ladder 25feet long is leaning against the wall of a house. The base of the ladder is pulled away from the wall at a rate of 2feet per second. Find
How fast is the top moving down the wall when the base of the ladder is 7feet, 15feet and 24feet from the wall
To solve this problem, we will use the Pythagorean theorem, which states that in a right triangle, the square of the length of the hypotenuse (the ladder in this case) is equal to the sum of the squares of the other two sides.
Let's denote the distance from the wall to the base of the ladder as x (in feet) and the distance from the top of the ladder to the ground as y (also in feet). We can then express the relationship between x, y, and the length of the ladder using the Pythagorean theorem:
x^2 + y^2 = 25^2
Differentiating both sides of this equation with respect to time t (in seconds), we get:
2x(dx/dt) + 2y(dy/dt) = 0
Since we know that dx/dt (the rate at which the base of the ladder is moving away from the wall) is 2 feet per second, we can substitute this value into the equation and solve for dy/dt:
2x(2) + 2y(dy/dt) = 0
4x + 2y(dy/dt) = 0
2y(dy/dt) = -4x
dy/dt = -4x / (2y)
Simplifying further:
dy/dt = -2x/y
Now we can substitute the values of x and y for the given scenarios:
1) When x = 7 feet:
dy/dt = -2(7) / y
2) When x = 15 feet:
dy/dt = -2(15) / y
3) When x = 24 feet:
dy/dt = -2(24) / y
To find the value of y in each case, we can use the Pythagorean theorem again:
1) When x = 7, y^2 = 25^2 - 7^2 = 576, so y = 24 feet.
2) When x = 15, y^2 = 25^2 - 15^2 = 200, so y = 10√2 feet.
3) When x = 24, y^2 = 25^2 - 24^2 = 49, so y = 7 feet.
Now we can substitute these values back into the equations above to find the rate at which the top of the ladder is moving down the wall in each case:
1) When x = 7:
dy/dt = -2(7) / 24 = -7/12 feet per second.
2) When x = 15:
dy/dt = -2(15) / (10√2) = -3√2 feet per second.
3) When x = 24:
dy/dt = -2(24) / 7 = -48/7 feet per second.
To find the rate at which the top of the ladder is moving down the wall, we can use the concept of related rates.
Let's denote the distance between the base 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 dx/dt (the rate at which x is changing) is 2 ft/s.
We are asked to find dy/dt (the rate at which y is changing) when x = 7 ft, x = 15 ft, and x = 24 ft.
Using the Pythagorean theorem, we can relate x, y, and the length of the ladder (25 ft):
x^2 + y^2 = 25^2
Differentiating implicitly with respect to time, we get:
2x(dx/dt) + 2y(dy/dt) = 0
Simplifying, we have:
x(dx/dt) + y(dy/dt) = 0
Now, let's substitute dx/dt = 2 ft/s:
2x + y(dy/dt) = 0
To solve for dy/dt, we rearrange the equation:
y(dy/dt) = -2x
dy/dt = (-2x) / y
Now, let's substitute the given values of x:
When x = 7 ft:
y(dy/dt) = (-2 * 7) / y
= -14 / y
When x = 15 ft:
y(dy/dt) = (-2 * 15) / y
= -30 / y
When x = 24 ft:
y(dy/dt) = (-2 * 24) / y
= -48 / y
Therefore, the rate at which the top of the ladder is moving down the wall when the base of the ladder is 7 ft, 15 ft, and 24 ft from the wall is (-14 / y) ft/s, (-30 / y) ft/s, and (-48 / y) ft/s, respectively.