SOLUTION: A ladder 25 feet long is leaning against the wall of a house. The base of the ladder is pulled away from the wall at a rate of 2 feet per second. What is the velocity of the top of

Question 1123366: A ladder 25 feet long is leaning against the wall of a house. The base of the ladder is pulled away from the wall at a rate of 2 feet per second. What is the velocity of the top of the ladder when the base is given below?
a.) 20 feet away from the wall
Answer by rothauserc(4718) (Show Source): You can put this solution on YOUR website!
Let the ladder be y feet above the ground and x feet from the wall
:
use Pythagorean theorem
:
x^2 + y^2 = 25^2
:
take first derivative of each vaiable with respect to time
:
1) 2x * dx/dt + 2y * dy/dt = 0
:
we are given dx/dt = 2 feet per second
:
we want to find dy/dt when the ladder is 20 feet from the wall(x=20)
:
when x=20
:
20^2 + y^2 = 25^2
:
400 + y^2 = 625
:
y^2 = 225
:
y = 15 feet (we ignore the negative root, -15)
:
now use equation 1
:
(2 * 20 * 2) + (2 * 15 * dy/dt) = 0
:
80 + (30 * dy/dt) = 0
:
dy/dt = -80/30 is approximately -2.67
:
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the velocity of the top of the ladder is -2.67 feet per second
:
the velocity is negative because the top of the ladder is sliding down the wall
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