University:
University of Michigan
Course:
MATH 116 | Calculus II
Academic year:

2024
Views:

48

Pages:

6
Author:

banenetta0l

Question 1 Whether the statement "If a system of two linear equations in two variables is dependent, then it has infinitely many solutions" is true or false. Answer: Definition used: If a system of equations has at least one solution, it is known as consistent. For a system of two linear equations in two variables, if one equation is a constant multiple of the other equation, the systems are dependent. Calculation: A consistent system is dependent if the equations have the same slope and the same y-intercepts. In this case, the lines coincide, representing infinitely many solutions. Hence, a dependent system of two linear equations in two variables always represents infinitely many solutions. Conclusion: The statement is true. Question 2 The Cartesian coordinates of a point are given: a) ( 2 , − 2 ) (2,−2) b) ( − 1 , 3 ) (−1, 3 ) Find the polar coordinates ( 𝑟 , 𝜃 ) (r,θ) of the point where: 𝑟 > 0 r>0, 0 ≤ 𝜃 < 2 𝜋 0≤θ<2π 𝑟 < 0 r<0, 0 ≤ 𝜃 < 2 𝜋 0≤θ<2π Answer: Definition: Polar coordinates ( 𝑟 , 𝜃 ) (r,θ): 𝑟 2 = 𝑥 2 + 𝑦 2 , tan ⁡ 𝜃 = 𝑦 𝑥 r 2 =x 2 +y 2 ,tanθ= x y a) For ( 2 , − 2 ) (2,−2): 𝑟 2 = 2 2 + ( − 2 ) 2 = 8 , 𝑟 = ± 2 2 r 2 =2 2 +(−2) 2 =8,r=±2 2 tan ⁡ 𝜃 = − 2 2 = − 1 tanθ= 2 −2 =−1 The angle 𝜃 θ is 7 𝜋 4 4 7π for 𝑟 > 0 r>0 and 3 𝜋 4 4 3π for 𝑟 < 0 r<0. Polar Coordinates: For 𝑟 > 0 r>0: ( 2 2 , 7 𝜋 4 ) (2 2 , 4 7π ) For 𝑟 < 0 r<0: ( − 2 2 , 3 𝜋 4 ) (−2 2 , 4 3π ) b) For ( − 1 , 3 ) (−1, 3 ): 𝑟 2 = ( − 1 ) 2 + ( 3 ) 2 = 4 , 𝑟 = ± 2 r 2 =(−1) 2 +( 3 ) 2 =4,r=±2 tan ⁡ 𝜃 = 3 − 1 = − 3 tanθ= −1 3 =− 3 The angle 𝜃 θ is 2 𝜋 3 3 2π for 𝑟 > 0 r>0 and 5 𝜋 3 3 5π for 𝑟 < 0 r<0. Polar Coordinates: For 𝑟 > 0 r>0: ( 2 , 2 𝜋 3 ) (2, 3 2π ) For 𝑟 < 0 r<0: ( − 2 , 5 𝜋 3 ) (−2, 3 5π ) Question 3 Find or create an example of a system of equations with one solution. Graph and label the lines on a coordinate plane. Provide their equations. State the accurate solution to the system. Answer: Equations: 𝑥 − 𝑦 = 2 (1) x−y=2(1) 𝑥 + 𝑦 = 4 (2) x+y=4(2) Graph: Two lines plotted intersect at ( 3 , 1 ) (3,1). Solution: From (1) and (2): 𝑥 = 3 , 𝑦 = 1 x=3,y=1 Final Answer: The solution to the system is ( 3 , 1 ) (3,1). Question 4 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 − 𝑥 2 y<9−x 2 𝑦 ≥ 𝑥 + 3 y≥x+3 Answer: Plot the inequality 𝑦 < 9 − 𝑥 2 y<9−x 2 . Plot the inequality 𝑦 ≥ 𝑥 + 3 y≥x+3. Find the vertices of the shaded region (intersection). Solve 9 − 𝑥 2 = 𝑥 + 3 9−x 2 =x+3: 𝑥 2 + 𝑥 − 6 = 0 x 2 +x−6=0 ( 𝑥 + 3 ) ( 𝑥 − 2 ) = 0 , 𝑥 = − 3 , 2 (x+3)(x−2)=0,x=−3,2 Substituting 𝑥 = − 3 x=−3 and 𝑥 = 2 x=2 gives 𝑦 = 0 y=0 and 𝑦 = 5 y=5. Vertices: ( − 3 , 0 ) , ( 2 , 5 ) (−3,0),(2,5) Conclusion: The solution set is bounded. Question 5 Plot the given point ( − 4 , 0 ) (−4,0) in a rectangular coordinate system. Answer: Step 1: Locate ( − 4 , 0 ) (−4,0) on the graph (4 units to the left of the origin). Step 2: Draw the point and label it as ( − 4 , 0 ) (−4,0). Conclusion: Point 𝐴 A is shown in the graph.