Questions and Answers #1 Parametric Equations, Polar Coordinates, and Vector-V

Questions and Answers Sheet 1 Parametric Equations, Polar Coordinates, and Vector-V Question #1 If a curve has the property that the position vector r(t) is always perpendicular to the tangent vector r'(t), show that the curve lies on a sphere with center the origin. Answer: It is given that 𝑟(𝑡) × 𝑟 ′ (𝑡) = 0 for all 𝑡 ∈ 𝑅 Now, let us consider, d (r (t )  r (t )) = r (t )  r (t ) + r (t )  r (t ) dt = 2 × 𝑟(𝑡) × 𝑟 ′ (𝑡) (given 𝑟(𝑡) × 𝑟′(𝑡) = 0 ) =0 2 This implies that 2 ⇒ ||𝑟(𝑡)|| || r(t)||=K 𝑑(||𝑟(𝑡)|| ) 𝑑𝑡 =0 is constant and therefore, || r(t)|| should also be a constant. Let K be the constant such that This implies that, || r (t ) || = K 2 2 Let us say x, y, z are the components of vector function r(t), this implies that, we have, 𝑥 2 + 𝑦2 + 𝑧2 = 𝐾 2 which is a sphere centered at (0, 0, 0) and with radius K Thus, we have 𝑟(𝑡) =≪ 𝑥(𝑡), 𝑦(𝑡), 𝑧(𝑡) ≫ such that, ||𝑟(𝑡)|| = √𝑥 2 + 𝑦 2 + 𝑧 2 2 ⇒ ||𝑟(𝑡)|| = 𝑥 2 + 𝑦 2 + 𝑧 2 Hence, the curve r(t) must be a sphere centered at origin. Question #2 Find two unit vectors orthogonal to both j - k and i + j. Answer: Step 1 Consider the vectors, j-k and i+j The objective is to find two vectors orthogonal to both the vectors. From the vectors, Let, a=j-k =≪ 0, 1, − 1 ≫ b=i+j =≪ 1, 1, 0 ≫ Definition: Let a and b two vectors, the cross product (a\times b) of a and b is orthogonal to both the vector a and b Find the cross product: i j k a  b =| 0 1 −1| 1 1 0 =(0-(-1)0i-(0-(-1))j+(0-1)k =i-j-k The orthogonal vector for the given two vectors is i-j-k And ||𝑎 × 𝑏|| = √(1)2 + (−1)2 + (−1)2 = √1 + 1 + 1 = √3 Step 2 Find two vectors orthogonal to both the vectors as follows: The desired unit vector, then can be obtained by, 𝑎×𝑏 1) ± ( ||𝑎×𝑏|| ) Substitute 𝑎 × 𝑏 = 𝑖 − 𝑗 − 𝑘 and ||𝑎 × 𝑏|| = √3 in equation (1) ±( 𝑎×𝑏 ) = ±( ||𝑎×𝑏|| 1 = ±( = +( = 1 √3 =≪ √3 1 √3 𝑖− 1 √3 𝑖− 𝑖− 1 1 √3 1 √3 𝑗− √3 , − 1 √3 𝑖−𝑗−𝑘 𝑗− 𝑗− 1 √3 √3 1 √3 1 √3 ) 𝑘) 𝑘) or − ( 𝑘 or − , − 1 √3 1 √3 𝑖+ 1 √3 1 √3 1 ≫ or ≪ − 1 𝑖− 𝑗+ √3 , √3 1 𝑗− , 𝑘) 𝑘 √3 1 1 √3 1 √3 √3 ≫ Thus, the required vectors orthogonal to both the vectors are, 1 1 1 1 1 1 =≪ , − , − ≫ 𝑜𝑟 ≪ − , , ≫ √3 √3 √3 √3 √3 √3 Question #3 Find parametric equations for the path a particle that moves along the circle 𝑥 2 + (𝑦 − 1)2 = 4 Find parametric equations for the path a particle that moves along the circle 𝑥 2 + (𝑦 − 1)2 = 4. In the manner describe a) One around clockwise starting at (2,1) b) Three times around counterclockwise starting at (2,1) c) halfway around counterclockwise starting at (0,3) Answer: HINT : Compare 𝑥 ( )+( 2 𝑦−1 2 2 2 ) =1 with 𝑐𝑜𝑠 𝑡 + sin2 𝑡 = 1. EDIT: Drawing the circle will help. In your case, since one around clockwise, y-coordinate of the point has to be 𝜋 𝜋 decreasing first. This is the reason why the sin t is with minus. Setting 𝑡 = 0, , , ... will also help. 6 4 Question #4 Find the area of the largest rectangle that can be inscribed in the ellipse 𝑥2 𝑎2 + 𝑦2 𝑏2 =1 Answer: Question #5 Find the area of the region that lies inside both curves. 𝑟 = √3 𝑐𝑜𝑠 𝜃, 𝑟 = 𝑠𝑖𝑛 𝜃 Answer: Question #6 Sketch the vector field F by drawing a diagram. 𝐹(𝑥, 𝑦) = Answer: Question #7 Find a Cartesian equation for the curve and identify it. 𝑟 = 7 cos(𝜃) 𝑦𝑖+𝑥𝑗 1 (𝑥 2 +𝑦 2 )2 Answer: Question #8 Describe in words the surface whose equation is given. 𝜙 = 𝜋 4 (select the correct answer) 1) the top half of the right circular cone with vertex at the origin and axis the positive z-axis 2) the plane perpendicular to the xz-plane passing through z = x, where 𝑥 ≥ 0 3) the plane perpendicular to the xy-plane passing through y = x, where 𝑥 ≥ 0 4) the base of the right circular cone with vertex at the origin and axis the positive z-axis 5) the plane perpendicular to the yz-plane passing through z = y, where 𝑦 ≥ 0 Answer: Question #9 Consider an object moving along the parametrized curve with equations: 𝑥(𝑡) = 𝑒 𝑡 + 𝑒 −𝑡 𝑦(𝑡) = 𝑒 −𝑡 where t is in the time interval [0,4] seconds. Fill in the bla below. The maximum speed of the object on the time interval is ____ at time ____. The minimum speed of the object on the time interval is ____ at time ____. Answer: Question #10 Use polar coordinates to find the volume of the given solid. Inside the sphere 𝑥 2 + 𝑦 2 + 𝑧 2 = 16 and outside the cylinder 𝑥 2 + 𝑦 2 = 1 Answer: To find the volume of the region that is inside the sphere and outside the cylinder is same as finding the volume of the whole sphere and then subtract the volume of the region that is inside the sphere and the cylinder. Let's find the volume of the whole sphere first, but we are going to use a non-calculus formula to do so. 4 Recall that the volume of the sphere is 𝑉 = 𝜋𝑟 3 and here r=4, so the volume of the sphere is 𝑉 = 4 3 3 𝜋(4) = 256 3 3 𝜋. Now let's find the volume of the region that is inside the sphere and the cylinder using 𝑉 = ∫ ∫𝐷 𝑧1 − 𝑧2 𝑑𝐴 where 𝑧1 is the upper z-function and 𝑧2 is the lower z-function (in terms of x and y) over some region of integration D So we have to solve z in terms of x and y 𝑥 2 + 𝑦 2 + 𝑧 2 = 16 ⇒= 16 − 𝑥 2 − 𝑦 2 ⇒ 𝑧 = ±√16 − 𝑥 2 − 𝑦 2 Therefore, the upper z-function is vz_1=\sqrt{16-x^2-y^2} and the lower z-function is 𝑧2 = −√16 − 𝑥 2 − 𝑦 2 As for dA ,we can use either dxdy or dydx since the region of integration D is the circular base of the cylinder 𝑥 2 + 𝑦 2 = 4 on the xy-plane. Let’s use the order dydx. So we have to solve y in terms of x. 𝑥 2 + 𝑦 2 = 4 ⇒ 𝑦 2 = 4 − 𝑥 2 ⇒ 𝑦 = ±√4 − 𝑥 2 Therefore, the lower limit for y is −√4 − 𝑥 2 and the upper limit for y is √4 − 𝑥 2 Lastly the limits for x are -2 as the lower limit and 2 as the upper limit. Therefore, the volume of the region S that is both inside the sphere and the cylinder is: 2 √4−𝑥 2 𝑉 = ∫ ∫𝐷 𝑧1 − 𝑧2 𝑑𝐴 = ∫−2 ∫−√4−𝑥 2 (√16 − 𝑥 2 − 𝑦 2 − (−√16 − 𝑥 2 − 𝑦 2 )) 𝑑𝑦𝑑𝑥 However, this triple integral is not easy to do except for the innermost one. After that and for the outer two integrals, we have to use polar coordinates to make our integration easier. Calculation is in the next cell. √4−𝑥 2 2 𝑉 = ∫−2 ∫−√4−𝑥 2 (√16 − 𝑥 2 − 𝑦 2 − (−√16 − 𝑥 2 − 𝑦 2 )) 𝑑𝑦𝑑𝑥 2 √4−𝑥 2 2𝜋 2 = ∫−2 ∫−√4−𝑥 2 2√16 − 𝑥 2 − 𝑦 2 𝑑𝑦𝑑𝑥 = ∫0 ∫0 2√16 − 𝑟 2 ⋅ 𝑟𝑑𝑟𝑑𝜃(1) 2 2𝜋 = ∫0 2𝑟√16 − 𝑟 2 𝑑𝑡 ⋅ ∫0 𝑑𝜃 (2) 2 2 = [− (16 − 𝑥 2 )3/2 ] ⋅ [𝜃]2𝜋 0 (3) =( = 3 128 3 256 3 0 − 16√3) (2𝜋) 𝜋 − 32𝜋√3 Notes (1) Polar-coordinate substitution: Recall that 𝑥 2 + 𝑦 2 = 𝑟 2 , so 16 − 𝑥 2 − 𝑦 2 = 16 − 𝑟 2 . The extra factor of r is the Jacobian, the limits of r are 0 to 2 an those for \theta are 0 to 2𝜋 when the region is entire circular disk with radius 2. 𝑑 𝑏 𝑏 𝑑 (2) We are using the formula ∫𝑐 ∫𝑎 𝑓(𝑟)𝑔(𝜃)𝑑𝑟𝑑𝜃 = ∫𝑎 𝑓(𝑟)𝑑𝑟 ⋅ ∫𝑐 𝑔(𝜃)𝑑 𝜃 where a,b,c,d are constants. Here, we can assume 𝑔(𝜃) = 1 (3) The integral with respect to r can be obtained by using a u-substitution by letting 𝑢 = 16 − 𝑟 2 . Here, we limit the integration process. Notice that the volume in the above cell is the volume of the region that is inside the sphere and the cylinder. We need to subtract this from the volume of the whole sphere to get the volume of the region that is inside the sphere and outside the cylinder. Therefore, the volume sought here is 𝑉= 256 3 𝜋−( 256 3 𝜋 − 32𝜋√3) = 32𝜋√3