Questions and Answers #16 Parametric Equations, Polar Coordinates, and Vector-V

Questions and Answers Sheet 16 Parametric Equations, Polar Coordinates, and Vector-V Question #1 Find parametric equations and a parameter interval for the motion of a particle that starts at (a, 0) and traces the circle 𝑥 2 + 𝑦 2 = 𝑎2 once counterclockwise Answer: Step 1 a particle that starts at (a, 0) and traces the circle 𝑥 2 + 𝑦 2 = 𝑎2 once counterclockwise. The parametric equations for a circle 𝑥 2 + 𝑦 2 = 𝑎2 and stats (a, 0) are 𝑥 = 𝑎 𝑐𝑜𝑠 𝑦 = 𝑎 sin It requires going around a circle once. Step 2 Therefore required parametric equations are 𝑥 = 𝑎 cos 𝑡 , 𝑡 = 𝑎𝑠𝑖𝑛 and d 0 ≤ 𝑡 ≤ 2π. Question #2 Determine whether the statement '' The graph of the parametric equations 𝑥 = 𝑡 2 and 𝑦 = 𝑡 2 is the line y=x' '', is true or false. If it is false, explain why or give an example that shows it is false. Answer: Step 1 Given: The graph of the parametric equations 𝑥 = 𝑡 2 and 𝑦 = 𝑡 2 is the line y=x Step 2 The objective is to determine that the given statement is true or false On considering the statement, ''The graph of the parametric equations 𝑥 = 𝑡 2 and 𝑦 = 𝑡 2 is the line y=x'' The given statement is false The graph of the parametric equation 𝑥 = 𝑡 2 and 𝑦 = 𝑡 2 is the portion of the line y=x only when 𝑥 ≥ 0 Question #3 Find two different sets of parametric equations for a rectangular equation 𝑦 − 𝑥2 − 3 Answer: We have 1) 𝑦 = 𝑥 2 − 3 Let x=t and put it in equation (1), we get 𝑦 = 𝑡2 − 3 Again, let 𝑥 = 𝑡 3 and put it in equation (1), we get ⇒ 𝑦 = (𝑡 3 )2 − 3 ⇒ 𝑦 = 𝑡6 − 3 Hence, the two different sets of parametric equations are 𝑦 = 𝑡2 − 3 𝑦 = 𝑡6 − 3 Question #4 Find parametric equations for the line in which the planes x+2y+z=1 and x-y+2z=-8 intersect. Answer: Step 1 To find the parametric equations for the line in which the planes x+2y+z=1 and x-y+2z-8 intersect Step 2 We have given the equations of the planes: 1) x+2y+z=1 and 2) x-y+2z=-8 we have to find the parametric equations for the line in which the planes intersect. Firstly we find the point on the line by setting z=0 From (1) we get: 3) x+2y=1 and from (2) 4) x-y=-8 From (3) and (4) we get: x=-5 and y=3 So we get: (-5, 3, 0) is a point on the line. Now the Normal vectors of the planes are 𝑛1 = (1, 2, 1) ⃗⃗⃗⃗ 𝑛2 = (1, − 1, 2) ⃗⃗⃗⃗ Now the normal vector perpendicular to both is: 𝑎 = ⃗⃗⃗⃗ 𝑛1 × ⃗⃗⃗⃗ 𝑛2 iˆ ˆj kˆ =| 1 2 1 | 1 −1 2 = 𝑖̂(5) − 𝑗̂(1) + 𝑘̂ (−3) =≪ 5, − 1, − 3 ≫ Therefore the line passes through (−5, 3, 0) and parallel to the vector 𝑎 =≪ 5, − 1, − 3 ≫ As we know the parametric equation of a line passing through the point (𝑥0 , 𝑦0 , 𝑧0 ) and parallel to the vector << a, b, c>> is given by 𝑥 = 𝑥0 + 𝑎𝑡, 𝑦 = 𝑦0 + 𝑏𝑡, 𝑧 = 𝑧0 = 𝑐𝑡 Here (𝑥0 , 𝑦0 , 𝑧0 ) = (−5, 3, 0) and << a, b, c>>=<<5, -1, -3>> So the parametric equation of the line is x=-5+5t, y=3-t, z=-3t Therefore the parametric equations for the line in which planes x+2y+z=1 and x-y+2z=-8 intersect is: x=-5+5t, y=3-t, z=-3t Question #5 Find a set of parametric equations of the line with the given characteristics The line passes through the point (−2, 4, 7) and is perpendicular to the plane given by -x+2y+z=4 Answer: The parametric equations of a line through (𝑥0 , 𝑦0 , 𝑧0 ) and perpendicular to a vector << a, b, c>> is, 𝑥 = 𝑥0 + 𝑡𝑎 𝑦 = 𝑦0 + 𝑡𝑏 𝑧 = 𝑧0 + 𝑡𝑐 A vector perpendicular to the plane -x+2y+z-4=0 is << -1, 2, 4>> and (𝑥0 , 𝑦0 , 𝑧0 ) = (−2, 4, 7) Therefore the parametric equations are, x=-2-t y=4+2t z=7+t Question #6 Find a set of parametric equations of the line that passes through the point (5, 6, 7) and is parallel to the xz-plane and the yz-plane. Answer: Step 1 The given point is (5, 6, 7). to find the set of parametric equations of line passing through the given point and it is parallel to z-plane and yz-plane. Step 2 The parametric equations of given point (5, 6, 7) are 𝑥 = 5 + 0𝑡, 𝑦 = 6 + 0𝑡, 𝑧 = 7 + 0𝑡 If a line is parallel to both xz-plane and yz-plane, it has to be parallel to the intersection of the two planes that is the z-axis. Therefore we take the unit vector that on the z-axis. 𝑣 =≪ 0, 0, 1 ≫ Question #7 Represent the line segment from P to Q by a vector-valued function. (P corresponds to t = 0 . Q corresponds to t = 1 .) P(-4, -1, -1), Q(-1, -7, -3) r(t)=? Represent the line segment from P to Q by a set of parametric equations. (Enter your answers as a comma-separated list of equations.) Answer: Given, point 𝑃 = (−4, − 1, − 1) and 𝑄(−1, − 7, − 3) We know that line segment r(t) is given by 𝑟(𝑡) = (1 − 𝑡)𝑟0 + 𝑡𝑟1 , 0 ≤ 𝑡 ≤ 1 Where 𝑟0 =≪ −4, − 1, − 1 ≫ and 𝑟1 =≪ −1, − 7, − 3 ≫ put this value in eq.(1) we get , 𝑟(𝑡) = (1 − 𝑡) ≪ −4, − 1, − 1 ≫ +𝑡 ≪ −1, − 7, − 3 ≫ =≪ −4(1 − 𝑡), − 1(1 − 𝑡), − 1(1 − 𝑡) ≫ +≪ −1𝑡, − 7𝑡, − 3𝑡 ≫ =≪ −4 + 4𝑡, − 1 + 𝑡, − 1 + 𝑡 ⟩+≪ −𝑡, − 7𝑡, − 3𝑡 ≫ =≪ −4 + 4𝑡 − 𝑡, − 1 + 𝑡 − 7𝑡, − 1 + 𝑡 − 3𝑡 ≫ =≪ −4 + 3𝑡, − 1 − 6𝑡, − 1 − 2𝑡 ≫ Hence the line segment is 𝑟(𝑡) =≪ −4 + 3𝑡, −1 − 6𝑡, −1 − 2𝑡 ≫this is the set of parametric equations 𝑟(𝑡) = (−4 + 3𝑡)𝑖 + (−1 − 6𝑡)𝑗 + (−1 − 2𝑡)𝑘 this is the vector valued function Question #8 Find a set of parametric equations of the line with the given characteristics. The line passes through the point (−4, 5, 2) and is parallel to the xy-plane and the yz-plane Answer: Step 1 Given: The line passes through the point (-4, 5, 2) and is parallel to the xy-plane and the yz-plane. Since the line is parallel to xy-plane and yz-plane, therefore it must be parallel to y-axis (Intersection of xy-plane and yz-plane). Now to find a set of parametric equations of the line, take any point on the y-axis. Let us take, (0, 2, 0) Step 2 Now the set of parametric equations is, 𝑥 = −4 + 0 × 𝑡, 𝑦 = 5 + 2 × 𝑡, 𝑧 = 2 + 0 × ⇒ 𝑥 = −4, 𝑦 = 5 + 2𝑡, 𝑧 = 2, 𝑡 ∈ 𝑅 Question #9 Find two different sets of parametric equations for the rectangular equation y=6x-5 Answer: Step 1 Given: The rectangular equation is: y=6x-5 We have to find the two different sets of parametric equations for a given rectangular equation. Step 2 Now Consider, t is a parameter for the given rectangular equation. 𝑥 = 𝑡 ⇒ 𝑦 = 6𝑡 − 5 Then the set is: {(𝑥, 𝑦)|𝑥 = 𝑡 and 𝑦 = 6𝑡 − 5} Consider, t is a parameter for the given rectangular equation. x=t+2 ⇒ 𝑦 = 6(𝑡 + 2) − 5 ⇒ 𝑦 = 6𝑡 + 12 − 5 ⇒ 𝑦 = 6𝑡 + 7 Then the set is: {(𝑥, 𝑦)|𝑥 = 𝑡 + 2 and 𝑦 = 6𝑡 + 7} Hence, the two different sets of parametric equations for a given rectangular equation are: {(𝑡, 6𝑡 − 5)|𝑥 = 𝑡 and 𝑦 = 6𝑡 − 5} and {((𝑡 + 2), (6𝑡 + 5))|𝑥 = 𝑡 + 2 and 𝑦 = 6𝑡 + 7} Question #10 Find parametric equations for the given curve. (𝑥 + 9)2 + (𝑦 − 4)2 = 49 Answer: Step 1 To find the parametric equation of the given curve (𝑥 + 9)2 + (𝑦 − 4)2 = 49 Given equation is in the form of equation of a circle given by (𝑥 − ℎ)2 + (𝑦 − 𝑘)2 = 𝑟 2 with center (h, k) and radius r. The parametric equations of a circle is given by s(t)= 𝑥(𝑡) = 𝑟 cos(𝑡) + ℎ 𝑦(𝑡) = 𝑟 sin(𝑡) + 𝑘 From the given equation (𝑥 + 9)2 + (𝑦 − 4)2 = 49, we have ℎ = −9, 𝑘 = 4 and 𝑟 2 = 49 ⇒ 𝑟 = 7 Step 2 Then the parametric equations of the given circle is s(t)= 𝑥(𝑡) = 𝑟 cos(𝑡) + ℎ 𝑦(𝑡) = 𝑟 sin(𝑡) + 𝑘, [0,2𝜋] s(t)= 𝑥(𝑡) = 7 cos(𝑡) − 9 𝑦(𝑡) = 7 sin(𝑡) + 4, [0, 2𝜋]