Question 1 Find the point(s) of intersection of the following two parametric curves by first eliminating the parameter, then solving the system of equations: 𝑥 = 𝑡 + 5 , 𝑦 = 𝑡 2 and 𝑥 = 1 16 𝑡 , 𝑦 = 𝑡 x=t+5,y=t 2 andx= 16 1 t,y=t Answer 1 From the first equation, 𝑡 = 𝑥 − 5 t=x−5. Substituting into 𝑦 = 𝑡 2 y=t 2 : 𝑦 = ( 𝑥 − 5 ) 2 y=(x−5) 2 For the second curve, 𝑡 = 16 𝑥 t=16x. Substituting into 𝑦 = 𝑡 y=t: 𝑦 = 16 𝑥 y=16x Equating the 𝑦 y-values: 16 𝑥 = ( 𝑥 − 5 ) 2 16x=(x−5) 2 16 𝑥 = 𝑥 2 − 10 𝑥 + 25 16x=x 2 −10x+25 𝑥 2 − 26 𝑥 + 25 = 0 x 2 −26x+25=0 ( 𝑥 − 1 ) ( 𝑥 − 25 ) = 0 (x−1)(x−25)=0 𝑥 = 1 or 𝑥 = 25 x=1orx=25 For 𝑥 = 1 x=1: 𝑦 = 16 𝑥 = 16 ( 1 ) = 16 y=16x=16(1)=16 For 𝑥 = 25 x=25: 𝑦 = 16 𝑥 = 16 ( 25 ) = 400 y=16x=16(25)=400 Points of intersection: ( 1 , 16 ) and ( 25 , 400 ) (1,16)and(25,400) Question 2 Sketch the vector field 𝐹 ( 𝑥 , 𝑦 ) = 𝑦 𝑖 + 𝑥 𝑗 𝑥 2 + 𝑦 2 F(x,y)= x 2 +y 2 yi+xj . Answer 1 The vector field 𝐹 ( 𝑥 , 𝑦 ) F(x,y) describes unit vectors pointing radially outward. At ( 1 , 0 ) (1,0): 𝐹 ( 1 , 0 ) = ⟨ 0 , 1 ⟩ F(1,0)=⟨0,1⟩. At ( 0 , 1 ) (0,1): 𝐹 ( 0 , 1 ) = ⟨ 1 , 0 ⟩ F(0,1)=⟨1,0⟩. At ( − 1 , 0 ) (−1,0): 𝐹 ( − 1 , 0 ) = ⟨ 0 , − 1 ⟩ F(−1,0)=⟨0,−1⟩. At ( 0 , − 1 ) (0,−1): 𝐹 ( 0 , − 1 ) = ⟨ − 1 , 0 ⟩ F(0,−1)=⟨−1,0⟩. These vectors form a circular, counterclockwise pattern, representing the radial symmetry of the field. Question 3 Take the graph of 𝑓 f over the interval ( 0 , 7 ) (0,7) and find: a) Intervals where 𝑓 f is increasing. b) Intervals where 𝑓 f is decreasing. c) Intervals where 𝑓 f is concave upward. d) Intervals where 𝑓 f is concave downward. e) Coordinates of inflection points. Answer 1 a) 𝑓 f is increasing: ( 0 , 1 ) ∪ ( 3 , 7 ) (0,1)∪(3,7). b) 𝑓 f is decreasing: ( 1 , 3 ) (1,3). c) 𝑓 f is concave upward: ( 2 , 4 ) ∪ ( 5 , 7 ) (2,4)∪(5,7). d) 𝑓 f is concave downward: ( 0 , 2 ) ∪ ( 4 , 5 ) (0,2)∪(4,5). e) Inflection points: ( 2 , 𝑦 2 ) , ( 4 , 𝑦 4 ) , ( 5 , 𝑦 5 ) (2,y 2 ),(4,y 4 ),(5,y 5 ). Question 4 Convert the polar equation 𝜃 = − 𝜋 6 θ=− 6 π to rectangular form. Answer 1 For 𝜃 = − 𝜋 6 θ=− 6 π : tan ⁡ 𝜃 = 𝑦 𝑥 , tan ⁡ ( − 𝜋 6 ) = − 3 3 tanθ= x y ,tan(− 6 π )=− 3 3 Substitute: 𝑦 𝑥 = − 3 3 x y =− 3 3 𝑦 = − 3 3 𝑥 y=− 3 3 x Rectangular form: 𝑦 = − 3 3 𝑥 y=− 3 3 x Question 5 Find the equation of the tangent plane to the parametric surface: 𝑥 = 𝑢 2 + 1 , 𝑦 = 𝑣 3 + 1 , 𝑧 = 𝑢 + 𝑣 x=u 2 +1,y=v 3 +1,z=u+v at ( 𝑥 , 𝑦 , 𝑧 ) = ( 5 , 2 , 3 ) (x,y,z)=(5,2,3). Answer 1 Step 1: Parametric derivatives: ∂ 𝑥 ∂ 𝑢 = 2 𝑢 , ∂ 𝑥 ∂ 𝑣 = 0 , ∂ 𝑦 ∂ 𝑢 = 0 , ∂ 𝑦 ∂ 𝑣 = 3 𝑣 2 , ∂ 𝑧 ∂ 𝑢 = 1 , ∂ 𝑧 ∂ 𝑣 = 1 ∂u ∂x =2u, ∂v ∂x =0, ∂u ∂y =0, ∂v ∂y =3v 2 , ∂u ∂z =1, ∂v ∂z =1 At 𝑢 = 2 , 𝑣 = 1 u=2,v=1: ∂ 𝑥 ∂ 𝑢 = 4 , ∂ 𝑦 ∂ 𝑣 = 3 , ∂ 𝑧 ∂ 𝑢 = ∂ 𝑧 ∂ 𝑣 = 1 ∂u ∂x =4, ∂v ∂y =3, ∂u ∂z = ∂v ∂z =1 Step 2: Tangent vectors: 𝑟 𝑢 = ⟨ 4 , 0 , 1 ⟩ , 𝑟 𝑣 = ⟨ 0 , 3 , 1 ⟩ r u =⟨4,0,1⟩,r v =⟨0,3,1⟩ Normal vector: 𝑟 𝑢 × 𝑟 𝑣 = ∣ 𝑖 𝑗 𝑘 4 0 1 0 3 1 ∣ = 𝑖 ( 0 − 3 ) − 𝑗 ( 4 − 0 ) + 𝑘 ( 12 − 0 ) = ⟨ − 3 , − 4 , 12 ⟩ r u ×r v = i 4 0 j 0 3 k 1 1 =i(0−3)−j(4−0)+k(12−0)=⟨−3,−4,12⟩ Step 3: Equation of tangent plane: − 3 ( 𝑥 − 5 ) − 4 ( 𝑦 − 2 ) + 12 ( 𝑧 − 3 ) = 0 −3(x−5)−4(y−2)+12(z−3)=0 Simplify: − 3 𝑥 − 4 𝑦 + 12 𝑧 = − 53 −3x−4y+12z=−53 Tangent plane equation: 3 𝑥 + 4 𝑦 − 12 𝑧 + 53 = 0 3x+4y−12z+53=0