Q&A #17: Parametric Equations & Polar Coordinates

Questions and Answers Sheet 17 Parametric Equations, Polar Coordinates, and Vector-V Question #1 Find a set of parametric equations that represents the graph of 𝑦 = 1 − 𝑥 2 using each of the following parameters. a) t=x b) The slope 𝑚 = 𝑑𝑦 𝑑𝑥 at the point (x, y) Answer: Step 1 Given: 𝑦 = 1 − 𝑥2 To find: Find a set of parametric equations that represents the above graph using each of the following parameters a) t=x b) The slope 𝑚 = 𝑑𝑦 𝑑𝑥 at the point (x, y) Step 2 Explanation: Rewrite the given expression, 1) 𝑦 = 1 − 𝑥 2 a) As we have to find a set of parametric equations in the parameter t=x , we get, 𝑦 = 1 − (𝑡)2 𝑦 = 1 − 𝑡2 b) As we have to find the slope, so differentiating the above equation w.r.t. x, we get, 𝑑𝑦 𝑑𝑥 𝑑𝑦 = 0 − 2𝑥 | 𝑑𝑥 (𝑥,𝑦) = −2𝑥 So, the slope of the above equation at point (x,y) is -2x. Hence, a set of parametric equations in the parameter t=x is 𝑦 = 1 − 𝑡 2 and the slope of the function 𝑦 = 1 − 𝑥 2 at point (x, y) is -2x. Question #2 Find parametric equation of the line passing through point (4, 2) parallel to vector 𝑣⃗ = −𝑖̂ + 5𝑗̂ Answer: Step 1 Given: A line passes through the point (4, 2) The line is parallel to the direction vector 𝑣⃗ = −𝑖̂ + 5𝑗̂ We want to find the parametric equations of the line passing through (4, 2) and parallel to the vector 𝑣⃗ = −𝑖̂ + 5𝑗̂ The parametric equations for a line though the point (𝑥0 , 𝑦0 ) and parallel to direction vector << a, b>> is given by 𝑥 = 𝑥0 + 𝑎𝑡 and 𝑦 = 𝑦0 + 𝑏𝑡 Step 2 In the given problem, we have (𝑥0 , 𝑦0 ) = (4, 2) and << a, b>>=<< -1, 5>> Since 𝑥 = 𝑥0 + 𝑎𝑡 is a parametric equation, we can replace values of 𝑥0 and a in the parametric equation. Hence 𝑥 = 𝑥0 + 𝑎𝑡 ⇒ 𝑥 = 4 + (−1)𝑡 ⇒𝑥 =4−𝑡 Now 𝑦 = 𝑦0 + 𝑏𝑡 is a parametric equation, we can replace values of 𝑦0 and b in the parametric equation. Hence 𝑦 = 𝑦0 + 𝑏𝑡 ⇒ 𝑦 = 2 + (5)𝑡 ⇒ 𝑦 = 2 + 5𝑡 Hence x=4-t and y=2+5t are parametric equations for the line passing through the point (4, 2) and parallel to the vector 𝑣⃗ = −𝑖̂ + 5𝑗̂ Question #3 Find parametric equations the line (Use the parameter t.) The line through the origin and the point (9, 3, − 1) (𝑥(𝑡), 𝑦(𝑡), 𝑧(𝑡)) =? Find the symmetric equations. 𝑥 𝑦 a) = = 𝑧 9 3 b) x+9=y+3=z-1 c) x-9=y-3=z+1 𝑥 𝑦 d) = = −𝑧 e) 9 𝑥 3 = 3 𝑦 9 = −𝑧 Answer: 𝑥 = 𝑥0 + 𝑎𝑡, 𝑦 = 𝑦0 + 𝑏𝑡, 𝑧 = 𝑧0 + 𝑐𝑡 𝑥 = 0 + 9𝑡, 𝑦 = 0 + 3𝑡, 𝑧 = 0 − 𝑡 For symmetric equations solve each of the parametric equations for t and set equal to each other 𝑥 𝑦 𝑡 = , 𝑡 = , 𝑡 = −𝑧 𝑥 9 = 9 𝑦 3 = 𝑧 3 −1 Thus parametric equations are 𝑥 = 9𝑡, 𝑦 = 3𝑡, 𝑧 = −𝑡 𝑥 𝑦 and symmetric equations are = = −𝑧 9 3 Question #4 Represent the line segment from 𝑃(−2, − 3, 8), 𝑄(5, 1, − 2)by a vector-valued function and by a set of parametric equations. Answer: We have, points (−2, − 3, 8) and 𝑄(5, 1, − 2) Now, the component vector from P to Q , 𝑣(𝑡) =≪ 5 − (−2), 1 − (−3), − 2 − 8 ≫ ⇒ 𝑣(𝑡) =≪ 7, 4, − 10 ≫ So, the direction number are a=7, b=4 and c=-10 For initial point on the line, use point 𝑃(−2, − 3, 8) We get, 𝑥0 = −2, 𝑦0 = −3 and 𝑧0 = 8 Now, to form the parametric equation: 1) 𝑥 = 𝑥0 + 𝑎𝑡, 𝑦 = 𝑦0 + 𝑏𝑡 and 𝑧 = 𝑧0 + 𝑐𝑡 Putting the value of a=7, b=4, c=-10 and 𝑥0 = −2, 𝑦0 = −3, 𝑧0 = 8 in eq. (1), we get x=-2+7t, y=-3+4t and z=9-10t Now, to form a vector-valued function: r(t)=x(t)i+y(t)j+z(t)k We have, x=-2+7t, y=-3+4t and z=8-10t So, 𝑟(𝑡) = (7𝑡 − 2)𝑖 + (4𝑡 − 3)𝑗 + (8 − 10𝑡)𝑘 Hence, set of parametric equations be, x=-2+7t, y=-3+4t and z=8-10t and vector-valued function be, r(t)=(7t-2)i+(4t-3)j+(8-10t)k Question #5 Find a set of parametric equations for the rectangular equation a) 𝑦 = 2𝑥 6 + 6 b) y=9x+1 Answer: a) 𝑦 = 2𝑥 6 + 6 Now let x=t, then 𝑦 = 2𝑡 6 + 6 Therefore the set of parametric equations is x=t & 𝑦 = 2𝑡 6 + 6 b) y=9x+1 Now let x=t-1, then y=9(t-1)+1 =9t-9+1 9t-8 Therefore the set of parametric equations is x=t-1 & y=9t-8 Question #6 Find the arc length of the curve on the given interval. Parametric Equations 𝑥 = √𝑡, 𝑦 = 3𝑡 − 1 Interval 0 ≤ 𝑡 ≤ 1 Answer: Step 1 Given: Parametric Equations 𝑥 = √𝑡 y=3t-1 Interval 0 ≤ 𝑡 ≤ 1 Step 2 Explanation: Formula: Arc length in parametric form: 𝑏 Arc length = ∫𝑎 𝑑𝑠 where 𝑑𝑥 2 𝑑𝑦 2 𝑑𝑡 𝑑𝑡 𝑑𝑠 = √( ) + ( ) Step 3 𝑥 = √𝑡 Differentiate with respect to t 𝑑𝑥 𝑑𝑡 = 1 2√𝑡 y=3t-1 Differentiate with respect to t 𝑑𝑦 𝑑𝑡 =3 Interval ≤ 𝑡 ≤ 1 Therefore use above values in the formula to determine the arc length, Step 4 𝑏 Arc length = ∫𝑎 𝑑𝑠 2 1 𝑑𝑥 𝑑𝑦 Arc length = ∫0 √( ) + ( ) 𝑑𝑡 Arc length = 1 ∫0 √( 1 2 𝑑𝑡 2 1 2√𝑡 ) + (3)2 𝑑𝑡 1 Arc length = ∫0 √ + 9𝑑𝑡 4𝑡 1 1+36𝑡 Arc length = ∫0 √ 4𝑡 1 √1+36𝑡 Arc length = ∫0 Arc length = 𝑑𝑡 𝑑𝑡 √4𝑡 1 √1+36𝑡 ∫0 2√𝑡 𝑑𝑡 1 1 √1+36𝑡 Arc length = ∫0 2 √𝑡 Put 𝑢 = 6√𝑡 ⇒ 𝑑𝑢 = 3 √𝑡 𝑑𝑡 𝑑𝑡 ⇒ 𝑑𝑡 = √𝑡𝑑𝑢 3 and 𝑢2 = 36𝑡 Put 𝑡=0⇒𝑢=0 𝑡 = 1 ⇒ 𝑢2 = 36 ⇒ 𝑢 = 6 Therefore, 1 6 √1+𝑢2 Arc length = ∫0 2 √𝑡 × √𝑡𝑑𝑢 3 1 6 1 (ln(√1 + 𝑢2 + 𝑢) + 𝑢√1 + 𝑢2 )0 Arc length = ∫0 √1 + 𝑢2 × 𝑑𝑢 6 Arc length = Arc length = 12 1 12 6 (ln(√37 + 6) + 6√37) Arc length ≈ 3.249 This is the required arc length of the given parametric equations over the interval [0, 1] Hence the solution. Question #7 Find parametric equations for the line that passes through the point (5, 2) and is parallel to the line −7 𝑦= 𝑥−2 2 Answer: Step 1 The given equation of line −7 𝑦= 𝑥−2 2 comparing with the y=mx +b , we get −7 𝑚= and b=-2 2 The slope of the line is −7 2 Given that the line passes through the point (5, 2) is parallel to the line 𝑦 = We know that the slope of parallel line is same −7 So, the line passes through the point (5, 2) has slope 𝑚 = 2 −7 2 𝑥−2 Step 2 using point slope formula, 𝑦 − 𝑦1 = 𝑚(𝑥 − 𝑥1 ) −7 Here 𝑚 = , (𝑥1 , 𝑦1 ) = (5, 2) ⇒𝑦−2= ⇒ 𝑦−2 −7 Now, = 2 𝑥−5 𝑦−2 −7 2 −7 2 (𝑥 − 5) = 𝑡(𝑙𝑒𝑡) =𝑡 ⇒ 𝑦 − 2 = −7𝑡 ⇒ 𝑦 = 2 − 7𝑡 and 𝑥−5 2 =𝑡 ⇒ 𝑥 = 2𝑡 ⇒ 𝑥 = 5 + 2𝑡 so, parametric equations for the line that passes through the point (5, 2) is 𝑦 = 2 − 7𝑡, 𝑥 = 5 + 2𝑡 This is the required answer. Question #8 Represent the line segment from P to Q by a vector-valued function and by a set of parametric equations 𝑃(0, 0, 0), 𝑄(5, 2, 2) Answer: Step 1 Given information: The point are 𝑃(0, 0, 0), 𝑄(5, 2, 2) Calculation: Find the vector valued function. 𝑣(𝑡) =≪ (5 − 0), (2 − 0), (2 − 0) ≫ ≪ 3, 2, 2 ≫ The direction numbers are ≪ 5, 2, 2 ≫ a=5, b=2, c=2 The value of (0, 0, 0) that is the initial points on the line 𝑥0 = 0, 𝑦0 = 0, 𝑧0 = 0 Step 2 The form of the parametric equations 𝑥 = 𝑥0 + 𝑎𝑡, 𝑦 = 𝑦0 + 𝑏𝑡, 𝑧 = 𝑧0 + 𝑐𝑡 Substitute 𝑥0 = 0, 𝑦0 = 0, 𝑧0 = 0, a=5, b=2, c=2 in the parametric equations x=0+5t y=0+2t z=0+2t We get x=5t y=2t z=2t Step 3 Find the vector valued function 𝑟(𝑡) = 𝑥(𝑡)𝑖 + 𝑦(𝑡)𝑗 + 𝑧(𝑡)𝑘 = (5𝑡)𝑖 + (2𝑡)𝑗 + (2𝑡)𝑘 Answer: 𝑟(𝑡) = (5𝑡)𝑖 + (2𝑡)𝑘 Question #9 What are the parametric equations for the intersection of the planes x-y-z=1 and 2x3y+z=2 ? Answer: Step 1 Given, the two planes x-y-z=1 and 2x+3y+z=2, we have to find the parametric equations for the intersections of these two planes. Step 2 To find a point on this line we can for instance set z = 0 and then use the given equations to solve for x and y. x-y=1 2x+3y=2 On solving these two equations we get x = 1, y = 0. So the point on the line is (1, 0, 0) Also, the direction of the line lives in both the planes and so, in particular, is perpendicular to both normal vectors, therefore a vector which is parallel to the line is given by i j k (1, − 1, − 1)  (2, 3, 1) =| 1 −1 −1| 2 3 1 =i(-1+3)-j(1+2)+k(3+2) =2i-3j+5k = (2, − 3, 5) Thus the equation of the line is given by the vector equation (𝑥, 𝑦, 𝑧) = (1, 00) + 𝑡(2, − 3, 5) =(1+2t,\ -3t,\ 5t) Hence the parametric equations are x=1+2t, y=-3t and z=5t Question #10 Find parametric equations and a parameter interval for the motion of a particle that starts at the point (-2, 0) in the xy-plane and traces the circle 𝑥 2 + 𝑦 2 = 4 three times clockwise. Answer: Step 1 Concept used: Parametric equation: 𝑥 = 𝑟 × 𝑐𝑜𝑠 𝜃 and 𝑦 = 𝑟 × 𝑠𝑖𝑛 𝜃 Equation of circle is 𝑥 2 + 𝑦 2 = 𝑟 2 Here, r is radius Step 2 The equation x 2 + 𝑦 2 = 4 can be written as, 𝑥 2 + 𝑦 2 = 22 Compare 𝑥 2 + 𝑦 2 = 22 with 𝑥 2 + 𝑦 2 = 𝑟 2 to obtain r r=2 Therefore, the parametric equations form is, 𝑥 = 2 × 𝑐𝑜𝑠 𝜃 and 𝑦 = 2 × 𝑠𝑖𝑛 𝜃 The particle that starts at the point (-2, 0), therefore, the parametric equation is 𝑥 = −2 × 𝑐𝑜𝑠 𝜃 and 𝑦 = 2 × sin 𝜃. The three times clockwise means, 3(2𝜋) = 6𝜋. Thus, the parametric equation is 𝑥 = −2 × 𝑐𝑜𝑠 𝜃 and 𝑦 = 2 × 𝑠𝑖𝑛 𝜃, and interval is 0 ≤ 𝜃 ≤ 6𝜋.