A board 5 feet long slides down a wall. At the instant the bottom end is 4 feet from the wall, the other end is moving down the wall at a rate of 2 feet per second. At the moment, how fast is the bottom end sliding along the ground?
well, you just said math, not calculus, which is why I completed the square to find the vertex of the parabola in the other problem.
3 4 5 triangle
x = 4
y = 3
hypotenuse = 5
dy/dt = -2
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x^2 + y^2 = 25
2 x dx/dt + 2 y dy/dt = 0
dx/dy = -(y/x)dy/dt
= -(3/4)(-2) = +3/2
= +1.5 feet/second
Well, with the board sliding down the wall, it seems like the situation is really falling into place. Now, let's get down to business and calculate the rate at which the bottom end is sliding along the ground.
We can start by calling the distance between the bottom end of the board and the wall 'x'. So, at the given moment, x = 4 feet. We're also given that the other end of the board is moving down the wall at a rate of 2 feet per second.
Now, let's use some good old Pythagoras to establish a relationship between the different lengths involved. We can say that the hypotenuse of the right triangle formed by the board, the wall, and the ground is the length of the board, which is 5 feet.
Using the Pythagorean theorem, we have:
(4)^2 + x^2 = 5^2
16 + x^2 = 25
x^2 = 9
x = 3 feet
So, the bottom end of the board is sliding along the ground at a rate of 3 feet per second. And just like that, we've nailed it!
To solve this problem, we can use related rates, where we relate the rates of change of different quantities involved in the problem.
Let's assume the distance from the bottom end of the board to the ground is represented by y, and the distance from the top end of the board to the ground is represented by x.
Given:
- dx/dt = 2 ft/s (the rate at which the top end is moving down the wall)
- x = 4 ft (the distance from the top end to the wall)
- y is the distance from the bottom end to the ground (the quantity we want to find)
We can use the Pythagorean theorem to relate x, y, and the length of the board (5 ft):
x^2 + y^2 = 5^2
Differentiating both sides of the equation with respect to time t, we get:
2x(dx/dt) + 2y(dy/dt) = 0
Substituting the given values, we have:
2(4)(2) + 2y(dy/dt) = 0
Simplifying:
16 + 2y(dy/dt) = 0
Now, we can solve for dy/dt (the rate at which the bottom end is sliding along the ground):
2y(dy/dt) = -16
dy/dt = -16 / (2y)
At the moment, y = 3 ft (from the Pythagorean theorem).
Substituting this value into the equation:
dy/dt = -16 / (2*3)
= -8/3 ft/s
Therefore, at the moment, the bottom end of the board is sliding along the ground at a rate of -8/3 ft/s. Note that the negative sign indicates that it is sliding in the opposite direction, away from the wall.
To find the rate at which the bottom end of the board is sliding along the ground, we can use related rates.
Let's label the distance between the bottom end of the board and the wall as x, and the distance between the bottom end of the board and the ground as y. We are given the rates of change of both x and y.
We want to find dx/dt, the rate at which x is changing with respect to time. We are given dy/dt, the rate at which y is changing with respect to time.
First, we need to relate x and y using the given information. Since the board is sliding down the wall, these two distances are related by the Pythagorean theorem:
x^2 + y^2 = 5^2 (because the board is 5 feet long)
Differentiating both sides of the equation with respect to time t, we get:
2x(dx/dt) + 2y(dy/dt) = 0
We want to find dx/dt, so let's solve the equation for dx/dt:
dx/dt = -y(dy/dt) / x
Now we can plug in the given values: x = 4 feet, dy/dt = 2 feet per second.
We still need to find the value of y. We can use the Pythagorean theorem to find it:
4^2 + y^2 = 5^2
Simplifying, we get: y^2 = 25 - 16 = 9
Taking the square root, y = 3 feet.
Now, we can substitute the values into the equation for dx/dt:
dx/dt = -3(2) / 4 = -6/4 = -3/2 feet per second
Therefore, at that moment, the bottom end of the board is sliding along the ground at a rate of -3/2 feet per second.