a ladder 25ft long leans against a wall with its foot on level ground 7ft from the base of the wall. If the foot is pulled away from the wall at the rate 2 ft/s express the distance (y) of the top of the ladder above the ground as a function of the time, t seconds in moving
x^2 + y^2 = 625
x = 2t
y^2 = 625 - 4t^2
so,
y = √(625-4t^2)
To find the distance, y, of the top of the ladder above the ground as a function of time, we can use the Pythagorean theorem. Let's break down the problem and solve it step by step.
Step 1: Visualize the scenario
We have a ladder leaning against a wall, with its foot on level ground. The length of the ladder, L, is 25ft, and the distance from the foot of the ladder to the base of the wall is 7ft.
|
|
|
| L y
Wall --------------
7ft
Step 2: Understand the problem and determine the given and unknown values
Given:
- The length of the ladder (L) = 25ft
- The distance from the foot of the ladder to the base of the wall (x) = 7ft
- The rate at which the foot of the ladder is pulled away from the wall (dx/dt) = 2ft/s
Unknown:
- The distance of the top of the ladder above the ground (y)
Step 3: Formulate an equation using the Pythagorean theorem
According to the Pythagorean theorem, the sum of the squares of the two sides of a right triangle is equal to the square of the hypotenuse.
In this case, we have:
(x + dx)^2 + y^2 = L^2
Step 4: Express the equation in terms of variables
Substitute the given values and express the equation in terms of variables:
(7 + 2t)^2 + y^2 = 25^2
Step 5: Simplify the equation
Expand and simplify the equation:
49 + 28t + 4t^2 + y^2 = 625
Step 6: Rearrange the equation to solve for y
Rearrange the equation to solve for y:
y^2 = 625 - 49 - 28t - 4t^2
Simplify further:
y^2 = 576 - 28t - 4t^2
Take the square root of both sides:
y = √(576 - 28t - 4t^2)
Therefore, the distance of the top of the ladder above the ground as a function of time (t) is y = √(576 - 28t - 4t^2).
x^2 + y^2 = 625 (Path. theorem)
x = 7+2t
(7+2t)^2 + y^2 = 625
y^2 = 625 -(7+2t)^2
y=√625 -(7+2t)^2