A 16 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 15 feet above the ground?
Well, it sounds like this ladder is really trying to escape from doing any work! Alright, let's help it out. We're given that the top of the ladder is slipping down the wall at a rate of 4 ft/s, and we want to find out how fast the foot of the ladder is moving away from the wall when the top is 15 feet above the ground.
To solve this problem, we'll use the Pythagorean theorem. The ladder is like a right triangle, with the wall being the vertical side, the ground being the horizontal side, and the ladder itself being the hypotenuse.
The Pythagorean theorem states that in a right triangle, the square of the length of the hypotenuse (ladder) is equal to the sum of the squares of the lengths of the other two sides (wall and ground).
So, let's call the distance between the foot of the ladder and the wall "x". We have an equation:
x^2 + 15^2 = 16^2
Simplifying, we get:
x^2 + 225 = 256
Subtracting 225 from both sides:
x^2 = 31
Now, let's differentiate both sides of the equation with respect to time (t). We get:
2x * dx/dt = 0
Dividing both sides by 2x:
dx/dt = 0 / (2x)
Since x = sqrt(31), we can plug this back into the equation:
dx/dt = 0 / (2 * sqrt(31))
Now, I'm not sure if the ladder really wants to move away from the wall or not, but if it really insists, then the foot of the ladder will be moving away from the wall at a whopping rate of 0 ft/s. It seems like the ladder is just too lazy to make a move.
To solve this problem, we can use related rates. Let's assign variables to the different parts of the ladder:
Let x be the distance from the wall to the foot of the ladder.
Let y be the distance from the ground to the top of the ladder.
We are given the following information:
dx/dt = ? (rate at which the foot is moving away from the wall, what we need to find)
dy/dt = -4 ft/s (rate at which the top is slipping down the wall)
x = ? (distance from the wall to the foot of the ladder, what we need to find)
y = 15 ft (distance from the ground to the top of the ladder, given)
Now, we need to find a relationship between x and y. We can use the Pythagorean theorem:
x^2 + y^2 = 16^2
Differentiating both sides with respect to time t:
2x(dx/dt) + 2y(dy/dt) = 0
Now, substitute the given values and solve for dx/dt:
2x(dx/dt) + 2(15 ft)(-4 ft/s) = 0
2x(dx/dt) = 2(15 ft)(4 ft/s)
2x(dx/dt) = 120 ft^2/s
dx/dt = 120 ft^2/s / (2x)
To find dx/dt when y = 15 ft, we need to find the corresponding value of x from the Pythagorean theorem:
x^2 + 15^2 = 16^2
x^2 + 225 = 256
x^2 = 256 - 225
x^2 = 31
x = sqrt(31)
Now we can substitute this value into the equation for dx/dt:
dx/dt = 120 ft^2/s / (2 * sqrt(31) ft)
dx/dt = 60 / sqrt(31) ft/s
So, when the top is 15 feet above the ground, the foot will be moving away from the wall at a rate of 60 / sqrt(31) ft/s.
To find how fast the foot of the ladder is moving away from the wall, we need to use related rates. Let's denote the distance between the foot of the ladder and the wall as "x" and the distance between the top of the ladder and the ground as "y".
We are given that the ladder is leaning against the wall, so we have a right triangle formed by the ladder, the wall, and the ground. Using the Pythagorean theorem, we have the equation x^2 + y^2 = 16^2.
Taking the derivative with respect to time (t) of both sides of the equation, we get:
2x(dx/dt) + 2y(dy/dt) = 0
We are asked to find how fast the foot of the ladder is moving away from the wall when y = 15 ft and given that dy/dt = 4ft/s. We need to find dx/dt.
We can rearrange the equation and substitute the known values to solve for dx/dt:
dx/dt = -xy / y = -x(y / y)
Using the Pythagorean theorem, we know that x = sqrt(16^2 - y^2). Substituting this expression for x, we get:
dx/dt = -sqrt(16^2 - y^2) * (y / y)
When the ladder is 15 feet above the ground (y = 15), we can plug this into the equation:
dx/dt = -sqrt(16^2 - 15^2) * (15 / 15)
Simplifying this, we find:
dx/dt = -sqrt(256 - 225) = -sqrt(31)
Therefore, when the top of the ladder is 15 feet above the ground, the foot of the ladder is moving away from the wall at a rate of -sqrt(31) ft/s.
x^2 + y^2 = 16^2 = 256
2 x dx/dt + 2 y dy/dt = d 256 /dt = 0
at start y = 15 so x = sqrt (256-225) = sqrt 31 = 5.568
11.14 dx/dt + 30 dy/dt = 0
dy/dt = -4 given so
11.14 dx/dt = -30 * -4 = +120 ft/s
dx/dt = 120/11.14
about 10.8 ft/s