Need to typed the work using math tools || copy paste diagrams as it's is/n Solution Solution Where ·Z N HW Q #1 (3.1) އ 1/2 Зл Xp =-31 cos %, Yp =^ Zp = 32 Sin%2 The co-ordinates of the position vector P Pp = -3 ^ Cosq₂ C₁+^ C₂+3^sing &₂ Р For velocity of Paint P wirit c P Cup = capp at CS CamScanner c₁ = & C₂) 2+ (-31 cost/2 & +^2 +31 Sin P C√ = 3^%₂ (Sin V/2 C₁ + Cas &/2 (3) - -Анд For acceleration of point P wirt c P cα = "dvp dt c₁ = d (31% (sing/ ε, + cos V₂ (3)) dt C₁₂² = 3^ [ ( ~ ₂ Sin 1/2 + For the velocity of point p wirit L L₁ = copp + w`x pp dt Sin Anz LvP = df (-3rcosan & tre₂+3a Sing (s) P alt +122 +(9), c₁) x (-31 cosa, &, + 3 ^ sings (3) L√ = R (3 sin v₂ (V₂ Ĉ, - V, ^2) + (2), +3% cost) ^3]/ % Am CS CamScanner and for acceleration of pount P wirt L LP L P L d = dv α dt P d {R (35in v2 (2 c₁ - v, ĉ2) = dt os + (%, +39% (as %) ĉ z}]} α = ^ { 3 (0%) Six V2 + va casar) & - (3°v, Sin 92 +69², ov₂ casq√2 +9, ²)^2 + [ª, +3% cos q− 3 (q;,² + V²) sin v ] } } / Анг Sin' CS CamScanner Q#2 (3.2) Solution: Show that V = vb₁ and a α = ib₁ + 2x²+ b ₁₂ The velocity v of paint P оп Сихне с Can be expressed as √ = 25, Let vector P on current) Wheek d^(+) V = खं The unit tangent vector b, is given by, bi = 리 21 || ducts || Thus =reb, Anz and Whore = ā = dt d at (61) (vbi) = da bi +a +2 of 5, dt CS CamScanner where, by 318 ख ds. chain rule = - db ds 25 det dre (b,) 2 + = य 154 Р ♥(금토리) = ā = at dre b₁ + bi dt ā = ا کافه لل 212 + 26. Wher थे 15 ā مل = dv वत = i b₁ + حل Am CS CamScanner

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Need to typed the work using math tools || copy paste diagrams as it's is/n Solution Solution Where ·Z N HW Q #1 (3.1) އ 1/2 Зл Xp =-31 cos %, Yp =^ Zp = 32 Sin%2 The co-ordinates of the position vector P Pp = -3 ^ Cosq₂ C₁+^ C₂+3^sing &₂ Р For velocity of Paint P wirit c P Cup = capp at CS CamScanner c₁ = & C₂) 2+ (-3