What is the rate of change in the following problem:
A 13 ft ladder is leaning against a house when its base starts to slide away. At a particular moment of time when the base is 12 ft from the house, the base is moving at the rate of 5 ft/sec.
How fast is the top of the ladder sliding down the wall at that moment and at what rate is the area of the triangle formed by the ladder, wall, and ground changing at that moment.
1 Answer | Add Yours
The 13 ft ladder is leaning against the house and its base starts to slide away from the wall. This reduces the height at which the top of the ladder touches the wall.
Let the height at which the top of the ladder touches the wall be y and the distance of the bottom of the ladder from the wall be x.
x^2 + y^2 = 13^2
Using implicit differentiation
`2*x*(dx/dt) + 2*y*(dy/dt) = 0`
=> `x*(dx/dt) = -y*(dy/dt)`
At a moment when the base is 12 ft from the wall, it is moving at the rate of 5 ft/sec.
`x = 12, dx/dt = 5`
y = `sqrt(13^2 - 12^2)` = 5
`dy/dt` = -12*5/5 = -12 ft/sec
The top of the wall is sliding down at the given moment at a rate of 12 ft/sec.
The area of the triangle formed by the ladder consisting of the ground, the wall and the ladder is equal to `(x*y)/2`
The rate of change of the area is `(1/2)*(x*(dy/dt) + y*(dx/dt))` = `(1/2)*(12*-12 + 5*5) = (1/2)*(-119)`
The area of the triangle at the given moment is reducing at `119/2` ft^2/sec.
We’ve answered 319,863 questions. We can answer yours, too.Ask a question