Parametric equations, polar coordinates, and vector-valued functions #1 - Questions and Answers

Question 1 Find the area of the parallelogram with vertices 𝐴 ( − 3 , 0 ) A(−3,0), 𝐵 ( − 1 , 6 ) B(−1,6), 𝐶 ( 8 , 5 ) C(8,5), and 𝐷 ( 6 , − 1 ) D(6,−1). Answer 1 Vertices are 𝐴 ( − 3 , 0 ) A(−3,0), 𝐵 ( − 1 , 6 ) B(−1,6), 𝐶 ( 8 , 5 ) C(8,5), and 𝐷 ( 6 , − 1 ) D(6,−1). So, 𝐴 𝐵 → = ⟨ − 1 − ( − 3 ) , 6 − 0 ⟩ = ⟨ 2 , 6 ⟩ , 𝐴 𝐷 → = ⟨ 6 − ( − 3 ) , − 1 − 0 ⟩ = ⟨ 9 , − 1 ⟩ . AB =⟨−1−(−3),6−0⟩=⟨2,6⟩, AD =⟨6−(−3),−1−0⟩=⟨9,−1⟩. Hence, area of parallelogram: Area = ∣ 𝐴 𝐵 → × 𝐴 𝐷 → ∣ . Area=∣ AB × AD ∣. Using the determinant: 𝐴 𝐵 → × 𝐴 𝐷 → = ∣ 𝑖 𝑗 𝑘 2 6 0 9 − 1 0 ∣ = 𝑖 ( 0 ) − 𝑗 ( 0 ) + 𝑘 ( − 2 ⋅ − 1 − 6 ⋅ 9 ) = − 56 𝑘 . AB × AD = ​ i 2 9 ​ j 6 −1 ​ k 0 0 ​ ​ =i(0)−j(0)+k(−2⋅−1−6⋅9)=−56k. Area = ( − 56 ) 2 = 56. Area= (−56) 2 ​ =56. Question 2 Evaluate the following integral in cylindrical coordinates (triple integral): ∫ 𝑥 = − 1 𝑥 = 1 ∫ 𝑦 = 0 1 − 𝑥 2 ∫ 𝑧 = 0 2 1 1 + 𝑥 2 + 𝑦 2   𝑑 𝑧   𝑑 𝑦   𝑑 𝑥 . ∫ x=−1 x=1 ​ ∫ y=0 1−x 2 ​ ​ ∫ z=0 2 ​ 1+x 2 +y 2 1 ​ dzdydx. Answer: Convert to cylindrical coordinates: ∫ ∫ ∫ 𝑓 ( 𝑥 , 𝑦 , 𝑧 )   𝑑 𝑥   𝑑 𝑦   𝑑 𝑧 = ∫ ∫ ∫ 𝑓 ( 𝑟 cos ⁡ 𝜃 , 𝑟 sin ⁡ 𝜃 , 𝑧 )   𝑟   𝑑 𝑧   𝑑 𝑟   𝑑 𝜃 . ∫∫∫f(x,y,z)dxdydz=∫∫∫f(rcosθ,rsinθ,z)rdzdrdθ. The region is: 𝐸 = { ( 𝑟 , 𝜃 , 𝑧 )   ∣   0 ≤ 𝜃 ≤ 2 𝜋 ,   0 ≤ 𝑟 ≤ 1 ,   0 ≤ 𝑧 ≤ 2 } . E={(r,θ,z)∣0≤θ≤2π,0≤r≤1,0≤z≤2}. ∫ 𝑥 = − 1 𝑥 = 1 ∫ 𝑦 = 0 1 − 𝑥 2 ∫ 𝑧 = 0 2 1 1 + 𝑥 2 + 𝑦 2   𝑑 𝑧   𝑑 𝑦   𝑑 𝑥 = ∫ 0 2 𝜋 ∫ 0 1 ∫ 0 2 𝑟 1 + 𝑟 2   𝑑 𝑧   𝑑 𝑟   𝑑 𝜃 . ∫ x=−1 x=1 ​ ∫ y=0 1−x 2 ​ ​ ∫ z=0 2 ​ 1+x 2 +y 2 1 ​ dzdydx=∫ 0 2π ​ ∫ 0 1 ​ ∫ 0 2 ​ 1+r 2 r ​ dzdrdθ. Integrate: ∫ 0 2 𝜋 𝑑 𝜃 ∫ 0 1 ∫ 0 2 𝑟 1 + 𝑟 2 𝑑 𝑧   𝑑 𝑟 = 2 𝜋 ∫ 0 1 𝑟 1 + 𝑟 2 ⋅ 2   𝑑 𝑟 = 4 𝜋 ∫ 0 1 𝑟 1 + 𝑟 2   𝑑 𝑟 . ∫ 0 2π ​ dθ∫ 0 1 ​ ∫ 0 2 ​ 1+r 2 r ​ dzdr=2π∫ 0 1 ​ 1+r 2 r ​ ⋅2dr=4π∫ 0 1 ​ 1+r 2 r ​ dr. Substitute 𝑢 = 1 + 𝑟 2 ,   𝑑 𝑢 = 2 𝑟   𝑑 𝑟 u=1+r 2 ,du=2rdr: 4 𝜋 ∫ 0 1 𝑟 1 + 𝑟 2   𝑑 𝑟 = 4 𝜋 ∫ 1 2 1 𝑢 ⋅ 𝑑 𝑢 2 = 2 𝜋 ln ⁡ 𝑢 ∣ 1 2 = 2 𝜋 ln ⁡ 2. 4π∫ 0 1 ​ 1+r 2 r ​ dr=4π∫ 1 2 ​ u 1 ​ ⋅ 2 du ​ =2πlnu ​ 1 2 ​ =2πln2. Question 3 Find an equation of the plane through the points ( 4 , 1 , 4 ) (4,1,4), ( 5 , − 8 , 6 ) (5,−8,6), and ( − 4 , − 5 , 1 ) (−4,−5,1). Answer 1 Step 1: Find two vectors on the plane: 𝑃 1 𝑃 2 → = ( 5 − 4 , − 8 − 1 , 6 − 4 ) = ( 1 , − 9 , 2 ) , 𝑃 1 𝑃 3 → = ( − 4 − 4 , − 5 − 1 , 1 − 4 ) = ( − 8 , − 6 , − 3 ) . P 1 ​ P 2 ​ ​ =(5−4,−8−1,6−4)=(1,−9,2), P 1 ​ P 3 ​ ​ =(−4−4,−5−1,1−4)=(−8,−6,−3). Step 2: Find the normal vector to the plane: 𝑛 ⃗ = 𝑃 1 𝑃 2 → × 𝑃 1 𝑃 3 → = ∣ 𝑖 𝑗 𝑘 1 − 9 2 − 8 − 6 − 3 ∣ = 𝑖 ( 27 − ( − 12 ) ) − 𝑗 ( − 3 − ( − 16 ) ) + 𝑘 ( − 6 − 72 ) . n = P 1 ​ P 2 ​ ​ × P 1 ​ P 3 ​ ​ = ​ i 1 −8 ​ j −9 −6 ​ k 2 −3 ​ ​ =i(27−(−12))−j(−3−(−16))+k(−6−72). 𝑛 ⃗ = 39 𝑖 − 13 𝑗 − 78 𝑘 . n =39i−13j−78k. Step 3: Write the equation of the plane: 39 ( 𝑥 − 4 ) − 13 ( 𝑦 − 1 ) − 78 ( 𝑧 − 4 ) = 0. 39(x−4)−13(y−1)−78(z−4)=0. Simplify: 39 𝑥 − 13 𝑦 − 78 𝑧 + 255 = 0. 39x−13y−78z+255=0. Question 4 Find the points on the ellipse 4 𝑥 2 + 𝑦 2 = 4 4x 2 +y 2 =4 that are farthest away from the point ( − 1 , 0 ) (−1,0). Answer: Substitute 𝑥 = 1 2 x= 2 1 ​ : The points are ( 7 2 , ± 15 2 ) ( 2 7 ​ ​ ,± 2 15 ​ ​ ). Question 5 Two components of a minicomputer have the following joint PDF for their useful lifetimes 𝑋 X and 𝑌 Y: 𝑓 ( 𝑥 , 𝑦 ) = 𝑥 𝑒 − 𝑥 ( 1 + 𝑦 ) , 𝑥 ≥ 0 , 𝑦 ≥ 0 , 0  otherwise. f(x,y)=xe −x(1+y) ,x≥0,y≥0,0 otherwise. (a) What is the probability that the lifetime 𝑋 X of the first component exceeds 3? (b) What are the marginal PDFs? Answer: (a) Find 𝑃 ( 𝑋 > 3 ) P(X>3): 𝑃 ( 𝑋 > 3 ) = ∫ 3 ∞ ∫ 0 ∞ 𝑥 𝑒 − 𝑥 ( 1 + 𝑦 )   𝑑 𝑦   𝑑 𝑥 = ∫ 3 ∞ 𝑥 𝑒 − 𝑥 ∫ 0 ∞ 𝑒 − 𝑥 𝑦   𝑑 𝑦   𝑑 𝑥 . P(X>3)=∫ 3 ∞ ​ ∫ 0 ∞ ​ xe −x(1+y) dydx=∫ 3 ∞ ​ xe −x ∫ 0 ∞ ​ e −xy dydx. Evaluate inner integral: ∫ 0 ∞ 𝑒 − 𝑥 𝑦   𝑑 𝑦 = 1 𝑥 . ∫ 0 ∞ ​ e −xy dy= x 1 ​ . So, 𝑃 ( 𝑋 > 3 ) = ∫ 3 ∞ 𝑒 − 𝑥   𝑑 𝑥 = 𝑒 − 3 . P(X>3)=∫ 3 ∞ ​ e −x dx=e −3 .