Please answer the three questions An 8-foot ladder is leaning against a building
ID: 3341516 • Letter: P
Question
Please answer the three questions
An 8-foot ladder is leaning against a building. If the bottom of the ladder is sliding along the pavement directly away from the building at 1 ft/sec, how fast is the top of the ladder moving down when the foot of the ladder is 2 feet from the wall? The top of the ladder is moving down at a rate of ft/sec when the foot of the ladder is 2 feet from the wall. (Round to the nearest thousandth as needed.) A pulley is on the edge of a dock, 6 feet above the water level. A rope is being used to pull in a boat. The rope is attached to the boat at water level. The rope is being pulled in at the rate of 1 foot per second. Find the rate at which the boat is approaching the dock at the instant the boat is 4 feet from the dock. The boat is approaching the dock at a rate of feet per second when the boat is 4 feet from the dock. (Round to the nearest thousandth as needed.) An inverted conical water tank with a height of 16 ft and a radius of 8 ft is drained through a hole in the vertex at a rate of 8 ft3 / s (see figure). What is the rate of change of the water depth when the water is 2 ft? (Hint. Use similar triangles.) dh / dt = ft / s (Round to the nearest hundredth as needed.)Explanation / Answer
1)
applying pythogorous equation x^2 + y^2 = z^2 ==> at x = 2,z = 8 y = sqrt(64 - 4) = sqrt (60) 2*x* dx/dt + 2*y* dy/dt = 0 since length of thh ladder is constant 2*2(-1) + 2*y*dy/dt = 0 = 2/sqrt(60) = 0.2581 ft/sec
2)
here same as in prev question instead of z, y is constant as pulley is at a constant height z = sqrt(6^2 + 4^2) = sqrt(52) = 7.211 2x dx/dt = 2z dz/dt ==> dx/dt = 7.211*1/4 = 1.8027 foot/sec
3)sintheta = 8/16 = 1/2 ==> angle between vertical height and slanted height is 30 degrees.therefore h = 2r
when height is 2, radius is 1
volume is pi r^2h/3 = pi*h^3/12
solving dV/dt = 3/12 * pi* h^2* dh/dt = 8 ==>pih^2/4 *dh/dt = 8 ==>dh/dt = 2.545 ft/s