A 12 foot ladder is leaning against a wall. If the top slips down the wall at a rate of 4 ft/s, how fast will the foot be moving away from the wall when the top is 8 feet above the ground?
i havent posted 7 questions...
To solve this problem, we can use related rates, specifically the Pythagorean theorem.
Let's denote the distance between the top of the ladder and the ground as 'x' and the distance between the foot of the ladder and the wall as 'y'.
According to the Pythagorean theorem, we have x^2 + y^2 = 12^2 (using the length of the ladder).
Now, let's differentiate the equation with respect to time (t):
2x(dx/dt) + 2y(dy/dt) = 0
We are given that dx/dt = -4 ft/s because the top of the ladder is sliding down the wall. Also, we want to find dy/dt when x = 8 ft.
Substituting these values in the equation, we get:
2(8)(-4) + 2y(dy/dt) = 0
Simplifying the equation:
-64 + 2y(dy/dt) = 0
Rearranging the terms:
2y(dy/dt) = 64
Finally, solving for dy/dt:
dy/dt = 64 / (2y)
To find the value of y, we can use the Pythagorean theorem again. Substitute x = 8 in x^2 + y^2 = 12^2, we get:
(8)^2 + y^2 = 12^2
64 + y^2 = 144
y^2 = 144 - 64
y^2 = 80
y = sqrt(80)
Now substitute this value of y in dy/dt = 64 / (2y):
dy/dt = 64 / (2 * sqrt(80))
Simplifying further:
dy/dt = 64 / (2 * 4 * sqrt(5))
dy/dt = 8 / sqrt(5)
Therefore, the foot of the ladder will be moving away from the wall at a rate of 8 / sqrt(5) ft/s when the top is 8 feet above the ground.
Ok Joe, you have now posted 7 Calculus problems in the last 1/2 hour.
Do you really expect us to do your assignment for you?
Exactly what difficulties are you having with these rather basic Calculus questions?
Show that what efforts you have made and I will gladly assist you with those difficulties.