Q&A #13: Parametric Equations & Polar Coordinates

Questions and Answers Sheet 13 Parametric Equations, Polar Coordinates, and Vector-V Question #1 Find equations fot the osculation, normal, and rectifying planes of the curve 𝑟(𝑡) = 𝑡𝑖 + 𝑡3𝑘 at the point (1, 1, 1). Answer: Step 1 Given: The curve 𝑟(𝑡) = 𝑡𝑖 + 𝑡 2 𝑗 + 𝑡 3 𝑘. The point (1, 1, 1). Step 2 To find: Osculating, normal, and rectifying planes of the curve 𝑟(𝑡) = 𝑡𝑖 + 𝑡 2 𝑗 + 𝑡 3 𝑘. If 𝑟(𝑡) = (𝑥(𝑡), 𝑦(𝑡), 𝑧(𝑡)) be vector-valued function and 𝑝0 (𝑥0 , 𝑦0 , 𝑧0 ) be a point on a curve c generated by vector-valued function ⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗ 𝑟(𝑡0 ). Then The normal plane of P is perpendicular to ⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗ 𝑟 3 (𝑡0 ).. 3 (𝑡 ) × ⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗ {𝑟 ′′ (𝑡0 )]} × ⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗ The rectifying plane of P is perpendicular to [𝑟⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗ 𝑟 3 (𝑡0 ). 0 The osculating plane P is perpendicular to 𝑟 3 (𝑡0 ) × 𝑟 ′′ (𝑡0 ). Now, if ∇𝑓(𝑥0 , 𝑦0 , 𝑧0 ) is perpendicular to the tangent vector at point P ( corresponding to 𝑡0 and passes though the point 𝑃(𝑥0 , 𝑦0 , 𝑧0 ) then the equation of tangent plane at point P is, ∇𝑓(𝑥0 , 𝑦0 , 𝑧0 ). (𝑥 − 𝑥0 , 𝑦 − 𝑦0 , 𝑧 − 𝑧0 ) = 0. Step 3 To find a normal plane corresponding to the point (1,1,1) that is at t=1 first find the vector perpendicular to these planes 3 (𝑡) at t=1. Find 𝑟⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗ Here ⃗⃗⃗⃗⃗⃗⃗⃗ 𝑟(𝑡) = 𝑡𝑖 + 𝑡 2 𝑗 + 𝑡 3 𝑘 ⇒ ⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗ 𝑟 3 (𝑡) = 𝑖 + 2𝑡𝑗 + 3𝑡 2 𝑘 ⇒ ⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗ 𝑟 3 (𝑡) = (1,2𝑡, 3𝑡 2 ) ⇒ ⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗ 𝑟 3 (𝑡) = (1,2,3) Therefore the equation of the tangent plane is. 1(𝑥 − 1) + 2(𝑦 − 1) + 3(𝑧 − 1) = 0 𝑥 − 1 + 2𝑦 − 2 + 3𝑧 − 3 − 0 𝑥 + 2𝑦 + 3𝑧 − 6 = 0 Step 4 To find the osculating plane corresponding to the point (1,1,1) that is at t=1 need to find a vector perpendicular to these planes. To find an osculating curve need to find 𝑟 3 (𝑡) × 𝑟 ′′ (𝑡) at t=1. Now 𝑟1 (𝑡) − 𝑖 + 2𝑡𝑗 + 3𝑡 2 𝑘 𝑟 ′′ (𝑡) = 𝑜𝑖 + 2𝑗 + 6𝑡𝑘 ⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗ 𝑟 3 (𝑡) × ⃗⃗⃗⃗ 𝑟 ′′ = 𝑖𝑗𝑘 1 2𝑡 3𝑡 2 0 2 6𝑡 = ((12𝑡 2 − 6𝑡 2 )𝑖 − 6𝑡𝑖 + 2𝑘) = 6𝑡 2 − 6𝑡𝑖 + 2𝑘 ⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗ 𝑟 3 (𝑡) × ⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗ 𝑟 ′′ (𝑡) = (6𝑡 2 , −6𝑡, 2) ⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗ 𝑟 3 (1) × ⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗ 𝑟 ′′ (1) = (6(1)2 , −6(1), 2) = (6, −6,2) Therefore, the equation of the osculating plane is given as 6(𝑥 − 1) − 6(𝑦 − 1) + 2(𝑧 − 1) = 0 6𝑥 − 6 − 6𝑦 + 6 + 2𝑧 − 2 = 0 6𝑥 − 6𝑦 + 2𝑧 − 2 = 0 Step 5 3 (𝑡 ) × ⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗ To find a rectifying plane need to find [𝑟⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗ 𝑟 ′′ (𝑡0 )] × ⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗ 𝑟 3 (𝑡0 ) at t=1 0 Now 𝑟1 (𝑡) = 𝑖 + 2𝑡𝑗 + 3𝑡 2 𝑘 𝑟 ′′ (𝑡) = 0𝑖 + 2𝑗 + 6𝑡𝑘 3 (𝑡 ) × 𝑟 ′′ (𝑡 )] × 𝑟 1 (1) = 𝑖𝑗𝑘 ⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗ ⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗ [𝑟⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗ 0 0 6 −62 123 = (−18 − 4)𝑖 − (18 − 2)𝑗 + (12 + 6)𝑘 = −22𝑖 − 16𝑗 + 18𝑘 Therefore, the equation of the rectifying plane is given as −22(𝑥 − 1) − 16(𝑦 − 1) + 18(𝑧 − 1) = 4 −22𝑥 + 22 − 16𝑦 + 16 + 18𝑧 − 18 = 0 −22𝑥 − 16𝑦 + 18𝑧 − 20 = 0 That is 11𝑥 + 8𝑦 − 9𝑧 + 10 = 0. Question #2 Suppose that the vector-valued function 𝑟(𝑡) =< 3𝑡, cos(5𝑡) , sin(5𝑡) represents the position of an object at time t. Part a) Find a vector-valued function 𝑎(𝑡) representing the object's acceleration at time t. Part b) What is the distance travelled by the object between time t=0 and time t=1 ? Answer: Step 1 Given is 𝑟(𝑡) =≪ 3𝑡, cos 5 𝑡, sin 5 𝑡 ≫ Represent the position of an object at time t a) Acceleration at time t 𝑎(+) = = 𝑑 𝑑𝑡 𝑑2𝑟 𝑑𝑡 2 = 𝑑 ( 𝑑 𝑑𝑡 𝑑𝑡 ≪ 3𝑡, cos 5 𝑡, sin 5 𝑡 ≫) (≪ 3, − 5 sin 5 𝑡, 5 cos 5 𝑡 ≫) =≪ 0, − 25 cos 5 𝑡, − 25 sin 5 𝑡 ≫ Acceleration at time t 𝑎(𝑡) ⇒> 0, −25 cos 5 𝑡, − 25 sin 5 𝑡 ≫ Ans. b) Distance travelled by the object between time t=0 & t=1 Distance 2 2 2 2 1 𝑑𝑥 𝑑𝑦 𝑑 𝑑 = ∫0 √( ) + ( ) + ( ) ∵ 𝑥 = 3𝑡 𝑑𝑡 𝑑𝑥 𝑑𝑡 𝑦 = cos 5 𝑡 𝑧 = sin 5 𝑡 𝑑𝑡 𝑑𝑡 =3 𝑑𝑦 𝑑𝑡 = −5 sin(5𝑡) 𝑑𝑧 = 5 cos 5 𝑡 𝑑𝑡 Step 2 distance 1 𝑑 = ∫0 √(3)2 + (−5 sin 5 𝑡)2 + (5 cos 5 𝑡)2 𝑑𝑡 1 = ∫0 √9 + 25(sin2 5𝑡 + cos 2 5𝑡)𝑑𝑡 ∵ sin2 0 + cos 2 0 = 1 0 ∈ 𝑅 1 𝑑 = ∫0 √9 + 25 𝑑𝑡 1 = ∫0 √34𝑑𝑡 1 = [√34𝑡]0 = √34 − 0 = 5.83 Distance travelled by the object is 5.83 unit. Question #3 Let r(t) and u(t) be vector-valued functions whose limits exist as t c. Prove that lim[𝑟(𝑡) × 𝑢(𝑡)] = lim × lim 𝑢 (𝑡) 𝑡⇒𝑐 𝑡⇒𝑐 𝑡⇒𝑐 Answer: Step 1 Given r(t) and u(t) be vector valued function whose lint exist as 𝑡 ⇒ 𝑐 To prove lim[𝑟(𝑡) × 𝑢(𝑡)] = lim 𝑟(𝑡) × lim 𝑢(𝑡) 𝑡⇒𝑐 𝑡⇒𝑐 𝑡⇒𝑐 Step 2 Dot product If 𝑎 = 𝑎1 𝑖 + 𝑎2 𝑗 + 𝑎3 𝑘 and 𝑏 = 𝑏[1} 𝑖 + 𝑏2 𝑗 + 𝑏3 𝑘 𝑎 × 𝑏 = 𝑎1 𝑏1 + 𝑎2 𝑏2 + 𝑎3 𝑏3 Step 3 Let 𝑟(𝑡) = 𝑎(𝑡)𝑖 + 𝑏(𝑡)𝑗 + 𝑐(𝑡)𝑘 𝑢(𝑡) = 𝑚(𝑡)𝑖 + 𝑛(𝑡)𝑗 + 𝑝(𝑡)𝑘 On taking L.H.S lim[𝑟(𝑡) × 𝑢(𝑡)] = lim(𝑎(𝑡)𝑖 + 𝑏(𝑡)𝑗 + 𝑐(𝑡)𝑘) × (𝑚(𝑡)𝑖 + 𝑛(𝑡)𝑗 + 𝑝(𝑡)𝑘) 𝑡⇒𝑐 𝑡⇒𝑐 = lim(𝑎(𝑡)𝑚(𝑡) + 𝑏(𝑡)𝑛(𝑡) + 𝑐(𝑡)𝑝(𝑡)) 𝑡⇒𝑐 = lim(𝑎(𝑡)𝑚(𝑡)) + lim(𝑏(𝑡)𝑛(𝑡)) + lim(𝑐(𝑡)𝑝(𝑡)) 𝑡⇒𝑐 𝑡⇒𝑐 𝑡⇒𝑐 Since limit exist as t tends to c. = 𝑎(𝑐)𝑚(𝑐) + 𝑏(𝑐)𝑛(𝑐) + 𝑐(𝑐)𝑝(𝑐) lim 𝑝 𝑟(𝑡) × 𝑢(𝑡) = 𝑎(𝑐)𝑚(𝑐) + 𝑏(𝑐)𝑛(𝑐) + 𝑐(𝑐)𝑝(𝑐) 𝑡⇒𝑐 Now taking R.H.S lim 𝑟(𝑡) × lim 𝑢(𝑡) = lim(𝑎(𝑡)𝑖 + 𝑏(𝑡)𝑗 + 𝑐(𝑡)𝑘) × lim 𝑡⇒𝑐 𝑡⇒𝑐 𝑡⇒𝑐 𝑡⇒𝑐 𝑚(𝑡)𝑖 + 𝑛(𝑡)𝑗 + 𝑝(𝑡)𝑘 Since limit as t tends to c = (𝑎(𝑐)𝑖 + 𝑏(𝑐)𝑗 + 𝑐(𝑐)𝑘) × (𝑚(𝑐)𝑖 + 𝑛(𝑐)𝑗 + 𝑝(𝑐)𝑘) = 𝑎(𝑐)𝑚(𝑐) + 𝑏(𝑐)𝑛(𝑐) + 𝑐(𝑐)𝑝(𝑐) lim 𝑟(𝑡) × lim 𝑢(𝑡) = 𝑎(𝑐)𝑚(𝑐) + 𝑏(𝑐)𝑛(𝑐) + 𝑐(𝑐)𝑝(𝑐) 𝑡⇒𝑐 𝑡⇒𝑐 Step 4 L.H.S=R.H.S lim[𝑟(𝑡) × 𝑢(𝑡)] = lim × lim 𝑢(𝑡) 𝑡⇒𝑐 𝑡⇒𝑐 𝑡⇒𝑐 Hence proved. Question #4 The acceleration of an object after t seconds is given by the vector-valued function 𝑎(𝑡) =≪ cos(8𝑡) , sin(8𝑡) , − 5 ≫ If the object's initial position is (−9, − 3, 10) , and the object's initial velocity is ≪ 1, 0, 1 ≫ , find a vector-valued function 𝑟(𝑡) representing the object's position at time t. Answer: Step 1 𝑎(𝑡) = (cos(8𝑡) , sin(8𝑡) . − 5) to find the velocity equation, take integration 𝑣(𝑡) = ∫ 𝑎(𝑡)𝑑𝑡 𝑣(𝑡) = ∫(cos(8𝑡) , sin(8𝑡) , − 5)𝑑𝑡 1 1 8 8 1) 𝑣(𝑡) = ( sin(8𝑡) + 𝑥, − cos(8𝑡) + 𝑦, − 5𝑡 + 𝑧) Step 2 here initial velocity is 𝑣(0) = (1, 0, 1) 1 1 (1, 0, 1) = ( sin(8 × 0) + 𝑥, − cos(8 × 0) + 𝑦𝑚 − 5 × 0 + 𝑧) 8 8 1 (1, 0, 1) = (0 + 𝑥, − + 𝑦, 𝑧) 8 1 (1, 0, 1) = (𝑥, − + 𝑦, 𝑧) 8 1 (𝑥, − + 𝑦, 𝑧) = (1,0,1) 8 1 (𝑥, 𝑦, 𝑧) = (1, , 1) put it back in equation 1 8 Step 3 1 1 8 1 8 1 1 8 8 8 𝑣(𝑡) = ( sin(8𝑡) + 𝑥, − cos(8𝑡) + 𝑦, −5𝑡 + 𝑧) 𝑣(𝑡) = ( sin(8𝑡) + 1, − cos(8𝑡) + , − 5𝑡 + 1) to find the position equation, take integration 𝑟(𝑡) = ∫ 𝑣(𝑡)𝑑𝑡 1 1 8 8 1 𝑟(𝑡) = ∫ ( sin(8𝑡) + 1, − cos(8𝑡) + , − 5𝑡 + 1) 2) 𝑟(𝑡) = (− 1 64 8 cos(8𝑡) + 𝑡 + 𝑥, − 1 64 1 5𝑡 2 8 2 sin(8𝑡) + 𝑡 + 𝑦, − + 𝑡 + 𝑧) here initial position is 𝑟(0) = (−9, − 3, 10) Step 4 (−9, − 3, 10) = (− (−9, − 3, 10) = (− (−9 + (− 575 64 1 64 1 64 1 64 cos(8 × 0) + 0 + 𝑥, − cos(0) + 0 + 𝑥, − 1 64 1 64 1 5×02 8 2 sin(8 × 0) + × 0 + 𝑦, − sin(0) + 0 + 𝑦, 𝑧) , − 3, 10) = (𝑥, 𝑦, 𝑧) , − 3, 10) = (𝑥, 𝑦, 𝑧) (𝑥, 𝑦, 𝑧) = (− 𝑟(𝑡) = (− 1 64 575 64 , −,10) put it back in equation 2 cos(8𝑡) + 𝑡 ± 575 64 , − 1 64 1 5𝑡 2 8 2 sin(8𝑡) + 𝑡 ± 3, − + 𝑡 + 10) Question #5 Find the domain of the vector-valued function 𝑟(𝑡) = √(4 − 𝑡)2 𝑖 + 𝑡 {2] 𝑗 − 6𝑡𝑘 Answer: Step 1 Given vector-valued function 𝑟(𝑡) = √(4 − 𝑡)3 𝑖 + 𝑡 2 𝑗 − 6𝑡𝑘 + 0 + 𝑧) Step 2 For the domain of given function Here (4 − 𝑡)2 is non negative for every real number ⇒ √(4 − 𝑡)2 is also non negative for every real number ⇒ √(4 − 𝑡)2 is defined for every real number And 𝑡 2 and -6t is polynomial function The domain of polynomial function is every real number. So, Domain of 𝑟(𝑡) = 𝑅 or Domain of 𝑟(𝑡) = (−∞, ∞) Question #6 Determine the interval(s) on which the vector-valued function is continuous 𝑟(𝑡) = 2𝑒 −𝑡 𝑖 + 𝑒 −𝑡 𝑗 + ln(𝑡 − 1) 𝑘 Answer: Step 1 We have to determine the interval on which the vector valued function is continuous. 1) 𝑟(𝑡) = 2𝑒 −𝑡 𝑖 + 𝑒 −𝑡 𝑗 + ln(𝑡 − 1) 𝑘 Let us break r(t) in parametric equations. 2) 𝑘𝑟(𝑡) = 𝑎(𝑡)𝑖 + 𝑏(𝑡)𝑗 + 𝑐(𝑡)𝑘 Step 2 On comparing equation (10 with (2) we get 𝑎(𝑡) = 2𝑒 −𝑡 , 𝑏(𝑡) = 𝑒 −𝑡 , 𝑐(𝑡) = ln(𝑡 − 1) for 𝑎(𝑡) = (−∞, ∞) 𝑏(𝑡) = (−∞, ∞) 𝑐(𝑡) = ln(𝑡 − 1) t-1>0 ⇒𝑡>1 ∴ 𝑐(𝑡) = (1, ∞) ∴ for 𝑟(𝑡) = (1, ∞) Question #7 Explain how the rectangular equation y=5x can have infinitely many sets of parametric equations. Answer: Step 1 Given rectangular equation is, y=5x It represents a line passing through the origin. Step 2 We find the set of parametric equations by choosing arbitrary parametric equation for x and then substituting it in the given equation. For example, Let x=t, then y=5t Therefore set of parametric equations are x=t and y=5t Now let x=t+1, then y=5(t+1) =5t+5 Therefore set of parametric equations are x=t+1 and y=5t+5 Similarly by choosing different parametric equation for x, we will get different set of parametric equations. Hence the rectangular equation y=5x can have infinitely many sets of parametric equations. Question #8 Find parametric equations for the tangent line to the curve with the parametric equations 𝑥 = √𝑡 2 + 3, 𝑦 = ln(𝑡 2 + 3) , 𝑧 = 𝑡 at the point (2, ln 4 , 1) Answer: Step 1 Given: The parametric equations, 𝑥 = √𝑡 2 + 3, 𝑦 = ln(𝑡 2 + 3) , 𝑧 = 𝑡 and the point (2, ln(4) , 1) To find: The parametric equations for the tangent line to the curve with given parametric equations at the given point. Step 2 Let, 𝑥 = √𝑡 2 + 3, 𝑦 = ln(𝑡 2 + 3) , 𝑧 = 𝑡 and the point is (2, ln(4) , 1) ⇒ 2 = √𝑡 2 + 3, ln(4) = ln(𝑡 2 + 3) , 1 = 𝑡 Now we will be lookin for the vector parallel to the tangent, and calculate r'(1) Let, 𝑟(𝑡) =≪ √𝑡 2 + 3, ln(𝑡 2 + 3) , 𝑡 ≫ 2𝑡 2𝑡 ⇒ 𝑟 ′ (𝑡) =≪ , (𝑡 2 , 1 ≫ 2 𝑟 ′ (𝑡) =≪ +3) 2√𝑡 +3 𝑡 2𝑡 √𝑡 2 +3 , (𝑡 2+3) ,1≫ Step 3 Put t=1 ⇒ 𝑟 ′ (1) =≪ 𝑟 ′ (1) 1 1 √1+3 1 2 2 =≪ , , 2 4 , 1≫ ,1≫ We can use our given starting point (2, ln(4) , 1) and the vector r'(1) to define our line. 1 1 𝑥 = 2 + 𝑡, 𝑦 = ln(4) + 𝑡, 𝑧 = 1 + 𝑡 2 2 Result: The parametric equations of the required tangent line is, 1 1 𝑥 = 2 + 𝑡, 𝑦 = ln(4) + 𝑡, 𝑧 = 1 + 𝑡 2 2 Question #9 Which of the following vector valued functions represent the same graph? a) 𝑟(𝑡) = (−3 cos 𝑡 + 1)𝑖 + (5 sin 𝑡 + 2)𝑗 + 4𝑘 b) 𝑟(𝑡) = 4𝑖 + (−3 cos 𝑡 + 1)𝑗 + (5 sin 𝑡 + 2)𝑘 c) 𝑟(𝑡) = (3 cos 𝑡 − 1)𝑖 + (−5 sin 𝑡 − 2)𝑗 + 4𝑘 d) 𝑟(𝑡) = (−3 cos 2 𝑡 + 1)𝑖 + (5 sin 2 𝑡 + 2)𝑗 + 4𝑘 Answer: Step 1 Since 𝑟(𝑡) = 𝑥𝑖 + 𝑦𝑖 + 𝑧𝑘 a) 𝑟(𝑡) = (−3 cos 𝑡 + 1)𝑖 + (5 sin 𝑡 + 2)𝑗 + 4𝑘 Parametric equations, 𝑥 = −3 cos 𝑡 + 1 𝑦 = 5 sin 𝑡 + 2 z=4 Then, 1−𝑥 cos 𝑡 = 3 𝑦−2 sin 𝑡 = 5 And cos 2 𝑡 + sin2 𝑡 = 1 ( 1−𝑥 2 3 ) +( 𝑦−2 2 ) =1 5 2 𝑦−2 1 − 2𝑥 + 𝑥 2 ( ) = 1− 5 9 ( 𝑦−2 2 5 ) = 8+2𝑥−𝑥 2 9 5 𝑦 = √8 + 2𝑥 − 𝑥 2 + 2 3 b) 𝑟(𝑡) = 4𝑖 + (−3 cos 𝑡 + 1)𝑗 + (5 sin 𝑡 + 2)𝑘 Parametric equations, x=4 𝑦 = −3 cos 𝑡 + 1 𝑧 = 5 sin 𝑡 + 2 Hence, 1−𝑦 cos 𝑡 = sin 𝑡 = 3 𝑧−2 5 And cos 2 𝑡 + sin2 𝑡 = 1 ( 1−𝑦 2 ) +( 3 [1−𝑦}2 ( 5 =1− 3 1−𝑦 2 3 𝑧−2 2 ) = ) =1 4−4𝑧+𝑧 2 25 25+4𝑧−𝑧 2 25 3 𝑦 = 1 − √21 + 4𝑧 − 𝑧 2 5 Step 2 c) 𝑟(𝑡) = (3 cos 𝑡 − 1)𝑖 + (−5 sin 𝑡 − 2)𝑗 + 4𝑘 Parametric equations, 𝑥 = 3 cos 𝑡 − 1 𝑦 = −5 sin 𝑡 − 2 z=4 Then 1+𝑥 cos 𝑡 = sin 𝑡 = 3 −2−𝑦 5 And cos 2 𝑡 + sin2 𝑡 = 1 ( ( −2−𝑦 2 ) =1− 5 −2−𝑦 2 5 ) = 1+2𝑥+𝑥 2 9 8−2𝑥−𝑥 2 9 5 𝑦 = − √8 − 2𝑥 − 𝑥 2 − 2 3 d) 𝑟(𝑡) = (−3 cos 2 𝑡 + 1)𝑖 + (5 sin 2 𝑡 + 2)𝑗 + 4𝑘 Parametric equations, 𝑥 = −3 cos 2 𝑡 + 1 𝑦 = 5 sin 2 𝑡 + 2 z=4 Then, 1−𝑥 cos 2 𝑡 = sin 2 𝑡 = 3 𝑦−2 5 And cos 2 2𝑡 + sin2 2𝑡 = 1 1−𝑥 2 𝑦−2 2 ( ) +( ) =1 3 5 ( 𝑦−2 2 5 ) =1− 1−2𝑥+𝑥 2 9 𝑦 − 2 2 8 + 2𝑥 − 𝑥 2 ( ) = 5 9 5 𝑦 = √8 + 2𝑥 − 𝑥 2 + 2 3 Here, 𝑟(𝑡) = (−3 cos 𝑡 + 1)𝑖 + (5 sin 𝑡 + 2)𝑗 + 4𝑘 And 𝑟(𝑡) = (−3 cos 2 𝑡 + 1)𝑖 + (5 sin 2 𝑡 + 2)𝑗 + 4𝑘 Represent the same graph Question #10 Find parametric equations for the following curves. Include an interval for the parameter values. Answers are not unique. The line segment starting at 𝑃(−1, − 3) and ending at 𝑄(6, − 16) Answer: Step 1 Given, The line segment starting at 𝑃(−1, − 3) and ending at 𝑄(6, − 16) . We have to find the parametric equations for this line segment. Step 2 Concept Used The parametric equations for a line segment starting from P and ending at Q P+tQ Step 3 Calculation The parametric equations of the line segment starting at 𝑃(−1, − 3) and ending at 𝑄(6, − 16) is 𝑃 + 𝑡𝑄 = (−1, − 3) + 𝑡(6, − 16) = (−1 + 6𝑡, − 3 − 16𝑡) ∴ 𝑥 = −1 + 6𝑡, 𝑦 = −3 − 16𝑡