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 7 feet above the ground?
Well, well, well, it seems we have a ladder in a bit of a pickle! Let's see if we can help it out with a dash of math and a sprinkle of humor.
We have ourselves a 12-foot ladder leaning against a wall. Now, if I've learned anything from my ladder friends, it's that they like to slide down at a constant rate. In this case, the top of the ladder is slipping down the wall at a rate of 4 ft/s.
Now, we want to know how fast the foot of the ladder is moving away from the wall when the top is 7 feet above the ground. Ah, the suspense is killing me! But fear not, my friend, for we shall find the answer.
Let's call the distance between the foot of the ladder and the wall "x." And since the ladder is 12 feet long, we can call the height from the foot to the top of the ladder "y." Now, we know that x and y are related by a neat little equation called the Pythagorean theorem: x^2 + y^2 = 12^2.
Now, let's take the derivative of both sides of this equation with respect to time. Remember, we want to find how fast the foot of the ladder is moving away from the wall, so we need to find dx/dt. As for dy/dt, we know that it's a constant 4 ft/s.
Differentiating our equation, we get 2x(dx/dt) + 2y(dy/dt) = 0. Substituting in the values we know, which are x = 12 - 7 = 5 and y = 7, we can plug them into the equation and get 2(5)(dx/dt) + 2(7)(4) = 0. Simplifying, we find that 10(dx/dt) + 56 = 0.
And now, my friend, it's time to reveal the answer! Solving for dx/dt, we find that 10(dx/dt) = -56, which gives us dx/dt = -56/10 = -5.6 ft/s.
So, when the top of the ladder is 7 feet above the ground, the foot of the ladder is moving away from the wall at a rate of 5.6 ft/s. Oh ladders, always causing trouble, but at least we can give them some speed limits!
To solve this problem, we can use related rates. Let's assign variables to the different quantities:
Let H be the height of the ladder above the ground (in feet).
Let x be the distance between the base of the ladder and the wall (in feet).
We are given:
H = 7 ft (the height of the ladder above the ground)
dH/dt = -4 ft/s (the rate at which the top of the ladder is sliding down the wall)
We need to find:
dx/dt (the rate at which the foot of the ladder is moving away from the wall)
Now, let's use the Pythagorean theorem:
x^2 + H^2 = 12^2
Differentiating both sides of the equation with respect to time t, we get:
2x (dx/dt) + 2H (dH/dt) = 0
Substituting the given values into the equation, we have:
2x (dx/dt) + 2(7)(-4) = 0
Simplifying the equation, we have:
2x (dx/dt) - 56 = 0
Now, let's solve for dx/dt:
2x (dx/dt) = 56
dx/dt = 28/x
To find the value of dx/dt, we need to determine the value of x when H = 7 ft. We can use the Pythagorean theorem:
x^2 + 7^2 = 12^2
x^2 + 49 = 144
x^2 = 95
x = √95 ≈ 9.75 ft
Now, we can find the rate at which the foot of the ladder is moving away from the wall when the top is 7 feet above the ground:
dx/dt = 28/9.75 ≈ 2.87 ft/s
Therefore, the foot of the ladder is moving away from the wall at a rate of approximately 2.87 ft/s when the top is 7 feet above the ground.
To solve this problem, we can use the Pythagorean Theorem to relate the movement of the top of the ladder and the foot of the ladder with respect to time.
Let's label the distance of the top of the ladder from the ground as "y" and the distance of the foot of the ladder from the wall as "x".
According to the Pythagorean Theorem, we have:
x^2 + y^2 = 12^2
Differentiating both sides of the equation implicitly with respect to time (t), we get:
2x(dx/dt) + 2y(dy/dt) = 0
Simplifying, we have:
x(dx/dt) + y(dy/dt) = 0
Now, we need to find dx/dt, which is the speed at which the foot of the ladder is moving away from the wall when the top is 7 feet above the ground.
Given that the top slips down the wall at a rate of 4 ft/s, we are given that dy/dt = -4 ft/s (negative because the top is moving down).
Substituting this into the equation above, we have:
7(dx/dt) -4y = 0
Since we are looking for dx/dt, we can solve for it:
7(dx/dt) = 4y
dx/dt = (4y) / 7
We know that, for our particular scenario, y = 7 feet (since the top is 7 feet above the ground). Substituting this value into the equation, we have:
dx/dt = (4 * 7) / 7
dx/dt = 4 ft/s
Therefore, the foot of the ladder will be moving away from the wall at a speed of 4 ft/s when the top is 7 feet above the ground.
w^2 + g^2 = 12^2
... 7^2 + g^2 = 12^2
... g = √95
2 w dw/dt + 2 g dg/dt = 0
(2 * 7 * -4) + (2 * √95 * dg/dt) = 0
solve for dg/dt