a 12 foot ladder is leaning against a building. If the bottom of the ladder is sliding along the pavement directly away from the building at 1 feet/second, how fast is the top of the ladder moving down when the foor of the ladder is 2 feet from the wall?
This is the classic example used by most textbooks in the introduction to Related Rates in Calculus.
let the foot of the ladder be x ft from the wall, let the ladder reach y ft above the ground
x^2 + y^2 = 12^2
2x dx/dt + 2y dy/dt = 0
or
x dx/dt + y dy/dt = 0
given: dx/dt = 1 ft/s
find: dy/dt when x = 2
when x = 2
4 + y^2 = 144
y^2 = 140
y = √140
2(1) + √140 dy/dt = 0
dy/dt = -√140/2 ft/s
= ....
notice the dy/dt is negative, indicating that y is decreasing
Correct answer is -.169
you are right, I got my fraction upside down, sorry
dy/dx = -2/√140 = appr -.169
To solve this problem, we can use the concepts of similar triangles and related rates from calculus. Let's assign some variables to the given information:
Let "x" represent the distance between the bottom of the ladder and the wall (in feet).
Let "y" represent the height between the pavement and the top of the ladder (in feet).
We're given that the bottom of the ladder is sliding along the pavement directly away from the building at a rate of 1 feet/second. This means that dx/dt = 1 (the rate of change of x with respect to time is 1).
We need to find dy/dt, the rate at which the top of the ladder is moving down with respect to time.
Using similar triangles, we can see that the height between the pavement and the top of the ladder (y) and the distance between the bottom of the ladder and the wall (x) are also in a constant ratio.
Therefore, we have the following relationship:
x/y = constant
Since the ladder has a length of 12 feet, we know that when the bottom of the ladder is 2 feet from the wall (x = 2), the height between the pavement and the top of the ladder (y) will be 10 feet (12 - 2 = 10).
Plugging these values into the relationship, we have:
2/10 = constant
Simplifying, we find:
1/5 = constant
Now, we can differentiate both sides of the equation with respect to time (t) using implicit differentiation:
d(x/y) / dt = d(constant) / dt
To find the derivative of the left side, we'll use the quotient rule:
(x * dy/dt - y * dx/dt) / (y^2) = 0
Since dx/dt = 1, we can substitute it into the equation:
(x * dy/dt - y * 1) / (y^2) = 0
Rearranging the equation, we get:
x * dy/dt = y
Now we can substitute the known values:
2 * dy/dt = 10
Simplifying, we find:
dy/dt = 10/2
Therefore, the top of the ladder is moving down at a rate of 5 feet/second when the bottom of the ladder is 2 feet from the wall.
So, the answer is: The top of the ladder is moving down at a rate of 5 feet/second.