Questions and Answers #19 Parametric Equations, Polar Coordinates, and Vector-V

Questions and Answers Sheet 19 Parametric Equations, Polar Coordinates, and Vector-V Question #1 Consider the following planes: 5x - 4y + z = 1 4x + y - 5z = 5 a) Find parametric equations for the line of intersection of the planes. (Use the parameter t.) b) Find the angle between the planes. (Round your answer to one decimal place.) Answer: Step 1 Consider the two given planes. 5x-4y+z=1 4x+y-5z=5 a) The objective of this part is to find the parametric equations for the line of intersection of the planes. To find the point of intersection, set z= 0 in the two given equations and solve for x and y. Step 2 1) 5x-4y=1 2) 4x+y=5 Multiply equation (1) by 1 and add with equation (2) multiplied by 4. x − 4y =1 16 x + 4 y = 20 21x = 21 x =1 Substitute x=1 in equation (2) to find y. 4x+y=5 4(1)+y=5 y=5-4 y=1 So, one point of intersection is ≪ 1, 1, 0 ≫ 𝑎1 𝑥 + 𝑏1 𝑦 + 𝑐𝑧 − 𝑑1 = 0 and 𝑎2 𝑥 + 𝑏2 𝑦 + 𝑐𝑧 = 𝑑2 = 0 Step 3 Find the cross product of the normal to the planes and it is a vector which is parallel to the line of intersection. ≪ 𝑎, 𝑏, 𝑥 ≫×≪ 𝑑, 𝑒, 𝑓 ≫=≪ 𝑏𝑓 − 𝑐𝑒, 𝑐𝑑 − 𝑎𝑓, 𝑎𝑒 − 𝑏𝑑 ≫ The given planes can be written as ≪ 5, − 4, 1 ≫ and ≪ 4, 1, − 5 ≫ So, the cross product is calculated as, ≪ 5, − 4, 1 ≫×≪ 4, 1, − 5 ≫=≪ −4(−5) − 1(1), 1(4) − 5(−5), 5(1) − (−4(4)) ≫ =≪ 19, 29, 21 ≫ Use the point of intersection (𝑥0 , 𝑦0 , 𝑧0 ) and the vector obtained << a, b, c>> to find the parametric equations. 𝑥 = 𝑥0 + 𝑎𝑡 𝑦 = 𝑦0 + 𝑏𝑡 𝑧 = 𝑧0 + 𝑐𝑡 So, the parametric equations are obtained as: x=1+19t y=1+29t z=21t Step 4 b) The objective is to find the angle between the two given planes. Let 𝜃 be the angle between the planes. The formula for angle between planes 𝑎1 𝑥 + 𝑏1 𝑦 + 𝑐1 𝑧 − 𝑑1 = 0 and 𝑎2 𝑥 + 𝑏2 𝑦 + 𝑐2 𝑧 − 𝑑2 = 0 is defined as 𝑎1 𝑎2 +𝑏1 𝑏2 +𝑐1 𝑐2 cos 𝜃 = || || √𝑎12 +𝑏12 +𝑐12 √𝑎22 +𝑏22 +𝑐22 Substitute the values from the given planes, the angle is calculated as, 5(4)+(−4)1+1(−5) cos 𝜃 = | | √52 +(−4)2 +12 √4 2 +12 +(−5)2 =| = 20−4−5 √42√42 11 | 42 So, the angle 𝜃 is calculated as, 11 𝜃 = cos −1 ( ) 42 = 74.8∘ Thus, the angle between the planes is 74.8∘ Question #2 Let 𝐴(2, 1, 1), 𝐵(1, 2, 1), and 𝐶(1, 1, 2) be three given points. Find 1) parametric and symmetric equations of the line through the point A and parallel to the line through the points B and C; Answer: Step 1 Let 𝐴(2, 1, 1), 𝐵(1, 2, 1, ) and 𝐶(1, 1, 2)be three given points. To find parametric and symmetric equations of the line through the point A and parallel to the line through the points B and C Step 2 Parametric equations of the line through the point (𝑥1 , 𝑦1 , 𝑧1 ) and parallel to the line through the points (𝑥2 , 𝑦2 , 𝑧2 ) and (𝑥3 , 𝑦3 , 𝑧3 ) 𝑥 = 𝑥1 + 𝑡(𝑥3 − 𝑥2 ) 𝑦 = 𝑦1 + 𝑡(𝑦3 − 𝑦2 ) 𝑧 = 𝑧1 + 𝑡(𝑧3 − 𝑧2 ) Parametric equations of the line through the point 𝐴(2, 1, 1) and parallel to the line through the points 𝐵(1, 2, 1), and 𝐶(1, 1, 2) x = 2 + t (1 − 1) x=2 y = 1 + t (1 − 2) z = 1 + t (2 − 1) y = 1− t z = 1+ t Step 3 Symmetric equations of the line through the point (𝑥1 , 𝑦1 , 𝑧1 ) and parallel to the line through the points (𝑥2 , 𝑦2 , 𝑧2 ) and (𝑥3 , 𝑦3 , 𝑧3 ) 𝑥 − 𝑥1 𝑦 − 𝑦1 𝑧 − 𝑧1 = = 𝑥3 − 𝑥2 𝑦3 − 𝑦2 𝑧3 − 𝑧2 Symmetric equations of the line through the point 𝐴(2, 1, 1) and parallel to the line through the points 𝐵(1, 2, 1), and 𝐶(1, 1, 2) 𝑥−2 𝑦−1 = 1−1 1−2𝑧 −1 2−1 𝑥−2 𝑦−1 𝑧−1 = = 0 −1 1 Question #3 Find the second derivative of y with respect to x from the parametric equations given. 𝑥= 1 𝑡2 , 𝑦 = 𝑡 2 − 4𝑡 + 1 Answer: Step 1 The given parametric equations are: 1 𝑥 = 2 , 𝑦 = 𝑡 2 − 4𝑡 + 1 𝑡 To find the second derivative of y with respect to x, i.e. 𝑑𝑦 𝑑𝑥 Differentiate the given y and x with respect to t. That is, 𝑑𝑥 𝑑𝑡 1) 𝑑𝑦 𝑑𝑡 2) = 𝑑 1 ( ) 𝑑𝑡 𝑡 2 𝑑𝑥 2 =− 𝑑𝑡 = 𝑑 𝑑𝑡 𝑑𝑦 (𝑡 2 𝑡3 − 4𝑡 + 1) = 2𝑡 − 4 𝑑𝑡 Step 2 From equations (1) and (2). 𝑑𝑦 𝑑𝑥 𝑑𝑦 𝑑𝑥 = 𝑑𝑦 𝑑𝑡 𝑑𝑥 𝑑𝑡 = 2𝑡−4 2 − 3 𝑡 = −𝑡 3 (𝑡 − 2) Question #4 Equation of a line and a line segment Let l be the line that passes through the points 𝑃0 (−3, 5, 8) and 𝑃1 (4, 2, − 1). a. Find an equation of l. b. Find parametric equations of the line segment that extends from 𝑃0 to 𝑃1 . Answer: Step 1 Formula: The equation of line passing through the points 𝐴(𝑥1 , 𝑦1 , 𝑧1 ) & 𝐵(𝑥2 , 𝑦2 , 𝑧2 ) is given by: 𝑥−𝑥1 𝑦−𝑦1 𝑧−𝑧1 = = 𝑥2 −𝑥1 𝑦2 −𝑦1 𝑧2 −𝑧1 Step 2 The equation of line passing through the points 𝑃0 = (−3, 5, 8) & 𝑃1 = (4, 2, − 1) is given by: 𝑥−(−3) 4−(−3) 𝑥+3 ⇒ 7 = = 𝑦−5 2−5 𝑦−5 −3 = = 𝑧−8 −1−8 𝑧−8 −9 Hence, the equation of line is 𝑥+3 7 = 𝑦−5 −3 = 𝑧−8 −9 Step 3 The parametric equations of line segment is given by: 𝑥+3 𝑦−5 𝑧−8 = = =𝑡 7 −3 −9 𝑥+3 𝑦−5 𝑧−8 ⇒ = 𝑡, = 𝑡, =𝑡 7 −3 −9 ⇒ 𝑥 = 7𝑡 − 3, 𝑦 = −3𝑡 + 5, 𝑧 = −9𝑡 + 8 Hence, the parametric equations are: 𝑥 = 8𝑡 − 3, 𝑦 = −3𝑡 + 5, 𝑧 = −9𝑡 + 8 Question #5 Suppose parametric equations for the line segment between (6, 7) and (4, 5) have the form:  x(t ) = a = bt   y (t ) = c + dt If the parametric curve starts at (6, 7) when t=0 and ends at (4, 5) at t=1, then find a, b, c and d. Answer: The parametric equations for the line segment between (6, 7) and (4, 5) have the form  x(t ) = a = bt   y (t ) = c + dt The parametric curve straits at (6, 7) when t=0 x(0)=6 and t(0)=7 a=6 and c=7 and ends at (4, 5) at t=1 x(1)=4 and y(1)=5 a+b=4 and c+d=5 6+b=4 and 7+d=5 b=-2 and d=-2 The parametric equation is,  x(t ) = 6 − 2t   y (t ) = 7 − 2t Answer: The values are a=6 b=-2 c=7 d=-2 Question #6 Find a path that traces the circle in the plane y=5 with radius r=2 and center (−1, 5, − 5) with constant speed 8. 𝑟1 (𝑠) =≪ ≫ Answer: The radius if circle =2 and center (−1, 5, − 4) The equation of such circle is given by: 2 2 (𝑥 − (−1)) + (𝑧 − (−4)) = 22 , 𝑦 = 5 (𝑥 + 1)2 + (𝑧 + 4)2 = 4, 𝑦 = 5 The parametric equations of the circle are given by: 𝑥 + 1 = 2 cos 𝜔 𝑡, 𝑦 = 5, 𝑧 + 4 = 2 sin 𝜔 𝑥 = −1 + 2 cos 𝜔 𝑡, 𝑦 = 5, 𝑧 = −4 + 2 sin 𝜔. Speed is given by: 𝑑𝑥 2 𝑑𝑦 2 𝑑𝑧 2 𝑑𝑡 𝑑𝑡 𝑑𝑡 𝑣 = √( ) + ( ) + ( ) Compute the derivatives: 𝑑𝑥 = 0 + 2𝜔(− sin 𝜔 𝑡) = −2𝜔 sin 𝜔 𝑑𝑡 𝑑𝑦 =0 𝑑𝑡 𝑑𝑧 = 0 + 2𝜔(cos 𝜔 𝑡) = 2𝜔 cos 𝜔. 𝑑𝑡 In given question we have, speed = 8 Substitute these derivatives and speed value in the speed equation above: 8 = √(−2𝜔 sin 𝜔 𝑡)2 + (0)2 + (2𝜔 cos 𝜔 𝑡)2 8 = √4𝜔 2 sin2 𝜔𝑡 + 4𝜔 2 cos {2] 𝜔𝑡 8 = √4𝜔 2 (sin2 𝜔𝑡 + cos 2 𝜔𝑡) 9 = √4𝜔 2 (1) 8 = 2𝜔 8 =𝜔 2 4 = 𝜔. Therefore, the parametric equations are : 𝑥 = −1 + 2 cos 4 𝑡, 𝑦 = 5, 𝑧 = −4 + 2 sin 4 𝑡 The path is given by: 𝑟(𝑡) =≪ −1 + 2 cos 4 𝑡, 5, −4 + 2 sin 4 𝑡 ≫ Question #7 Use polar coordinates to find the volume of the given solid. Above the cone 𝑧 = 𝑥 2 + 𝑦 2 and below the sphere 𝑥 2 + 𝑦 2 + 𝑧 2 = 49 Answer: To calculate the volume, we can use ∫ ∫𝐷(√49 − 𝑥 2 − 𝑦 2 − √𝑥 2 + 𝑦 2 )𝑑𝐴 Where D is the region specified (above the cone, below the sphere) The boundary of the region D must be the projection of the curve of intersection of the two surface. Switching both surface to polar coordinates we have the cone given by z = r and the sphere by 𝑧 = √49 − 𝑟 2 To find the curve of intersection we must solve the equations simultaneously. √49 − 𝑟 2 = 𝑟 7 𝑟= √2 Hence we describe the radius as 7 𝐷 = (𝑟, 𝜃)|0 ≤ 𝜃 ≤ 2𝜋, 0 ≤ 𝑟 ≤ √2 Therefore the required volume is 7 𝑉 2𝜋 = ∫0 ∫0√2(√49 7 2𝜋 √2 =∫ ∫ 0 0 2𝜋 =∫ 0 2𝜋 =∫ 0 =( − 𝑥 2 − 𝑦 2 − √𝑥 2 + 𝑦 2 )𝑑𝐴 (√49 − 𝑟 2 − 𝑟) 𝑟𝑑𝑟𝑑 𝜃 7 3 1 1 √2 (− (49 − 𝑟 2 )2 − 𝑟 3 ) 𝑑 𝜃 3 3 0 3 3 1 49 2 1 7 3 1 (− (49 − ) ) − ( ) + (492 ) 𝑑𝜃 3 2 3 √2 3 343 343 − ) 2𝜋 3 3√2 = 686𝜋 3 (1 − 1 √2 ) =210,4 cubic units Question #8 𝜋 𝜋 𝑟(𝑡) = cos (𝑡 − ) 𝑖 + 2 sin (𝑡 − ) 𝑗: 𝐴 = [0, 𝜋); 4 4 1. 𝑟(𝑡) = (1 + 2 sin 𝑡)𝑖 + (2 − 2 cos 𝑡)𝑗: 𝐴 = [0,2𝜋); Sketch the curves, r(t), that are the ranges of the following vector-valued functions i j l Answer: Consider the given vector-valued functions: 𝜋 𝜋 4 4 1. 𝑟(𝑡) = cos (𝑡 − ) 𝑖 + 2 sin (𝑡 − ) 𝑗: 𝐴 = [0. 𝜋) 2. 𝑟𝑡( ) = (1 + 2 sin 𝑡)𝑖 + (2 − 2 cos 𝑡)𝑗: 𝐴 = [0,2𝜋) Here, the objective is to sketch the curves represented by vector-valued function r(t).. The given vector-valued function is, 𝜋 𝜋 𝑟(𝑡) = cos (𝑡 − ) 𝑖 + 2 sin (𝑡 − ) 𝑗: 𝐴 = [0, 𝜋)(1) 4 4 It can also be written as: 𝑟(𝑡) = 𝑥𝑖 + 𝑦𝑗. (2) Compare (1) and (2) 𝜋 𝑥 = cos (𝑡 − ) 4 𝜋 𝑦 4 2 𝑦 = 2 sin (𝑡 − ) ⇒ 𝜋 = sin (𝑡 − ) 4 Use Trigonometric identity sin2 𝜃 + cos 2 𝜃 = 1 𝜋 Substitute 𝜃 = (𝑡 − ) 4 𝜋 2 2 𝜋 sin (𝑡 − ) + cos (𝑡 − ) = 1 4 4 𝜋 𝑦 4 2 Substitute 𝑥 = cos (𝑡 − ) 𝑎𝑛𝑑 𝜋 = sin (𝑡 − ) 4 y ( )2 + ( x 2 ) = 1 2 𝑥2 1 + 𝑦2 4 =1 Hence, the given vector-valued function represents a curve which is an ellipse. The sketch of curves represented by vector-valued function r(t) is shown below: The given vector-valued function is, |𝑟(𝑡) = (1 + 2 𝑠𝑖𝑛 𝑡)𝑖 + (2 — 2 𝑐𝑜𝑠 𝑡)𝑗 ∶ 𝐴 = [0,2𝜋) (1) It can also be written as: 𝑟(𝑡) = 𝑥𝑖 + 𝑦𝑗(2) Compare (1) and (2) 𝑥 = 1 + 2 𝑠𝑖𝑛 |𝑥 − 1 = 2𝑠𝑖 𝑛 (𝑥 − 1) = 𝑠𝑖𝑛 2 And 𝑦 = 2 − 2𝑐𝑜 𝑠 2 − 𝑦 = 2 𝑐𝑜𝑠 (2 − 𝑦) = 𝑐𝑜𝑠 2 Use Trigonometric identity sin2 𝑡 + 𝑐𝑜𝑠 2 𝑡 = 1 Substitute [( (𝑥−1) 2 = sin 𝑡 and x −1 2 2 − y 2 ) +( ) =1 2 2 (𝑥 − 1)2 + (𝑦 − 2)2 = 4 (2−𝑦) 2 = 𝑐𝑜𝑠 Hence, the given vector-valued function represents a curve which is a circle with a center at (1,2) and a radius equal to 2 units. Question #9 Find the Domain of the vector-valued function R Let 𝑅⃗ be a vector-valued function defined by sin 3t  , In(t + 1)  , ift  0  4 − t R(t ) =  t  , ift = 0  2,3, 0  Give dom 𝑅⃗ Determine whether 𝑅⃗ is a continuous at t= 0 . Answer: sin 3t  , In(t + 1)  , if t  0  4 − t R(t ) =  t  2,3, 0  , if t = 0 Domain of √4 − 𝑡 ⇒ 4 − 𝑡 ≥ 0 ⇒ 𝑡 ≤ 4 ⇒ 𝑡𝜖(−∞, 4] Domain of lim 3𝑡 𝑡 ⇒ 𝑡 ≠ 0 ⇒ 𝑡𝜖(−∞, 0) ∪ (−, ∞) Domain of ≤ 𝑔(𝑡 + 1) ⇒ 𝑡 + 1 > 0 ⇒ 𝑡 ≻ 1 ⇒ 𝑡𝜖(−1, ∞) lim 3𝑡 Domain of √4 − 𝑡, ≠ log(𝑡 + 1) 𝑡 = (−∞, 4] ∩ {(−∞, 0) ∩ ∞)} ∩ (−1, +∞) = (−1,0) ∩ (0,4] but the 𝑅⃗ (𝑡) is defined at t=0 ∴ dom 𝑅⃗(𝑡) = (−1,4] For continuity of 𝑅⃗ at t=0 we have to check the continuity of component of 𝑅⃗ . sin 3𝑡 lim 𝑅⃗ (𝑡) = lim ≪ √4 − 𝑡, , log(𝑡 + 1) ≫ 𝑡⇒0 =≪ 𝑡 𝑡⇒0 lim 𝑡⇒0√4−𝑡 , lim 𝑡⇒0 =≪ √4 − 0, 3 lim sin 3𝑡 , lim log(𝑡 + 1) ≫ 𝑡 𝑡⇒0 sin 3𝑡 𝑡⇒0 3𝑡, , log(0 + 1) ≫ =≪ √4, 3 × 1, log 1 ≫ {∴ lim 𝑡⇒0 sin 𝑥 𝑥 = 1} =<<2,3,0>> = 𝑅⃗ (0) ⇒ lim 𝑅⃗(𝑡) = 𝑅⃗(0) 𝑡⇒0 ∴ 𝑅⃗ is continuous at t=0 Question #10 Find the maximum and minimum values attained by the function f along the path c(t). (a) 𝑓(𝑥, 𝑦) = 𝑥𝑦; 𝑐(𝑡) = (cos(𝑡) , sin(𝑡)); 0 ≤ 𝑡 ≤ 2𝜋. maximum value__________ minimum value__________ (b) 𝑓(𝑥, 𝑦) = 𝑥 2 + 𝑦 2 ; 𝑐(𝑡) = (cos(𝑡) , 8 sin(𝑡)); 0 ≤ 𝑡 ≤ 2𝜋. maximum value__________ minimum value__________ Answer: (a) Given that, 𝑓 = (𝑥, 𝑦) = 𝑥𝑦; 𝑐(𝑡) = (cos 𝑡 , sin 𝑡); 0 ≤ 𝑡 ≤ 2𝜋. 𝑐(𝑡) = 𝑥𝑗̂ + 𝑦𝑗̂ = (cos 𝑡)𝑗̂ + (sin 𝑡)𝑗̂ 𝑓(𝑥, 𝑦) = cos 𝑡 × 𝑠𝑖𝑛 sin(2𝑡) = 2 (∵ sin 2 𝛼 = 2 sin 𝛼 cos 𝛼) It is known that, the range of the sine functions is -1 to 1. −1 ≤ sin(2𝑡) ≤ 1 −1 sin(2𝑡) 1 ≤ ≤ 2 2 2 −1 1 ≤ 𝑓(𝑥, 𝑦) ≤ 2 2 Thus, the function has the maximum value 1 2 1 and minimum value − . 2