Find the rate of change of the angle of elevation when the balloon is 9 feet above the ground.

Question

Question 1. A balloon rises at the rate of 8 feet per second from a point on the ground 12 feet from an observer. To 2 decimal places in radians per second, find the rate of change of the angle of elevation when the balloon is 9 feet above the ground.

Question 2. A water tank has the shape of an inverted right circular cone with base radius 3 meters and height 6 meters. Water is being pumped into the tank at the rate of 12 meters³/sec. Find the rate, in meters/sec, at which the water level is rising when the water is 2 meters deep. 

Could you explain to me how you got the answer?

Answer 1

1. Hite of balloonn at time t H(t) = 12tanθ, where θ is angle of elevation.

H'(t) = 12sec²θ·θ', from here θ' = H'(t)cos²θ/12,

H^'(t) = 8ft/sec, cosθ = 12/sqrt(12² + 9²) = 12/15 = 0.8≈.

θ' = 8·0.8²/12 = 0.43 rad/sec

2 Volume of water is V = 1/3πr²h, where h is hite of cone and level of water. Using similyarity of triangle

h/r = 6/3 = 2, r = h/2; V = 1/3 π(h/2)²h = 1/12πh³. Now derivative: V' = 1/12π·3h²·h' = 1/4πh²h'; from her

h' = 4V'/(πh²) = 4·12m³/sec/(π·4m²) = 12/π m/sec ≈ 3.82 m/sec