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© Prep101 MATH 2ZZ3 Solutions Page 1 of 20 Chapter 1 – Vector Calculus Q1. 𝒓𝒓 ( 𝑡𝑡 ) = (2 𝒊𝒊 − 3 𝑡𝑡𝒋𝒋 + 4 𝑡𝑡 2 𝒌𝒌 ) 𝑑𝑑𝑡𝑡 = (2 𝑡𝑡 + 𝑐𝑐 1 ) 𝒊𝒊 − � 3 𝑡𝑡 2 2 + 𝑐𝑐 2 � 𝒋𝒋 + 4 𝑡𝑡 3 3 + 𝑐𝑐 3 � 𝒌𝒌 𝒓𝒓 (0) = 𝑐𝑐 1 𝒊𝒊 − 𝑐𝑐 2 𝑗𝑗 + 𝑐𝑐 3 𝒌𝒌 = −𝒊𝒊 + 2 𝒋𝒋 − 3 𝒌𝒌 ⇒ 𝑐𝑐 1 = 1, 𝑐𝑐 2 = 2, 𝑐𝑐 3 = 3 ⇒ 𝒓𝒓 ( 𝑡𝑡 ) = (2 𝑡𝑡 − 1) 𝒊𝒊 − � 3 𝑡𝑡 2 2 2 � 𝒋𝒋 + 4 𝑡𝑡 3 3 3 � 𝒌𝒌 Q2. We know that Speed = ‖𝒓𝒓 ( 𝑡𝑡 ) , 𝒓𝒓 ( 𝑡𝑡 ) = 1, 6 𝑡𝑡 , 3 𝑡𝑡 2 , Therefore ‖𝒓𝒓 ( 𝑡𝑡 ) = 1 + 6 𝑡𝑡 2 + 9 𝑡𝑡 4 = (3 𝑡𝑡 2 + 1) 2 = 3 𝑡𝑡 2 + 1 So, ‖𝒓𝒓 (2) = 13 Answer – B. Q3. The tangential and normal components of acceleration are 𝒂𝒂 𝑻𝑻 = 𝒓𝒓 ( 𝑡𝑡 ) ⋅ 𝒓𝒓 ′′ ( 𝑡𝑡 ) ‖𝒓𝒓 ( 𝑡𝑡 ) 𝒂𝒂 𝑵𝑵 = ‖𝒓𝒓 ( 𝑡𝑡 ) × 𝒓𝒓 ′′ ( 𝑡𝑡 ) ‖𝒓𝒓 ( 𝑡𝑡 ) So we have, 𝒓𝒓 ( 𝑡𝑡 ) = 2,2 𝑡𝑡 , 0 𝒓𝒓′ ( 𝑡𝑡 ) = 0,2,0 ‖𝒓𝒓 ( 𝑡𝑡 ) = 2 𝒓𝒓 ( 𝑡𝑡 ) ⋅ 𝒓𝒓 ′′ ( 𝑡𝑡 ) = 2,2 𝑡𝑡 , 0 〉 ⋅ 〈 0,2,0 = 4 𝑡𝑡 𝒓𝒓 ( 𝑡𝑡 ) × 𝒓𝒓 ′′ ( 𝑡𝑡 ) = 𝒊𝒊 𝒋𝒋 𝒌𝒌 2 2 𝑡𝑡 0 0 2 0 = 0,0,4 ‖𝒓𝒓 ( 𝑡𝑡 ) × 𝒓𝒓 ′′ ( 𝑡𝑡 ) = 4
© Prep101 MATH 2ZZ3 Solutions Page 2 of 20 So, 𝑎𝑎 𝑇𝑇 = 4𝑡𝑡 2 = 2 𝑡𝑡 and 𝑎𝑎 𝑁𝑁 = 4 2 = 2 . Q4. The curvature of a function is 𝜅𝜅 ( 𝑡𝑡 ) = ‖𝒓𝒓 ( 𝑡𝑡 ) × 𝒓𝒓 ′′ ( 𝑡𝑡 ) ‖𝒓𝒓 ( 𝑡𝑡 ) 3 𝒓𝒓 ( 𝑡𝑡 ) = 1 cos( 𝑡𝑡 ) , 1 + sin( 𝑡𝑡 ) 𝒓𝒓′ ( 𝑡𝑡 ) = sin( 𝑡𝑡 ) , cos( 𝑡𝑡 ) 𝒓𝒓 ( 𝑡𝑡 ) × 𝒓𝒓 ′′ ( 𝑡𝑡 ) = 𝒊𝒊 𝒋𝒋 𝒌𝒌 1 cos( 𝑡𝑡 ) 1 + sin( 𝑡𝑡 ) 0 sin( 𝑡𝑡 ) cos( 𝑡𝑡 ) 0 = 0,0, cos( 𝑡𝑡 ) cos 2 ( 𝑡𝑡 ) sin( 𝑡𝑡 ) sin 2 ( 𝑡𝑡 ) = 0,0, cos( 𝑡𝑡 ) sin( 𝑡𝑡 ) 1 ‖𝒓𝒓 ( 𝑡𝑡 ) × 𝒓𝒓 ′′ ( 𝑡𝑡 ) = cos( 𝑡𝑡 ) sin( 𝑡𝑡 ) 1 ‖𝒓𝒓 ( 𝑡𝑡 ) 𝟑𝟑 = [(1 cos( 𝑡𝑡 )) 2 + (1 + sin( 𝑡𝑡 )) 2 ] 3 2 = [3 + 2 sin( 𝑡𝑡 ) 2 cos( 𝑡𝑡 )] 3 2 So, 𝜅𝜅 ( 𝑡𝑡 ) = cos( 𝑡𝑡 ) sin( 𝑡𝑡 ) 1 [3 + 2 sin( 𝑡𝑡 ) 2 cos( 𝑡𝑡 )] 3 2 This gives 𝜅𝜅 � 𝜋𝜋 2 = 0 1 1 (3 + 2) 3 / 2 = 2 5 3 / 2 Q5. a) 𝛁𝛁𝑓𝑓 = 𝜕𝜕𝜕𝜕 𝜕𝜕𝜕𝜕 , 𝜕𝜕𝜕𝜕 𝜕𝜕𝜕𝜕 , 𝜕𝜕𝜕𝜕 𝜕𝜕𝜕𝜕 = 2 𝑥𝑥 + 𝑦𝑦𝑒𝑒 𝜕𝜕𝜕𝜕 , 2 𝑦𝑦 + 𝑥𝑥𝑒𝑒 𝜕𝜕𝜕𝜕 , 2 𝑧𝑧〉 b) 𝑑𝑑𝑑𝑑𝑑𝑑 ( 𝛁𝛁𝑓𝑓 ) = 𝜕𝜕 𝜕𝜕𝜕𝜕 (2 𝑥𝑥 + 𝑦𝑦𝑒𝑒 𝜕𝜕𝜕𝜕 ) + 𝜕𝜕 𝜕𝜕𝜕𝜕 (2 𝑦𝑦 + 𝑥𝑥𝑒𝑒 𝜕𝜕𝜕𝜕 ) + 𝜕𝜕 𝜕𝜕𝜕𝜕 (2 𝑧𝑧 ) = 2 + 𝑦𝑦 2 𝑒𝑒 𝜕𝜕𝜕𝜕 + 2 + 𝑥𝑥 2 𝑒𝑒 𝜕𝜕𝜕𝜕 + 2 = 6 + ( 𝑥𝑥 2 + 𝑦𝑦 2 ) 𝑒𝑒 𝜕𝜕𝜕𝜕 c) 𝛁𝛁𝑓𝑓 ( 1,1) = 〈− 2 + 𝑒𝑒 −1 , 2 − 𝑒𝑒 −1 , 0 𝒖𝒖 = 𝒊𝒊 − 2 𝒋𝒋 ‖𝒊𝒊 − 2 𝒋𝒋‖ = 1 5 , 2 5 , 0 So, 𝛁𝛁𝑓𝑓 ( 1,1) ⋅ 𝒖𝒖 = 1 5 1 𝑒𝑒 2, 2 1 𝑒𝑒 2 �〉
© Prep101 MATH 2ZZ3 Solutions Page 3 of 20 d) −‖𝛁𝛁𝑓𝑓 ( 1,1) = 1 √5 1 𝑒𝑒 2 2 + 4 1 𝑒𝑒 2 2 = 1 5 5 1 𝑒𝑒 2 2 = − � 1 𝑒𝑒 2 2 Q6. 𝐹𝐹 ( 𝑥𝑥 , 𝑦𝑦 , 𝑧𝑧 ) = 𝑥𝑥 3 + 𝑦𝑦 3 2 𝑧𝑧 3 3 , So 𝐹𝐹 𝜕𝜕 ( 𝑥𝑥 , 𝑦𝑦 , 𝑧𝑧 ) = 3 𝑥𝑥 2 , 𝐹𝐹 𝜕𝜕 ( 2,3,2) = 12 𝐹𝐹 𝜕𝜕 ( 𝑥𝑥 , 𝑦𝑦 , 𝑧𝑧 ) = 3 𝑦𝑦 2 , 𝐹𝐹 𝜕𝜕 ( 2,3,2) = 27 𝐹𝐹 𝜕𝜕 ( 𝑥𝑥 , 𝑦𝑦 , 𝑧𝑧 ) = 6 𝑧𝑧 2 , 𝐹𝐹 𝜕𝜕 ( 2,3,2) = 24 So, the equation of the tangent plane is 12( 𝑥𝑥 + 2) + 27( 𝑦𝑦 − 3) 24( 𝑧𝑧 − 2) = 0 12 𝑥𝑥 + 27 𝑦𝑦 − 24 𝑧𝑧 = 9 Answer – C. Q7. 𝛁𝛁𝑓𝑓 = 2 𝑥𝑥 , 2 𝑦𝑦 , 2 𝑧𝑧〉 𝛁𝛁𝑓𝑓 × 𝑭𝑭 = 𝒊𝒊 𝒋𝒋 𝒌𝒌 2 𝑥𝑥 2 𝑦𝑦 2 𝑧𝑧 𝑥𝑥𝑦𝑦 𝑦𝑦𝑥𝑥 𝑥𝑥𝑧𝑧 = 2 𝑥𝑥𝑦𝑦𝑧𝑧 − 2 𝑦𝑦𝑧𝑧 2 , 2 𝑥𝑥𝑦𝑦𝑧𝑧 − 2 𝑥𝑥 2 𝑧𝑧 , 2 𝑥𝑥𝑦𝑦𝑧𝑧 − 2 𝑥𝑥𝑦𝑦 2 𝛁𝛁 × ( 𝛁𝛁𝑓𝑓 × 𝑭𝑭 ) = 𝒊𝒊 𝒋𝒋 𝒌𝒌 𝜕𝜕 𝜕𝜕𝑥𝑥 𝜕𝜕 𝜕𝜕𝑦𝑦 𝜕𝜕 𝜕𝜕𝑧𝑧 2 𝑥𝑥𝑦𝑦𝑧𝑧 − 2 𝑦𝑦𝑧𝑧 2 2 𝑥𝑥𝑦𝑦𝑧𝑧 − 2 𝑥𝑥 2 𝑧𝑧 2 𝑥𝑥𝑦𝑦𝑧𝑧 − 2 𝑥𝑥𝑦𝑦 2 = 2 𝑥𝑥𝑧𝑧 − 4 𝑥𝑥𝑦𝑦 − 2 𝑥𝑥𝑦𝑦 + 2 𝑥𝑥 2 , 2 𝑥𝑥𝑦𝑦 − 4 𝑥𝑥𝑦𝑦 − 2 𝑦𝑦𝑧𝑧 + 2 𝑦𝑦 2 , 2 𝑦𝑦𝑧𝑧 − 4 𝑥𝑥𝑧𝑧 − 2 𝑥𝑥𝑧𝑧 + 2 𝑧𝑧 2 = 2 𝑥𝑥 2 6 𝑥𝑥𝑦𝑦 + 2 𝑥𝑥𝑧𝑧 , 2 𝑦𝑦 2 2 𝑥𝑥𝑦𝑦 − 2 𝑦𝑦𝑧𝑧 , 2 𝑧𝑧 2 6 𝑥𝑥𝑧𝑧 + 2 𝑦𝑦𝑧𝑧〉
© Prep101 MATH 2ZZ3 Solutions Page 4 of 20 Q8.

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