Question 1 Find the lengths of the sides of the triangle 𝑃 𝑄 𝑅 PQR. Is it a right triangle? Is it an isosceles triangle? 𝑃 ( 2 , − 1 , 0 ) , 𝑄 ( 4 , 1 , 1 ) , 𝑅 ( 4 , − 5 , 4 ) P(2,−1,0),Q(4,1,1),R(4,−5,4) Answer 1 Step 1: Use the distance formula: 𝑑 ( 𝑃 1 , 𝑃 2 ) = ( 𝑥 2 − 𝑥 1 ) 2 + ( 𝑦 2 − 𝑦 1 ) 2 + ( 𝑧 2 − 𝑧 1 ) 2 d(P 1 ,P 2 )= (x 2 −x 1 ) 2 +(y 2 −y 1 ) 2 +(z 2 −z 1 ) 2 Lengths of the sides: 𝑃 𝑄 PQ: 𝑃 𝑄 = ( 4 − 2 ) 2 + ( 1 − ( − 1 ) ) 2 + ( 1 − 0 ) 2 = 2 2 + 2 2 + 1 2 = 4 + 4 + 1 = 9 = 3 PQ= (4−2) 2 +(1−(−1)) 2 +(1−0) 2 = 2 2 +2 2 +1 2 = 4+4+1 = 9 =3 𝑄 𝑅 QR: 𝑄 𝑅 = ( 4 − 4 ) 2 + ( − 5 − 1 ) 2 + ( 4 − 1 ) 2 = 0 2 + ( − 6 ) 2 + 3 2 = 0 + 36 + 9 = 45 = 3 5 QR= (4−4) 2 +(−5−1) 2 +(4−1) 2 = 0 2 +(−6) 2 +3 2 = 0+36+9 = 45 =3 5 𝑅 𝑃 RP: 𝑅 𝑃 = ( 4 − 2 ) 2 + ( − 5 − ( − 1 ) ) 2 + ( 4 − 0 ) 2 = 2 2 + ( − 4 ) 2 + 4 2 = 4 + 16 + 16 = 36 = 6 RP= (4−2) 2 +(−5−(−1)) 2 +(4−0) 2 = 2 2 +(−4) 2 +4 2 = 4+16+16 = 36 =6 Step 2: Check for a right triangle: 𝑃 𝑄 2 + 𝑄 𝑅 2 = 3 2 + ( 3 5 ) 2 = 9 + 45 = 54 , 𝑅 𝑃 2 = 6 2 = 36 PQ 2 +QR 2 =3 2 +(3 5 ) 2 =9+45=54,RP 2 =6 2 =36 Since 𝑃 𝑄 2 + 𝑄 𝑅 2 ≠ 𝑅 𝑃 2 PQ 2 +QR 2  =RP 2 , it is not a right triangle. Step 3: Check for an isosceles triangle: Compare the lengths: 𝑃 𝑄 = 3 , 𝑄 𝑅 = 3 5 , 𝑅 𝑃 = 6 PQ=3,QR=3 5 ,RP=6 Since no two sides are equal, it is not an isosceles triangle. Conclusion: Lengths of sides: 𝑃 𝑄 = 3 PQ=3, 𝑄 𝑅 = 3 5 QR=3 5 , 𝑅 𝑃 = 6 RP=6. Not a right triangle. Not an isosceles triangle. Question 2 Replace the Cartesian equation 𝑥 = 𝑦 x=y with an equivalent polar equation. Answer 1 Using polar coordinates: 𝑥 = 𝑟 cos ⁡ 𝜃 , 𝑦 = 𝑟 sin ⁡ 𝜃 x=rcosθ,y=rsinθ Substitute into 𝑥 = 𝑦 x=y: 𝑟 cos ⁡ 𝜃 = 𝑟 sin ⁡ 𝜃 rcosθ=rsinθ Divide by 𝑟 r (assuming 𝑟 ≠ 0 r  =0): cos ⁡ 𝜃 = sin ⁡ 𝜃 cosθ=sinθ tan ⁡ 𝜃 = 1 tanθ=1 𝜃 = 𝜋 4 θ= 4 π Polar equation: 𝜃 = 𝜋 4 θ= 4 π Question 3 Match the parametric equations with the correct graph: 𝑥 = 𝑡 cos ⁡ 𝑡 , 𝑦 = 𝑡 sin ⁡ 𝑡 , 𝑧 = 𝑡 x=tcost,y=tsint,z=t Answer 1 These parametric equations describe a helix in three dimensions, spiraling around the z-axis. The graph corresponding to this is the 5th graph (a helical curve). Question 4 Give an example of a vector-valued function that is defined but not continuous at 𝑡 = 3 t=3. Answer 1 Let: 𝑟 ( 𝑡 ) = ⟨ 2 𝑡 + 1 , 𝑡 2 − 9 𝑡 − 3 , 𝑡 2 ⟩ r(t)=⟨2t+1, t−3 t 2 −9 ,t 2 ⟩ Domain: 𝑡 ≠ 3 t  =3, because the second component 𝑡 2 − 9 𝑡 − 3 t−3 t 2 −9 is undefined at 𝑡 = 3 t=3. The function simplifies to: 𝑟 ( 𝑡 ) = ⟨ 2 𝑡 + 1 , 𝑡 + 3 , 𝑡 2 ⟩ ( 𝑡 ≠ 3 ) r(t)=⟨2t+1,t+3,t 2 ⟩(t  =3) At 𝑡 = 3 t=3, 𝑟 ( 𝑡 ) r(t) is undefined, so it is not continuous. Conclusion: The function is defined for 𝑡 ≠ 3 t  =3 but is not continuous at 𝑡 = 3 t=3. Question 5 Find a vector-valued function whose graph is the surface of the cone 𝑥 = 16 𝑦 2 + 𝑧 2 x= 16y 2 +z 2 . Answer 1 Given 𝑥 = 16 𝑦 2 + 𝑧 2 x= 16y 2 +z 2 : Let 𝑦 = 𝑢 y=u and 𝑧 = 𝑣 z=v. Then: 𝑥 = 16 𝑢 2 + 𝑣 2 x= 16u 2 +v 2 The vector-valued function is: 𝑟 ( 𝑢 , 𝑣 ) = 16 𝑢 2 + 𝑣 2 𝑖 + 𝑢 𝑗 + 𝑣 𝑘 r(u,v)= 16u 2 +v 2 i+uj+vk