A ladder 12 meters long is leaning against the wall of a building. The bottom of the ladder is sliding away from the wall at the rate of 1.5 m/s. How fast is the top of the ladder sliding down when it is 3 meters above the ground?
Express your answer in square root form. Include unit of measurement.
Almost every Calculus textbook I have seen has a variation of this
question as one of the examples to introduce "Rates Of Change"
foot of ladder from wall ---- x m
height of ladder up the wall ---- y m
x^2 + y^2 = 12^2
2x dx/dt + 2y dy/dt = 0
given: dx/dt = 1.5 m/s
find : dy/dt when y = 3
x^2 + 3^2 = 12^2
x = √135
sub into 2x dx/dt + 2y dy/dt = 0
x dx/dt + y dy/dt = 0
√135(1.5) + 3dy/dt = 0
dy/dt = -1.5√135/3 = ..... m/s
Well, well, well, it seems we have a ladder-slash-wall-slash-sliding situation going on here! Let's break it down, shall we?
We have a ladder that's 12 meters long and it's leaning against a wall. The bottom of the ladder is sliding away from the wall at a rate of 1.5 m/s. We need to figure out how fast the top of the ladder is sliding down when it's 3 meters above the ground.
To solve this, we can use good ol' Pythagoras. You see, we have a right triangle formed by the ladder, the ground, and the distance from the wall to the ladder. The length of the ladder is the hypotenuse, and the distance from the wall is one side of the triangle.
Let's denote the distance from the wall as x, and the height of the ladder (from the ground to the top) as y. As the bottom of the ladder slides away, x is increasing, while y is decreasing.
Now, we can use the Pythagorean theorem: x^2 + y^2 = 12^2. If we differentiate this equation with respect to time, we get:
2x (dx/dt) + 2y (dy/dt) = 0.
We want to find dy/dt when y = 3 meters, so we can plug in these values and solve for dx/dt:
2x (dx/dt) + 2y (dy/dt) = 0,
2(3)(0) + 2(3)(dy/dt) = 0,
6(dy/dt) = 0,
dy/dt = 0.
Oh, no! It seems like when the top of the ladder is 3 meters above the ground, it's not sliding down at all! That's quite a balancing act! The top of the ladder remains at the same height, without any vertical movement.
So, the answer is 0. And if you're feeling extra fancy, you can express it as the square root of zero, or √0, with the unit of measurement being meters per second.
Keep those ladders in check, my friend, and remember, when in doubt, call in the clowns!
To solve this problem, we can use the Pythagorean theorem, which states that for a right triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.
Let's assume that the distance between the bottom of the ladder and the wall is x meters. According to the problem, the bottom of the ladder is sliding away from the wall at a rate of 1.5 m/s. That means dx/dt (the rate at which the distance is changing) is equal to 1.5 m/s.
We can now draw a triangle to represent the situation:
|\
| \
| \
| \
| \
| \
| \
| \
|________\
In this triangle, the ladder is the hypotenuse, the distance between the bottom of the ladder and the wall is x, and the height of the ladder (the vertical distance from the ground to the top of the ladder) is y. We want to find dy/dt (the rate at which the height is changing) when y is equal to 3 meters.
Using the Pythagorean theorem, we know that:
x^2 + y^2 = 12^2 (since the ladder is 12 meters long)
Differentiating this equation with respect to time (t) gives us:
2x*dx/dt + 2y*dy/dt = 0
Now we can substitute the given values into the equation:
2( x )( 1.5 ) + 2( y )( dy/dt ) = 0
Simplifying the equation:
3x + 2y(dy/dt) = 0
To find dy/dt, we need to solve for it. We know that x is changing at a rate of 1.5 m/s, and we want to find dy/dt when y is equal to 3 meters.
Let's set x as the variable:
x^2 + y^2 = 12^2
x^2 + 3^2 = 12^2
x^2 + 9 = 144
x^2 = 135
x = √135 (approximately 11.62 meters)
Now, substitute the values into the equation:
3(√135) + 2(3)(dy/dt) = 0
Simplifying the equation:
3√135 + 6(dy/dt) = 0
To find dy/dt, isolate it:
6(dy/dt) = -3√135
dy/dt = (-3√135) / 6
dy/dt = -√135 / 2
So, the top of the ladder is sliding down at a rate of -√135 / 2 meters per second when it is 3 meters above the ground.
To solve this problem, we can use related rates. Let's label the length of the ladder as "L", the rate at which the bottom of the ladder is sliding away from the wall as "dx/dt", the height of the ladder on the wall as "y", and the rate at which the top of the ladder is sliding down as "dy/dt".
We know that the ladder has a fixed length of 12 meters (L = 12 m), and the bottom of the ladder is sliding away from the wall at a rate of 1.5 m/s (dx/dt = -1.5 m/s, negative because it is moving away from the wall).
We need to find the rate at which the top of the ladder is sliding down (dy/dt) when it is 3 meters above the ground (y = 3 m).
We can use the Pythagorean theorem to establish a relationship between x, y, and L:
x^2 + y^2 = L^2
Now, taking the derivative of both sides with respect to time (t), we get:
2x(dx/dt) + 2y(dy/dt) = 0
Rearranging this equation, we have:
dy/dt = -x(dx/dt)/y
To find the value of x when y = 3 m, we can use the Pythagorean theorem:
x^2 + 3^2 = 12^2
x^2 + 9 = 144
x^2 = 144 - 9
x^2 = 135
x ≈ √135
Now we can substitute these values into our equation:
dy/dt = -√135(-1.5) / 3
Simplifying this expression, we get:
dy/dt = √135 / 2
So, the rate at which the top of the ladder is sliding down when it is 3 meters above the ground is approximately √135 / 2 m/s.