Questions and Answers #1 Alternate Coordinate Systems

Questions and Answers Sheet Alternate Coordinate Systems Question #1 Find Maximum Volume of a rectangular box that is inscribed in a sphere of radius r. The part i do not get the solutions manual got 2𝑥, 2𝑦 , and 2𝑧 as the volume. Can you explain how the volume of the cube for this problem is 𝑣 = 8𝑥𝑦𝑧? Answer: Recall that the volume of a box is 𝑉 = 𝑙 × 𝑤 × ℎ . The volume of the box inscribed inside a sphere where (𝑥, 𝑦, 𝑧) is the point in octant 1 where the box touches, the sphere will be 𝑉 = (2𝑥)(2𝑦)(2𝑧) = 8𝑥𝑦𝑧 . Since this point lies on a sphere, it must satisfy 𝑥 2 + 𝑦 2 + 𝑧 2 = 𝑟 2 . We have two equations in which we can get rid of one variable. 𝑉 = 8𝑥𝑦𝑧 𝑧= 𝑉 8𝑥𝑦 𝑥 2 + 𝑦2 + 𝑧2 = 𝑟2 𝑧 = √(𝑟 2 − 𝑥 2 − 𝑦 2 ) (no ± because in octant 1, z is positive) 𝑉 8𝑥𝑦 = √(𝑟 2 − 𝑥 2 − 𝑦 2 ) 𝑉 = 8𝑥𝑦√(𝑟 2 − 𝑥 2 − 𝑦 2 ) This is just an optimization problem now. (𝑥 2 )𝑦 ∂𝑓 = 8 (𝑦√(𝑟 2 − 𝑥 2 − 𝑦 2 ) − )=0 ∂𝑥 √(𝑟 2 − 𝑥 2 𝑦 2 ) (𝑦 2 )𝑥 ∂𝑓 = 8 (𝑦√(𝑟 2 − 𝑥 2 − 𝑦 2 ) − )=0 ∂𝑦 √(𝑟 2 − 𝑥 2 𝑦 2 ) The solution to these sets of equations are ( 𝑟 √3 , 𝑟 √3 ) (there's also (0, 0) but that's obvious not a maximum point) The maximum volume therefore is: 𝑉 = 8𝑥𝑦𝑧 𝑟 𝑟 𝑟 √3 √3 √3 = 8( )( )( ) = 8𝑟 3 (3√3) Question #2 The Cartesian coordinates of a point are given. a) (2, −2) b) (−1, √3) Find the polar coordinates (𝑟, θ) of the point, where r is greater than 0 and 0 is less than or equal to θ , which is less than 2π Answer: Consider the Cartesian coordinates be (𝑥, 𝑦). The polar coordinates (𝑟, θ) is defined as, 𝑟 2 = 𝑥 2 + 𝑦 2 𝑥 = 𝑟 cos θ 𝑦 = 𝑟 sin θ This implies that, tan θ = 𝑦 𝑥 a) The Cartesian coordinates be (2, −2) . Calculate r and θ as follows: 𝑟 2 = 22 + (−2)2 =8 𝑟 = ±2√2 tan θ = = 𝑦 𝑥 −2 2 = −1 Thus, the angle θ is 3π 7π 4 , 4 as the angle 0 ≤ θ2π The polar coordinates of (2, −2) are, (2√2, 7π 4 ), where 𝑟 > 0 (−2√2, 3π 4 ) , where 𝑟 < 0 b) The Cartesian coordinates be (−1, √3). Calculate r and θ as follows: 2 𝑟 2 = (−1)2 + (√3) =4 𝑟 = ±√2 tan θ = = 𝑦 𝑥 √3 −1 = −√3 Thus, the angle θ is 2π 5π 3 , 3 as the angle 0 ≤ θ ≤ 2π The polar coordinates of (−1, √3) are, (2, 2π 5π 3 3 ) , where 𝑟 > 0 (−2, ) , where 𝑟 < 0 Question #3 For the polar point (−4, − 3π ) 4 a) Give alternate coordinates with 𝑟 > 0 and −2 π ≤ θ < 0 . b) Give alternate coordinates with 𝑟 < 0 and 0 ≤ θ < 2 π. c) Give alternate coordinates of your choice. Answer: Step 1 Given: 𝑃 (−4, − 3π 4 ) a) (−4, − 3π 4 ) = (4, − 3π 4 + π) π = (4, ) 4 So, (−4, − 3π π 4 4 ) = (4, ) Step 2 b) (−4, − = (−4, − 4 ) = (−4, − 3π 4 + 2π) 5π ) 4 So, (−4, − 3π 5π 4 4 ) = (−4, c) (−4, − = (4, 3π 3π 4 ) ) = (4, − 3π 4 + 3π) 9π ) 4 So, (−4, − 3π 9π 4 4 ) = (4, ) Question #4 Given point P(-2,6,3) and vector 𝐵 = 𝑦𝑎𝑥 + (𝑥 + 𝑧)𝑎𝑦 , Express P and B in cylindrical and spherical coordinates. Evaluate A at P in the Cartesian, cylindrical and spherical systems. Answer: Step 1 The objective is to express the point P-2,6,3 and vector B=yax+x+zay in cylindrical and spherical coordinates. Step 2 To convert Cartesian coordinates x,y,z to cylindrical coordinates 𝑟, θ, 𝑧 𝑟 = 𝑥2 + 𝑦2 θ = 𝑡𝑎𝑛 − 1𝑦𝑥𝑧 = 𝑧 The point P-2,6,3 in cylindrical coordinates is, 𝑟 = 4 + 36 = 40 = 210 And θ = 𝑡𝑎𝑛 − 16 − 2 = −1.25 Cylindrical coordinates is 210,-1.25,3 To convert Cartesian coordinates x,y,z to spherical coordinates ρθϕ ρ = 𝑥2 + 𝑦2 + 𝑧2 θ = 𝑡𝑎𝑛 − 1𝑦𝑥 ϕ = 𝑡𝑎𝑛 − 1𝑥2 + 𝑦2𝑧 The point P-2,6,3 in spherical coordinates is, ρ = 𝑥2 + 𝑦2 + 𝑧2 = 4 + 36 + 9 = 49 = 7 And θ = 𝑡𝑎𝑛 − 16 − 2 = −1.25 And ϕ = 𝑡𝑎𝑛 − 14 + 363 = 𝑡𝑎𝑛 − 1403 = 1.3034 Spherical coordinates is 7, −1.25,1.3034 Step 3 The vector B=yax+x+zay in cylindrical and spherical coordinates. In the cartesian system B at P is B=6ax+ay For vector B, Bx=yBy=x+zBz=0 In cylindrical system 𝐴𝑟 = 𝑦 cos θ + 𝑥 + 𝑧 sin θ 𝐴θ = −𝑦 sin θ + 𝑥 + 𝑧 cos θ 𝐴𝑧 = 0 From step 2 we have substituting values of θ, we get 𝐴 = −0.9487𝑎𝑟 − 6.008𝑎 θ 𝑆imilarly in spherical system 𝐴ρ = 𝑦 sin θ cos ϕ + 𝑥 + 𝑧 sin θ sin ϕ 𝐴θ = 𝑦 cos θ cos ϕ + 𝑥 + 𝑧 cos θ sin ϕ 𝐴ϕ = −𝑦 sin ϕ + (𝑥 + 𝑧) cos ϕ. Since, 𝑥 = ρ sin θ cos ϕ 𝑦 = ρ sin θ sin ϕ 𝑧 = ρ cos θ. From step 2, substitute ρ, θ, ϕ we get, 𝐴 = −0.8571𝑎 ρ − 0.4066𝑎 θ − 6.008𝑎 ϕ Question #5 a) What coordinate system is suggested to be used when seeing the integrand (𝑥 2 + 𝑦 2 )5/2 , cylindrical or Cartesian? b) List the conversion factors for rectangular to cylindrical coordinate systems. Answer: Step 1 We determine which coordinate system is used Step 2 a) (𝑥 2 + 𝑦 2 )5/2 cylindrical b) we know that 𝑥 2 + 𝑦 2 = γ2 (𝑥 2 + 𝑦 2 )5/2 = (γ2 )5/2 = γ5 Question #6 Below are various vectors in cartesian, cylindrical and spherical coordinates. Express the given vectors in two other coordinate systems outside the coordinate system in which they are expressed a) 𝐴⃗(𝑥, 𝑦, 𝑧) = ⃗⃗⃗⃗ 𝑒𝑥 ⃗ d) 𝐴(ρ, ϕ, 𝑧) = ⃗⃗⃗⃗ 𝑒ρ ⃗ g) 𝐴(𝑟, θ, ϕ) = ⃗⃗⃗⃗ 𝑒θ ⃗⃗⃗⃗⃗+𝑥𝑒 ⃗⃗⃗⃗⃗ −𝑦𝑒 j) 𝐴⃗(𝑥, 𝑦, 𝑧) = 2𝑥 2 𝑦 𝑥 +𝑦 Answer: Step 1 Given: a) 𝐴⃗(𝑥, 𝑦, 𝑧) = ⃗⃗⃗⃗ 𝑒𝑥 To express it in other two coordinate systems: ie., in terms of cartesian, cylindrical and spherical coordinates Step 2 Given: a) 𝐴⃗(𝑥, 𝑦, 𝑧) = ⃗⃗⃗⃗ 𝑒𝑥 spherical coordinates can be expressed as (𝑟, θ, ϕ) = 𝑦 𝑧 𝑥 √𝑥 2 +𝑦 2 +𝑧 2 (√𝑥 2 + 𝑦 2 + 𝑧 2 , tan−1 ( ) , cos −1 ( π )) Therefore, (𝑟, θ, ϕ) = (1,0, ) 2 Question #7 Convert between the coordinate systems. Use the conversion formulas and show work. π π spherical: (8, , ) 3 6 Change to cylindrical. Answer: π π Given: spherical coordinates: (8, , ). To find: convert into, cylindrical coordinates. Solution: The 3 6 π π 3 6 spherical coordinates are given by (𝑠, ϕ, ψ) So here 𝑠 = 8; ϕ = ; ψ = These coordinates can be π π 6 3 converted into cylindrical coordinates by: γ = 𝑠 sin ψ ϕ = ϕ𝑧 = 𝑠 cos ψ. So; γ = 8 sin ( ) ; ϕ = ; 𝑧 = π 1 π √3 6 2 3 2 8 cos ( ) . Thus γ = 8 ⋅ ; ϕ = ; 𝑧 = 8 π Thus; γ = 4; ϕ = ; 𝑧 = 4√3 3 Question #8 This question has to do with binary star systems, where 'i' is the angle of inclination of the system. Calculate the mean expectation value of the factor sin3 i, i.e., the mean value it would have among an ensemble of binaries with random inclinations. Find the masses of the two stars, if sin3 i has its mean value. Hint: In spherical coordinates, (θ, ϕ) , integrate over the solid angle of a sphere where the observer is in the direction of the z-axis, with each solid angle element weighted by sin 3 (θ). 𝑣1 = 100𝑘𝑚/𝑠 𝑣2 = 200𝑘𝑚/𝑠 Orbital period =2 days 𝑀1 = 5.74𝑒33𝑔 𝑀2 = 2.87𝑒33𝑔 Answer: Step 1 For a binary star system, the ratio of their velocities and their masses are related as, 𝑚1 𝑚2 = 𝑣2 𝑣1 = 200𝑘𝑚/𝑠 100𝑘𝑚/𝑠 = 2𝑚1 = 2𝑚2 … (1) Step 2 The Kepler equation for a binary star system is, 𝑚1 + 𝑚2 = 𝑃 2π𝐺 (v1 + v2 )3 (2  24  3600s) (105 ms −1 + 2 105 ms −1 )3 = sin 3 (i) 2 (6.67 10−11 Nm2 kg −2 sin 3 (i) (1.1110 = kg )  M d    30 sin 3 ( i )  2 10 kg  31 = 5.556M sin 3 (i) Step 3 π/2 The mean expectation value of sin3 i be given as, ≪ sin3 (𝑖) ≫= ∫0 3𝑖 sin(2𝑖) 8 4 [ − + π sin(4𝑖) 2 32 ] = 0 3π 16 π/2 sin3 (𝑖) {sin 𝑖 𝑑𝑖} ∫0 sin4 (𝑖) 𝑑𝑖 = = 0.59 Step 4 From equation (1), 𝑚1 = 2𝑚2 Hence, m1 = 6.3M . Here M 3m2 = 5.556M 5.556M m = = 3.1389M 2 sin 3 (i) 3(0.59) defined as the solar mass. Solar mass equal to approximately 2 × 1030 𝑘𝑔. Step 5 Hence, m1 = 6.3M Similarly, = (6.3)(2 1030 )kg = (12.6 1030 kg )( 1000 g ) = 12.6 1033 g 1kg And m2 = 3.13M = (3.13)(2 1030 )kg = (6.26 1030 kg )( 1000 g ) = 6.26 1033 g 1kg Question #9 Solve the following systems of inequalities graphically on the set of axes below.The state the following coordinates of a point in the solution say 𝑦 ≤𝑥+6 3 𝑦≻ 𝑥−4 2 Answer: Step 1 The graph is shown below: FSP196104013522.jpgFSZ Step 2 The solutions are (1,0),(2,0),(1,1),(-3,0),(0,0) Question #10 4.(a) Sketch the region of solutions of the following systems of inequalities: 2? + 3? ≤ 12 -? + 2? < 4 ?≥0 ?≥0 b. List the coordinates of all of the corner points: c Which corner point would maximize the equation ? = 5? + 3? ? Answer: Step 1 Inequalities Step 2 x 0 6 x 0 −4 2 x + 3 y  12 − x + 2 y  4 x  0 y  0 4 4 0 4 2 0 Intersection pt: 2x+3y=12 -2x+4y=8 7y=20 y=20/7 x=12/7 b) Comer points: (12/7, 20/7), (6,0), (0,2), (0,0) c) z=5x+3y A(6,0), z=30 (Max) B(0,2), z=6 C(0,0), z=0 D(12/7, 20/7)m z=120/7 Question #11 Systems of Inequalities Graph the solution set of the system if inequalities. Find the coordinates of all vertices, and determine whether the solution set is bounded.  y  9 − x2  y  x +3 Answer: To graph: The given system of inequality. Also find the coordinates of all vertices, and check whether the solution set is bounded. Graph: The given system of the inequalities is, 𝑦 < 9 − 𝑥 2 .....(1) 𝑦 ≥ 𝑥 + 3 .....(2) The corresponding equation of the inequality (1) is, 𝑦 = 9 − 𝑥 2 Since, the inequality 𝑦 < 9 − 𝑥 2 involves the strictly less than sign, then the points on the parabola y=9-x^{2} does not satisfy the inequality 𝑦 < 9 − 𝑥 2 . Therefore, the parabola 𝑦 = 9 − 𝑥 2 itself is not the part of the solution set. The corresponding equation of the inequality (2) is, 𝑦 = 𝑥 + 3 Since, the inequality 𝑦 ≥ 𝑥 + 3 involves the greater than or equal to sign, then the points on the line y=x+3 satisfies the inequality 𝑦 ≥ 𝑥 + 3 Therefore, the line y=x+3 itself is the part of the solution set. Consider the test points (0,4) to check whether the solution satisfies each inequality of the given system. Substitute 0 for x and 4 for y in the inequality 𝑦 < 9 − 𝑥 2 , 𝑦 < 9 − 𝑥 2, 4<9-0 4<9. The point (0,4) is inside the parabola 𝑦 = 9 − 𝑥 2 and satisfies the inequality 𝑦 < 9 − 𝑥 2 . So it is the part of the solution set. Substitute 0 for x and 4 for y in the inequality 𝑦 ≥ 𝑥 + 3 . 𝑦 ≥ 𝑥 + 3, 4 ≥ 0 + 3, 4 ≥ 3. The points (0,4) is below the line y=x+3 and satisfies the inequality 𝑦 ≥ 𝑥 + 3 . So it is the part of the solution set. Therefore, the test point(0,4) satisfies each inequality of the given system. The solution set of the given system of inequalities is the intersection of the solutions of each of the given inequality. Therefore, the solution set is shown as shaded region. The vertices occur at the points of intersection of the corresponding equation of the given system of inequalities. 𝑦 = 9 − 𝑥 2 .....(3) 𝑦 = 𝑥 + 3 .....(4) It is observed from Figure 1 that the parabola 𝑦 = 9 − 𝑥 2 and line y=x+3 intersect each other. Substitute x+3 for y in equation (3). 𝑦 = 9 − 𝑥 2 , 𝑥 + 3 = 9 − 𝑥 2 , 𝑥 2 + 𝑥 = 9 − 3, 𝑥 2 + 𝑥 − 6 = 0. Further solve the above equation for the value of x. 𝑥 2 + 3𝑥 − 2𝑥 − 6 = 0, (𝑥 − 2)(𝑥 + 3) = 0, x+3=0 or x-2=0 x=-3 or x=2. Therefore, the x-coordinate of vertex are -3 and 2. Substitute -3 for x in equation (3). 𝑦 = 9 − 𝑥 2 , 𝑦 = 9 − (−3)2 , y=0 Substitute 2 for x in exuation (3). 𝑦 = 9 − 𝑥 2 , 𝑦 = 9 − (2)2, y=9-4 y=5 Therefore, the y-coordinate of vertex are 0 and 5. Therefore, the vertices of the shaded region are (-3,0) and (2,5). It is observed from Figure 1 that the shaded region is enclosed by the boundary lines of the given system of inequalities. Therefore, the shaded region is bounded. Interpretation: The solution set of the given system of inequality lies in 1 and 2 quadrant as shown in Figure 1. Conclusion: Thus, the vertices of the given system of inequalities are (-3,0) and (2,5), the solution set is bounded.'' Question #12 Consider the following system.  y = 30 − x  2 2 y = − x + 16 x − 12 Answer: 𝑦 = 30 − 𝑥 ⇒ (1) 2𝑦 = −𝑥 2 + 16𝑥 − 12 ⇒ (2) from 1 and 2 we have 2(30 − 𝑥) = −𝑥 2 + 16𝑥 − 12 60 − 2𝑥 = −𝑥 2 + 16𝑥 − 12 𝑥 2 − 16𝑥 − 2𝑥 + 12 + 60 = 0 𝑥 2 − 18𝑥 + 72 = 0 𝑥 2 − 6𝑥 − 12𝑥 + 72 = 0 𝑥(𝑥 − 6) − 12(𝑥 − 6) = 0(𝑥 − 6)(𝑥 − 12) = 0 ∵ 𝑥 − 6 = 0, 𝑥 − 12 = 0 ∵ x=6, x=12 when x=6, y=30-6=24 [from 1] when x=12, y=30-12=18 [from 1]. Hence the requined solution (6,24) and (12,18)