Questions and Answers #7 Parametric Equations, Polar Coordinates, and Vector-V

Questions and Answers Sheet 7 Parametric Equations, Polar Coordinates, and Vector-V Question #1 Represent the plane curve by a vector-valued function. (There are many correct answers.) 3𝑥 + 4𝑦 − 12 = 0 Answer: Step 1 Represents the plane curve by a vector valued function We have an equation 3x+4y-12=0 Explanation: We have an equation 3x+4y-12=0 We know that the vector form is given by r(t)=x(t)i+y(t)j Now simplify the equation and find the value of y as follows. Step 2 3x+4y-12=0 4y=-3x+12 −3 12 𝑦= 𝑥+ 𝑦= 4 −3 4 4 𝑥+3 3 𝑦 = 3− 𝑥 4 So let us consider x=t then y becomes 3 𝑦 = 3− 𝑡 4 So the vector valued function becomes 3 𝑟(𝑡) = 𝑡𝑖 + (3 − 𝑡) 𝑗 4 Question #2 Represent the plane curve by a vector-valued function 𝑦 = 9 − 𝑥 2 Answer: Step 1 Given: 𝑦 = 9 − 𝑥2 The vector valued function for a plane curve: 𝑟(𝑡) = 𝑥(𝑡)𝑖 + 𝑦(𝑡)𝑗 Step 2 𝑥(𝑡) = 𝑡 𝑦(𝑡) = 9 − 𝑡 2 𝑟(𝑡) = 𝑡𝑖 + (9 − 𝑡 2 )𝑗 Question #3 Find the limit of the following vector-valued functions at the indicated value of t. lim ≪ √𝑡 − 3, 𝑡⇒4 √𝑡−2 , 𝑡−4 𝜋 tan ( ) ≫ 𝑡 Answer: To find the limit , we evaluate the limit of each component 𝜋 √𝑡 − 2 lim 𝑟(𝑡) = lim [√𝑡 − 3, , tan ] 𝑡⇒4 𝑡⇒4 𝑡−4 𝑡 𝜋 √𝑡 − 2 = [ lim , lim , lim tan ] 𝑡⇒4 𝑡⇒4 𝑡 − 4 𝑡 𝑡⇒4√𝑡−3 since second term when applied limit is of the form √𝑡−2 𝑡−4 1 − = 0 0 we use L hospital rule. So 1 2√𝑡 1 2√𝑡 So r(t) changes to = [ lim lim 1 𝑡⇒4√𝑡−3 𝑡⇒4 2√𝑡 1 = [√4 − 3, = [1, = [1, 1 2×2 1 4 2√4 , lim tan 𝑡⇒4 𝜋 , tan 4 𝜋 𝑡 ] ] , 1] , 1] Question #4 Find the domain of the vector-valued function 𝑟(𝑡) = ln(𝑡 − 1) 𝑖 Square root 4-tj Answer: Step 1 This vector valued function is a combination of two functions: 1. Logarithmic function and 2. Square root function Step 2 For, ln(𝑡 − 1) to be defined 𝑡−1>0⇒𝑡 >1 For Square root function to be defined 4−𝑡 ≥0⇒𝑡 ≤4 So, the domain is: (1, 4] Question #5 Represent the plane curve by a vector-valued function 𝑦 = (𝑥 − 2)2 Answer: Step 1 𝑦 = (𝑥 − 2)2 Now let x(t)=t Now put x(t)=t in 𝑦 = (𝑥 − 2)2 , we get 𝑦 = (𝑡 − 2)2 √𝑡−2 𝑡−4 will be equal to Step 2 𝑟(𝑡) = 𝑥(𝑡)𝑖 + 𝑦(𝑡)𝑗 ⇒ 𝑟(𝑡) = 𝑡𝑖 + (𝑡 − 2)2 𝑗 Question #6 2 Compute the derivative of the vector-valued function 𝑟(𝑡) =≪ sin 2 𝑡, 𝑒 𝑡 ≫ Answer: Step 1 If 𝑟(𝑡) =≪ 𝑓(𝑡), 𝑔(𝑡), ℎ(𝑡) ≫ where f, g, h are differentiable function, then 𝑟 ′ (𝑡) =≪ 𝑓 ′ (𝑡), 𝑔′ (𝑡), ℎ′ (𝑡) ≫ Step 2 2 Given, 𝑟(𝑡) =≪ sin 2 𝑡, 𝑒 𝑡 ≫ 𝑑 𝑑 (sin 2 𝑡) = cos 2 𝑡 (2𝑡) 𝑑𝑡 𝑑𝑡 = 2 cos 2 𝑡 𝑑 𝑡2 2 𝑑 (𝑡 2 ) (𝑒 ) = 𝑒 𝑡 𝑑𝑡 𝑑𝑡 2 = 2𝑡𝑒 𝑡 2 So, 𝑟 ′ (𝑡) =≪ 2 cos 2 𝑡, 2𝑡𝑒 𝑡 ≫ Question #7 Represent the plane curve by a vector-valued function. 𝑥2 16 − 𝑦2 4 =1 Answer: Step 1 𝑥2 𝑦2 − 16 =1 4 let x=t 𝑡2 ∴ 𝑦2 16 4 𝑦2 4 − = = 𝑦2 𝑡2 4 =1 −1 16 𝑡 2 −16 16 2 𝑦 =4× 2 𝑦 = 𝑡 2 −16 𝑡 2 −16 16 4 𝑡 2 −16 𝑦=√ 4 1 𝑦 = √𝑡 2 − 16 2 Step 2 let r(t) be a vector field function 1 ∴ 𝑟(𝑡) = (𝑡)𝑖 + ( √𝑡 2 − 16) 𝑗 2 Question #8 Represent the plane curve by a vector-valued function. 𝑥 2 + 𝑦 2 = 25 Answer: Step 1 It is given that, 𝑥 2 + 𝑦 2 = 25 Let 𝑥 = 5 𝑐𝑜𝑠 and 𝑦 = 𝑠𝑖𝑛 , Then (5 cos 𝑡)2 + (5 sin 𝑡)2 = 25(cos 2 (𝑡) + sin2 (𝑡)) = 25 Step 2 The function of the form 𝑟(𝑡) = 𝑥(𝑡)𝑖 + 𝑦(𝑡)𝑗 is a vector valued function. Substitute x and y in the above equation, to get 𝑟(𝑡) = (5 cos 𝑡)𝑖 + (5 sin 𝑡)𝑗 Question #9 Derivatives of vector-valued functions Differentiate the following function. 𝑟(𝑡) =≪ 𝑡𝑒 −𝑡 , 𝑡 ln 𝑡 , 𝑡 cos 𝑡 ≫ Answer: Step 1 Given, 𝑟(𝑡) =≪ 𝑡𝑒 −𝑡 , 𝑡𝑒𝑛 𝑡, 𝑡 cos 𝑡 ≫ So, obtain differentiate individually 𝑑 (𝑡𝑒 −𝑡 ) = 𝑒 −𝑡 + 𝑡(𝑒 −𝑡 )(−1) 𝑑𝑥 = (1 − 𝑡)𝑒 −𝑡 Step 2 Similarly, 𝑑 𝑡 (𝑡 ln 𝑡) = + 1 × 𝑙𝑛 𝑑𝑥 𝑡 = ln 𝑡 + 1 𝑑 (𝑡 cos 𝑡) = cos 𝑡 + 𝑡(− sin 𝑡) 𝑑𝑥 = cos 𝑡 − 𝑡 sin ∴ 𝑟 ′ (𝑡) =≪ (1 − 𝑡)𝑒 −𝑡 , (1 + ln 𝑡), (cos 𝑡 − 𝑡 sin 𝑡) ≫ Question #10 Represent the plane curve by a vector-valued function. 2𝑥 − 3𝑦 + 5 = 0 Answer: Step 1 Given: 2x-3y+5=0 For finding given equation in vector form, we assume x=t and find value of y in terms of t So, 2t-3y(t)+5=0 -3y(t)=-2t-5 3y(t)=2t+5 𝑦(𝑡) = 2𝑡+5 3 Step 2 So, vector equation of line is given by 2𝑡+5 𝑟(𝑡) = 𝑡𝑖 + 𝑗 3 Hence, vector equation of given line is 𝑟(𝑡) = 𝑡𝑖 + 2𝑡+5 3 𝑗