1) A kite 100feet above the ground moves horizontally at a speed of 2 feet/secon
ID: 2882880 • Letter: 1
Question
1) A kite 100feet above the ground moves horizontally at a speed of 2 feet/second At what rate is the angle between the string and the horizontal decreasing when 300 feet of string has been let out?
Answer (in radians per second): ___________
2) Water is leaking out of an inverted conical tank at a rate of 0.0095 meter^3/minute. At the same time water is being pumped into the tank at a constant rate. The tank has height 6 meters and the diameter at the top is 5 meters. If the water level is rising at a rate of 0.21 m/min when the height of the water is 4.5 meters, find the rate at which water is being pumped into the tank.
Water is being pumped in at _____________- meter^3/minute
Explanation / Answer
Solution:(1)
tan() = 100/x
sec²() (d/dt) = (-100/x²) (dx/dt) ........... chain rule: d/dt = d/dx * dx/dt
d/dt = (-100/x²) cos²() (dx/dt) ........ dx/dt = 2 ft/s
{let given cos²() = x²/(300)² => cos²() / x² = 1/(300)²}
d/dt = -100/(300)² (2 ft/s)
= -0.0022 radians/s
Answer: d/dt = -0.0022 radians/s
Solution:(2)
V = (1/3)r²h
height = 6 m, diameter = 5 m, radius = 5/2 m
h/r = 6/(5/2)
r/h = 5/12
r = 5h/12
V = (1/3)r2h = (1/3)(5h/12)2h = (25/432)h3
dV/dh = (25/144)h2
When h = 4.5m, dh/dt = 0.21m/min
dV/dt = (dV/dh)(dh/dt) = (25/144)*(4.5)2 * (0.21) = 2.3194 m3/min
Water is leaking at the rate of 0.0095m³/min
Rate at which water is being pumped into the tank
= 2.3194 m3/min + 0.0095m3/min
2.3289 m3/min
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