A 13 foot ladder is leaning against a house when its base starts to slide away. By the time the base is 12 ft. from the house, the base is moving at the rate of 10ft./s. How fast is the top of the ladder sliding down the wall then?
To find the rate at which the top of the ladder is sliding down the wall, we can apply related rates.
Let's denote the distance between the ladder base and the house as x (in feet) and the distance between the top of the ladder and the ground as y (in feet). According to the problem, we have the following information:
x = 12 ft (distance of the base from the house)
dx/dt = 10 ft/s (rate at which the base is moving away from the house)
We need to find dy/dt (rate at which the top of the ladder is sliding down the wall).
From the problem, we have a right triangle formed by the ladder, the wall, and the ground. The length of the ladder is given as 13 ft.
Using the Pythagorean theorem, we can express the relationship between x and y as:
x^2 + y^2 = 13^2
Differentiating both sides with respect to t (time):
2x(dx/dt) + 2y(dy/dt) = 0
Substituting the known values:
2(12 ft)(10 ft/s) + 2y(dy/dt) = 0
Rearranging the equation:
20(12 ft) + 2y(dy/dt) = 0
240 ft + 2y(dy/dt) = 0
2y(dy/dt) = -240 ft
Now we can solve for dy/dt:
dy/dt = -240 ft / (2y)
To find y, we can use the Pythagorean theorem:
x^2 + y^2 = 13^2
12^2 + y^2 = 169
144 + y^2 = 169
y^2 = 169 - 144
y^2 = 25
y = 5 ft (using the positive value since y represents a distance)
Substituting y = 5 ft into the equation:
dy/dt = -240 ft / (2*5 ft)
dy/dt = -240 ft / 10 ft
dy/dt = -24 ft/s
Therefore, the top of the ladder is sliding down the wall at a rate of 24 ft/s.
looks like a homework dump.
Five questions without showing us any of your starting steps or any indication where your problems lie with these questions.