y = x2 Correct x 2 – 3b derivative with = (x 2)2x x2 = 0 for stat pts yꞌ – – correct solutions. (x 2)2 – So 2x2 4x x2 = 0 – – x2 4x = 0 – x(x 4) = 0 – x = 0 or 4 3c Correct y = xln(x) 2x – derivative = x× 1 + lnx 2 yꞌ – for u x = 1 + lnx 2 – = lnx 1 – Decreasing if < 0 yꞌ x < e So lnx 1 < 0 – is needed for r lnx < 1 x < e1 M 3d h dh = - 90 θ dt Q 800 C h = tan θ dh = - 90 h = tan θ 800 dt 800 and dh = 800sec2 θ dh = 800sec2 = 800 θ dθ dθ cos2 θ for u d = d × dh θ θ dt dh d t d = cos2 ×(–90) θ θ dt 800 If h = 500 then tan = 5 so = 320 θ θ all correct for r 8 accept decreasing d = cos2 ×–90 = cos232 ×–90 θ θ at 0.0809 rad/sec dt 800 800 d = rad/sec θ –0.0809 dt
3b y = x2 x – 2 yꞌ = (x – 2)2x – x2 = 0 for stat pts (x – 2)2 So 2x2 – 4x – x2 = 0 x2 – 4x = 0 x(x – 4) = 0 x = 0 or 4 Correct derivative with correct solutions. 3c y = xln(x) – 2x yꞌ = x× 1 + lnx – 2 x = 1 + lnx – 2 = lnx – 1 Decreasing if yꞌ < 0 So lnx – 1 < 0 lnx < 1 x < e1 Correct derivative for u x < e is needed for r 3d M h θ Q 800 C dh = - 90 h = tan θ dt 800 dh = 800sec2 θ = 800 dθ cos2 θ d θ = d θ × dh dt dh d t d θ = cos2 θ ×(–90) dt 800 If h = 500 then tan θ = 5 so θ = 320 8 d θ = cos2 θ ×–90 = cos232 ×–90 dt 800 800 d θ = –0.0809 rad/sec dt dh = - 90 dt h = tan θ 800 and dh = 800sec2 θ dθ for u all correct for r accept decreasing at 0.0809 rad/sec