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. How fast is the top of the ladder moving down the wall when its base is 7 feet?
x^2+y^2 = 25
x dx/dt + y dy/dt = 0
So, find y when x=7, and then just use your numbers to find dy/dt
-14/24 = dy/dt?
Looks good to me, but I'd write it as
-7/12 ft/s
To find how fast the top of the ladder is moving down the wall, we can use related rates. Let's use the variables:
- y: the distance between the top of the ladder and the ground (in feet)
- x: the distance between the base of the ladder and the house wall (in feet)
We are given that the ladder is 25 feet long, so we can write an equation relating x and y using the Pythagorean theorem:
x^2 + y^2 = 25^2
Differentiating implicitly with respect to time (t) gives us:
2x(dx/dt) + 2y(dy/dt) = 0
Since we want to find dy/dt (the rate at which the top of the ladder is moving down the wall), we can rearrange the equation and solve for dy/dt:
dy/dt = - (x/y) * (dx/dt)
Now, we can use the information we have to find dx/dt when x = 7 ft and dy/dt when y = 24 ft:
Given dx/dt = 2 ft/s, x = 7 ft, and y = 24 ft, we can substitute these values into the equation:
dy/dt = - (7/24) * 2
Simplifying, we get:
dy/dt = - 14/24 ft/s
So, when the base of the ladder is 7 feet away from the wall, the top of the ladder is moving down the wall at a rate of -14/24 ft/s (or approximately -0.58 ft/s). The negative sign indicates that the top of the ladder is moving downward.