Questions and Answers #2 Parametric Equations, Polar Coordinates, and Vector-V

Questions and Answers Sheet 2 Parametric Equations, Polar Coordinates, and Vector-V Question #1 The plane x+y+2z=18 intersects the paraboloid 𝑧 = 𝑥 2 + 𝑦 2 in an ellipse. Find the points on this ellipse that are nearest to and farthest from the origin. Answer: It suffices to find the extreme values of the square of the distance to the origin 𝐷 = 𝑥 2 + 𝑦 2 + 𝑧 2 , subject to the constraints g = x + y + 2z = 18 and ℎ = 𝑥 2 + 𝑦 2 − 𝑧 = 0. 𝜇 By Lagrange Multipliers, ∇𝐷 = 𝜆∇𝑔 + 𝜇∇ℎ. =⇒ < 2𝑥, 2𝑦, 2𝑧 > = 𝜆 < 1, 1, 2 > + 𝜇 < 2𝑥, 2𝑦, −1 >. Equating like entries, 2𝑥 = 𝜆 + 2𝜇𝑥 2𝑦 = 𝜆 + 2𝜇𝑦 2𝑧 = 2𝜆 − 𝜇 Since the last equation can be rewritten as 𝜇 = 2𝜆 − 2𝑧 , we now have 2𝑥 = 𝜆 + 2𝑥(2𝜆 − 2𝑧) =⇒ 2𝑥 = 𝜆(4𝑥 + 1) − 4𝑥𝑧 2𝑦 = 𝜆 + 2𝑦(2𝜆 − 2𝑧) =⇒ 2𝑦 = 𝜆(4𝑦 + 1) − 4𝑦𝑧 So, (2𝑥 + 4𝑥𝑧)/(4𝑥 + 1) = 𝜆 = (2𝑦 + 4𝑦𝑧)/(4𝑦 + 1) =⇒ 2𝑥(1 + 2𝑧)/(4𝑥 + 1) = 2𝑦(1 + 2𝑧)/(4𝑦 + 1). *If 1 + 2𝑧 = 0 ⇔ 𝑧 = −1/2, then ℎ = 𝑥 2 + 𝑦 2 = −1/2 has no solution. So, we can ignore this case. Now, we can freely divide both sides by (1 + 2z) : 2𝑥/(4𝑥 + 1) = 2𝑦/(4𝑦 + 1) =⇒ 2𝑥(4𝑦 + 1) = 2𝑦(4𝑥 + 1) =⇒ 𝑥 = 𝑦. Substitute this into the constraint equations: 𝑥 + 𝑥 + 2𝑧 = 18 =⇒ 𝑥 + 𝑧 = 9 𝑥 2 + 𝑥 2 − 𝑧 = 0 =⇒ 𝑧 = 2𝑥 2 So, 𝑥 + 2𝑥 2 = 9. =⇒ 2𝑥 2 + 𝑥 − 9 = 0 =⇒ 𝑥 =. [−1 + √73]/4 or [−1 − √73]/4 So, (𝑥, 𝑦) = ([−1 + √73]/4, [−1 + √73]/4) or ([−1 − √73]/4, [−1 − √73]/4), since y = x =⇒ (𝑥, 𝑦, 𝑧) = ([−1 + √73]/4, [−1 + √73]/4, [37 − √73]/4) 𝑜𝑟([−1 − √73]/4, [−1 − √73]/4, [37 + √73]/ 4), using either constraint. Finally, substitute this into D: 𝐷([−1 + √73]/4, [−1 + √73]/4, [37 − √73]/4) = [1590 − 78√3]/16 <---Closest 𝐷([−1 − √73]/4, [−1 − √73]/4, [37 + √73]/4) = [1590 + 78√3]/16 <---Farthest. Question #2 The part of the surface 2𝑦 + 4𝑧 − 𝑥 2 = 5 that lies above the triangle with vertices (0, 0), (2, 0), and (2, 4) Find the area of the surface. Answer: 2𝑦 + 4𝑧 − 𝑥 2 = 5 5 𝑥2 4 4 1 𝑓(𝑥, 𝑦) = + 𝑥 − 2𝑦 4 𝑓𝑥 = ; 𝑓𝑥 = − 0 ≤ 𝑥 ≤ 2 2 2 𝑆 = ∫ √1 + 𝑓𝑥 2 + 𝑓𝑦 𝑑𝐴 0 ≤ 𝑦 ≤ 2𝑥 𝑑𝐴 = 𝑑𝑦𝑑𝑥 2 𝑆=∫ ∫ 2𝑥 √1 + 0 𝑆= 𝑆= 𝑥2 1 + 𝑑𝑦𝑑𝑥 4 4 0 2 2𝑥 ∫ ∫ √5 + 𝑥 2 𝑑𝑦𝑑𝑥 2 0 0 1 2 ∫ 2𝑥√5 + 𝑥 2 𝑑𝑥 2 0 2 1 Let 5 + 𝑥 = 𝑢 2𝑥𝑑𝑥 = 𝑑𝑢 1 2 𝑆 = ∫0 √𝑢 𝑑𝑢 2 1 32 𝑆 = (5 + 𝑋 2 )2|0 3 1 𝑆 = (27 − 5√5) 3 Question #3 Find the area of the surface. The part of the paraboloid 𝑧 = 1 − 𝑥 2 − 𝑦2 that lies above the plane 𝑧 = −6 Answer: Question #4 Change from rectangular to cylindrical coordinates. (Let 𝑟 ≥ 0 and 0 ≤ 𝜃 ≤ 2𝜋.) a) (-2, 2, 2) b) (−9,9√3,6) c) Use cylindrical coordinates. Evaluate ∫ ∫ ∫𝐸 𝑥𝑑𝑉 where E is enclosed by the planes z=0 and z=x+y+10 and by the cylinders 𝑥 2 + 𝑦 2 = 16 and 𝑥 2 + 𝑦 2 = 36 d) Use cylindrical coordinates. Find the volume of the solid that is enclosed by the cone 𝑧 = √𝑥 2 + 𝑦 2 and the sphere 𝑥 2 + 𝑦 2 + 𝑧 2 = 8. Answer: Step 1 Change from rectangular to cylindrical coordinates. Let 𝑟 ≥ 0 and 0 ≤ 𝜃 ≤ 2𝜋. a) (x, y, z)=(-2, 2, 2) Use cylindrical coordinates. 𝑟 = √𝑥 2 + 𝑦 2 = √(−2)3 + 23 = 2√2 𝑦 2 3𝜋 𝑥 −2 4 𝜃 = arctan ( ) = arctan ( ) (−1) = 𝑧=𝑧=2 Therefore, the required coordinates is, (𝑟, 𝜃, 𝑧) = (2√2, 2𝜋 4 , 2) Step 2 b) (𝑥, 𝑦, 𝑧) = (−9.9√3.6) Use cylindrical coordinates. 2 𝑟 = √𝑥 2 + 𝑦 2 = √(−9)2 + (9√3) = 18 𝑦 9√3 𝑥 −9 𝜃 = arctan ( ) = arctan ( ) = arctan(−√3) = 2𝜋 3 𝑧=𝑧=6 Therefore, the required coordinates is, (𝑟, 𝜃, 𝑧) = (18, 2𝜋 3 , 6) Step 3 c) Use cylindrical coordinates, to evaluate ∫ ∫𝐸 ∫ 𝑥𝑑𝑉 Where E is enclosed by the planes z=0 and z=x+y+10 and by the cylinders 𝑥 2 + 𝑦 2 = 16 and 𝑥 2 + 𝑦 2 = 36 16 ≤ 𝑥 2 + 𝑦 2 ≤ 36 ⇒ 16 ≤ 𝑟 2 ≤ 36 As 𝑥 2 + 𝑦 2 = 𝑟 2 ⇒4≤𝑟≤6 and 0 ≤ 𝜃 ≤ 2𝜋. and 0 ≤ 𝑧 ≤ 𝑥 + 𝑦 + 10 ⇒ 0 ≤ 𝑧 ≤ 𝑟 cos 𝜃 + 𝑟 sin 𝜃 + 10 As 𝑥 = 𝑟 cos 𝜃 , 𝑦 = 𝑟 sin 𝜃. and 𝑑𝑉 = 𝑑𝑥𝑑𝑦𝑑𝑧 = 𝑟𝑑𝑟𝑑𝜃𝑑𝑧 2𝑎 6 𝑟 cos 𝜃+𝑟 sin 𝜃+10 So, ∫ ∫𝐸 ∫ 𝑥𝑑𝑉 = ∫0 ∫4 ∫0 𝑟 cos 𝜃 (𝑟𝑑𝑟𝑑𝜃𝑑𝑧) 2𝜋 =∫ 6 𝑟 cos 𝜃+𝑟 sin 𝜃+10 𝑟 2 cos 𝜃 𝑑𝑟𝑑𝜃𝑑𝑧 ∫ ∫ 0 4 0 2𝜋 6 2 (𝑟 =∫ ∫ 𝑟 cos 𝜃 + 𝑟 sin 𝜃 + 10) cos 𝜃 𝑑𝑟𝑑𝜃. 0 4 2𝜋 6 ∫ (𝑟 3 (cos 2 𝜃 + sin 𝜃 cos 𝜃) + 10𝑟 2 cos 𝜃)𝑑𝑟𝑑𝜃 . =∫ 0 4 2𝜋 6 1 10 3 ( 𝑟 4 (cos 2 𝜃 + sin 𝜃 cos 𝜃) + 𝑟 cos 𝜃) 𝑑𝜃. 4 3 4 0 2𝜋 1 4 10 (63 − 43 ) cos 𝜃) 𝑑𝜃 . = ∫ ( (6 − 44 )(cos 2 𝜃 + sin 𝜃 cos 𝜃) + 4 3 0 2𝜋 1520 = ∫ (260(cos 2 𝜃 + sin 𝜃 cos 𝜃) + cos 𝜃) 𝑑𝜃. 3 0 = 260𝜋. =∫ Step 4 d) Use cylindrical coordinates. Find the volume of the solid that is enclosed by the cone 𝑧 = √𝑥 2 + 𝑦 2 and the sphere 𝑥 2 + 𝑦 2 + 𝑧 2 = 8 Use cylindrical polar coordinates, 𝑥 = 𝑟 cos 𝜃 , 𝑦 = 𝑟 sin 𝜃 , 𝑧 = 𝑧 ⇒ 𝑥 2 + 𝑦 2 = 𝑟 2 , 𝑑𝑉 = 𝑑𝑥𝑑𝑦𝑑𝑧 = 𝑟𝑑𝑟𝑑𝜃𝑑𝑧 Limits: As 𝑧 = √𝑥 2 + 𝑦 2 and 𝑥 2 + 𝑦 2 + 𝑧 2 = 8 ⇒ 2(𝑥 2 + 𝑦 2 ) = 8 ⇒ 𝑥 2 + 𝑦2 = 4 𝑟2 = 4 ⇒𝑟=2 So, 𝐸 = {(𝑟, 𝜃, 𝑧): 0 ≤ 𝑟 ≤ 2, 0 ≤ 𝜃 ≤ 2𝜋, 𝑟 ≤ 𝑧 ≤ √8 − 𝑟 2 } Volume = ∫ ∫𝐸 ∫ 𝑑𝑥𝑑𝑦𝑑𝑧 2 2𝜋 √8−𝑟 2 = ∫𝑟=0 ∫𝜃=0 ∫𝑧=𝑟 2 =∫ 𝑟𝑑𝑟𝑑𝜃𝑑𝑧 2𝜋 ∫ 𝑟 (√8 − 𝑟 2 − 𝑟) 𝑑𝑟𝑑𝜃 . 𝑟=0 𝜃=0 2 2𝜋 (𝑟√8 − 𝑟 2 − 𝑟 2 ) 𝑑𝑟 ) (∫ = (∫ 𝑟=0 𝑑𝜃) 𝜃=0 2 1 = [ (−(8 − 𝑟 2 )3/2 − 𝑟 3 ]0 [𝜃]2𝜋 0 3 1 = ( (−(8 − 4)3/2 − 8 + (8)3/2 )) (2𝜋) 3 16 = ((√2 − 1) ) (2𝜋) 3 = 32(√2−1)𝜋 3 Question #5 Find a polar equation for the curve represented by the given Cartesian equation. 𝑥 2 + 𝑦 2 + 2𝑐𝑥 Answer: 𝑥 2 + 𝑦 2 = 2𝑐𝑥 Recall that when converting from cartesian to polar, we substitute: 𝑥 = 𝑟 cos 0 And 𝑦 = 𝑟 sin 0 (𝑟 cos 0)2 + (𝑟 sin 0)2 = 2𝑐(𝑟 cos 0) 𝑟 2 cos 2 0 + 𝑟 2 sin2 0 = 2𝑐𝑟 cos 0 𝑟 2 (cos 2 0 + sin2 0) = 2𝑐𝑟 cos 0 Recall that: sin2 0 + cos 2 0 = 1 𝑟 2 = 2𝑐𝑟 cos 0 Divide both sides by r 𝑟 = 2𝑐 ∗ cos 0 Result: 𝑟 = 2𝑐 ∗ cos 0 Question #6 Find the length of the curve. 1 𝑟(𝑡) =≪ 6𝑡, 𝑡 2 , 𝑡 3 ≫ ,0 ≤ 𝑡 ≤ 1 9 Answer: 𝑟(𝑡) =≪ 6𝑡, 𝑡 2 , 𝑡3 𝑡2 ≫⇒ 𝑟 ′ (𝑡) =≪ 6,2𝑡, ≫ 9 3 |𝑟 ′ (𝑡)| = √36 + 4𝑡 2 + 𝑡4 9 1 = √𝑡 4 + 36𝑡 2 + 324 3 1 𝑡 2 +18 3 3 = √(𝑡 2 + 18)2 + 1 11 Length of the curve: 𝐿 = ∫0 |𝑟̅′ (𝑡)|𝑑𝑡 = ∫0 (𝑡 2 + 18)𝑑𝑡 3 1 𝑡2 = ∫0 ( + 𝑏) 𝑑𝑡 3 𝑡3 1 9 0 = [ + 𝑏𝑡 ] 1 1 55 9 9 9 = +𝑏−0 = +6 = Question #7 Find a unit vector that is orthogonal to both i+j and i+k, Answer: Let a=i+j, b=i+k The cross product 𝑎 × 𝑏 is arthogonal to both a and b i j k a  b =|1 1 0 | 1 0 1 = 𝑖(1 − 0) − 𝑗(1 − 0) + 𝑘(0 − 1) 𝑎×𝑏 =𝑖−𝑗−𝑘 Since the vector we found had magnitude √3 i.e. |𝑎 × 𝑏| = √1̌ + 1̌ + 1̌ = √3 A unit vector that is arthogonal to both i+j and i+k is 𝑢= (𝑖−𝑗−𝑘) √3 or 𝑢= √3(𝑖−𝑗−𝑘) 3 Question #8 1 √1−𝑥 2 Evaluate the following integral in cylindrical coordinates triple 𝑖𝑛𝑡𝑒𝑔𝑟𝑎𝑙 ∫𝑥=−1 ∫𝑦=0 2 1 ∫𝑧=0 (1+𝑥 2+𝑦2) 𝑑𝑧𝑑𝑥𝑑𝑦 Answer: Step 1 Evaluate integral by changing into cylindrical coordinates. The formula for triple integration in cylindrical coordinates is ℎ2 (𝜃) 𝛽 ∫ ∫ ∫ 𝑓(𝑥, 𝑦, 𝑧)𝑑𝑉 = ∫ ∫ 𝐸 ℎ1 (𝜃) 𝛼 𝑢2 (𝑟 cos 𝜃,𝑟 sin 𝜃) ∫ 𝑓(𝑟 cos 𝜃 , 𝑟 sin 𝜃 , 𝑧)𝑟𝑑𝑧𝑑𝑟𝑑𝜃 . 𝑢1 (𝑟 cos 𝜃,𝑟 sin 𝜃) Step 2 Here, the region is 𝐸 = {(𝑟, 𝜃, 𝑧)|0 ≤ 𝜃 ≤ 2𝜋, 0 ≤ 𝑟 ≤ 1,0 ≤ 𝑧 ≤ 2} √1−𝑥 2 1 ∫𝑥=−1 ∫𝑦=0 2𝜋 2 1 ∫𝑧=0 (1+𝑥 2+𝑦 2 ) 𝑑𝑧𝑑𝑥𝑑𝑦 1 2 1 ( ) 𝑟𝑑𝑧𝑑𝑟𝑑𝜃 . 2 𝜃=0 𝑟=0 𝑧=0 1 + 𝑟 2𝜋 1 𝑟 [𝑧]20 𝑑𝑟𝑑𝜃. =∫ ∫ 2 𝜃=0 𝑟=0 1 + 𝑟 2𝜋 1 2𝑟 =∫ ∫ 𝑑𝑟𝑑𝜃 . 2 𝜃=0 𝑟=0 1 + 𝑟 =∫ ∫ ∫ 2𝜋 [ln(1 + 𝑟 2 )]10 𝑑𝜃. =∫ 𝜃=0 2𝜋 = ln(2) ∫ 𝑑𝜃 . 𝜃=0 = 2𝜋 ln(2) 1 √1−𝑥 2 Thus, ∫𝑥=−1 ∫𝑦=0 2 1 ∫𝑧=0 (1+𝑥 2+𝑦2) 𝑑𝑧𝑑𝑥𝑑𝑦 = 2𝜋 ln(2) Question #9 Two components of a minicomputer have the following joint pdf for their useful lifetimes X and Y:  xe − x (1+ y ) f ( x, y ) =  0 x  0 and  0 otherwise a) What is the probability that the lifetime X of the first component exceeds 3? b) What are the marginal pdf's of X and Y? Are the two lifetimes independent? Explain. c) What is the probability that the lifetime of at least one component exceeds 3? Answer: Step 1 Consider two random variables X and Y which indicates the lifetime of the two components of the minicomputers. The joint probability density function is provided as:  xe − x (1+ y ) f ( x, y ) =  0 x  0 and  0 otherwise The required probability does not depend on Y values. Hence consider all the possible values of Y and the values of the X ranges from 3 to ∞ as the first component exceeds 3. So, the required probability is obtained as: ∞ ∞ 𝑃(𝑥 > 3) = ∫3 ∫0 𝑥𝑒 −𝑥(1+𝑦) 𝑑𝑦𝑑𝑥 ∞ = ∫3 ∞ ∞ ([−𝑒 −𝑥(1+𝑦) ]0 ) 𝑑𝑥 = ∫3 𝑒 𝑥 𝑑𝑥 =[ 𝑒 −𝑥 ∞ ] −1 3 = 0.049787 ≈ 0.05 The probability indicates that there is 5% chance that the lifetime X of the first component exceeds 3. Step 2 Substitute the provided probability density function and integrates with respect to y to obtain the marginal probability density function of X. Hence, the marginal distribution of X can be obtained as: ∞ 𝑓𝑥 (𝑥) = ∫∞ 𝑓(𝑥, 𝑦)𝑑𝑦 for −∞ < 𝑥 < ∞ ∞ = ∫0 𝑥𝑒 −𝑥(1+𝑦) 𝑑𝑦 ∞ = [−𝑒 −𝑥(1+𝑦) ]0 Substitute the given probability density function and integrates with respect to x to obtain the marginal probability density function of Y. Hence, the marginal distribution of Y can be obtained as: ∞ 𝑓𝑦 (𝑦) = ∫0 𝑥𝑒 −𝑥(1+𝑦) 𝑑𝑥 =[ =[ 𝑥𝑒 −𝑥(1+𝑦) −(1+𝑦) ∞ ∞ 𝑒 −𝑥(1+𝑦) ] − ∫0 −(1+𝑦) 0 ∞ −𝑦𝑥−𝑧 ((𝑦−1)𝑥+1)𝑒 𝑦 2 +2𝑦−1 1 𝑑𝑥 ] 0 = (1+𝑦)2 The marginal probability distribution of the random variable X and Y indicates the individual probability of occurrence of the random variable without considering the occurrence of the other variable. Substitute the obtained marginal probability density functions and the given joint probability density function in the condition for independence to find whether the two lifetimes are independent. 𝑓(𝑥, 𝑦) = 𝑓𝑥 (𝑥) × 𝑓𝑦 (𝑦) 1 𝑥𝑒 −𝑥(1+𝑦) ≠ (𝑒 −𝑥 ) ((1+𝑦)2 ) Since the product of marginal probability density function is not same as the given joint probability density function, the two lifetimes are not independent. As the joint probability distribution of the random variables does not re[presented in the form of the product of two marginal distribution so the random variable are not independent of each other that is the occurrence of one event effects the occurrence of other. Step 3 The probability that the lifetime of at least one component exceeds 3 is given by: 𝑃(𝑋 > 3 𝑜𝑟 𝑌 > 3) = 1 − 𝑃(𝑋, 𝑌 ≤ 3) 3 3 = 1 − ∫ ∫ 𝑥𝑒 −𝑥(1+𝑦) 𝑑𝑦𝑑𝑥 0 0 3 3 = 1 − ∫ ([−𝑒 −𝑥(1+𝑦) ]0 ) 𝑑𝑥 3 0 = 1 ∫0 (𝑒 −4𝑥 (𝑒 3𝑥 − 1))𝑑𝑥 The probability can be simplified as: 𝑃(𝑋 > 3 𝑜𝑟 𝑌 > 3) = 1 − [ 𝑒 −4𝑥 4 − 𝑒 −𝑥 ] 3 0 = 1 − 0.700 = 0.3000 The probability indicates that there is 30% chance that the lifetime of at least one of the components will exceed 3. Question #10 Determine whether the given vectors are orthogonal, parallel, or neither: 𝑢 = (−3, 9, 6) 𝑣 = (4, − 12, − 8) Answer: We first check the dot product of ''u'' and ''v'' to determine if they are orthogonal ( dot product will be 0 ): 𝑢 × 𝑣 = (−3) × (4) ÷ (9) × (−12) ÷ (6) × (−8) = −168 ≠ 0 so that these vectors are NOT orthogonal. We next determine if ''u'' and ''v'' are linearly dependent: (−3) (4) (9) =− (−12) (6) (−8) 3 4 =− =− −3 4 3 4 Hence, these vectors are linearly dependent and are thus parallel: Vectors ''u'' and ''v'' Are PARALLEL