 3 = 0 because these vectors are mutually perpendicular. Substituting these relations into the above equation and solving for c 1 yields Similar results hold for v 2 and v 3. c 1 = (v v 1 )/ v 1 2. 2

OHSx XM521 Multivariable Differential Calculus: Homework Solutions 13.4 (26) Show that in 3-space the distance d from a point P to the line L through points A and B can be expressed as d = AP AB AB. Solution: The distance d is equal to the length of the component of the vector AP orthogonal to the vector AB. With θ as the angle between these two vectors, d can be expressed as: d = AP sin θ = AB AP sin θ AB = AP AB AB. (31) What can you say about the angle between nonzero vectors u and v if u v = u v? Solution: Let θ be the angle between the nonzero vectors u and v. Using u v = u v cos θ and u v = u v sin θ, the condition u v = u v becomes cos θ = sin θ, or tan θ = 1. This is satisfied if and only if θ = π/4. (Recall that the angle between two vectors must be no greater than π.) (32) Show that if u and v are vectors in 3-space, then u v 2 = u 2 v 2 (u v) 2. [Note: This result is sometimes called Lagrange s identity.] Solution: If either u or v is zero, then clearly each side of the above equation is zero. So we may assume u and v are nonzero. Using u v = u v sin θ, sin 2 θ = 1 cos 2 θ, and cos θ = u v/ u v, we can write: ( u v ) ) 2 u v 2 = u 2 v 2 sin 2 θ = u 2 v 2 (1 cos 2 θ) = u 2 v (1 2. u v Multiplying out the right-most portion of the above gives: as desired. u v 2 = u 2 v 2 (u v) 2, (38) Prove part (b) of Theorem 13.4.1 for 3 3 determinants. [Just give the proof for the first two rows.] Then use (b) to prove (a). 1

Solution: Part (b) of Theorem 13.4.1 states that Interchanging two rows of a determinant multiplies its value by 1. We first show that interchanging the first two rows multiplies the determinant by 1. a 1 a 2 a 3 b 1 b 2 b 3 b = a 2 b 3 1 c 1 c 2 c 3 c 2 c 3 a 2 b 1 b 3 c 1 c 3 + a 3 b 1 b 2 c 1 c 2 = a 1 (b 2 c 3 b 3 c 2 ) a 2 (b 1 c 3 b 3 c 1 ) + a 3 (b 1 c 2 b 2 c 1 ) = a 1 b 2 c 3 a 1 b 3 c 2 a 2 b 1 c 3 + a 2 b 3 c 1 + a 3 b 1 c 2 a 3 b 2 c 1 ( ) = b 1 (a 2 c 3 a 3 c 2 ) b 2 (a 1 c 3 a 3 c 1 ) + b 3 (a 1 c 2 a 2 c 1 ) ( a = b 2 a 3 1 c 2 c 3 b 2 a 1 a 3 c 1 c 3 + b 3 a ) 1 a 2 c 1 c 2 b 1 b 2 b 3 = a 1 a 2 a 3 c 1 c 2 c 3. Now we consider the case of interchanging the second and third rows. We use the fact that interchanging the two rows of a 2 2 matrix multiplies its determinant by 1, a fact that is trivial to check. Expanding in terms of its 2 2 determinants, a 1 a 2 a 3 b 1 b 2 b 3 b = a 2 b 3 1 c 1 c 2 c 3 c 2 c 3 a 2 b 1 b 3 c 1 c 3 + a 3 b 1 b 2 c 1 c 2 c = a 2 c 3 1 b 2 b 3 + a 2 c 1 c 3 b 1 b 3 a 3 c 1 c 2 b 1 b 2 ( c = a 2 c 3 1 b 2 b 3 a 2 c 1 c 3 b 1 b 3 + a 3 c ) 1 c 2 b 1 b 2 a 1 a 2 a 3 = c 1 c 2 c 3 b 1 b 2 b 3. Finally, notice that interchanging the first and third rows of a matrix can be accomplished by interchanging rows 1 and 2, then interchanging rows 2 and 3, then interchanging rows 1 and 2. Each interchange multiplies the determinant of the original matrix by 1. So three interchanges multiplies the determinant by ( 1) 3 = 1. Part (a) of Theorem 13.4.1 states If two rows of a determinant are the same, then the value of the determinant is 0.. To prove this, let M be a matrix having two rows that are the same. Let M be the matrix obtained by interchanging these two rows. Clearly, M = M, so However, by part (b), det(m) = det(m ). det(m ) = det(m). These two equations imply that det(m) = det(m), which can happen only if det(m) = 0. (41) Prove: If a, b, c, and d lie in the same plane when positioned with a common initial point, then 2

(a b) (c d) = 0. Solution: Let P be the plane containing a, b, c, and d when positioned with a common initial point. Let x = a b and y = c d. x must be orthogonal to the plane P containing a and b. Similarly, y is orthogonal to the plane P containing c and d. In order for x and y to both be orthogonal to this same plane P, they must be scalar multiples of one another, so that x = ky for some scalar k. Therefore, x y = x (kx) = k(x x) = 0, since the cross product of any vector with itself is 0. Hence (a b) (c d) = 0. 3

OHSx XM521 Multivariable Differential Calculus: Homework Solutions 13.5 (23) Where does the line x = 1 + t, y = 3 t, z = 2t intersect the cylinder x 2 + y 2 = 16? Solution: Substitute the parametric equations for x and y into the equation for the cylinder to find the values for t at the intersection point: x 2 + y 2 = (1 + t) 2 + (3 t) 2 = 2t 2 4t + 10 = 16. This implies that t must satisfy t 2 2t 3 = (t 3)(t + 1) = 0, so t = 3 or t = 1 at the intersection points. Substituting these t values into the equation for the line, we see that the points of intersection are (4, 0, 6) and (0, 4, 2). (Note: We can check our result by directly verifying that these two points do satisfy the equation for the cylinder.) (31) Determine whether the points P 1 (6, 9, 7), P 2 (9, 2, 0), and P 3 (0, 5, 3) lie on the same line. Solution: Points P 1, P 2, and P 3 lie on the same line if and only if the directed line segments P 1 P 2 and P 1 P 3 are parallel vectors, which is true if and only if P 1 P 2 = cp 1 P 3 for some scalar c. Since P 1 P 2 = 3, 7, 7 and P 1 P 3 = 6, 14, 10, there is no c that satisfies P 1 P 2 = cp 1 P 3, so the points P 1, P 2, and P 3 do not all lie on the same line. (33) Show that the lines are the same. L 1 : x = 3 t, y = 1 + 2t L 2 : x = 1 + 3t, y = 9 6t. Solution: Represent the lines L 1 and L 2 via the following vector equations: L 1 : r 1 = r o1 + v 1 t = 3, 1 + 1, 2 t L 2 : r 2 = r o2 + v 2 t = 1, 9 + 3, 6 t. Since v 2 = 3v 1, the lines L 1 and L 2 are parallel. L 1 and L 2 also share the common point 1, 9 (when t = 4 in L 1 and t = 0 in L 2 ). Because parallel lines can share a common point only if they are the same line, L 1 and L 2 must be the same line. (39) Show that the lines L 1 : x = 2 t, y = 2t, z = 1 + t L 2 : x = 1 + 2t, y = 3 4t, z = 5 2t. are parallel, and find the distance between them. Solution: Represent the lines L 1 and L 2 via the following vector equations: L 1 : r 1 = r o1 + v 1 t = 2, 0, 1 + 1, 2, 1 t L 2 : r 2 = r o2 + v 2 t = 1, 3, 5 + 2, 4, 2 t. Since v 2 = 2v 1, the lines L 1 and L 2 are parallel. To find the distance between the lines, use the result from Exercise 26 in 13.4 which states that in 3-space the distance d from a point P to the line L through points A and B can be expressed as 1

d = AP AB AB. Let A = (2, 0, 1), B = (1, 2, 2) (this corresponds to t = 1 in L 1 ), and P = (1, 3, 5), so that AB = 1, 2, 1 and AP = 1, 3, 4. This gives AP AB = 5, 3, 1, so that AP AB = 5 2 + 3 2 + 1 2 = 35. AB = 6; so d = 35/6. (42) Let L be the line that passes through the point (x 0, y 0, z 0 ) and is parallel to the vector v = a, b, c where a, b, and c are nonzero. Show that a point (x, y, z) lies on the line L if and only if x x 0 a = y y 0 b = z z 0. c These equations, which are called the symmetric equations of L, provide a nonparametric representation of L. Solution: The line L can be represented parametrically as L : x = x 0 + at, y = y 0 + bt, z = z 0 + ct. Suppose (x, y, z) lies on this line. Then there exists some t 1 such that x = x 0 + at 1, y = y 0 + bt 1, and z = z 0 + ct 1. Solving for t 1 in each of these equations gives In short, t 1 = x x 0 a x x 0 a ; t 1 = y y 0 b = y y 0 b ; t 1 = z z 0. c = z z 0, c (and all are equal to t 1 ). This completes the first part of the proof. To prove the other direction of the if and only if, suppose that (x, y, z) satisfies x x 0 a Denote this common value by t 0. Then = y y 0 b = z z 0. c t 0 = x x 0 a ; t 0 = y y 0 b ; t 0 = z z 0. c Solving for (x, y, z) gives x = x 0 + at 0, y = y 0 + bt 0, z = z 0 + ct 0. Therefore, by letting t = t 0, we see that (x, y, z) lies on the line L : x = x 0 + at, y = y 0 + bt, z = z 0 + ct. (45) Let L 1 and L 2 be the lines whose parametric equations are L 1 : x = 1 + 2t, y = 2 t, z = 4 2t L 2 : x = 9 + t, y = 5 + 3t, z = 4 t. (a) Show that L 1 and L 2 intersect at the point (7, 1, 2). Solution: To determine if L 1 passes through (7, 1, 2), we look for a t value that satisfies the equations for L 1 with (x, y, z) replaced by (7, 1, 2). It s easy to see that t = 3 does satisfy these equations, so L 1 passes through (7, 1, 2) at t = 3. Similarly, we find that L 2 passes through (7, 1, 2) at t = 2. 2

(b) Find, to the nearest degree, the acute angle between L 1 and L 2 at their intersection. Solution: L 1 is parallel to the vector v 1 = 2, 1, 2 and L 2 is parallel to the vector v 2 = 1, 3, 1. (Note that if the angle θ between v 1 and v 2 is acute, then θ is the acute angle between L 1 and L 2. If θ is obtuse, then the acute angle between L 1 and L 2 is 180 θ.) Using the dot product we find v 1 v 2 = 2, 1, 2 1, 3, 1 = 1. Therefore, Solving for θ gives 1 = v 1 v 2 cos θ = 3 11 cos θ. θ = cos 1 (1/3 11) 84 o. (c) Find parametric equations for the line that is perpendicular to L 1 and L 2 and passes through their point of intersection. Solution: The desired line must be perpendicular to v 1 and v 2 as given above. So this line is parallel to v 1 v 2 = 7, 0, 7, which is parallel to 1, 0, 1. Since the desired line passes through the point (7, 1, 2) (which is the point of intersection of L 1 and L 2 ), it has parametric equations x = 7 + 1 t, y = 1 + 0 t, z = 2 + 1 t. 3

OHSx XM521 Multivariable Differential Calculus: Homework Solutions 13.6 (16) Determine whether the line and plane are parallel, perpendicular, or neither. (a) x = 3 t, y = 2 + t, z = 1 3t; 2x + 2y 5 = 0. Solution: v = 1, 1, 3 is a vector parallel to the given line, and n = 2, 2, 0 is a normal vector to the given plane. Since v n = 0, the line is parallel to the given plane. The line is not contained in the plane because there is no value of t that results in values for x, y, z that also satisfy the plane equation. (b) x = 1 2t, y = t, z = t; 6x 3y + 3z = 1. Solution: v = 2, 1, 1 is a vector parallel to the given line, and n = 6, 3, 3 is a normal vector to the given plane. Since n = 3v, the line is parallel to the normal of the given plane, and so the line is perpendicular to the plane. (c) x = t, y = 1 t, z = 2 + t; x + y + z = 1. Solution: v = 1, 1, 1 is a vector parallel to the given line, and n = 1, 1, 1 is a normal vector to the given plane. Since v n 0 and n is not some multiple of v, the line is not parallel to the plane and it is not perpendicular to the plane. (25) Find an equation of a plane that satisfies the stated conditions. The plane through (1, 2, 1) that is perpendicular to the line of intersection of the planes 2x + y + z = 2 and x + 2y + z = 3. Solution: Let P 1 : 2x + y + z = 2 and P 2 : x + 2y + z = 3 represent the two given planes. The line of intersection of the planes must be perpendicular to their normals n 1 = 2, 1, 1 and n 2 = 1, 2, 1, and therefore parallel to n = n 1 n 2 = 1, 1, 3, which must be normal to the desired plane. (Alternatively, the line of intersection can be found by solving the simultaneous equations given by the two planes.) In general, a plane passing through (x 0, y 0, z 0 ) with normal vector a, b, c has equation a(x x 0 ) + b(y y 0 ) + c(z z 0 ) = 0. In this case the desired plane passes through (1, 2, 1) and has normal vector 1, 1, 3. Therefore, 1(x 1) 1(y 2) + 3(z + 1) = 0, or x + y 3z = 6, is an equation for the desired plane. 1

(26) Find an equation of a plane that satisfies the stated conditions. The plane through the points P 1 ( 2, 1, 4), P 2 (1, 0, 3) that is perpendicular to the plane 4x y + 3z = 2. Solution: Because P 1 and P 2 lie in the desired plane, P 1 P 2 = 3, 1, 1 is parallel to that plane. Let n 1 = 4, 1, 3 be a normal vector to the given plane. Then a normal to the desired plane is any (nonzero) vector perpendicular to both P 1 P 2 and n 1. Therefore, n = P 1 P 2 n 1 = 3, 1, 1 4, 1, 3 = 4, 13, 1 is a normal to the desired plane. Since this plane passes through the point (1, 0, 3), it has equation 4(x 1) 13y + (z 3) = 0, or 4x + 13y z = 1. (29) Find an equation of a plane that satisfies the stated conditions. The plane whose points are equidistant from (2, 1, 1) and (3, 1, 5). Solution: The midpoint m between the two given points, P 1 (2, 1, 1) and P 2 (3, 1, 5), must lie in the desired plane and is given by m(x, y, z) = 1 2 (x 1 + x 2, y 1 + y 2, z 1 + z 2 ) = 1 2 (2 + 3, 1 + 1, 1 + 5) = ( 5 2, 0, 3). The directed line segment P 1 P 2 = 1, 2, 4 must be perpendicular to the desired plane and can be used as a normal vector n for the desired plane. Therefore, an equation for the desired plane is (x 5 2 ) + 2y + 4(z 3) = 0, or x + 2y + 4z = 29/2. (30) Find an equation of a plane that satisfies the stated conditions. The plane that contains the line x = 3t, y = 1 + t, z = 2t and is parallel to the intersection of the planes 2x y + z = 0 and y + z + 1 = 0. Solution: Let L be the given line, which is parallel to v = 3, 1, 2. Let P 1 : 2x y+z = 0 and P 2 : y + z + 1 = 0 represent the two given planes with normals n 1 = 2, 1, 1 and n 2 = 0, 1, 1. The line of intersection of P 1 and P 2 is parallel to n 1 n 2 = 2, 2, 2, which is parallel to w = 1, 1, 1. So a normal n to the desired plane is any (nonzero) vector perpendicular to both v and w. One such vector is n = v w = 3, 1, 2 1, 1, 1 = 3, 5, 2. Since the desired plane contains the line x = 3t, y = 1 + t, z = 2t, which contains the point (0, 1, 0), the desired plane passes through this point. Therefore, 3(x 0) + 5(y 1) + 2(z 0) = 0, or 3x 5y 2z = 5, is an equation for the desired plane. 2

OHSx XM521 Multivariable Differential Calculus: Homework Solutions 13.7 (18) Identify and sketch the quadric surface y 2 x2 4 z2 9 = 1. Solution: This is a hyperboloid of two sheets. It has no trace in the xz-plane since setting y = 0 results in x2 4 z2 9 = 1, which has no real solutions. The traces in planes parallel to the xz-plane with y 1 are ellipses. For example, when y = 2 the traces are given by the following equation: x 2 12 + z2 27 = 1. The two sheets of the hyperboloid open in the positve and negative y-directions with y-intercepts at (0, 1, 0) and (0, 1, 0). (25) The given equation represents a quadric surface whose orientation is different from those in Table 13.7.1. Identify and sketch the surface 2y 2 x 2 + 2z 2 = 8. Solution: in This is a hyperboloid of one sheet. Dividing the above equation by 8 results y 2 4 x2 8 + z2 4 = 1. This hyperboloid opens in the positive and negative x-directions. The trace in the yzplane (x = 0) is the circle y 2 + z 2 = 4. The traces in the planes x = 2 2 (parallel to the yz-plane) are circles given by y 2 + z 2 = 8. (35) Identify the surface 9x 2 + y 2 + 4z 2 18x + 2y + 16z = 10, and make a rough sketch that shows its position and orientation. 1

Solution: Completing the squares, we obtain which simplifies to or 9(x 2 2x + 1) + (y 2 + 2y + 1) + 4(z 2 + 4z + 4) = 10 + 9 + 1 + 16, 9(x 1) 2 + (y + 1) 2 + 4(z + 2) 2 = 36, (x 1) 2 + 4 (y + 1)2 36 + (z + 2)2 9 This is the equation of an ellipsoid centered at (1, 1, 2) and with axes of length 4 in the x-direction, length 12 in the y-direction, and length 6 in the z-direction. = 1. (37) Exercises 37 and 38 are concerned with the ellipsoid 4x 2 + 9y 2 + 18z 2 = 72. (a) Find an equation of the elliptical trace in the plane z = 2. Solution: Substituting z = 2 into the above equation, subtracting 36 from both sides, and then dividing both sides by 36 gives x2 9 + y2 4 = 1 as the equation of the elliptical trace. (b) Find the length of the major and minor axes of the ellipse in part (a). Solution: The major axis is along the x-direction and of length 2 3 = 6. The minor axis is along the y-direction and of length 2 2 = 4. 2

(c) Find the coordinates of the foci of the ellipse in part (a). Solution: The ellipse x 2 /a 2 +y 2 /b 2 = 1 with a > b has foci at coordinates (±c, 0) where c = a 2 b 2. In this case, c = 9 4 = 5. So the coordinates (in 3-space) of the foci are ( 5, 0, 2) and ( 5, 0, 2). (d) Describe the orientation of the focal axis of the ellipse in part (a) relative to the coordinate axes. Solution: The focal axis is parallel to the x-axis and in the plane z = 2. (39) Exercises 39-42 refer to the hyperbolic parabloid z = y 2 x 2. (a) Find an equation of the hyperbolic trace in the plane z = 4. Solution: Substituting z = 4 into the above equation and dividing both sides of equation by 4 results in y2 4 x2 4 = 1 for the hyperbolic trace. (b) Find the vertices of the hyperbola in part (a). Solution : The vertices are at y = 2 and y = 2, so (in 3-space) the coordinates are (0, 2, 4) and (0, 2, 4). (c) Find the foci of the hyperbola in part (a). Solution : The hyperbola y 2 /b 2 x 2 /a 2 = 1 has foci at coordinates (0, ±c) where c = a 2 + b 2. So in this case the coordinates (in 3-space) of the foci are (0, ±c, 4) where c = a 2 + b 2 = 4 + 4 = 2 2. (d) Describe the orientation of the focal axis of the hyperbola in part (a) relative to the coordinate axes. Solution: The focal axis is parallel to the y-axis and in the plane z = 4. (45) In Exercises 45 and 46, find an equation for the surface generated by revolving the curve about the axis. y = 4x 2 (z = 0) about the y-axis. Solution: Revolving any point (x 0, y 0 ) on the given curve about the y-axis results in a circle in the plane y = y 0 with radius equal to x 0 = y 0 /4, and centered at (0, y 0, 0). This circle consists of the set of points {(x, y, z) y = y 0, x 2 + z 2 = ( y 0 /4) 2 = y 0 /4}. The generated surface consists of all such circles where y 0 0. So the generated surface consists of the set of points {(x, y, z) x 2 + z 2 = y/4, y 0}, which (because squares are 3

always nonnegative) can be more concisely written {(x, y, z) 4x 2 + 4z 2 = y}. In short, the surface has equation 4x 2 + 4z 2 = y. (49) If a sphere x 2 a 2 + y2 a 2 + z2 a 2 = 1 of radius a is compressed in the z-direction, then the resulting surface, called an oblate spheroid, has an equation of the form x 2 a 2 + y2 a 2 + z2 c 2 = 1 where c < a. Show that the oblate spheroid has a circular radius a in the xy-plane and an elliptical trace in the xz-plane with major axis of length 2a along the x-axis and minor axis of length 2c along the z-axis. Solution: or simply In the xy-plane z = 0, so the spheroid equation becomes: x 2 a 2 + y2 a 2 = 1, x 2 + y 2 = a 2. 4

This is the standard equation of a circle with radius a in the xy-plane. In the xz-plane y = 0, so the spheroid equation becomes: x 2 a 2 + z2 c 2 = 1. This is the standard equation for an ellipse in the xz-plane with major axis of length 2a along the x-axis and minor axis of length 2c along the z-axis. 5

OHSx XM521 Multivariable Differential Calculus: Homework Solutions 13.8 (18) In Exercises 15-22, an equation is given in cylindrical coordinates. Express the equation in rectangular coordinates and sketch the graph. z = r cos θ. Solution: Since x = r cos θ, the equation becomes z = x. In the xz-plane, this is the line z = x. In 3-space, this is a plane defined by x z = 0 with normal vector n = 1, 0, 1. (20) In Exercises 15-22, an equation is given in cylindrical coordinates. Express the equation in rectangular coordinates and sketch the graph. r = 2 sec θ. Solution: r = 2 sec θ = 2/ cos θ, which can be rewritten as r cos θ = 2. Since x = r cos θ, the equation in rectangular coordinates becomes x = 2, which is obviously a plane. (28) In Exercises 23-30, an equation is given in spherical coordinates. Express the equation in rectangular coordinates and sketch the graph. ρ sin φ = 1. 1

Solution: ρ sin φ = r = x 2 + y 2 so the equation becomes x 2 + y 2 = 1, which is equivalent to x 2 + y 2 = 1. In the xy-plane, this is an equation of the unit circle, which in 3-space extrudes along the z-direction to a cylinder of radius 1. (30) In Exercises 23-30, an equation is given in spherical coordinates. Express the equation in rectangular coordinates and sketch the graph. ρ 2 sin φ cos θ = 0. Solution: Using the fact that sin φ cos θ = x/ρ, this equation becomes ρ 2x ρ = 0. Multiplying by ρ and using ρ 2 = x 2 + y 2 + z 2, this becomes x 2 + y 2 + z 2 2x = 0. Completing the square for x, this becomes (x 1) 2 + y 2 + z 2 = 1. This is the equation of a sphere with radius 1 centered at (1, 0, 0). (40) In Exercises 31-42, an equation of a surface is given in rectangular coordinates. Find an equation of the surface in (a) cylindrical coordinates and (b) spherical coordinates. x 2 + y 2 z 2 = 1. Solution (a): Since r 2 = x 2 + y 2, the equation becomes r 2 z 2 = 1 in cylindrical coordinates. This is an equation for a hyperboloid of one sheet. 2

Solution (b): Since r 2 = ρ 2 sin 2 φ and z = ρ cos φ, the equation becomes ρ 2 sin 2 φ ρ 2 cos 2 φ = 1. Using cos 2φ = cos 2 φ sin 2 φ, this simplifies to ρ 2 cos 2φ = 1. (42) In Exercises 31-42, an equation of a surface is given in rectangular coordinates. Find an equation of the surface in (a) cylindrical coordinates and (b) spherical coordinates. x 2 + y 2 + z 2 = 2z. Solution (a): Since r 2 = x 2 + y 2, the equation becomes r 2 + z 2 = 2z. Subtracting 2z from both sides and completing the square for z gives r 2 + (z 1) 2 = 1. This is a sphere of radius 1 centered at (0, 0, 1) (this coordinate is valid either in cylindrical or rectangular coordinates). Solution (b): Since ρ 2 = x 2 +y 2 +z 2 and z = ρ cos φ, this equation becomes ρ 2 = 2ρ cos φ, which simplifies to ρ = 2 cos φ. (43) In Exercises 43-46, describe the region in 3-space that satisfies the given inequalities. r 2 z 4. Solution: The constraint r 2 z implies that the points must be on or above (within) the circular parabloid z = r 2 = x 2 + y 2. Therefore the region in 3-space consists of those points that lie on or directly above (within) the circular parabloid z = r 2 = x 2 + y 2 and on or below the plane z = 4. The projection of this region onto the xy-plane is the disk given by the inequality x 2 + y 2 4. (44) In Exercises 43-46, describe the region in 3-space that satisfies the given inequalities. 0 r 2 sin θ, 0 z 3. Solution: Using sin θ = y/r, the equation r = 2 sin θ becomes r = 2y/r. Multiplying both sides by r, using r 2 = x 2 + y 2, subtracting 2y from both sides, and completing the square for the y term, this becomes x 2 + (y 1) 2 = 1, which is a circle of radius 1 centered at (0, 1) in the xy-plane. Since r 2 sin θ, the region in 3-space must be those points (x, y, z) such that (x, y) lies on or within the circle x 2 + (y 1) 2 = 1 and 0 z 3. This 3

is a cylinder of radius 1 and height 3, parallel to the z-axis and with base centered at (0, 1) in the xy-plane. (49) The accompanying figure shows a right circular cylinder of radius 10 cm spinning at 3 revolutions per minute about the z-axis. At time t = 0, a bug at the point (0, 10, 0) begins walking straight up the face of the cylinder at the rate of 0.5 cm/min. (a) Find the cylindrical coordinates of the bug after 2 min. Solution: Since the bug will stay on the surface of the cylinder, r = 10 cm for the bug even when the cylinder rotates. After 2 min., the cylinder will undergo (3rev/min)(2min)=6 revolutions, so the θ coordinate will be the same as its initial value. Since the bug starts on the y-axis, θ = π 2 radians. Since the bug moves in the vertical direction at a rate of 0.5 cm/min., z =(0.5cm/min)(2 min)=1 cm. Therefore the cylindrical coordinates are (10 cm, π 2 rad,1 cm). (b) Find the rectangular coordinates of the bug after 2 min. Solution: Since the 6 revolutions will bring the bug back to the same x and y coordinates as it initially starts from, x = 0 cm and y = 10 cm. The z coordinate will be the same as in (a): z = 1 cm. Therefore the rectangular coordinates are (0, 10, 1) cm. (c) Find the spherical coordinates of the bug after 2 min. Solution: Since ρ = r 2 + z 2, the value of ρ can be calculated by converting the cylindrical coordinates from (a): ρ = (10) 2 + 1 2 = 101 cm. The θ value will be the same in spherical and cylindrical coordinates: θ = π 2 radians. The value of φ can be calculated by converting from cylindrical coordinates using the formula tan φ = r/z = 10/1 = 10; so φ = tan 1 (10) 1.47 radians. Therefore, the spherical coordinates are ( 101 cm, π 2 rad, 1.47 rad). 4

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