Questions and Answers #5 Parametric Equations, Polar Coordinates, and Vector-V

Questions and Answers Sheet 5 Parametric Equations, Polar Coordinates, and Vector-V Question #1 Determine the interval (s) on which the vector-value function is continious. (Enter your answer using interval notation.) 𝑟(𝑡) = √𝑡𝑖 + √𝑡 − 1𝑘 Answer: Step 1 𝑟(𝑡) = (√𝑡)𝑖 + (√𝑡 − 1)𝑘 we know for a function to be continuous in Interval then it should be defined at every point in that interval. we know that inside a root function there can't be a negative number so t should greater than or equal to 0 and t - 3 should greater than or equal to 0 by taking the intersection of both t should greater than or equal to 3 But at 0 and 3 limit does not exist as the left-hand side limit is not equal to the right-hand side limit. Step 2 Therefore, the interval is (1, co) on which the vector-valued function is continuous. Question #2 Vector-valued functions Find a function r(t) that describes the following curve. Anellipse in te plane x=2 satisfying the equation 𝑦2 9 + 𝑧2 16 =1 Answer: Step 1 Given, An ellipse in the plane x=2 satisfying the equation 𝑦2 9 + 𝑧2 16 = 1. We have to find a function r(t) that describes the above curve. Step 2 Calculation The parametric equations of an ellipse Hence the parametric equations for 𝑦2 9 𝑥2 𝑎2 + + 𝑧2 16 𝑦2 𝑏2 = 1 are 𝑥 = 𝑎 cos , 𝑦 = 𝑏 sin . = 1 are 𝑦 = 3 𝑐𝑜𝑠 and 𝑧 = 4 sin . Therefore the vector function for the given curve is 𝑟⃗(𝑡) = 2𝑖̂ + 3 cos 𝑡𝑗̂ + 4 sin 𝑡𝑘̂ Question #3 Find a vector-valued function whose graph is the indicated surface. The cylinder 4𝑥 2 + 𝑦 2 = 16 Answer: Step 1 In polar coordinates, 𝑥 = 𝑟 cos 𝜃 , 𝑦 = 𝑟 sin 𝜃. Using this rule, using appropriate substitutions: 𝑥 = 2 cos 𝑢 , 𝑦 = 4 sin 𝑢 Step 2 Considering z=v, the vector form is given by: 𝑟(𝑢, 𝑣) = 2 cos 𝑢 𝑖 + 4 sin 𝑢 𝑗 + 𝑣𝑘 Step 3 Answer: 𝑟(𝑢, 𝑣) = 2 cos 𝑢 𝑖 + 4 sin 𝑢 𝑗 + 𝑣𝑘 Question #4 Consider the vector-valued function 𝑟(𝑡)) = 𝑡^2𝑖 + (𝑡 − 3)𝑗 + 𝑡𝑘 . Write a vector-valued function s(t) that is the specified transformation of r. i. a vertical traation 3 units upward Answer: Step 1 𝑟(𝑡) = 𝑡 2 𝑖 + (𝑡 − 3)𝑗 + 𝑡𝑘 we have to find vector valued function s(t) Step 2 s(t) is the specified transformation of r(t) That is a vertical traation 3 units upward 𝑥(𝑡) = 𝑡 2 𝑦(𝑡) = 𝑡 − 3 2(𝑡) = 𝑡 + 3 𝑠(𝑡) = 𝑡 2 𝑖 + (𝑡 − 3)𝑗 + (𝑡 + 3)𝑘 Question #5 3 Determine the interval(s) on which the vector-valued function is continuous 𝑟(𝑡) =≪ 8, √𝑡, √𝑡 ≫. Answer: Step 1 To determine: The interval(s) on which the given vector values function is continuous 3 𝑟(𝑡) =≪ 8, √𝑡, √𝑡 ≫. Step 2 Explanation: The vector-valued function, 𝑟(𝑡) =≪ 𝑓(𝑡), 𝑔(𝑡), ℎ(𝑡) ≫ is continuous on the interval if each component function, f(t), g(t), and h(t) are continuous on that interval. 3 Here 𝑟(𝑡) =≪ 8, √𝑡, √𝑡 ≫ 3 So, the function r(t) is continuous when the component functions, 8, √𝑡, √𝑡 are continuous. f(t)=8 is the constant function so it is continuous on the interval −∝, ∝ The function 𝑔(𝑡) = √𝑡 is the radical function so itis continuous on the interval [0, ∝) 3 And the function ℎ(𝑡) = √𝑡 ≫ is the cube root function continuous on the interval −∝, ∝ So the common interval is [0, ∝) Therefore the given vector values function r/t) is continuous on the interval [0, ∝) Question #6 Sketch the conic sections whose polar coordinate equations are given in. Give polar coordinates for the vertices and, in the case of ellipses, for the centers as well. 𝑟= 2 1+cos 𝜃 Answer: Given that, the conic equation is 𝑟 = Compare this equation 𝑟 = 2 1+cos 𝜃 2 1+cos 𝜃 with 𝑟 = 𝑒𝑘 1+𝑒 cos 𝜃 Here, the value of eccentricity e is 1. The value of directrix k is 2. 2 From above statement, it is observed that the equation 𝑟 = 1+cos 𝜃 represents parabola with k = 2 directrix to the right side of the pole. Due to positive sign in denominator the parabola opens to the left along the polar axis and p=1. From above figure, it is observed that the vertex is (1,0). Question #7 Consider the vector-valued function 𝑟(𝑡) = 3𝑡 2 𝑖 + (𝑡 − 1)𝑗 + 𝑡𝑘. Write a vector-valued function u(t) that is the specified transformation of r. A vertical traation two units upward Answer: A vector-valued function in the different directions of the x-axis, the y-axis, and the z-axis can be written as: 𝑣⃗ = 𝑥𝑖̂ + 𝑦𝑗̂ + 𝑧𝑘̂ Since it is given to us to give a vertical traation of two units upward i.e. in the positive direction of the yaxis will be y=t-1+2 y=t+1 So the unit vector 𝑢(𝑡) = 3𝑡 2 𝑖̂ + (𝑡 + 1)𝑗̂ + 𝑡𝑘̂ Question #8 Consider the vector-valued function 𝑟(𝑡) = 𝑡 2 𝑖̂ + (𝑡 − 3)𝑗̂ + 𝑡𝑘̂. Write a vector-valued function s(t) that is the specified transformation of r. iii. a horizontal traation 5 units in the direction of the positive y-axis Answer: Consider the vector valued function 𝑟(𝑡) = 𝑡 2 𝑖̂ + (𝑡 − 3)𝑗̂ + 𝑡𝑘̂ The x axis components is 𝑡 3 (horizontal) The y axis Components is (t-3) The z axis Components is t (vertica) So, horizontally traate 5 units in the positive y axis. The y axis component will be 𝑡 − 3 + 5 = 𝑡 + 2 Hence the horizontal traation 5 units in the direction of the positive y axis 𝑟⃗(𝑡) = 𝑟⃗(𝑡) + 5𝑗 = 𝑡 3 𝑖̂ + (𝑡 − 3 + 5)𝑗̂ + 𝑡𝑘̂ 𝑟⃗(𝑡) = 𝑡 3 𝑖̂ + (𝑡 + 2)𝑗̂ + 𝑡𝑘̂ Question #9 1 Evaluate, don't forget the constant vector, ∫ (𝑡 2 𝑖 − 2𝑡𝑗 + 𝑘) 𝑑𝑡 𝑡 Answer: Step 1 1 We have given vector-valued function as 𝑡 2 𝑖 − 2𝑡𝑗 + 𝑘. While evaluating the indefinite integration of 𝑡 vector valued, the constant term will be considered with every component of the vector and at last, combined the constant term with their respective component. Step 2 1 Now, find the integration of the vector function 𝑡 2 𝑖 − 2𝑡𝑗 + 𝑘. Use the sum, difference and constant multiple rules. Also, use the power rule ∫  = 𝑑𝑢 = 𝑢𝑛+1 𝑛+1 𝑡 1 + 𝐶 and ∫ 𝑑𝑢 = ln|𝑢| + 𝐶 𝑢 1 1 (t 2i − 2tj + k )dt = (  t 2 dt )i − (2  tdt ) j + (  dt )k t t 𝑡3 𝑡2 3 2 = ( + 𝐶1 ) 𝑖 − (2 ⋅ = 𝑢𝑛 𝑡3 3 𝑡3 3 + 𝐶2 ) 𝑗 + (ln|𝑡| + 𝐶3 )𝑘 𝑖 − 𝑡 2 𝑗 + ln|𝑡| 𝑘 + (𝐶1 𝑖 + 𝐶2 𝑗 + 𝐶3 𝑘) 𝑖 − 𝑡 2 𝑗 + ln|𝑡| 𝑘 + 𝐶 Hence,  1 t3 (t 2i − 2tj + k )dt = i − t 2 j + ln | t | k + C , where 𝐶 = 𝐶1 𝑖 + 𝐶2 𝑗 + 𝐶3 𝑘 is a constant vector. t 3 Question #10 Find the open interval(s) on which the curve given by the vector valued function 𝑟( ) = 2 cos 3 3 sin3 𝑗, 0 ≤≤ 2 is smooth. 𝑖+ Answer: Consider the given vector-valued function: 𝑟(𝑡) = (2 cos 3 𝑡)𝑖̂ + (3 sin3 𝑡)𝑗̂ Here the objective is to find the open intervals on which the curve given by the vector-valued function is smooth. Let 𝑟(𝑡) = 𝑓(𝑡)𝑖̂ + 𝑔(𝑡)𝑗̂ + ℎ(𝑡)𝑘̂ be the parameterization of the curve that is differentiable (i.e 𝑓 ′ (𝑡), 𝑔′ (𝑡) and ℎ′ (𝑡) is continuous) on an open interval I. Then r(t) is smooth on the open interval I, if 𝑟 ′ (𝑡) ≠ 0̅. for any value of t in the interval I. The given vector-valued function is 𝑟(𝑡) = (2 cos 3 𝑡)𝑖̂ + (3 sin3 𝑡)𝑗̂ Differentiate r(t) with respect to t 𝑟 ′ (𝑡) = (−6 sin2 𝑡)𝑖̂ + (9 cos 2 𝑡)𝑗̂ To find the open interval sin 2 t = 0 r (t ) = 0   2 cos t = 0 Here required such value of t for which sin2 𝑡 = 0 and cos 2 𝑡 = 0 simultaneously. To solve the above equation, draw the graph of sin2 𝑡 = 0 and cos 2 𝑡 = 0 From the graph, it is clear that there is no such value of t for which sin2 𝑡 = 0 and cos 2 𝑡 = 0 simultaneously. Therefore 𝑟 ′ (𝑡) ≠ 0 in (−∞, ∞) Hence the curve is smooth on (−∞, ∞)