A 10 foot ladder leans against a 20 foot wall. Someone begins pushing the base of the ladder toward the wall at the rate of one foot per second. How quickly is the top of the ladder moving up the wall after 2 seconds?
Let y be the wall height
let x be the horizontal distance
x^2+y^2=100
2x dx/dt+2y dy/dt=0
dy/dt=-x dx/dt
so calculate the x,y positions after 2 seconds, I have no idea the starting position.
This problem is the most popular lead-in question to "rate of change" questions for almost any Calculus text I have seen
Let the ladder be x ft from the wall, and y m up the wall
we know x^2 + y^2 = 100
2x(dx/dt) + 2y(dy/dt) = 0
given: dx/dt = 1 ft/s
find dy/dt when t = 2 or
in other words, when x = 2 ft since we know in 2 s it moved 2 ft.
also when x = 2
2^2 + y^2 = 100
y = √96
2(2)(1) + 2√96(dy/dt) = 0
dy/dt = -4/(2√96
= -.204 ft/s
at that moment the top of the ladder is dropping (note the - sign) at .204 ft/s
Just realized I read the question as the foot of the ladder moving away from the wall, so just change
dx/dt to -1,
the result will change to dy/dt = +.204
how can a 10 foot ladder lean on a 20 foot wall?
To solve this problem, we can use related rates, which involves finding the rate at which one quantity is changing with respect to another quantity. In this case, we want to find the rate at which the top of the ladder is moving up the wall, given that the base is being pushed towards the wall.
Let's assign some variables to the quantities involved:
- Let x be the distance between the base of the ladder and the wall.
- Let y be the height of the ladder on the wall.
- Let t be the time in seconds.
Given that the base of the ladder is being pushed towards the wall at a rate of 1 foot per second, we have dx/dt = 1. We want to find dy/dt, the rate at which the top of the ladder (y) is moving up the wall.
Using the Pythagorean theorem, we know that x^2 + y^2 = 10^2 (since the ladder is 10 feet long). Differentiating both sides of the equation with respect to time (t) will give us a related rates equation:
2x(dx/dt) + 2y(dy/dt) = 0
Plugging in the known values, we have:
2(20)(1) + 2y(dy/dt) = 0
Simplifying:
40 + 2y(dy/dt) = 0
Now, let's solve for dy/dt, the rate at which the top of the ladder is moving up the wall:
2y(dy/dt) = -40
dy/dt = -40 / (2y)
We are interested in finding the rate after 2 seconds, so let's substitute t = 2 into the equation:
dy/dt = -40 / (2y) = -20 / y
Now, we need to find the height of the ladder at t = 2. Plugging t = 2 into the original equation x^2 + y^2 = 10^2, we can solve for y:
x^2 + y^2 = 100
(2)^2 + y^2 = 100
4 + y^2 = 100
y^2 = 96
y = sqrt(96) ≈ 9.80
Now we can substitute y = 9.80 into dy/dt = -20 / y to find the rate at which the top of the ladder is moving up the wall after 2 seconds:
dy/dt = -20/9.8 ≈ -2.04 feet per second
Therefore, after 2 seconds, the top of the ladder is moving up the wall at approximately -2.04 feet per second. The negative sign indicates that the height is decreasing.