Related rates multiple choice questions. i dont understand any of them, trying to review before an exam. please help!

\fA rectangle with area A : bh is expanding with time such that the length of the base, b (t) : t2 + 2, and the height of the rectangle is constant at h = 4. What is the formula for the rate of change of the area of the rectangle with respect to time? o %=b(t)=t2+2 0 %=(%)(%)=0 0 %=(%)=2t 0%:M%=% Dan is blowing up a balloon. Assume that the volume of the balloon is V : 4?: , r(t) = t3 + 1. What is the formula for the rate of change of the volume of the balloon, with respect to time? 0 %:12w(t3+1)2(t2) o %=%(t3+1)2(t2) o %=W(t3+1)2(t2) 0 %:127r(t3+1)2 Assume that a helicopter is landing perpendicular to you at a constant distance x away from you. In the triangle below, h(t) is the height of the helicopter. Assume that $4, is how quickly the helicopter is approaching the ground. When the helicopter is h miles high, what is the rate of change of 0? tan (0) = h O de h2 d h d t x2 + h2 dt O de h d h dt = x2 + h2 dt O de x2 d h d t x2+ h2 dt O X dh d t x2 + h2 dt\fA cone is expanding at the rate of 6 cm3a'sec. The formula for the volume of a cone is: V = \"Sh. Let h = 2r. What is E when r = 2? dt 0 zcm/s O =cms O %:%0m/s dr_i O 4Wcm/s a. n The surface area of a sphere is expanding at the rate of 6 cm2/sec, where S = 4Tir2. What is the change in the radius when r = 2? O dr 1 dt 2 IT cm/s O dr 1 dt 4 T cm/s O dr 3 cm/s dt 8 7 dr 3 O cm/s dt 4 TA cylinder is being filled such that the height of the contents is increasing at a rate of 2 cm/sec. What is the rate of change of the volume if r =2 and V = Tir h? O d V dt =87 cm3/s O dv dt =97 cm3/s O d V dt =27 cm3/s O d V dt = 187 cm3 /sA ladder is 27 feet long and is leaning against the side of a house. The base is moved away at 3 ft per min. At what rate is the ladder moving down the wall when the base is 10 feet from the house? 0 % = -2.516ft!m o % =-'l.196ftlm 0 % = 0.797 lm 0 % = -0.418 lm Assume that a rocket is lifting off 10 miles away, with a position h(t) = 5t2. What is the rate of change of O, given t = 5 seconds? tan (0) = h O dt de = 0.25 degrees/second O dt de = 0.03 degrees/second O de = 0.5 degrees/second dt O dt de = 0.69 degrees/second