Question 1193964: A ladder inclined at 60 ° with the horizontal is leaning against a vertical wall. The foot of the ladder is 3 meters away from the foot of the wall. A boy climbs the ladder such that his distance z meters with respect to the foot of the ladder is given by z=6t, where t is the time in seconds. Find the rate at which his vertical distance from the ground changes with respect to t. Find the rate at which his distance from the foot of the wall is changing with respect to t when he is 3 meter away from the foot of the ladder.
Answer by greenestamps(13198)   (Show Source):
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The z=6t measures the boy's distance from the foot of the ladder at time t. Since the ladder is inclined at 60 degrees to the horizontal, his vertical distance from the ground is y=(3*sqrt(3))t and his horizontal distance from the base of the wall is x=A-3t, where A is the distance from the base of the wall to the foot of the ladder.
Both those functions are linear, so the rates of change of his vertical distance from the ground and of his horizontal distance from the foot of the wall are constants.
ANSWERS:
(1) rate at which his vertical distance from the ground is changing = dy/dt = 3*sqrt(3)
(2) rate at which his horizontal distance from the foot of the wall is changing = dx/dt = -3
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