Final answer:
The velocity of the top of the ladder at time t=3 seconds is approximately 3 ft/sec downward, as calculated using the Pythagorean theorem and related rates.
Step-by-step explanation:
The student is asking to find the velocity of the top of the ladder at time t=3 seconds, where the ladder is initially 3 ft away from the wall and is slipping away at a constant rate. Since the ladder's length does not change, we can use the Pythagorean theorem to relate the distance of the bottom of the ladder from the wall, the height of the top of the ladder on the wall, and the ladder's length.
Let x be the distance of the bottom from the wall, y be the height of the top of the ladder on the wall, and L be the length of the ladder which is 17 ft. Then:
L2 = x2 + y2
At time t=3 seconds, x=3 ft + 3 ft/sec × 3 sec = 12 ft. To find y, we can substitute x=12 ft into the equation and solve for y.
L2 = 122 + y2
172 = 144 + y2
289 = 144 + y2
y2 = 145
y = √145
Now we can calculate dy/dt using related rates. Since we are given dx/dt=3 ft/sec:
2x dx/dt + 2y dy/dt = 0
Plugging in the values we get:
2(12) × 3 + 2(√145) dy/dt = 0
dy/dt = -72 / (2√145)
dy/dt ≈ -3 ft/sec (since we are only concerned with the magnitude)
The velocity of the top of the ladder at time t=3 seconds is approximately 3 ft/sec downward.