A ladder 25ft long is leaning against the wall of a house. The ladder is pulled away from the wall at a rate of 2 ft per second. How fast is the top of the ladder moving down the wall when its base is 7ft from the wall?
This is the classic problem that almost every text in Calculus uses to introduce "rate of change"
Make a diagram
Let the ladder by y ft above the ground, and
x ft away from the wall
so you have a right angled triangle
x^2 + y^2 = 25
2x dx/dt + 2y dy/dt = 0 (#1)
given : dx/dt = 2 ft/s
find : dy/dt when x = 7
when x = 7
49 + y^2 = 625
y = √576 = 24
back in #1
2(7)(2) + 2(24)dy/dt = 0
dy/dt = -28/48 = -7/12 ft/s
the negative indicates that the value of y is decreasing, or in other words, the ladder is dropping along the wall
To solve this problem, we can use the related rates approach.
Let's label the distance between the base of the ladder and the wall as "x" (in feet) and the distance from the top of the ladder to the ground as "y" (in feet).
We are given that dx/dt = 2 ft/s, which represents the rate at which the base of the ladder is moving away from the wall. We need to find dy/dt, the rate at which the top of the ladder is moving down the wall.
Using the Pythagorean theorem, we have
x^2 + y^2 = 25^2.
Taking the derivative with respect to time t, we get
2x(dx/dt) + 2y(dy/dt) = 0.
Now, we can substitute the given values to solve for dy/dt when x = 7 ft.
2(7)(2) + 2y(dy/dt) = 0.
28 + 2y(dy/dt) = 0.
2y(dy/dt) = -28.
dy/dt = -28/(2y).
To find the value of y when x = 7 ft, we can plug x = 7 into the Pythagorean theorem:
7^2 + y^2 = 25^2.
49 + y^2 = 625.
y^2 = 576.
y = √576.
y = 24 ft.
Now we can substitute y = 24 ft into the rate equation to find dy/dt.
dy/dt = -28/(2 * 24).
dy/dt = -7/12 ft/s.
Therefore, the top of the ladder is moving down the wall at a rate of 7/12 ft/s when the base is 7 ft from the wall.
To find the rate at which the top of the ladder is moving down the wall, we need to use the related rates formula.
Let's denote the distance between the base of the ladder and the wall as x (in feet), and the distance between the top of the ladder and the ground as y (in feet). We want to find dy/dt (the rate at which y is changing) when x = 7ft.
Given:
dx/dt = -2ft/s (since the ladder is being pulled away from the wall, x is decreasing)
x = 7ft
We can start by using the Pythagorean theorem to relate x and y:
x^2 + y^2 = 25^2
Differentiating both sides of this equation with respect to time t:
2x(dx/dt) + 2y(dy/dt) = 0
Plugging in the values we know:
2(7)(-2) + 2y(dy/dt) = 0
Simplifying the equation:
-28 + 2y(dy/dt) = 0
2y(dy/dt) = 28
dy/dt = 28/(2y)
To find y, we can use the Pythagorean theorem with x = 7ft:
7^2 + y^2 = 25^2
49 + y^2 = 625
y^2 = 625 - 49
y^2 = 576
y = √576
y = 24ft
Now we can plug the value of y into our equation for dy/dt:
dy/dt = 28/(2 * 24)
dy/dt = 7/12 ft/s
Therefore, the top of the ladder is moving down the wall at a rate of 7/12 ft/s when its base is 7ft from the wall.