Questions and Answers #7 Equations

Questions and Answers Sheet 7 Equations Question #1 Find a polar equation for the curve represented by the given Cartesian equation. 𝑥 2 + 𝑦 2 = 6𝑥 − 4𝑦 Answer: Step 1 Given equation is : 𝑥 2 + 𝑦 2 = 6𝑥 − 4𝑦 Concept : To convert an equation in polar form, we substitue 𝑥 = 𝑟 cos 𝜃 and 𝑦 = 𝑟 sin 𝜃; where r = length of the pole and θ is the angle of the pole with +x axis Now, we know 𝑥 = 𝑟 cos 𝜃 𝑦 = 𝑟 sin 𝜃 Step 2 Substituting the value of x and y in the given equation 𝑥 2 + 𝑦 2 = 6𝑥 − 4𝑦 (𝑟 cos 𝜃)2 + (𝑟 sin 𝜃)2 = 6 ∗ 𝑟 cos 𝜃 − 4𝑟 sin 𝜃 𝑟 2 cos 2 𝜃 + 𝑟 2 sin2 𝜃 = 6𝑟 cos 𝜃 − 4𝑟 sin 𝜃 𝑟 2 (cos 2 𝜃 + sin2 𝜃) = 6𝑟 cos 𝜃 − 4𝑟 sin 𝜃 𝑟 2 = 𝑟(6 cos 0 − 4 sin 0) [∵ cos 2 0 + sin2 0 = 1] 𝑟 = 6 cos 0 − 4 sin 0 (cancelling out r from both side) 𝑟 = 6 cos 0 − 4 sin 0 Question #2 Find an equation of the tangent line to the graph of the equation 𝑥 + 𝑦 − 𝑒 𝑥−𝑦 = 3 at the point (2,2). Answer: Step 1 We have 𝑥 + 𝑦 − 𝑒 𝑥−𝑦 = 3 Now find the first derivative 𝐷(𝑥 + 𝑦 − 𝑒 𝑥−𝑦 ) = 𝐷(3) 1 + 𝑦 ′ − 𝑒 𝑥−𝑦 + 𝑒 𝑥−𝑦 𝑦 ′ = 0 𝑦 ′ + 𝑦 ′ 𝑒 𝑥−𝑦 = 𝑒 𝑥−𝑦 − 1 𝑦 ′ (1 + 𝑒 𝑥−𝑦 ) = 𝑒 𝑥−𝑦 − 1 𝑦′ = 𝑒 𝑥−𝑦 −1 𝑒 𝑥−𝑦 +1 Now put x=2, y=2 𝑦′ = ′ 𝑒 2−2 −1 𝑒 2−2 +1 0 𝑦 = =0 2 𝑚 = 𝑦′ = 0 Step 2 The tangent line is of the form y=mx+c put x=2,y=2, m=0 y=mx+c 2=(0)2+c ⇒𝑐=2 So the equation of the tangent line becomes. y=mx+c y=(0)x+2 y=2 Step 3 Answer: y=2 is the equation of tangent line. Question #3 Convert the given polar equation to a Cartesian equation. Write in the standard form of a conic if possible, and identify the conic section represented. 𝑟 = 2 sec 0 Answer: Step 1 Given: 𝑟 = 2 sec 0 2 𝑟= cos 0 𝑟 cos 0 = 2 Step 2 The polar coordinates r, 0 and cartesian coordinates x,y are related by 𝑥 = 𝑟 cos 0 Hence, the equation 𝑟 cos 0 = 2 in cartesian coordinates is x=2 The equation represents a straight line parallel to y-axis Question #4 Transform the given differential equation or system into an equivalent system of first-order differential equations. 𝑥 ′′ + 3𝑥 ′ + 7𝑥 = 𝑡 2 Answer: Step 1 Given: 𝑥 ′′ + 3𝑥 ′ + 7𝑥 = 𝑡 2 Step 2 Substituting 𝑦1 = 𝑥, 𝑦2 = 𝑥 ′ , we get 𝑦1 = 𝑥 ′ = 𝑦2 , 𝑦2′ = (𝑥 ′ )′ = 𝑥 ′′ But from the original differential equation 𝑥 ′′ + 3𝑥 ′ + 7𝑥 = 𝑡 2 , we have 𝑥 ′′ + 3𝑥 ′ + 7𝑥 = 𝑡 2 ⇒ 𝑥 ′′ = 𝑡 2 − 3𝑥 ′ − 7𝑥 Substituting 𝑥 = 𝑦1 , 𝑥 ′ = 𝑦2 we get 𝑦2′ = 𝑡 2 − 3𝑦2 − 7𝑦1 Therefore, our original differential equation is equivalent to 𝑦1′ = 𝑥 ′ = 𝑦2 , 𝑦2′ = 𝑥 ′′ = 𝑡 2 − 3𝑦2 − 7𝑦1 Therefore, we have the following system, of linear first order differential equations 𝑦1′ = 𝑦2 𝑦2′ = −7𝑦1 − 3𝑦2 + 𝑡 2 Question #5 Find the solution of the equation in the interval 00 ≤ 𝑥 < 3600 . 2 cos 2 𝑥 − 5 cos 𝑥 = 3 Answer: Step 1 It is given that, 2 cos 2 𝑥 − 5 cos 𝑥 = 3 We have to find the solution of the equation in the interval 00 ≤ 𝑥 < 3600 . Step 2 We have, 2 cos 2 𝑥 − 5 cos 𝑥 = 3 ⇒ 2 cos 2 𝑥 − (6 − 1) cos 𝑥 − 3 = 0 ⇒ 2 cos 2 𝑥 − 6 cos 𝑥 + 1 cos 𝑥 − 3 = 0 ⇒ 2 cos 𝑥 (cos 𝑥 − 3) + 1(cos 𝑥 − 3) = 0 ⇒ (cos 𝑥 − 3)(2 cos 𝑥 + 1) = 0 ⇒ (cos 𝑥 − 3) = 0 or (2 cos 𝑥 + 1) = 0 1 ⇒ cos 𝑥 = 3 or cos 𝑥 = − ⇒𝑥= 4𝜋 or 𝑥 = 3 2 2𝜋 3 ⇒ 𝑥 = 240∘ or 𝑥 = 120∘ Hence, 𝑥 = 120∘ , 240∘ Question #6 Solve the equation. (find all solutions of the equation in the interval (0,2𝜋). (7 sin(2𝑥) + 7 cos(2𝑥))2 = 49 Answer: Given (7 sin(2𝑥) + 7 cos(2𝑥))2 = 49 solution (7 sin(2𝑥) + 7 cos(2𝑥))2 = 49 72 (sin(2𝑥) + cos(2𝑥))2 = 49 (sin(2𝑥) + cos(2𝑥))2 = 1 sin2 (2𝑥) + cos 2 (2𝑥) + 2 sin(2𝑥) cos(2𝑥) = 1 1 + 2 sin(2𝑥) cos(2𝑥) = 1 2 sin(2𝑥) cos(2𝑥) = 0 sin(2(2𝑥)) = 0 (since, 2 sin 𝑥 cos 𝑥 = sin 2 𝑥 ) sin 4 𝑥 = 0 4𝑥 = sin−1 0 4𝑥 = 𝑛𝜋, 𝑛 ∈ 𝑍 𝜋 2𝜋 3𝜋 4𝜋 5𝜋 6𝜋 7𝜋 8𝜋 𝑥= , 4 4 , 4 , 4 , , 4 4 , 4 , 4 ,..... Therefore the solution are, 𝜋 𝜋 3𝜋 𝑥= , , 4 2 4 , 𝜋, 5𝜋 3𝜋 7𝜋 4 , 2 , 4 Question #7 Find the solutions of the equation. 𝑥 2 + 4𝑥 + 13 = 0 Answer: Step 1 The given equation is: 𝑥 2 + 4𝑥 + 13 = 0 … (1) As, the given equation is a quadratic equation. The general form f a quadratic equation is: 𝑎𝑥 2 + 𝑏𝑥 + 𝑐 = 0 … (2) So, finding the roots of any quadratic equation, we use: 𝑥= −𝑏±√𝑏 2 −4𝑎𝑐 2𝑎 On comparing equation (1) with the equation (2), we have: a=1 b=4 c=13 Step 2 Solving it, we get: 𝑥= 𝑥= 𝑥= −4±√(−4)2 −4(1)(13) 2(1) −4±√16−52 2 −4±√−36 2 𝑥 = −2 ± √−9 𝑥 = −2 ± 3√−1 Since, √−1 = 𝑖 So, 𝑥 = −2 ± 3𝑖 Hence, the quadratic equation have complex roots. Therefore, we have: 𝑥 = −2 + 3𝑖 𝑥 = −2 − 3𝑖 The required solution of the given quadratic equation. Question #8 Solve each equation for the other variable. 𝑣 = 8(2.5)𝑡 Answer: Step 1 Given equation: 𝑣 = 8(2.5)𝑡 Step 2 Now, Solving for variable 't' 𝑣 = 8(2.5)𝑡 𝑣 (2.5)𝑡 = 8 Taking natural logarithm both sides: 𝑣 ln[(2.5)𝑡 ] = ln [ ] 8 Apply log rule: ln(𝑥 𝑦 ) = 𝑦 ln(𝑥) 𝑣 So, ln[(2.5)𝑡 ] = ln [ ] 8 𝑣 𝑡 ln(2.5) = ln ( ) 8 𝑡= 𝑣 8 ln( ) ln(2.5) Question #9 A nonhomogeneous equation and a particular solution are given. Find a general solution for the equation. 𝑦 ′′ − 𝑦 = 10𝑡, 𝑦𝑝 (𝑡) = −10𝑡 𝑦(𝑡) =? Answer: Step 1 According to the given information, it is required to find the general solution of the given differential equation whose particular solution is given. 𝑦 ′′ − 𝑦 = 10𝑡, 𝑦𝑝 (𝑡) = −10𝑡 Step 2 The general solution to the non-homogeneous differential equation can be written as: 𝑦(𝑡) = 𝑦𝑐 (𝑡) + 𝑦𝑝 (𝑡) where 𝑦𝑐 (𝑡) is the solution to homogeneous differential equation and 𝑦𝑝 (𝑡) is the particular solution Step 3 Now, first find the solution to homogeneous differential equation. characteristic equation: 𝑟2 − 1 = 0 𝑟2 = 1 𝑟 = ±1 𝑦𝑐 (𝑡) = 𝑐1 𝑒 𝑡 + 𝑐2 𝑒 −𝑡 Step 4 Now, the general solution to differential equation is: 𝑦(𝑡) = 𝑦𝑐 (𝑡) + 𝑦𝑝 (𝑡) = 𝑐1 𝑒 𝑡 + 𝑐2 𝑒 −𝑡 − 10𝑡 Question #10 Find the parametric equation (linear). remove the parameter to find the equation. 𝑥 = 6 − 3𝑡 3 𝑦 = 4 + 2𝑡 3 Answer: Step 1 To eliminate the parameter, we need to solve one of the equation for the parameter (t) and substitute the value of parameter in other equation. Step 2 Given: 𝑥 = 6 − 3𝑡 3 … (1) 𝑦 = 4 + 2𝑡 3 … (2) From (1), we get −3𝑡 3 = 𝑥 − 6 𝑥 ⇒ 𝑡3 = 2 − 3 now, substituting the value of 𝑡 3 in equation (2), we get 𝑥 𝑦 = 4 + 2 (2 − ) 3 ⇒𝑦=8− 2𝑥 3 ⇒ 3𝑦 + 2𝑥 = 24 Question #11 Solve the equation. (Find all solutions of the equation in the interval (0,2𝜋) sin(2𝑥) − sin(𝑥) = 0 Answer: Step 1 given, sin(2𝑥) − sin(𝑥) = 0 Step 2 solution, sin(2𝑥) − sin(𝑥) = 0 2 sin 𝑥 cos 𝑥 − sin 𝑥 = 0 sin 𝑥 (2 cos 𝑥 − 1) = 0 sin 𝑥 = 0 𝑥 = 00 , or 2 cos 𝑥 − 1 = 0 1 cos 𝑥 = 2 𝜋 5𝜋 𝑥= , 3 3 Question #12 Find general solutions of the equation. First find a small integral root of the characteristic equation by inspection; then factor by division. 𝑦 3 + 3𝑦 ′′ − 4𝑦 = 0 Answer: Given: The given differential equation is 𝑦 (3) + 3𝑦 ′′ − 4𝑦 = 0 Calculation: The characteristic equation is 𝑟 3 + 3𝑟 − 4 = 0 ⇒ (𝑟 − 1)(𝑟 + 2)2 = 0 ⇒ 𝑟1 = 1, 𝑟2 = 𝑟3 = −2 This is combination of case 1 and case 2. Then the general solution of the equation is 𝑦(𝑥) = 𝑐1 𝑒 𝑥 + (𝑐2 + 𝑐3 𝑥)𝑒 −2𝑥 . Question #13 Determine whether each point lies on the graph of the equation. 𝑦 = √𝑥 + 4 (a)(0,2) (B)(5,3) Answer: Step 1 Solution: Given 𝑦 = √{𝑥 + 4} 𝑎)(0,2) 𝑏)(5,3) To determine whether each point lies on the graph of the equation. Step 2 𝑦 = √𝑥 + 4 a) At x=0 𝑦 = √4 = 2 The point (0,2) lie on the graph of the equation b) At x=5 𝑦 = √9 = 3 The point (5,3) lie on the graph of the equation. Question #14 State whether the equation or system of equation is linear. 𝑥 2 + 𝑦2 + 𝑧2 = 4 Answer: Step 1 Given equation is 𝑥 2 + 𝑦 2 + 𝑧 2 = 4. To determine whether equation is linear. Solution: A equation is a linear equation if its degree is 1. The form of the linear equation is: Ax+Bx=C The graph of the linear equation is a straight line. We have given equation 𝑥 2 + 𝑦 2 + 𝑧 2 = 4. Here, the term having highest power is 2. So, the degree of equation is 2. Step 2 Therefore, this equation is not a linear equation. Hence, the given equation is not a linear equation. Question #15 Solve the equation 2 9(𝑥 ) = 33𝑥+2 Answer: Step 1 2 The given equation is 9(𝑥 ) = 33𝑥+2 . This equation is with a single variable x. Any equation of the form 𝑎𝑥 2 + 𝑏𝑥 + 𝑐 is a quadratic equation The quadratic equation will have two solutions, hence there will be two values of x. Step 2 Solve the given equation to get a quadratic equation 2 9(𝑥 ) = 33𝑥+2 2 log(9𝑥 ) = log(33𝑥+2 ) (taking 𝑙𝑜𝑔 on both sides) 𝑥 2 log(9) = (3𝑥 + 2) log(3) (log 𝑏 𝑎 = 𝑎 log 𝑏) 𝑥 2 log(32 ) = (3𝑥 + 2) log(3) (𝑥 2 × 2) log(3) = (3𝑥 + 2) log(3) (log 𝑏 𝑎 = 𝑎 log 𝑏) 2𝑥 2 = 3𝑥 + 2 … (1) Equation (1) is a quadratic equation. Solving a quadratic equation will give 2 roots. Step 3 Solve equation (1) for x 2𝑥 2 − 3𝑥 − 2 = 0 2𝑥 2 − 4𝑥 + 𝑥 − 2 = 0 2𝑥(𝑥 − 2) + 1(𝑥 − 2) = 0 (2𝑥 + 1)(𝑥 − 2) = 0 1 𝑥 = 2, − 2 Therefore, value of x is 2 and − 1 2 Question #16 Find an equation in cylindrical coordinates for the surface represented by each rectangular equation. a. 𝑥 2 + 𝑦 2 = 4𝑧 2 b. 𝑦 2 = 𝑥 Answer: Step 1 a. Given 𝑥 2 + 𝑦 2 = 4𝑧 2 It can be compared with the equation of elliptic cone with its axis along z-axis. 𝑧2 ℎ2 = 𝑥2 𝑎2 + 𝑦2 𝑏2 Converting rectangular coordinates to cylindrical coordinates. 𝑟2 = 𝑥 2 + 𝑦2 𝑦 tan 0 = 𝑥 z=z Step 2 so here the equation becomes 𝑥 2 + 𝑦 2 = 4𝑧 2 ⇒ 𝑟 2 = 4𝑧 2 Step 3 b. Given 𝑦 2 = 𝑥 In conversion 𝑥 = 𝑟 cos 0 𝑦 = 𝑟 sin 0 z=z Step 4 Substitute values of x and y in equation. (𝑟 sin 0)2 = 𝑟 cos 0 𝑟 2 sin2 0 = 𝑟 cos 0 𝑟(𝑟 sin2 0 − cos 0) = 0 divide both sides by r 𝑟 sin2 0 − cos 0 = 0 cos 0 𝑟= 2 sin 0 𝑟 = cos 𝑒 𝑐𝜃 cot 𝜃 Question #17 Solve the equation. (Find all solutions of the equation in the interval (0,2𝜋). cos(2𝑥) − cos(𝑥) = 0 Answer: Step 1 The given equation is, cos(2𝑥) − cos(𝑥) = 0 Step 2 Use the trigonometric identity to solve the equation. cos(2𝑥) − cos(𝑥) = 0 2 cos 2 (𝑥) − 1 − cos 𝑥 = 0 2 cos 2 (𝑥) − cos 𝑥 − 1 = 0 2𝑢2 − 𝑢 − 1 = 0 ( ∵ Assume 𝑢 = cos 𝑥 ) 2𝑢2 − 2𝑢 + 𝑢 − 1 = 0 2𝑢(𝑢 − 1) + 1(𝑢 − 1) = 0 (2𝑢 + 1)(𝑢 − 1) = 0 1 𝑢 = − ,1 2 Step 3 Replace u by cos 𝑥. 1 cos 𝑥 = − and cos 𝑥 = 1 2 1 𝑥 = cos −1 (− ) and𝑥 = cos −1 (1) 2 𝑥= 2𝜋 4𝜋 3 𝑥 = 0, , and x=0 3 2𝜋 4𝜋 3 , 3 Question #18 1 Consider the parametric equations 𝑥 = √𝑡 − 2 and 𝑦 = 𝑡 + 1, 𝑡 ≥ 2 . What is implied about the domain of 2 the resulting rectangular equation? Answer: Step 1 Consider the given parametric equation as 1 𝑥 = √𝑡 − 2 and 𝑦 = 𝑡 + 1 2 Step 2 Now find the value of t 𝑥 = √𝑡 − 2 𝑥2 = 𝑡 − 2 𝑡 = 𝑥2 + 2 1 Substitute the value of t in the 𝑦 = 𝑡 + 1 1 2 𝑦 = (𝑥 2 + 2) + 1 2 Now find the domain as 𝑥2 + 2 ≥ 2 𝑥2 ≥ 0 𝑥≥0 Function is defined for all real numbers [0, ∞). Question #19 Solve the equation. (Find all solutions of the equation in the interval [0,2𝜋). 2 sin(2𝑥) sin(𝑥) = 2 cos(𝑥) Answer: Step 1 Given equation is, 2 sin(2𝑥) sin(𝑥) = 2 cos(𝑥) Step 2 Now, 2 sin 𝑥 cos 𝑥 sin 𝑥 cos 𝑥 2 =1 2 sin 𝑥 = 1 1 sin2 𝑥 = 2 1 sin 𝑥 = ± √2 Step 3 Then the values of x, 𝜋 𝑥 = 𝑛𝜋 + (−1)𝑛 4 at n=0 𝜋 𝑥= 4 at n=1 𝜋 3𝜋 𝑥=𝜋− = 4 4 Therefore, values of x in the interval [0,2𝜋)are, 𝜋 3𝜋 𝑥= , 4 4 Question #20 Use the discriminant to determine whether each quadratic equation has two unequal real solutions, a repeated real solution (a double root), or no real solution, without solving the equation. 25𝑥 2 − 20𝑥 + 4 = 0 Answer: Step 1 We know, for a quadratic equation of the form 𝑎𝑥 2 + 𝑏𝑥 + 𝑐 = 0, (𝑎 ≠ 0) the discriminant D is given by 𝐷 = 𝑏 2 − 4𝑎𝑐, and depending on the nature of the discriminant the equation the nature of the roots is identified as follows: If D>0,Then given quadratic equation has two real and unequal roots. If D<0,Then given quadratic equation has no real roots. If D=0,Then given quadratic equation has a repeated real root(a root with multiplicity 2). Given equation is: 25𝑥 2 − 20𝑥 + 4 = 0, ⇒ 𝑎 = 25, 𝑏 = −20, 𝑐 = 4. ⇒ 𝐷 = 𝑏 2 − 4𝑎𝑐 ⇒ 𝐷 = (−20)2 − 4(25)(4) ⇒ 𝐷 = 400 − 400 ⇒𝐷 =0. Hence given quadratic equation has a repeated real root (a real root with multiplicity 2). Step 2 Result: Given quadratic equation has a single real root with multiplicity 2.